mm-1053 > Consider a polynomial P(x), with n factors such that g_1(x) g_2(x)...g_n(x) = P(x) and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. That is, for an algebraic integer x, and natural number ÔaÕ where > 1<=a<=n, g_a(x) is an algebraic integer. Now consider setting x=0, and notice that gives g_1(0) g_2(0)...g_n(0) = P(0) and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thatÕs not a trick. Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have c_1 c_2...c_n = P(0) and the cÕs are necessarily algebraic integers and factors of the > constant term P(0). Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. Then I can substitute and have (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). Now let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 then P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) when the aÕs are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so let g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). Setting x = 0 with the cubic defining the aÕs gives a^3 - 3a^2 = 0, so two of the aÕs are 0, and one is 3. Since indices are arbitrary let the first two equal 0, and I have c_1 = 7, c_2 = 7, and c_3 = 22 which is consistent with the constant term of P(x), which is 1078. But dividing P(x) by 49, gives P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, Up to this point you are fine. so > necessarily 7 is divided from the factors where the constant terms are > 7. No. What you are saying here is that if 7 divides g(0) then > 7 must divide g(x) for all x. This is simply not true. > It follows algebraically. If you need me to explain in further detail > then I can, but simple denial is not going to work. Actually it does not follow, algebraically or otherwise. Simple assertion is not going to work. I have explained in great detail, why you are wrong. You have ignored these posts. The rest of your post in nonsense. - William Hughes > I can explain the algebra in detail to you, but if you simply choose > to claim itÕs nonsense, then no progress can be made, right? Good, start by explaining all the counterexamples I have provided. You have quite a backlog. As an easy first step explain why, with a(0) = 0 and w1(0) =1 the constant terms of (a(x) + 7) and (a(x)/w1(x) + 7/w1(x)) are different. > Now then, if you will accept algebra, and point out an area of > confusion then I can explain to you. But if you are simply going to > just deny then thereÕs little point. > Which is it? Will you follow algebra, and give a mathematical area > for me to explain, or are you just going to deny? Which is it? Will you reply to the above and give mathematical arguments, or are you just going to ignore? > James Harris === Subject: JSH: Does it really matter? It seemed to me that Usenet potentially had a future as an important tool in mathematical research. However, it has weird ßaws, so IÕve been diligently, and with great effort, checking it to see if it can be made to be useful. Mathematics is a discipline where the truth can be determined. When explained out in great detail, mathematical truths can be understood without confusion. However, IÕve seen that after repeated explanations I not only still see confusion, but lots of emotion, and people who stubbornly hold on to false beliefs and make false claims which others apparently believe. Now if Usenet is immune to process then it is useless, as typically in any research you start vaguely and with multiple errors, and if youÕre lucky over time you may get to correct and valuable results. The value of results--their worth--is determinable in terms of impact and new knowledge gained, as well as the potential new knowledge to be gained, as paths of inquiry are opened up. So far, IÕve seen a lot of evidence that Usenet is immune to process in that in the early stages, a person can get a negative reputation, which is not surprising, but later even with valuable and correct results people will read what they have--thatÕs important--and ignore whatÕs mathematically correct to hold on to the negative reputation! ItÕs a key feature of Usenet. The group holds on to negatives, and communicates them forward, while refusing to reassess. So that the group tries to *lock* a person into one place for all time. Now if people would just ignore that person so that they never could tell when they had now switched to having correct results, that would be one thing, and not necessarily bad. But the sadder truth is that they will keep looking at what a person has, and choose to ignore mathematics that is correct, in a willful manner. Why is this important? One of the things that happens to a major researcher is that people end up asking their opinions about certain things, and looking to them for guidance on how others might better their own research. My hope for a while had been that Usenet would be a part of the useful tools for the researcher as a place where you could work through ideas in their early phases and maybe over time win people over when you had correct and valuable results. Instead it has been a tool for holding back research, and like with the sci.math posters who got together to send emails against a paper of mine, itÕs clearly an area where group forces can make it dangerous for researchers. So, my opinion will be to keep the best and brightest off of Usenet, and to make something like it, in terms of the ability to express new ideas and get heated criticism, elsewhere. My own theory is that there are disgruntled people who basically live on Usenet, who feel badly for their own lack of research success, who will latch on to anyone who starts to show signs of good results, and they will do their best to undermine them psychologically, by repeated postings determined to obscure their message, and even by going outside of Usenet by putting up ßame webpages, or through emailings meant to dissuade them or block their research, like what happened to me. These people are very energetic in attacking individuals, and they act over not just days or months, but *years* as if they will not stop until they accomplish taking a person down. They act obsessively, with great malice, and use multiple tools both on Usenet and off to go after a researcher. Those disgruntled people destroy the usefulness of Usenet itself, but not the format, as in other areas they can be controlled simply by not allowing them to post. Those areas will be behind closed doors, hidden from most, as I can now see the wisdom of keeping the process of discovery locked behind closed doors, away from the public. I am opting for greater secrecy, in the interests of progress. And no, the irony is not lost on me that I probably wouldnÕt have been able to post in such areas myself! The world is not perfect. One can only try for the best solution. That emphasis on secrecy may be the greatest legacy of my time on Usenet, as I think I can argue for it rather well, and that it will come to dominate mathematical and even scientific research in at least the near term. The arguments will continue, but the best of them will be hidden behind closed doors. James Harris === Subject: Re: Does it really matter? > My hope for a while had been that Usenet would be a part of the useful > tools for the researcher ... Well that was a stupid hope you had. And why are you telling us that this was your hope? === Subject: Re: Does it really matter? > Mathematics is a discipline where the truth can be determined. Can you prove in mathematics that truth can always be determined? There have been some cases where it has been proven that truth cannot be determined. You are somethines such a romantic idealist goofball James! You worship mathematics and logic like a zealot, but yuo have no skill or talent in applying it in even very simple situations. === Subject: Re: JSH: Does it really matter? >[...] >That emphasis on secrecy may be the greatest legacy of my time on >Usenet, as I think I can argue for it rather well, and that it will >come to dominate mathematical and even scientific research in at least >the near term. >The arguments will continue, but the best of them will be hidden >behind closed doors. Good plan. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Does JHS really matter? NO! === Subject: Re: JSH: Does it really matter? > It seemed to me that Usenet potentially had a future as an important > tool in mathematical research. Usenet is an Internet message board system. There are very few professional mathematicians on sci.math or *.math.* in general, and those that are here, are not doing professional math here. Where did you get the idea that Usenet is a tool for mathematical research? === Subject: JSH: Operator ambiguity, Escultura I would like to pull out and highlight something interesting that E. E. Escultura posted a few days ago, which IÕd guess heÕs probably talked about many times before, but I just noticed it and think itÕs neat. First some more preamble as *by convention* as has been noted when I brought up the subject of operator ambiguity before, sqrt(x) is taken to be positive. So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, which is not good. Naively then, you may believe that you can just say, take the positive of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. You see, the ambiguity in the square root operator still remains, despite the convention. It doesnÕt work to just try and always take the positive as EsculturaÕs example shows so clearly. Who has the resolution? IÕm curious as to whether or not any of you think you can answer. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > You see, the ambiguity in the square root operator still remains, > despite the convention. Your insistence that it is impossible to define a function for sqrt(x), i.e. to restrict the possible solutions to a single value, invalidates your own prime count function and disqualifies you from further study of algebra. Analogous situations arise in evaluating periodic functions, in finding Ôprincipal valuesÕ, in dealing with Ôbranch cutsÕ and elsewhere. If you cannot understand how these issues are resolved, take up a different line of work. (Not sales, however. I doubt if you could sell manure to an organic farmer.) -- 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: JSH: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. > First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. Wait, youÕre not suggesting that sqrt(0) is positive are you? Just kidding. Of course you arenÕt. As you well know, the convention that sqrt(x) is positive applies only to positive numbers x. > So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. Hey! I remember thinking this was really cool. I did it with 5 = -5, but it was the same idea. I showed it to my teacher (this was in sixth grade), and he said I couldnÕt do that, but he wasnÕt able to give me a good reason why not. The reason it appealed to me so much was that once you have a result like that, you can prove anything. Wow! All you need to prove anything is a fuzzy definition like sqrt(n) is the thing that gives you n when you square it. Provided youÕre willing to put your fingers in your ears and say la la la la I am not lis en ing when mathematicians try to explain why such a definition is invalid. > Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. Oh, this is the more sophisticated version. Usually itÕs shown with a few more steps: i = sqrt(-1) = sqrt(1/-1) = sqrt(1)/sqrt(-1) = 1/i = -i. ^ ERROR! > You see, the ambiguity in the square root operator still remains, > despite the convention. Umm, no. > It doesnÕt work to just try and always take the positive as > EsculturaÕs example shows so clearly. > Who has the resolution? IÕm curious as to whether or not any of you > think you can answer. You ask this question like itÕs at the forefront of mathematical knowledge because (a) itÕs at the frontier of your mathematical knowledge, and (b) you believe that you are blazing new trails in mathematics. Assumption (b) is whatÕs holding you back from learning math. Before you can learn, you have to realize that you donÕt know. > James Harris === Subject: Re: JSH: Operator ambiguity, Escultura hey, then 0 = 5 - 5 = -5 - 5 = -10; you should turn that into a tutorial on Zero Divisors ... and watch out for those chaotic rounding-off errors! > Hey! I remember thinking this was really cool. I did it with 5 = -5, but --Advice 0.05; Free, if wrong, again! http://tarpley.net/bush6.htm === Subject: Re: JSH: Operator ambiguity, Escultura > Assumption (b) is whatÕs holding you back from learning math. Before you > can learn, you have to realize that you donÕt know. A Fundamental Truth. === Subject: Re: JSH: Operator ambiguity, Escultura > Assumption (b) is whatÕs holding you back from learning math. Before you > can learn, you have to realize that you donÕt know. > A Fundamental Truth. Oh please, the poster never answered the mathematical issue raised but instead turned to personal attacks. You deleted out all of the context and made a reply that had nothing to do with the issue. For those who missed it, I used EsculturaÕs example of i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, to show that itÕs naive to think that you can remove the ambiguity from the square root operator. Rather than give a cogent answer the sci.math poster Jim Ferry babbled about his childhood, and simply *claimed* as if thatÕs all it takes that thereÕs no problem. The regular sci.math poster Gib Bogle then came in to delete out all of the context and make his own reply. ThatÕs how sci.mathÕers managed to paint me as a crank. They do this consistently, and sci.math readers cheer them on! The sci.math readership is remarkably stupid and gullible. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura Assumption (b) is whatÕs holding you back from learning math. Before you > can learn, you have to realize that you donÕt know. A Fundamental Truth. > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. > For those who missed it, I used EsculturaÕs example of > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, > to show that itÕs naive to think that you can remove the ambiguity > from the square root operator. > Rather than give a cogent answer the sci.math poster Jim Ferry babbled > about his childhood, and simply *claimed* as if thatÕs all it takes > that thereÕs no problem. > The regular sci.math poster Gib Bogle then came in to delete out all > of the context and make his own reply. > ThatÕs how sci.mathÕers managed to paint me as a crank. They do this > consistently, and sci.math readers cheer them on! Can you give me an example showing sci.math readers cheering them on? > The sci.math readership is remarkably stupid and gullible. Is it? How do you know? Brian Chandler http://imaginatorium.org === Subject: Re: JSH: Operator ambiguity, Escultura Assumption (b) is whatÕs holding you back from learning math. Before you > can learn, you have to realize that you donÕt know. A Fundamental Truth. > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. For those who missed it, I used EsculturaÕs example of i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, to show that itÕs naive to think that you can remove the ambiguity > from the square root operator. Rather than give a cogent answer the sci.math poster Jim Ferry babbled > about his childhood, and simply *claimed* as if thatÕs all it takes > that thereÕs no problem. The regular sci.math poster Gib Bogle then came in to delete out all > of the context and make his own reply. ThatÕs how sci.mathÕers managed to paint me as a crank. They do this > consistently, and sci.math readers cheer them on! > Can you give me an example showing sci.math readers cheering them on? IÕve been posting for years. These people routinely get all kinds of support, which has convinced me that the sci.math readership is stupid. Just dig into the record, and youÕll find plenty of people getting on my case about me supposedly being wrong since all these people supposedly have counter examples, and such, versus very few, if any questioning posts asking if maybe these obsessive people replying to me werenÕt full of it! > The sci.math readership is remarkably stupid and gullible. > Is it? How do you know? > Brian Chandler I was pissed off yesterday so basically I was ranting. I know, itÕs better not to go around calling a lot of people stupid, but hey, if mathematics is explained to you in detail, where the conclusion is quite evident, and you listen to people who rather stupidly lie about the mathematics, and instead of properly questioning them, just swallow their b.s., then do you really think you get points? After all, eventually my research will come out on top, and what will the sci.math newsgroup have? A long record of rather amazing stupidity, where its most notable point, and what will probably define the newsgroup from now on, was the gang emailing to the Southwest Journal of Pure and Applied Mathematics to get my paper censored. Whether I call you stupid now or not, more than likely that and resisting my research, namecalling, and other childish behavior is ALL that the sci.math newsgroup will become known for. The sci.math newsgroup will have its place in history. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > A long record of rather amazing stupidity, where its most notable > point, and what will probably define the newsgroup from now on, was > the gang emailing to the Southwest Journal of Pure and Applied > Mathematics to get my paper censored. Who is in this gang, and how do you know they emailed SWJPA? Also, the word censored is a bit strong. Is criticism really censorship? As far as I can see, nobody here wanted your paper dropped. It would have been fun for us all if it were actually left published. It would have been a problem for SWJPA, but I for one would have enjoyed it if it were left up there. Perhaps your paper was dropped just because the editor realized that it was wrong and worthless. === Subject: Re: JSH: Operator ambiguity, Escultura ... > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. But some posters answered it. > For those who missed it, I used EsculturaÕs example of > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, The contradiction exists because the assumption is made that sqrt(a/b) = sqrt(a)/sqrt(b), which is false when in the complex numbers. Similarly log(a.b) = log(a) + log(b) is false when in the complex numbers. Read a bit about Riemann surfaces and you may understand. -- 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: Operator ambiguity, Escultura > Read a bit about Riemann surfaces and you may understand. He wonÕt. -- --Tim Smith === Subject: Re: JSH: Operator ambiguity, Escultura > ... > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, This is a reductio proof that sqrt(x/y) <> sqrt(x)/sqrt(y) . This is a well-known and not very exciting fact. === Subject: Re: JSH: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. > First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. > So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. > Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > You see, the ambiguity in the square root operator still remains, > despite the convention. > It doesnÕt work to just try and always take the positive as > EsculturaÕs example shows so clearly. > Who has the resolution? IÕm curious as to whether or not any of you > think you can answer. Typical mistake of a beginner: In the complex numbers, sqrt (a/b) = sqrt (a) / sqrt (b) is not usually true. The convention decides about the sign of the result. This decision cannot possible made to be consistent in all cases for sqrt (a), sqrt (b) and sqrt (a/b). Convert your complex numbers into polar coordinates, look at what the convention does to the phase angle, and it should be quite obvious where your mistake is. === Subject: Re: JSH: Operator ambiguity, Escultura > It doesnÕt work to just try and always take the positive as > EsculturaÕs example shows so clearly. In general a complex number has two square roots, both complex. So it doesnÕt make sense to try to take the positive. However, there are two cases i) x^2 - a = 0 has two roots, one with positive real part and one with negative real part. ii) x^2 - a = 0 has two roots both with a real part of zero So to get a unique square root. i) take the root with positive real part ii) take the root with positive imaginary part Now x^2 +1 =0 has two roots, i and -i. Both have 0 real part. So by ii) we choose i and say the (principlal value of) the square root of -1 is i. > Who has the resolution? Me, No one else. Nobody ever though about this before me. I deserve an Acadamy Award. > IÕm curious as to whether or not any of you > think you can answer. > James Harris Be curious no longer. -William Hughes === Subject: Re: JSH: Operator ambiguity, Escultura > ... > of the square root but as Escultura showed, that doesnÕt work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. You, or Escultura, whomever, thereby proves that sqrt(x/y) != sqrt(x)/sqrt(y) === Subject: Re: JSH: Operator ambiguity, Escultura >I would like to pull out and highlight something interesting that E. >E. Escultura posted a few days ago, which IÕd guess heÕs probably >talked about many times before, but I just noticed it and think itÕs >neat. >First some more preamble as *by convention* as has been noted when I >brought up the subject of operator ambiguity before, sqrt(x) is taken >to be positive. >So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so >if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, >which is not good. >Naively then, you may believe that you can just say, take the positive >of the square root but as Escultura showed, that doesnÕt work: >i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. >You see, the ambiguity in the square root operator still remains, >despite the convention. >It doesnÕt work to just try and always take the positive as >EsculturaÕs example shows so clearly. Of course you canÕt take the positive square root of a negative number - it doesnÕt _have_ a positive square root! You seem to be under the impression that there is such a convention - there is not, the convention about the meaning of sqrt(x) is only for x > 0. You also seem to be under the impression that i is positive, which is kind of funny - itÕs not. ItÕs only real numbers that are positive or negative. Finally, you seem to have missed the most important point in the contradiction above, which is that itÕs impossible to define sqrt(z) for complex z in such a way that sqrt(zw) = sqrt(z) sqrt(w). Except for totally missing the main points it was an excellent post. >Who has the resolution? IÕm curious as to whether or not any of you >think you can answer. Guffaw. ThereÕs no contradiction to be resolved, because the calculuation above is simply wrong, using properties of sqrt that nobody ever claimed were valid. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Operator ambiguity, Escultura ... >Naively then, you may believe that you can just say, take the positive >of the square root but as Escultura showed, that doesnÕt work: > >i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > >You see, the ambiguity in the square root operator still remains, >despite the convention. ... > Finally, you seem to have missed the most important point > in the contradiction above, which is that itÕs impossible > to define sqrt(z) for complex z in such a way that > sqrt(zw) = sqrt(z) sqrt(w). He is missing one more point. That you can not algebraically distinguish i and -i is also a result from Galois theory (you can not algebraically distinguish the various roots of a primitive polynomial). -- 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: Operator ambiguity, Escultura >... >Naively then, you may believe that you can just say, take the positive >of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. You see, the ambiguity in the square root operator still remains, >despite the convention. >... > Finally, you seem to have missed the most important point > in the contradiction above, which is that itÕs impossible > to define sqrt(z) for complex z in such a way that > sqrt(zw) = sqrt(z) sqrt(w). >He is missing one more point. That you can not algebraically distinguish >i and -i is also a result from Galois theory (you can not algebraically >distinguish the various roots of a primitive polynomial). Well, yes, although thatÕs at least related to one I did point out: You also seem to be under the impression that i is positive. ************************ David C. Ullrich === Subject: Re: JSH: Operator ambiguity, Escultura >... >Naively then, you may believe that you can just say, take the positive >of the square root but as Escultura showed, that doesnÕt work: > >i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > >You see, the ambiguity in the square root operator still remains, >despite the convention. >... > Finally, you seem to have missed the most important point > in the contradiction above, which is that itÕs impossible > to define sqrt(z) for complex z in such a way that > sqrt(zw) = sqrt(z) sqrt(w). > >He is missing one more point. That you can not algebraically distinguish >i and -i is also a result from Galois theory (you can not algebraically >distinguish the various roots of a primitive polynomial). > Well, yes, although thatÕs at least related to one I did point out: > You also seem to be under the impression that i is positive. But being positive is not an algebraic property. -- 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: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. > First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. > So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. > Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. If we make the convention that sqrt(-1)=i, then itÕs not true that sqrt(1/-1)=sqrt(1)/sqrt(-1). > You see, the ambiguity in the square root operator still remains, > despite the convention. > It doesnÕt work to just try and always take the positive as > EsculturaÕs example shows so clearly. > Who has the resolution? IÕm curious as to whether or not any of you > think you can answer. > James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. If we make the convention that sqrt(-1)=i, then itÕs not true that > sqrt(1/-1)=sqrt(1)/sqrt(-1). So your assertion is that the substitution 1/-1 = -1 cannot be made? The resolution is that the sqrt() operator gives TWO answers. It always does, and convention canÕt force it to give only one answer. So i = +/-sqrt(-1), where the +/- in front is a nod to the reality that the result of using the square root operator is two solutions. ThatÕs the operator ambiguity. Naively you may believe that if you simply say, take only ONE answer from the square root, and try to figure out some way to do it that you can succeed, but you will always have a contradiction lurking. The ambiguity extends to other operators like the cuberoot operator or an infinity of other operators where you get more than one answer. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. If we make the convention that sqrt(-1)=i, then itÕs not true that > sqrt(1/-1)=sqrt(1)/sqrt(-1). > So your assertion is that the substitution 1/-1 = -1 cannot be made? > The resolution is that the sqrt() operator gives TWO answers. It > always does, and convention canÕt force it to give only one answer. No, youÕre confusing the square root of a number (which has both a positive answer and itÕs negation) with the sqrt() operator, which is _defined_ to yield the positive root. This definition prohibits the implied step above: i = sqrt(1)/sqrt(-1), because this equation is false since 1/sqrt(-1) = 1/i = -i. > So i = +/-sqrt(-1), where the +/- in front is a nod to the reality > that the result of using the square root operator is two solutions. > ThatÕs the operator ambiguity. No, i does not = +/-sqrt(-1). YouÕre tilting at definitions again. The _definition_ of ÔiÕ is i = sqrt(-1). KeithK > Naively you may believe that if you simply say, take only ONE answer > from the square root, and try to figure out some way to do it that you > can succeed, but you will always have a contradiction lurking. > The ambiguity extends to other operators like the cuberoot operator or > an infinity of other operators where you get more than one answer. > James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > If we make the convention that sqrt(-1)=i, then itÕs not true that > sqrt(1/-1)=sqrt(1)/sqrt(-1). > So your assertion is that the substitution 1/-1 = -1 cannot be made? No. > The resolution is that the sqrt() operator gives TWO answers. It > always does, and convention canÕt force it to give only one answer. You can make a convention that it only gives one answer, but you may have to sacrifice identities like sqrt(a/b)=sqrt(a)/sqrt(b). === Subject: Re: JSH: Operator ambiguity, Escultura > The resolution is that the sqrt() operator gives TWO answers. It > always does, and convention canÕt force it to give only one answer. Sure it can. Consider a square surface with an area of 4 square units. Answer the question: What is the length of one side? Are you saying that -2 units is a correct answer? -- 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: JSH: Operator ambiguity, Escultura posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. If we make the convention that sqrt(-1)=i, then itÕs not true that > sqrt(1/-1)=sqrt(1)/sqrt(-1). > So your assertion is that the substitution 1/-1 = -1 cannot be made? sqrt(a/b) = sqrt(a)/sqrt(b) is not true for every real a, b: a <> b > The ambiguity extends to other operators like the cuberoot operator or > an infinity of other operators where you get more than one answer. Can you show few examples with the other operators? === Subject: Re: Operator ambiguity, Escultura I think that the error is in the following step: sqrt(1/-1) = 1/i Its a bit like saying sqrt((-3)(-3))=-3 And then proclaiming that you have taken the positive square root. Clearly this is incorrect. > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which IÕd guess heÕs probably > talked about many times before, but I just noticed it and think itÕs > neat. > First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. > So, by the convention, sqrt(4) = 2, and thatÕs good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. > Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesnÕt work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > You see, the ambiguity in the square root operator still remains, > despite the convention. > It doesnÕt work to just try and always take the positive as > EsculturaÕs example shows so clearly. > Who has the resolution? IÕm curious as to whether or not any of you > think you can answer. > James Harris === Subject: Re: Operator ambiguity, Escultura >I would like to pull out and highlight something interesting > at E.E. Escultura posted a few days ago, which IÕd guess > heÕs probably talked about many times before, but I just > noticed it and think itÕs neat. You just ache to show mathematics is not on firm foundations, donÕt you?! Because you are stumped at times does not mean that mathematics is contradictory. On the contrary, it means you are not working hard enough, do not know enough, or are not mathematically sophisticated enough to overcome your mathematical difficulties. This elementary stuff has been worked over many, many times before by those much smarter than you or I. It would be prudent to give up the ghost now. You have not a snowballÕs chance of ever winning in this forum. === Subject: Re: Operator ambiguity, Escultura > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > Who has the resolution? IÕm curious as to whether > or not any of you think you can answer. As this example shows, in the domain of complex numbers, the square root of the product is not equal to the product of the square roots and the same for quotients. What does work is DeMoivreÕs Theorem. Since -1 = cos(pi) + i*sin(pi), we have r1 = [cos(pi)+i*sin(pi)]^(1/2) = cos(pi/2)+i*sin(pi/2) = i r2 = [cos(3pi)+i*sin(3pi)]^(1/2) = cos(3pi/2)+i*sin(3pi/2) = -i As a consequence of the Fundamental Theorem of Algebra, every complex number has exactly two complex square roots. So, i and -i are the only square roots of -1. === Subject: Re: Operator ambiguity, Escultura > As this example shows, in the domain of complex numbers, > the square root of the product is not equal to the product of the > square roots and the same for quotients. And how do you define the square root in the domain of complex numbers? > As a consequence of the Fundamental Theorem of Algebra, > every complex number has exactly two complex square roots. In any field, every element has at most two square roots. Besides, you donÕt need the Fundamental Theorem of Algebra in order to prove that every complex complex number has a square root. Jose Carlos Santos === Subject: Re: Operator ambiguity, Escultura >> As this example shows, in the domain of complex numbers, >> the square root of the product is not equal to the product of the >> square roots and the same for quotients. > And how do you define the square root in the domain of complex > numbers? A square root of a complex number a+bi is a solution of the equation z^2 = a+ib. >> As a consequence of the Fundamental Theorem of Algebra, >> every complex number has exactly two complex square roots. > In any field, every element has at most two square roots. Besides, > you donÕt need the Fundamental Theorem of Algebra in order to prove > that every complex complex number has a square root. === Subject: Re: Operator ambiguity, Escultura >As this example shows, in the domain of complex numbers, >the square root of the product is not equal to the product of the >square roots and the same for quotients. >>And how do you define the square root in the domain of complex >>numbers? > A square root of a complex number a+bi is a solution of the equation > z^2 = a+ib. reproduced above) something about the square root. Now, I know very well what _a_ square root is, thank you very much, but when you use the expression the square root you are making a choice: you are choosing one among the two square roots of a complex number (different from 0). Only after that you can state that in the domain of complex numbers, the square root of the product is not equal to the product of the square roots. So, I repeat my question: how do you define the square root in the domain of complex numbers? Jose Carlos Santos === Subject: Re: Operator ambiguity, Escultura Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Only after that you can state that in the domain of complex numbers, >the square root of the product is not equal to the product of the square >roots. >So, I repeat my question: how do you define the square root in the >domain of complex numbers? Any definition of which root is *the* root will have the problem stated, so the statement is true regardless of which definition you use. -- Richard === Subject: Re: Operator ambiguity, Escultura >>Only after that you can state that in the domain of complex numbers, >>the square root of the product is not equal to the product of the square >>roots. >>So, I repeat my question: how do you define the square root in the >>domain of complex numbers? > Any definition of which root is *the* root will have the problem > stated, so the statement is true regardless of which definition you > use. I know that. I just wanted to know which definition was N. Silver the square root of the product is not equal to the product of the square roots. Note that he did not add regardless of which definition you use. Jose Carlos Santos === Subject: What does this mean? What does An eigenvector is a (representative member of a) fixed point of the map on the projective plane induced by a linear map. mean? I read it here: http://foldoc.doc.ic.ac.uk/foldoc/foldoc.cgi?query= eigenvector&action=Search . === Subject: Re: What does this mean? Well an eigenvector of a linear map A is a vector v such that: Av=mv (for some scalar m - the eigenvalue) Now, from the website you referenced, the projective plane is the space of equivalence classes of vectors under non-zero scalar multiplication. Each element of the space is a set of the form: {kv: k != 0, k scalar, v != O, v a vector} Given our linear map A, we may induce a map on this projective plane space. LetÕs pick the element of the projective plane space that contains our eigenvector: {kv: k != 0, k scalar, v != O, v a vector} Then every element of this set is also an eigenvector: A(kv) = kAv=kmv=m(kv). Vector v is therefore a representative of this set. (We could have picked any one). Moreover, since v is an eigenvector, the induced map leaves the set unchanged (a fixed point). So, our eigenvector is a representative member of a set that is in turn a fixed point of the induced map (that acts on the projective plane) > What does An eigenvector is a (representative member of a) fixed point > of the map on the projective plane induced by a linear map. mean? > I read it here: http://foldoc.doc.ic.ac.uk/foldoc/foldoc.cgi?query= eigenvector&action=Search . === Subject: Re: What does this mean? > === Subject: The Empty Set Over stocked, must sell at once for next to nothing prices, 5,000 unsold empty sets produced for Buy Nothing Day. ;-) === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) What is the warranty? Suppose I buy some and they are not all empty. Can I, for nothing, exchange something for nothing? Alex === Subject: Re: The Empty Set alt.algebra.help === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) > What is the warranty? Suppose I buy some and they are not all > empty. Can I, for nothing, exchange something for nothing? Infinitesimals we may have included are massless and free of charge. ---- === Subject: Re: The Empty Set > .... Can I, for nothing, exchange something for nothing? HowÕbout we do an experiment? You send me $100 (something), and in exchange I promise to send you nothing, and charge you nothing for the privilege. If I fail to fulfil my end of the agreement, IÕll refund your $100. How much out of pocket will you end up? -- rgds David Brownridge === Subject: Re: The Empty Set >> .... Can I, for nothing, exchange something for nothing? >HowÕbout we do an experiment? >You send me $100 (something), and in exchange I promise to send you nothing, >and charge you nothing for the privilege. If I fail to fulfil my end of the >agreement, IÕll refund your $100. >How much out of pocket will you end up? Is that a math question? Will it be on the exam? === Subject: Re: The Empty Set > .... Can I, for nothing, exchange something for nothing? > HowÕbout we do an experiment? > You send me $100 (something), and in exchange I promise to send you nothing, > and charge you nothing for the privilege. If I fail to fulfil my end of the > agreement, IÕll refund your $100. > How much out of pocket will you end up? > -- > rgds > David Brownridge Well, in context, I already received the non-nothing (the $100) from you so if you allow me to return it I am out nothing. Alex === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) Hey, Jon Giffen and James Harris! You should get a hold of this; the null set is FULL of examples where your methods work! Every example in the null set guaranteed to prove your method, or triple your money back! -- Christopher Heckman === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) What is the shipping cost? -- Oppie the Bear aka John ÔRemoveÕ MYWORRIES to email me! (My mother never saw the irony of calling me a son-of-a-bitch) === Subject: Re: The Empty Set alt.algebra.help === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) > What is the shipping cost? We expect it to be $.41 first class with the forth coming postal rate increase, but exact prices cannot be determined until we have calculated how many empty sets will dance in the ball of Radius Epsilon being held at the Origin. ---- === Subject: Re: The Empty Set >> Over stocked, must sell at once for next to nothing prices, >> 5,000 unsold empty sets produced for Buy Nothing Day. ;-) > What is the shipping cost? I think that it will depend on the packaging. For example, a naked {} is difficult to ship, so youÕll probably need at least {{}} , and there may be some bulk packages available, such as {{},{{}}} , and some even fancier gift-boxed versions, which you may find useful and attractive for the holiday season. (Check with the OP for availability.) All of these are very low in calories, too, I understand. -- Vincent Johns Please feel free to quote anything I say here. === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) >> What is the shipping cost? >I think that it will depend on the packaging. For example, a naked > {} >is difficult to ship { } MineÕs a lot bigger than yours, so I have even more difficulty. There is no commemorative stamp I know of, so IÕm proposing that one be printed with CantorÕs name and have a face value of 0 cents [or yen, or whatever currency for any country issuing the stamp.] There could be a fee of 1 cent to cover cost of production and distribution. === Subject: Re: The Empty Set 5,000 unsold empty sets produced for Buy Nothing Day. ;-) >> What is the shipping cost? >I think that it will depend on the packaging. For example, a naked > {} >is difficult to ship > { } Ooooh, thatÕs a big lot of nothing. > MineÕs a lot bigger than yours, so I have even more difficulty. === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) New or used? bob === Subject: Re: The Empty Set alt.algebra.help === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) > New or used? Brand new, undefiled virgin empty sets with nothing ever inserted. ---- === Subject: Re: The Empty Set > alt.algebra.help === >Subject: Re: The Empty Set >>Over stocked, must sell at once for next to nothing prices, >>5,000 unsold empty sets produced for Buy Nothing Day. ;-) >> New or used? >Brand new, undefiled virgin empty sets with nothing ever inserted. Sound like empty promises to me. CB Klein bottle for sale. Inquire within. === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) > New or used? >>Brand new, undefiled virgin empty sets with nothing ever inserted. >Sound like empty promises to me. >Klein bottle for sale. Inquire within. I think that these people simply have nothing to talk about. === Subject: Re: The Empty Set Are they all the same size? > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) === Subject: Re: The Empty Set > Are they all the same size? Identical except some may be emptier than others, however not one empty set has been known to top post, an accomplishment with which empty minds are more adept. I remember one time when the contents of BushÕs mind inadvertently got included in an empty set. But as that didnÕt even as much as cause the empty indigestion, it went unnoticed by quality control, to be bought later by a Buy Nothing customer who fortuitous happened to be a CIA undercover operative; and then there was another time by special order of the imperial tailor, when the Emperor BushÕs clothes were packaged in an empty set. Thus just twice to our knowledge, an empty set has been emptier than empty. > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) === Subject: Re: The Empty Set > Are they all the same size? > Identical except some may be emptier than others, > however not one empty set has been known to top post, > an accomplishment with which empty minds are more adept. > I remember one time when the contents of BushÕs mind inadvertently got > included in an empty set. But as that didnÕt even as much as cause the > empty indigestion, it went unnoticed by quality control, to be bought > later by a Buy Nothing customer who fortuitous happened to be a CIA > undercover operative; and then there was another time by special order of > the imperial tailor, when the Emperor BushÕs clothes were packaged in an > empty set. Thus just twice to our knowledge, an empty set has been > emptier than empty. > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) An empty set is not really empty because it contains emptiness which is more than just being empty. Also, an empty empty set is not really empty empty because it contains empty emptiness which is more than being empty empty. Then there are dynamic empty sets which ßip ßop between being just empty and being empty empty. (Maybe that is why Kerry lost because no one could figure out to what degree his brain was empty). Alex === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) You can buy a real Klein bottle right here ... http://www.kleinbottle.com/ === Subject: Re: The Empty Set > Over stocked, must sell at once for next to nothing prices, > 5,000 unsold empty sets produced for Buy Nothing Day. ;-) > You can buy a real Klein bottle right here ... > http://www.kleinbottle.com/ If you take a strip of the plane [0,1] x [0,10] with area 10 and make it a mobius strip tacking (0,0) thur (1,0) to (1,10) thru (0,10) resp., is itÕs area 10 or 20? === Subject: Re: The Empty Set > If you take a strip of the plane > [0,1] x [0,10] > with area 10 and make it a mobius strip tacking (0,0) thur (1,0) > to (1,10) thru (0,10) resp., is itÕs area 10 or 20? Try painting the Moebius strip and get back to us. === Subject: Re: The Empty Set alt.algebra.help === Subject: Re: The Empty Set > If you take a strip of the plane > [0,1] x [0,10] > with area 10 and make it a mobius strip tacking (0,0) thur (1,0) > to (1,10) thru (0,10) resp., is itÕs area 10 or 20? > Try painting the Moebius strip and get back to us. I twisted my slipped off the Moebius strip spilling the paint when opening Klien bottle requiring application with cross cap. ---- === Subject: Re: The Empty Set > If you take a strip of the plane > [0,1] x [0,10] > with area 10 and make it a mobius strip tacking (0,0) thur (1,0) > to (1,10) thru (0,10) resp., is itÕs area 10 or 20? > Try painting the Moebius strip and get back to us. > I twisted my slipped off the Moebius strip spilling the paint Right. ThatÕs my point. It takes more than twice as much paint. === Subject: Re: The Empty Set >> Over stocked, must sell at once for next to nothing prices, >> 5,000 unsold empty sets produced for Buy Nothing Day. ;-) >> You can buy a real Klein bottle right here ... >> http://www.kleinbottle.com/ > If you take a strip of the plane > [0,1] x [0,10] > with area 10 and make it a mobius strip tacking (0,0) thur (1,0) > to (1,10) thru (0,10) resp., is itÕs area 10 or 20? When you start with the strip, the surface area is 20, 10 on the front and 10 on the back. When yomake it a mobius strip, the surface area is still 20. === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) >>You can buy a real Klein bottle right here ... >>http://www.kleinbottle.com/ > If you take a strip of the plane > [0,1] x [0,10] > with area 10 and make it a mobius strip tacking (0,0) thur (1,0) > to (1,10) thru (0,10) resp., is itÕs area 10 or 20? Yes! -- Oppie the Bear aka John ÔRemoveÕ MYWORRIES to email me! (My mother never saw the irony of calling me a son-of-a-bitch) === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) criminal offence under the Digital Millenium Copyright Act (IANAL). There has only ever been one actual Empty Set produced, so these alleged ones being sold must be forgeries. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: The Empty Set alt.algebra.help === Subject: Re: The Empty Set >Over stocked, must sell at once for next to nothing prices, >5,000 unsold empty sets produced for Buy Nothing Day. ;-) > a criminal offence under the Digital Millenium Copyright Act > (IANAL). There has only ever been one actual Empty Set produced, so > these alleged ones being sold must be forgeries. They are all one of a kind artistic replicas of the original masterpiece. ---- === Subject: Re: The Empty Set > They are all one of a kind artistic replicas of the original masterpiece. ThatÕs a good point. How do we know that weÕd be getting an AUTHENTIC empty set and not a reproduction? -- Oppie the Bear aka John ÔRemoveÕ MYWORRIES to email me! Nom de Dieu de putain de bordel de merde de saloperie de connard dÕencul.8e de ta m.8fre! Now bring me my silk! === Subject: Re: The Empty Set > They are all one of a kind artistic replicas of the original masterpiece. > ThatÕs a good point. How do we know that weÕd be getting an AUTHENTIC > empty set and not a reproduction? You canÕt have the original masterpiece, itÕs being used by most everybody. If you donÕt want to be like one of them, using such a common set, then get one of these replicas, new and unused in mint condition as first wrought by the artisan. === Subject: Re: The Empty Set >> They are all one of a kind artistic replicas of the original masterpiece. >> ThatÕs a good point. How do we know that weÕd be getting an AUTHENTIC >> empty set and not a reproduction? > You canÕt have the original masterpiece, itÕs being used by most > everybody. If you donÕt want to be like one of them, using such a common > set, then get one of these replicas, new and unused in mint condition as > first wrought by the artisan. If I have two empty sets and accidentally push them together, will I lose half my investment? If so, they should come with warning labels. Bill === Subject: Re: The Empty Set > If I have two empty sets and accidentally push them together, will I > lose half my investment? If so, they should come with warning labels. Yes and donÕt put anything in them not even warning labels. Perhaps for the benefit of the unaware youÕd prepare warning advisoriers for empty sets. Below are efforts made by others for warnings on difficult items. Possibility their examples may assist your efforts. ---- Warning Labels Mandated by 20th Century Physics As safety experts and concerned citizens, we applaud the recent trend towards legislation that requires the prominent placing of warnings on products that present hazards to the general public. Yet we must also offer the cautionary thought that such warnings, however well-intentioned, merely scratch the surface of what is really necessary in this important area. This is especially true in light of the findings of 20th century physics. We are therefore proposing that, as responsible professionals and science enthusiasts, we join together in an intensive push for new laws that will mandate the conspicuous placement of suitably informative warnings on the packaging of every product in every category offered for sale. Our suggested list of required warnings follows. -- WARNING This Product Warps Space and Time in Its Vicinity. -- WARNING This Product Attracts Every Other Piece of Matter in the Universe, Including the Products of Other Manufacturers, with a Force Proportional to the Product of the Masses and Inversely Proportional to the Distance Between Them. -- CAUTION The Mass of This Product Contains the Energy Equivalent of 85 Million Tons of TNT per Net Ounce of Weight. -- HANDLE WITH EXTREME CARE Velocities in Excess of Five Hundred Million Miles per Hour. -- CONSUMER NOTICE Because of the Uncertainty Principle, It Is Impossible for the Consumer to Find Out at the Same Time Both Precisely Where This Product Is and How Fast It Is Moving. -- ADVISORY There is an Extremely Small but Nonzero Chance That, Through a Process Known as Tunneling, This Product May Spontaneously Disappear from Its Present Location and Reappear at Any Random Place in the Universe, Including Your NeighborÕs Domicile. The Manufacturer Will Not Be Responsible for Any Damages or Inconvenience That May Result. -- READ THIS BEFORE OPENING PACKAGE According to Certain Suggested Versions of a Grand Unified Theory, the Within the Next Four Hundred Million Years. -- THIS IS A 100% MATTER PRODUCT In the Unlikely Event That This Merchandise Should Contact Antimatter in Any Form, a Catastrophic Explosion Will Result. -- PUBLIC NOTICE AS REQUIRED BY LAW Any Use of This Product, In Any Manner Whatsoever, Will Increase the Amount of Disorder in the Universe. Although No Liability is Implied Herein, the Consumer Is Warned That This Process Will Ultimately Lead to the Heat Death of the Universe. -- NOTE Gluing Force About Which Little Is Currently Known and Whose Adhesive Power Can Therefore Not Be Permanently Guaranteed. -- ATTENTION Despite Any Other Listing of Product Contents Found Hereon, the Consumer Is Advised That, in Actuality, This Product Consists of 99.9999999999% Empty Space. -- NEW GRAND UNIFIED THEORY DISCLAIMER The Manufacturer May Technically Be Entitled To Claim That This Product Is Ten-Dimensional. However, the Consumer Is Reminded That This Confers No Legal Rights Above and Beyond Those Applicable to Three-Dimensional Objects, Since the Seven New Dimensions Are Rolled Up Into Such a Small Area That They Cannot Be Detected. -- PLEASE NOTE Some Quantum Physics Theories Suggest That When the Consumer Is Not Directly Observing This Product, It May Cease to Exist or Will Exist Only in a Vague and Undetermined State. -- COMPONENT EQUIVALENCY NOTICE Are Exactly the Same in Every Measurable Respect as Those Used in the Products of Other Manufacturers, and No Claim to the Contrary May Legitimately Be Expressed or Implied. -- IMPORTANT NOTICE TO PURCHASERS The Entire Physical Universe, Including This Product, May One Day Collapse Back into an Infinitesimally Small Space. Should Another Universe Subsequently Reemerge, the Existence of This Product in That Universe Cannot Be Guaranteed. ---- HAZARDOUS MATERIALS INFORMATION SHEET MATERIALS SAFETY DATA SHEET; WOMEN - A CHEMICAL ANALYSIS ELEMENT Women SYMBOL Wo DISCOVERER Adam ATOMIC MASS Accepted at 53.6kg, but known to vary from 40-200kg OCCURRENCES Copious quantities in all urban areas PHYSICAL PROPERTIES 1. Surface usually covered in painted film. 2. Boils at nothing; freezes without known reason. 3. Melts if given special treatment. 4. Bitter if incorrectly used. 5. Found in various states from virgin metal to common ore. 6. Yields if pressure applied in correct places. CHEMICAL PROPERTIES 1. Has great affinity for gold, silver, and a range of precious stones. 2. Absorbs great quantities of expensive substances. 3. May explode spontaneously without prior warning and for no known reason. 4. Insoluble in liquids, but activity increases greatly by saturation in alcohol. 5. Most powerful money reducing agent known to man. COMMON USES 1. Highly ornamental, especially in sports cars. 2. Can be a great aid to relaxation. 3. Very effective cleaning agent. TESTS 1. Pure specimen turns rosy pink when discovered in the natural state 2. Turns green when placed beside a better specimen. HAZARDS 1. Highly dangerous except in experienced hands. 2. Illegal to possess more than one, although several can be maintained at different locations as long as specimens do not come into direct contact. ---- === Subject: Re: The Empty Set > alt.algebra.help === >Subject: Re: The Empty Set >>Over stocked, must sell at once for next to nothing prices, >>5,000 unsold empty sets produced for Buy Nothing Day. ;-) >> a criminal offence under the Digital Millenium Copyright Act >> (IANAL). There has only ever been one actual Empty Set produced, so >> these alleged ones being sold must be forgeries. >They are all one of a kind artistic replicas of the original masterpiece. Sounds like Franklin Mint. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Marginal Cost -- Suppose the total cost (in dollars) of manufacturing x units of a certain commodity is C(x) = 4(x^2) +20x + 400. At what level of production is the avg. cost per unit the smallest? At what level of production is the avg. cost per unit equal to the marginal cost? -- That being the question posed; I know that Marginal Cost is C(x+1) - c(x), or cÕ(x). However, IÕve a couple of questions: a) How do we find the smallest average cost? If I set CÕ(x) = 0, I get x = -20/8. Assuming that I could produce something for free, itÕs still not the correct answer. If I try to find cÕÕ (x) = 0, so that I can minimize c Ô (x) [Note: I honestly do now know what theyÕre asking when they ask for the average cost], I end up with a constant 8 (which is not the correct answer.) b) When asking to set cost per unit equal to marginal cost, are they essentially asking us to set CÕx to c(x+1) - c(x)? === Subject: Re: Marginal Cost >How do we find the smallest average cost? By writing an average cost function and minimizing it. Hint: If the cost function for x units is C(x) = 14x, what is the average cost per unit? How did you find it? Now, if the cost function is C(x) = 4(x^2) +20x + 400, what is the average cost per unit at the level of x units? Once you have the average cost function, you find its minimum the way you find a minimum of any function. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Marginal Cost > -- > Suppose the total cost (in dollars) of manufacturing x units of a certain > commodity is C(x) = 4(x^2) +20x + 400. > At what level of production is the avg. cost per unit the smallest? > At what level of production is the avg. cost per unit equal to the marginal > cost? > -- > That being the question posed; > I know that Marginal Cost is C(x+1) - c(x), or cÕ(x). These two are not the same (except if C is a linear function which here it isnÕt). Marginal cost is literally the cost of production of one additional unit, or C(x+1) - C(x). CÕ(x) is an often-used approximation to this, which is reasonably close provided that the function C is reasonably linear around x. This is because, by definition CÕ(x) = lim{h -> 0: [C(x+h) - C(x)]/h} and if you set h = 1 (reasonably close to zero), you get CÕ(x) ~ C(x+1) - C(x). This approximation is, in a general mathematical sense, nonsense. But it works well enough in many practical financial situations, so marginal cost is often calculated (or even defined) as CÕ(x). > However, IÕve a couple of questions: > a) How do we find the smallest average cost? If I set CÕ(x) = 0, I get x > = -20/8. Assuming that I could produce something for free, itÕs still not > the correct answer. If I try to find cÕÕ (x) = 0, so that I can minimize c Ô > (x) [Note: I honestly do now know what theyÕre asking when they ask for the > average cost], I end up with a constant 8 (which is not the correct > answer.) If 200 units cost $1,400, then what is the average cost per unit? If x units cost C(x) dollars then what is the average cost per unit? Find the expression for average cost per unit, and then find the value of x to minimise that. > b) When asking to set cost per unit equal to marginal cost, are they > essentially asking us to set CÕx to c(x+1) - c(x)? No. Take the expression for average cost per unit and set it equal to marginal cost, which will be either CÕ(x) or C(x+1) - C(x), depending on how they expect you to calculate marginal cost (I suspect the former). Then solve for x. === Subject: Re: Marginal Cost > Suppose the total cost (in dollars) of manufacturing x units of a certain > commodity is C(x) = 4(x^2) +20x + 400. > At what level of production is the avg. cost per unit the smallest? > At what level of production is the avg. cost per unit equal to the marginal > cost? > -- > That being the question posed; > I know that Marginal Cost is C(x+1) - c(x), or cÕ(x). Yes. *Marginal cost* is dC/dx. But thatÕs *not* what average cost is. Average cost is (not surprisingly), total cost divided by number of units produced. In other words: Cav(x) = C(x)/x Now re-work your questions (a) and (b). -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Marginal Cost >> Suppose the total cost (in dollars) of manufacturing x units of a certain >> commodity is C(x) = 4(x^2) +20x + 400. >> At what level of production is the avg. cost per unit the smallest? >> At what level of production is the avg. cost per unit equal to the >> marginal >> cost? >> -- >> That being the question posed; >> I know that Marginal Cost is C(x+1) - c(x), or cÕ(x). > Yes. *Marginal cost* is dC/dx. > But thatÕs *not* what average cost is. > Average cost is (not surprisingly), total cost divided > by number of units produced. In other words: > Cav(x) = C(x)/x > Now re-work your questions (a) and (b). === Subject: Graph Coloring Properties Hello everyone, Suppose there is a simple, planar graph that requires more than four colors to color. Among all such graphs, there is one with the fewest number of vertices. Let G be a triangulation of this graph. Then G also has a minimal number of vertices and by Theorem 1 requires more than four colors to color. Theorem 1. If a triangulation GÕ of a simple, plnar graph G can be colored with n colors, so can G. Given the above information: - Show that G cannot have a vertex of degree 3. - Show that G cannot have a vertex of degree 4. - Show that G has a vertex of degree 5. I really donÕt see any reason why G canÕt have at least one vertex of degree 3 or 4. What gives? === Subject: Method of Images and GreenÕs function. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATJQKQ04627; Using the method of images, construct the GreenÕs function of the Neumann problem for Laplaces equation in the half-space, D = {(x,y,z)is an element of (the real numbers)^3 : z > 0}. Hence solve (Delta^2)u=0, ((delta u)/(delta z))=1-(x^2)-(y^2), (x^2)+(y^2)<1, z=0 ((delta u)/(delta z))=0, (x^2)+(y^2)>1, z=0 and evaluate the resulting integral on the z-axis. I have worked this out myself but I am not sure whther I am in the right direction. === Subject: Re: Method of Images and GreenÕs function. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAUMgpA19887; >Using the method of images, construct the GreenÕs function of the >Neumann problem for Laplaces equation in the half-space, >D = {(x,y,z)is an element of (the real numbers)^3 : z > 0}. >Hence solve (Delta^2)u=0, > ((delta u)/(delta z))=1-(x^2)-(y^2), (x^2)+(y^2)<1, z=0 > ((delta u)/(delta z))=0, (x^2)+(y^2)>1, z=0 >and evaluate the resulting integral on the z-axis. >I have worked this out myself but I am not sure whther I am in the >right direction. am not sure whether I am right. === Subject: f(x) = 2x + 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATNAVG24579; Help, I need to draw this graph of f(x) for 0 <= x <= 2. And explain it. Then I need to graph f(g) = f(x+3) - 2 and draw for g(x) for -3<= x <= -1. And then I need to explain. === Subject: Re: f(x) = 2x + 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAUItta31733; >Help, I need to draw this graph of f(x) for 0 <= x <= 2. And explain >it. Then I need to graph f(g) = f(x+3) - 2 and draw for g(x) for -3<= >x <= -1. And then I need to explain. Have a look: f(g) = f(x+3) - 2 -left side from f(x) = 2x + 1 f(g(x))=2g(x)+ 1 , (1) -right side f(x +3)=2(x+ 3)+ 1, f(x+3)-2 = ... (2) (1) =(2) => g(x) Alain. === Subject: JSH: Final exam In arguing with me posters continually push that x=0 is a special case. Well, hereÕs yet another answer and I want you to pay careful attention to what the sci.mathÕers do now. Remember 49(300125x^3 - 18375 x^2 - 360 x + 22) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) where the aÕs are the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and I find out what the constant terms are by using x=0, but now IÕll consider x = 7. Then I have a^3 + 3(342)a^2 - 49(816361) = 0 which is irreducible over Q, for the aÕs and 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) where youÕll notice that 102040002 is coprime to 7. So 7 divides TWO and only TWO of the factors. How do you know? Because when x=0, two of the aÕs equal 0, thatÕs why. x = 0, is equivalent to x = 0 mod 7. Notice though that the cubic defining the aÕs is still irreducible over Q, and irreducibility has no impact. However, in the ring of algebraic integers NONE of the aÕs have 7 as a factor, and each has a non-unit factor in common with 7. You will get the same result with x = 7k, where k is an integer, including k=0, which is not a special case after all. IÕm curious to see if any of the sci.mathÕers wish to argue over this point, as, of course, I can add more detail. You see, the sci.math readership is stupid, and will believe just about anything if certain people say, especially if they argue with me. So those posters are used to saying stupid things which are mathematically incorrect. I find it interesting to see what they will try now. They are stupid, after all. So they will try. James Harris === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now IÕll > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the aÕs and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youÕll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. How do you know? > Because when x=0, two of the aÕs equal 0, thatÕs why. Actually, thatÕs not how you know. > x = 0, is equivalent to x = 0 mod 7. Nope. It turns out that it is an error on my part to claim that you can say anything about what happens with x = 0 mod 7 from x=0 with the aÕs. > Notice though that the cubic defining the aÕs is still irreducible > over Q, and irreducibility has no impact. > However, in the ring of algebraic integers NONE of the aÕs have 7 as a > factor, and each has a non-unit factor in common with 7. > You will get the same result with x = 7k, where k is an integer, > including k=0, which is not a special case after all. > IÕm curious to see if any of the sci.mathÕers wish to argue over this > point, as, of course, I can add more detail. Nope. CanÕt do that as what I had was wrong. > You see, the sci.math readership is stupid, and will believe just > about anything if certain people say, especially if they argue with > me. So those posters are used to saying stupid things which are > mathematically incorrect. I find it interesting to see what they will > try now. > They are stupid, after all. So they will try. I was really ticked off yesterday and probably at least a little angry this morning and I notice that I talked about how stupid the sci.math readership is several times. Oh well. In any event, what I have in this thread was just wrong, as you canÕt say anything just from x= 0 mod 7. Now I can admit when IÕm wrong, and IÕve noticed that when I get upset I tend to leap to conclusions even more so than normally! And I was just feeling really pissed off yesterday. James Harris === Subject: Re: JSH: Final exam > Now I can admit when IÕm wrong, and IÕve noticed that when I get upset > I tend to leap to conclusions even more so than normally! > And I was just feeling really pissed off yesterday. projecting that weakness on everyone else. -- 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: JSH: Final exam ... > x = 0, is equivalent to x = 0 mod 7. > Nope. It turns out that it is an error on my part to claim that you > can say anything about what happens with x = 0 mod 7 from x=0 with the > aÕs. > Here readers can get a good dose of pseudo-math from one the most > obnoxious posters on sci.math as this guy not only lies about basic > mathematics, as you can see in his post, but he copied from my Usenet > posts without my permission on to his own webpage, where he added in > negative commentary! of the five pages I have devoted to your attempts, only two are your you. I keep them in, because they clearly show your method of discussion, your proofs to show the lack of logic. Which also has not changed a bit. -- 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: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now IÕll > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the aÕs and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youÕll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. How do you know? You donÕt. Since the polynomial is irreducible, 7 divides either NONE a_1(7), a_2(7), or a_3(7), or ALL of them. Suppose A is a solution of a^3 + 3(342)a^2 - 49(816361) = 0 and A is divisible by 7 in the algebraic integers. That is, A = 7 * B, where B is an algebraic integer. Then B satisfies the equation 7 B^3 + 3(342) B^2 - 816361 = 0, which is irreducible and non-monic and primitive. None of the roots of this latter equation are algebraic integers. Therefore none of a_1(7), a_2(7), or a_3(7) are divisible by 7. However it is equally true that each of them share a nonunit factor with 7 in the ring of algebraic integers. > Because when x=0, two of the aÕs equal 0, thatÕs why. > x = 0, is equivalent to x = 0 mod 7. x = 0 is indeed a special case, because then the polynomial that the aÕs satisfy is reducible. That is not true when x = 7 as you note above. > Notice though that the cubic defining the aÕs is still irreducible > over Q, and irreducibility has no impact. Wrong. See above. > However, in the ring of algebraic integers NONE of the aÕs have 7 as a > factor, and each has a non-unit factor in common with 7. True, as I just proved. But if that is true, how can you conclude that exactly TWO of (5 a_1(7) + 7), (5 a_2(7) + 7), and (5 a_3(7) + 7) are divisible by 7 ? [Of course you have slyly neglected to specify in what ring your original claims about divisibility were being made. The natural assumption is that it is the ring of algebraic integers. However, from what you say above, you evidently realize that this must not be true. This leads me to conclude that since you have not specified the ring, your statement is essentially vacuous.] [Note too here that the statement you just made is tantamount to conceding what we have said from the beginning about the main conclusion of ÔAdvanced Polynomial FactorizationÕ: i.e., the main conclusion is false. And it only took you a year and a half to get it.] > You will get the same result with x = 7k, where k is an integer, > including k=0, which is not a special case after all. > IÕm curious to see if any of the sci.mathÕers wish to argue over this > point, as, of course, I can add more detail. Go for it. > You see, the sci.math readership is stupid, and will believe just > about anything if certain people say, especially if they argue with > me. So those posters are used to saying stupid things which are > mathematically incorrect. I find it interesting to see what they will > try now. > They are stupid, after all. So they will try. It may turn out that you are wrong, as it has very often in the past and even now, as you are finally admitting regarding the conclusion of ÔAPFÕ. In that case, who is stupid? Who passes the final exam ? Nora B. > James Harris === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now IÕll > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the aÕs and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youÕll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. No. In this case, NONE of the factors are divisible by 7 in the algebraic integers. If any of 5a_i + 7 were divisible by 7, then for that factor, weÕd have to have a_i divisible by 7 so then b_i = a_i/7 would have to be an algebraic integer. Notice that since the aÕs satisfy a^3 + 3*342*a^2 - 49*816361 = 0 then, rearranging and factoring 816361, a^3 = -3*342*a^2 + 49*7*13*8971 and dividing through by 7^3 we have (a^3)/(7^3) = -(3*342/7)((a^2)/(7^2)) + 13*8971 or, using b = a/7 we would have b^3 + (3*342/7)b^2 - 13*8971 = 0 so the bÕs all satisfy 7b^3 + 3*342b^2 - 7*13*8971 = 0 The left side is a non-monic irreducible polynomial, showing that b cannot be an algebraic integer, from which we conclude that none of the aÕs are divisible by 7. Rick === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. Yup, I will argue. > Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now IÕll > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the aÕs and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youÕll notice that 102040002 is coprime to 7. Yup. > So 7 divides TWO and only TWO of the factors. How do you know? The question is *why*? > Because when x=0, two of the aÕs equal 0, thatÕs why. Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible by 7? It would be true if the aÕs were polynomials. They are not. > x = 0, is equivalent to x = 0 mod 7. Using equivalence classes mod 7 is only useful when you are doing additions and multiplications. With other functions it just does not work. You are actually saying that sqrt(7) should be 0 mod 7. Which means that sqrt(7) is divisible by 7, which means that 1/7 is an element of the ring you are in, and so 7 is a unit in that ring. > Notice though that the cubic defining the aÕs is still irreducible > over Q, and irreducibility has no impact. Yup, and so the aÕs are not polynomials and that has a terrific impact. > However, in the ring of algebraic integers NONE of the aÕs have 7 as a > factor, and each has a non-unit factor in common with 7. And the non-unit factors multiply together to give a multiple of 49. So, what is the problem? -- 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: Final exam > > In arguing with me posters continually push that x=0 is a special > > case. > Yup, I will argue. Here readers can get a good dose of pseudo-math from one the most obnoxious posters on sci.math as this guy not only lies about basic mathematics, as you can see in his post, but he copied from my Usenet posts without my permission on to his own webpage, where he added in negative commentary! And refused to quit using my Usenet posts in violation of both the spirit and letter of international copyright law. And he lies about math. > > Well, hereÕs yet another answer and I want you to pay careful > > attention to what the sci.mathÕers do now. > > Remember > > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > > where the aÕs are the roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > and I find out what the constant terms are by using x=0, but now IÕll > > consider x = 7. > > Then I have > > a^3 + 3(342)a^2 - 49(816361) = 0 > > which is irreducible over Q, for the aÕs and > > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > > where youÕll notice that 102040002 is coprime to 7. > Yup. > > So 7 divides TWO and only TWO of the factors. How do you know? > The question is *why*? > > Because when x=0, two of the aÕs equal 0, thatÕs why. Notice that you have 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) with 49 on the left side, and it has to divide through in some way. For two of the aÕs you know they result from functions that equal 0, when x = 0 mod 7 as they equal 0, when x = 0 and you canÕt get any better than that! The third equals 3 when x = 0, so it is blocked from havinng factors in common with 7, when x has 7 itself as a factor, like when x = 7. ItÕs basic. To believe otherwise you need to challenge algebra. > Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible > by 7? It would be true if the aÕs were polynomials. They are not. Now notice, the poster canÕt deny the result when the aÕs are polynomials as then you can physically SEE how it works, so he handles that right off the bat by asserting that whether or not they are polynomials matters. Also he tossed in a weird assertion that doesnÕt follow as thereÕs no reason to suggest that sqrt(7) is divisible by 7. HeÕs just trying to throw up smoke as one other tactic is to confuse readers to the point that they just give up and trust that if someone is disagreeing with me, then I must be wrong. > > x = 0, is equivalent to x = 0 mod 7. > Using equivalence classes mod 7 is only useful when you are doing > additions and multiplications. With other functions it just does So now congruences only matter with additions and multiplications, according to the sci.mathÕer, so isnÕt it interesting what math this person is teaching you? But, I didnÕt necessarily phrase my own sentence well, as my point is that x=0 is equivalent to x = 0 mod 7, with respect to having factors of 7 so my using x = 0 mod 7, is a neat way to show that the result at x=0 is NOT a special case at all, as the constant terms work as IÕve explained repeatedly, while the claims of people arguing with me fall ßat. If this werenÕt such a HUGE issue with massive repercussions for mathematicians worldwide, then maybe these sci.math-ers might finally give up and admit the truth, but wait, IÕm giving them too much credit. It could be a minor issue with few repercussions and these people, from what IÕve seen over the years would still lie to you. Lying is in their nature. > not work. You are actually saying that sqrt(7) should be 0 mod 7. No IÕm not. > Which means that sqrt(7) is divisible by 7, which means that 1/7 > is an element of the ring you are in, and so 7 is a unit in that ring. Desperation. > > Notice though that the cubic defining the aÕs is still irreducible > > over Q, and irreducibility has no impact. > Yup, and so the aÕs are not polynomials and that has a terrific impact. This poster hardly even tries, but why should he? These TACTICS WORK on sci.math, as IÕve noted for years now. The sci.math readership is like some kind of religious cult, willing to believe just about anything as long as posters disagree with me. They are true believers on sci.math, and itÕs freaky. > > However, in the ring of algebraic integers NONE of the aÕs have 7 as a > > factor, and each has a non-unit factor in common with 7. > And the non-unit factors multiply together to give a multiple of 49. So, > what is the problem? Oh, hey, itÕs all ok if it works out in the end, right? If you want the full picture you need to read my original post, and my other reply to the other sci.math-er. The story here is truly sad as the problem I found *was* just a historical one, where math professors today could simply point out that they followed what they were taught. But now as the days turn into months and the months move toward years, thereÕs little reason to believe that such a huge result hasnÕt traveled through the math community, so what was a fascinating historical error, is now a case of modern fraud. Math professors who go out today to teach ßawed algebraic number theory to their students may be criminals and it may be possible to prosecute them under anti-fraud laws, but that sounds so removed from reality that they will probably continue without real fear. But, for those of you listening to them, the impact is very real. You are being taught false information, and being forced to do homework, and take tests to regurgitate that false information and God help you if you come back saying that itÕs false based on what you learned on newsgroups from a guy called James Harris. I feel sorry for you, but if you look around you can see how often society does things that inside you feel are wrong, but people just go on about their business as thatÕs what theyÕve learned is necessary. And many of you, unfortunately, will be like them, today, and the next day. So, in a great field, where truth is paramount, you will knowingly learn false things, and maybe even pride yourself on the grades you get from professors who are compromised. But youÕre living in a mirage... James Harris === Subject: Re: JSH: Final exam Your cubic polynomial in ÔaÕ has coefficients which are functions of ÔxÕ: Q(a,x) = a^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x) Therefore its roots are functions of ÔxÕ. Your polynomial in ÔxÕ is: P(x) = 14706125x^2 - 900375x^2 - 17640x + 1078 which you have factored in terms of the roots of Q(a,x): P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) You claim that since the roots of Q(a,x) evaluated at x = 0, are 0, 0, and 3 that P(0) = (0 + 7)(0 + 7)(5*3 + 7) = (7)(7)(22) = 1078 from which you somehow conclude that two of the terms in parentheses are divisible by 7 for any value of ÔxÕ. Or have I misunderstood? If that is your claim, it is surely false. We have Q(a,x) @ x=0 : a^3 - 3a^2 Q(a,x) @ x=1 : a^3 + 144a^2 - 110593 Q(a,x) @ x=2 : a^3 + 291a^2 - 912674 Q(a,x) @ x=3 : a^3 + 438a^2 - 3112137 The roots of Q(a,x) are a_1(x), a_2(x) and a_3(x), so a_1(0) = 0 a_2(0) = 0 a_3(0) = 3 a_1(1) = -31.32993946314472877... a_2(1) = -138.2104344584112440... a_3(1) = 25.54037392155592731... a_1(2) = -63.31229381879031244... a_2(2) = -279.3003544412207553... a_3(2) = 51.61264826001106776... a_1(3) = -31.65614690939515622... a_2(3) = -420.3902390936727750... a_3(3) = 77.68498899747098938... Hence, Ôx = 0Õ is indeed a Ôspecial caseÕ from which you argue to a false generalization. -- 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: JSH: Final exam ... > > So 7 divides TWO and only TWO of the factors. How do you know? > > The question is *why*? > > > Because when x=0, two of the aÕs equal 0, thatÕs why. > > Notice that you have > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > with 49 on the left side, and it has to divide through in some way. > For two of the aÕs you know they result from functions that equal 0, > when > x = 0 mod 7 as they equal 0, when x = 0 This is wrong. They are 0 when x = 0, they are not 0 when x != 0, whether x = 0 mod 7 or not. Note that the aÕs are the roots of: > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) for any a to be 0, 2401 x^3 - 147 x^2 + 3x must be 0. This is (with integer x) only the case when x = 0. Talking about false math... > The third equals 3 when x = 0, so it is blocked from havinng factors > in common with 7, when x has 7 itself as a factor, like when x = 7. This is also wrong. Because if x has 7 as a factor a3(x) does not necessarily have 7 as a factor. You are at least consistent in your errors. You assume that if 1. f(0) = 0 2. x = 0 mod 7 you can conclude that f(x) = 0 mod 7. That conclusion is false in general and holds only when f(x) is a polynomial in x. > ItÕs basic. To believe otherwise you need to challenge algebra. Yup, it is basic, and to believe otherwise you need to challenge algebra. > Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible > by 7? It would be true if the aÕs were polynomials. They are not. > Now notice, the poster canÕt deny the result when the aÕs are > polynomials as then you can physically SEE how it works, so he handles > that right off the bat by asserting that whether or not they are > polynomials matters. > Also he tossed in a weird assertion that doesnÕt follow as thereÕs no > reason to suggest that sqrt(7) is divisible by 7. Lets check your assertion above with sqrt(x) as function: 1. sqrt(0) = 0? yes. 2. 7 = 0 mod 7? yes. sqrt(7) = 0 mod 7? no. So your conclusion above is not a valid conclusion. It is your logic that suggests that sqrt(7) is divisible by 7. > > x = 0, is equivalent to x = 0 mod 7. > > Using equivalence classes mod 7 is only useful when you are doing > additions and multiplications. With other functions it just does > So now congruences only matter with additions and multiplications, > according to the sci.mathÕer, so isnÕt it interesting what math this > person is teaching you? What other operations do you think it works with? Not with the sqrt function. Also not with division. > x=0 is equivalent to x = 0 mod 7, with respect to having factors of 7 > so my using x = 0 mod 7, is a neat way to show that the result at x=0 > is NOT a special case at all, as the constant terms work as IÕve > explained repeatedly, while the claims of people arguing with me fall > ßat. Can you please explain why your inference work *must* work for your function, but does not work for sqrt? -- 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: Final exam Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. To be pedantic, there are other classes of function that it holds for. For example, it holds for all functions that are zero at all the integers. -- Richard === Subject: Re: JSH: Final exam >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. > To be pedantic, there are other classes of function that it holds for. > For example, it holds for all functions that are zero at all the > integers. > -- Richard Yup. However, I *did* use it improperly and my example in my original post is misleading. I was wrong. It happens. I was kind of upset yesterday, and got emotional. James Harris === Subject: Re: JSH: Final exam >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >>-- Richard > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. > It happens. I was kind of upset yesterday, and got emotional. You are always emotional, and continually making errors. Why donÕt you save yourself some embarrassment and go away and learn the subject before proclaiming your mastery of it? === Subject: Re: JSH: Final exam > >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >>-- Richard > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. It happens. I was kind of upset yesterday, and got emotional. > You are always emotional, and continually making errors. Why donÕt you > save yourself some embarrassment and go away and learn the subject > before proclaiming your mastery of it? That doesnÕt make sense. I have a mathematical argument, which is correct in every detail. Problems arise when I try to find other arguments to explain what IÕve already proven one way. Trying to prove in different ways helps my own understanding, and itÕs fun. Besides, IÕve discovered is a problem in what is commonly taught, so it hardly makes sense to try and learn from what is commonly taught! Mistakes happen. They are MUCH more likely when doing research that is breaking new ground, and when trying to understand something that hasnÕt been well-worked out before. ItÕs what happens when youÕre in the lead, and donÕt have some book or reference to tell you the answer but have to figure it out yourself. I am in the lead position. Making mistakes isnÕt a problem. ItÕs human. Not being able to admit it when you make mistakes IS a problem. James Harris === Subject: Re: JSH: Final exam >You assume that if >1. f(0) = 0 >2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >>-- Richard >Yup. However, I *did* use it improperly and my example in my original >post is misleading. I was wrong. >It happens. I was kind of upset yesterday, and got emotional. >>You are always emotional, and continually making errors. Why donÕt you >>save yourself some embarrassment and go away and learn the subject >>before proclaiming your mastery of it? > That doesnÕt make sense. I have a mathematical argument, which is > correct in every detail. Problems arise when I try to find other > arguments to explain what IÕve already proven one way. Actually, what happens is enough complications disappear that you can see the error in your simplified explanations. What never happens is the logical step of realizing that the error in your simple cases *also* exists in your general argument. If your argument were correct, your simpler examples would be correct to. You admit they are not. Draw the natural conclusion. > Trying to prove in different ways helps my own understanding, and itÕs > fun. But your method of attack is always the same. All you have shown is that your primary argument does not work, in general. > Besides, IÕve discovered is a problem in what is commonly taught, so > it hardly makes sense to try and learn from what is commonly taught! > Mistakes happen. They are MUCH more likely when doing research that > is breaking new ground, and when trying to understand something that > hasnÕt been well-worked out before. > ItÕs what happens when youÕre in the lead, and donÕt have some book or > reference to tell you the answer but have to figure it out yourself. The answers to a *lot* of your questions are in books. > I am in the lead position. > Making mistakes isnÕt a problem. ItÕs human. Not being able to admit > it when you make mistakes IS a problem. When you find that you are making mistakes at a high rate, it is worth while to at least lose some of the arrogance. Actually, arrogance pisses people off, regardless of whether you are right or not. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Final exam > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. But you say things like There is no error. Then you say I was wrong. Do you wonder why everyone doubts you? > It happens. I was kind of upset yesterday, and got emotional. But you often say that you are just following the math, and things like emotion and feelings do not enter it. Which is it with you? === Subject: Re: Final exam > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now IÕll > consider x = 7. Where do you get this silly arithmetic? What are you choosing this particular equation as your starting point. Does it have any significance? Where do you get this silly arithmetic? What are you choosing this particular equation as your starting point. Does it have any significance? Did it come to you in a Kekule style benzene dream? Did god give it to you? Is it special, or is it random? === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = When x = 0, the above is 1078, which is divisible by 11 When x = 1, the above is 13789188, which is not divisible by 11. Certainly there are at least *some* properties where x=0 leads to different results than x=1. Of course, x=1 could be the special case, but either way, the properties are not the same. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. Well, hereÕs yet another answer and I want you to pay careful > attention to what the sci.mathÕers do now. Remember 49(300125x^3 - 18375 x^2 - 360 x + 22) = > When x = 0, the above is 1078, which is divisible by 11 > When x = 1, the above is 13789188, which is not divisible by 11. > Certainly there are at least *some* properties where x=0 leads to > different results than x=1. > Of course, x=1 could be the special case, but either way, the properties > are not the same. Told you theyÕd fight. Consider also that the poster deleted out the relevant information, an old sci.math tactic, and now consider the facts 49(300125x^3 - 18375 x^2 - 360 x + 22) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) where the aÕs are the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and setting x=0, letÕs you see that two of the aÕs go to 0, at that point as you have 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = (5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7) and a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0)) = 0 so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the aÕs equal 0, while one equals, 3, at x=0. Since I have the correct mathematics I can endlessly explain it in different variations which is fun for me, but a headache for those who lie about my work as notice that I simply went to using x = 0 mod 7, and in my original post specifically used x = 7 to show that you get an irreducible over Q cubic for the aÕs, which shows that in the ring of algebraic integers NONE of the aÕs have 7 as a factor. But, from the result at x=0, you know that one of the aÕs cannot have non-unit factors in common with 7 for x=0 mod 7, but they do in the ring of algebraic integers. It turns out what are non-unit factors in the ring of algebraic integers are actually unit factors that are not units in the ring of algebraic integers because they are not roots of some monic polynomial with integer coefficients. You see that definition focusing on roots of monic polynomials with integer coefficients gives a ring that has these quirks. You can figure out the quirks, if you trust algebra and use logic, but if youÕre a ßawed human being desperate to have an easy tool to prove things, then you will fight it. Much of the supposedly great accomplishments in algebraic number theory over the last hundred years plus are simply vapor. Now to believe the sci.mathÕers, you need to start fiddling with your understanding of congruence, and IÕm sure some of them will come forward to lie to you about it. ThatÕs just the fun of the game I play on sci.math and this newsgroup. WhatÕs not a game is that some of you probably will go to class today and some math professor is going to confidently teach you mathematical information that is wrong. When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory (given what youÕve learned here) I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. James Harris === Subject: Re: JSH: Final exam > When you go to class today, if your professor talks about algebraic > number theory, or misuses Galois Theory (given what youÕve learned > here) I want you to carefully notice how you feel. If any students think about you in class at all (which I doubt), IÕm certain theyÕll be telling their professors something like, Wait till you hear what that idiot Harris said *this* time! -- Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you improvise fwbrown@bellsouth.net | if youÕre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Final exam >In arguing with me posters continually push that x=0 is a special >case. >Well, hereÕs yet another answer and I want you to pay careful >attention to what the sci.mathÕers do now. >Remember >49(300125x^3 - 18375 x^2 - 360 x + 22) = >>When x = 0, the above is 1078, which is divisible by 11 >>When x = 1, the above is 13789188, which is not divisible by 11. >>Certainly there are at least *some* properties where x=0 leads to >>different results than x=1. >>Of course, x=1 could be the special case, but either way, the properties >>are not the same. > Told you theyÕd fight. > Consider also that the poster deleted out the relevant information, an > old sci.math tactic, and now consider the facts > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the aÕs are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and setting x=0, letÕs you see that two of the aÕs go to 0, at that > point as you have > 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = > (5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7) > and > a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0)) = 0 > so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the aÕs equal 0, > while one equals, 3, at x=0. > Since I have the correct mathematics I can endlessly explain it in > different variations which is fun for me, but a headache for those who > lie about my work as notice that I simply went to using x = 0 mod 7, > and in my original post specifically used x = 7 to show that you get > an irreducible over Q cubic for the aÕs, which shows that in the ring > of algebraic integers NONE of the aÕs have 7 as a factor. > But, from the result at x=0, you know that one of the aÕs cannot have > non-unit factors in common with 7 for x=0 mod 7, but they do in the > ring of algebraic integers. And it doesnÕt occur to you that you are making an argument that 7 is in the same category of special cases as 0 for the purposes of dividing by 7? Many people have shown that your arguments are simply false when x=1. They didnÕt do anything fancy, just cranked out the numbers and showed they donÕt have the properties you claim they do. Did I change what was being talked about? Yes. Did you miss the point? Yes. The burden of proof is always on you to show that x=0 is *not* a special case. The best you have done is to argue that x=0 mod 7 has a special property. That is *still* a special case. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Cubic Polynomial by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAUCnPR28302; here is the problem: find a polynomial p(x) of degree 3 with real coefficients satisfying the condition f(-1)=5 and whose roots include 0 and 2-3i. here is what I did and I`m not sure if it`s correct or not! Since complex solutions come in pairs:2+3i is also a solution: Sum (2-3i)+(2+3i)=4 Product (2-3i)(2+3i)=13 Therefore xî-4x+13=0 so the equation is: x(xî-4x+13)=0 xñ-4xî+13x=0 that`s all I came up to, is it complete? thank you for helping me! === Subject: Re: Cubic Polynomial > here is the problem: > find a polynomial p(x) of degree 3 with real coefficients satisfying > the condition f(-1)=5 and whose roots include 0 and 2-3i. > here is what I did and I`m not sure if it`s correct or not! > Since complex solutions come in pairs:2+3i is also a solution: > Sum (2-3i)+(2+3i)=4 > Product (2-3i)(2+3i)=13 > Therefore xî-4x+13=0 > so the equation is: x(xî-4x+13)=0 > xñ-4xî+13x=0 > that`s all I came up to, is it complete? > thank you for helping me! How does the f(-1) = 5 part fit in? Bill === Subject: Re: Cubic Polynomial Updated by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Wo31355; >it`s p(-1)=5 and not f, my bad! >I`m still stuck any help! f(x) = a(x-0)[x-(2-3i)][x-(2+3i)] = a(x)(x-2+3i)(x-2-3i) = a(x)(xî-2x-3ix-2x+4+6i+3ix-6i-9iî) = a(x)(xî-4x+13) = a(xñ-4xî+13x) This problem gave another condition, f(-1) = 5. This will help us find in this problem. f(x) = a(xñ-4xî+13x) f(-1) = a[(-1)ñ-4(-1)î+13(-1)] = 5. a(-1-4-13) = 5 -18a = 5 a = - 5/18 Putting it all together we get: f(x) = -5/18(xñ-4xî+13x) = -5/18xñ+10/9xî-65/18x that`s it! what do you think about? === Subject: Re: Cubic Polynomial > here is the problem: > find a polynomial p(x) of degree 3 with real coefficients satisfying > the condition f(-1)=5 and whose roots include 0 and 2-3i. > here is what I did and I`m not sure if it`s correct or not! > Since complex solutions come in pairs:2+3i is also a solution: > Sum (2-3i)+(2+3i)=4 > Product (2-3i)(2+3i)=13 > Therefore xî-4x+13=0 > so the equation is: x(xî-4x+13)=0 > xñ-4xî+13x=0 > that`s all I came up to, is it complete? > thank you for helping me! > How does the f(-1) = 5 part fit in? > Bill Also remember you were asked to find a polynomial, not an equation... Mike. === Subject: Please help with Parametric Equation! If anyone could give me a hint as to how to start this problem, IÕd greatly appreciate it! Directions: Transform the parametric equation to a Cartesian equation by eliminating the parameter. x=2^t + 2^-t y=2^t - 2^-t Christina === Subject: Re: Please help with Parametric Equation! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB12Wgo08114; >If anyone could give me a hint as to how to start this problem, IÕd >greatly appreciate it! >Directions: Transform the parametric equation to a Cartesian >equation by eliminating the parameter. >x=2^t + 2^-t >y=2^t - 2^-t >Christina x + y = 2*2^t x - y = 2*2^(-t) (x + y)*(x - y) = 4 x^2 - y^2 = 4 (x/2)^2 - (y/2)^2 = 1 Normal hyperbola with: the main axes a = b = 2, asymptotes y = x and y = -x, focal points at [0, 2*sqrt(2)] and [0, -2*sqrt(2)]. Since x > 0 for any real t, the parametric equations are for the right branch of the hyperbola. === Subject: Re: Please help with Parametric Equation! > If anyone could give me a hint as to how to start this problem, IÕd greatly > appreciate it! > Directions: Transform the parametric equation to a Cartesian equation by > eliminating the parameter. > x=2^t + 2^-t > y=2^t - 2^-t > Christina Put p = 2^t and write the equations as x = p + 1/p . . . . (1) y = p - 1/p . . . . (2) Note that (1) is p^2 - xp + 1 = 0 . . (3) Solve (3) for p using the quadratic equation and substitute the result into (2). ThatÕs the obvious approach. One could also manipulate (1) and (2) directly to get (x + y)(x - y) = 4. But if you want y = ... youÕll still have to solve a quadratic. === Subject: Re: Please help with Parametric Equation! > If anyone could give me a hint as to how to start this problem, IÕd greatly > appreciate it! > Directions: Transform the parametric equation to a Cartesian equation by > eliminating the parameter. > x=2^t + 2^-t > y=2^t - 2^-t Add the equations to get a new equation. Subtract the equations to get a 2nd new equation. Then: (a) remember that u^(-v) = 1/(u^v) (b) look at both of those new equations really closely. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: JSH: Point of logic A mathematical proof must be logical. What I aim to do here is elaborate on the logic, which proves IÕm correct, and letÕs see what happens. What I have is a polynomial P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 which has 49 as a multiple, as P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). I factor the polynomial as P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) when the aÕs are the three roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). I determine the constant terms of the three factors by setting x=0, which reveals that for two of them the constant term is 7, while for one it is 22. That corresponds with the constant term of P(x) being 1078, which is 7(7)(22). Now I note that dividing both sides by 49 gives me P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, which factors as P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 and I note that the constant term is now 22. Now what do I know about the constant terms? Well they are constant, and are specific numbers, specifically 7, 7 and 22, for the factors of P(x). But for P(x)/49, the constant term is 22. So, logically, two of the constant terms were divided by 7. That is the essential point on which everything hinges. Notice how simple it is, the constant terms are actual numbers. Numbers like 7 and 22 are NOT variables, and they remain the same without regard to the value of x. Dividing P(x) by 49 changes the constant terms. They go from being 7, 7 and 22 to being 1, 1 and 22. Therefore, exactly two of them were divided by 7. There is no other way to get from 7 to 1, and you can write it out algebraically. 7/w = 1, giving w = 7. By the distributive property for the constant terms of two of (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) to be divided by 7, the full factor has to be divided by 7, which gives P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) indicating that two of the aÕs have 7 as a factor, and IÕve arbitrarily selected a_1(x) and a_2(x). What can be shown is that in the ring of algebraic integers, if a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) is irreducible over Q, with integer coefficients, then the aÕs cannot have 7 as a factor in the ring of algebraic integers. So is there a problem with the logic in my argument above? How can there be? I simply assert that given 7 to get to 1, you must divide 7 by 7, and that by the distributive property, a factor that has 7 as a constant term must itself by divided by 7. The logic is trivial. If thatÕs not true, the the entire logical framework of mathematics itself is crap. Notice, I depend on the distributive property and 7/w = 1, meaning w=7. So why all the arguing? Well, my work shows that some people didnÕt properly understand algebraic integers, and they built a lot of proofs on their ßawed understanding. They did and people long dead did as well. That means that a lot of prestige is at stake, and egos and other boring things that help make the world a very nasty place. So these people fight the logic and the conclusion in order to try and keep the status quo. They are trying to maintain homeostasis in their environment. Trouble is theyÕre living a lie, and continuing to teach it to fresh students who deserve to be taught true mathematical ideas. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. > So is there a problem with the logic in my argument above? I thought it might be worthwhile revisiting this Harris argument, but with more of the detail filled in, as Harris provided in a post in another thread on November 14. This is quoted below: so I divide out the 49 to get >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22. >Now assume each of the aÕs has some non-unit factor in common with 7, >for a given x, which in fact they do in the ring of algebraic integers >whenever all the aÕs are irrational, which is a point that has been >brought up repeatedly by people arguing with me. For a while I >resisted that fact, but now as IÕve said before I concede that they >are in fact correct--each of the aÕs, in the ring of algebraic >integers does in fact share a non-unit factor with 7 when all of the >aÕs are irrational. >So let w_1(x) w_2(x) w_3(x) = 49, where the wÕs are those factors, so >a_1(x) = w_1(x) b_1(x), >a_2(x) = w_2(x) b_2(x), >and >a_3(x) = w_3(x) b_3(x) >and where >w_1(x) v_1(x) = 7, w_2(x) v_2(x) = 7, and w_3(x) v_3(x) = 7, >and divide through by 49 to get >P(x)/49 = > (5 b_1(x) + v_1(x))(5 b_2(x) + v_2(x))(5 b_3(x) + v_3(x)) >and if you allow that the factors are each factors of the constant >term as before, then you have > v_1(0) v_2(0)(15 + v_3(0)) = 22 >at x=0, and when x does not equal 0, you have > v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 >where I introduce u_3(x) to handle any further weirdness with how >a_3(x) behaves, where u_3(0) = 3, to agree with previous results, and >remember >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >and the constant term doesnÕt change, as, well, itÕs constant, as itÕs >22. >So, notice, I now have that v_1(x) and v_2(x) are factors of 22. >Provably, they cannot be units in the ring of algebraic integers when >a_1(x), a_2(x), and a_3(x) are all irrational as thatÕs what people >argued with me about for so long. >The full result was just so weird that it escaped even me for a while, >as following their arguments to their logical conclusion 22 and 7 must >share non-unit factors in the ring of algebraic integers. >But you can also appear to prove that 22 and 7 are coprime in the ring >of algebraic integers using various accepted definitions of >coprimeness, like you can find algebraic integers x, and y such that >22x + 7y = 1 >and claim to have proven that they are coprime. >So, in the ring of algebraic integers, using whatÕs commonly accepted >you can prove that 22 and 7 are both coprime and that they share >non-unit algebraic integer factors. >ThatÕs the full result. Actually I thought this argument looked pretty good! Harris assumes that there exist algebraic integers w_1(x), w_2(x), and w_3(x) such that (1) w_1(x)*w_2(x)*w_3(x) = 49 (2) a_1(x) is divisible by w_1(x), etc. - that is, a_1(x)/w_1(x) = b_1(x), where b_1(x) is an algebraic integer, and similarly for a_2(x) and a_3(x). (3) 7 is divisible by w_1(x), w_2(x), and w_3(x),; that is, 7/w_1(x) = v_1(x), 7/w_2(x) = v_2(x), and 7/w_3(x) = v_3(x), where v_1(x), v_2(x), and v_3(x) are algebraic integers. (4) v_1(0) * v_2(0) * (5 b_3(0) + v_3(0)) = 22. All of this is correct! Harris then goes on to say, >... and when x does not equal 0, you have > v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 >where I introduce u_3(x) to handle any further weirdness with how >a_3(x) behaves, where u_3(0) = 3, to agree with previous results. And he then concludes that >So, notice, I now have that v_1(x) and v_2(x) are factors of 22. And this is the crucial statement. If Harris is right about this, then he has justified his claims. Is he? We need to go back to his assertion that v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 Is this correct, for x <> 0 ? v_1(x) * v_2(x) * v_3(x) = 7 * 7 * 7 / 49 = 7. Thus easily, v_1(x) v_2(x) (5 u_3(x) + v_3(x)) = 7 + 5 u_3(x) v_1(x) v_2(x). Now, this will equal 22 only if u_3(x) * v_1(x) * v_2(x) = 3. This is known to be true when x = 0, because u_3(0) = 3 and v_1(0) = 1 and v_2(0) = 1. But unless you assume what you want to prove - that is, that v_1(x) and v_2(x) are constant functions, and in addition that u_3(x) is a constant function, there is no reason to conclude that u_3(x) * v_1(x) * v_2(x) = 3 for values of x other than 0. HarrisÕs statemtent that v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 is simply not justified. In fact it is false for x > 0. Again, this is where Harris abandons his own definition of constant term, which yields the correct statement v_1(0) v_2(0)(5u_3(0) + v_3(0)) = 22 and he assumes that the constant term is defined by its position in an expression rather than by a specific value. ------------------------------------------------------------- ------- Related to this: since w_1(x)*w_2(x)*w_3(x) = 49, it IS true that (7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22, for any x. This looks a great deal like exactly what Harris wants! Better yet, unlike what he claimed above, it is actually correct !!! But does it imply what he wants, i.e., does it imply that v_1(x) and v_2(x) are divisors of 22 ? Noting that v_1(x) = 7/w_1(x), v_2(x) = 7/w_2(x), and v_3(x) = 7/w_3(x), the equation above implies that [***] v_1(x) * v_2(x) * v_3(x) * 22/7 = 22, which also follows trivially from the equation noted previously, v_1(x) * v_2(x) * v_3(x) = 7. Because 22/7 is not an algebraic integer, equation [***] does NOT tell you that v_1(x) and v_2(x) are divisors of 22 in the algebraic integers. Clearly from the equation just quoted, v_1(x), v_2(x), and v_3(x) are all divisors of 7. That is not surprising or informative, and it does not lead to the conclusion that Harris wants. ------------------------------------------------------------- ---------- v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 is the following: If you are considering a product of functions of a variable x, then the product of the constant terms is the constant term of the product. And this is a true statement. However, it cannot be assumed, for example, that v_1(x) is the constant term of (5 b_1(x) + v_1(x)). By HarrisÕs own definition, the constant term of this expression is 5 b_1(0) + v_1(0) = 5 * 0 + v_1(0) = v_1(0) = 1. Assuming that v_1(x) = 1 for x nonzero is just assuming what you (or Harris, actually) want to prove. The correct value for the constant term is simply v_1(0), i.e., neither more nor less than what HarrisÕs definition of constant term specifies. ------------------------------------------------------------- -------- Another cut at the Harris reasoning. Harris notes correctly that P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22), where one may assume that a_1(x), a_2(x), and a_3(x) are functions that take their values in the ring of algebraic integers, and a_1(0) = 0, a_2(0) = 0, a_3(0) = 3, and b_3(0) = 0. [note that b_3(x) is defined to equal a_3(x) - 3 ]. So he has set up the factorization so that the constant terms are 7, 7, and 22, and of course P(0) = 49 * 22. Then Harris thinks: I must divide both sides by 49. The constant term of P(x)/49 is 22. The only way I can think of to divide 49 out of the three factors (which have constant terms 7, 7, and 22) is to divide 7 out of the first factor and second factor (giving constant terms of 1 in both cases), and divide the third term by 1 (giving constant term 22). The problem with that is, a_1(x) and a_2(x) are not divisible by 7. What you NEED to do is divide each of the three factors by numbers which DO divide a_1(x), a_2(x), and a_3(x), and also are divisors of 7. This *can* be done. There are three divisors which I will call w_1(x), w_2(x), and w_3(x), exactly as Harris describes above, and their product happens to be 49. They are functions of x. They HAVE to be, because when x = 0, a_1(x) and a_2(x) are divisible by 7, but when x > 0 , they are not. The result, however, is that 7/w_1(x) is NOT the constant term of (a_(x) + 7)/w_1(x). There is no reason it should be. The key thing here is that you divide both sides by 49 in such a way that all the numbers in sight are algebraic integers. Harris at this point has lost sight of what he wanted to do. He wanted to factor 49 out of both sides so that the factors are all algebraic integers. He has convinced himself that one of the objectives of the factoring was to preserve the constant terms. The key thing here is, you cannot have both. If you want to preserve the constant terms, then you will not end up with algebraic integers on both sides. If you want algebraic integers everywhere, then the constant terms cannot be preserved. Of these two goals, the essential one was the original one: to end up with algebraic integers on both sides after 49 is factored out. Preserving the constant terms amounts to nothing more than requiring that the functions w_1(x), etc., are constant, and there is no reason to do this. Nora B. === Subject: Re: Point of logic I still have a few questions... > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. If the aÕs are not polynomials, then how do you define constant term? Presumably you mean the value of each a_*(x) at the point x=0. In other words, the intercept of the functions a_*(x). Is that correct? > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. Well, you can certainly express each a_*(x) as a_*(x) = b_*(x) + const, such that b_*(0)=0. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). IÕm a bit confused here. When you talk about having 7 as a factor, this is usually applied to an integer value, i.e. (49 has 7 as a factor, etc). Are you saying that a_*(x) is integer valued for every x? It seems like you have only shown this for the case x=0? Maybe you can elaborate here. > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. So, are you setting x to be an integer value in order to get integer coefficients? For which x it is irreducible? Darren === Subject: Re: Point of logic > I still have a few questions... >>A mathematical proof must be logical. What I aim to do here is >>elaborate on the logic, which proves IÕm correct, and letÕs see what >>happens. >>What I have is a polynomial >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>which has 49 as a multiple, as >>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >>I factor the polynomial as >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >>when the aÕs are the three roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >>I determine the constant terms of the three factors by setting x=0, >>which reveals that for two of them the constant term is 7, while for >>one it is 22. > If the aÕs are not polynomials, then how do you define constant term? > Presumably you mean the value of each a_*(x) at the point x=0. In other > words, the intercept of the functions a_*(x). > Is that correct? >>That corresponds with the constant term of P(x) being 1078, which is >>7(7)(22). >>Now I note that dividing both sides by 49 gives me >>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, >>which factors as >>P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 >>and I note that the constant term is now 22. >>Now what do I know about the constant terms? >>Well they are constant, and are specific numbers, specifically 7, 7 >>and 22, for the factors of P(x). But for P(x)/49, the constant term >>is 22. >>So, logically, two of the constant terms were divided by 7. >>That is the essential point on which everything hinges. >>Notice how simple it is, the constant terms are actual numbers. >>Numbers like 7 and 22 are NOT variables, and they remain the same >>without regard to the value of x. > Well, you can certainly express each a_*(x) as a_*(x) = b_*(x) + const, such > that b_*(0)=0. >>Dividing P(x) by 49 changes the constant terms. >>They go from being 7, 7 and 22 to being 1, 1 and 22. >>Therefore, exactly two of them were divided by 7. >>There is no other way to get from 7 to 1, and you can write it out >>algebraically. >>7/w = 1, giving w = 7. >>By the distributive property for the constant terms of two of >>(5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) >>to be divided by 7, the full factor has to be divided by 7, which >>gives >>P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) >>indicating that two of the aÕs have 7 as a factor, and IÕve >>arbitrarily selected a_1(x) and a_2(x). > IÕm a bit confused here. When you talk about having 7 as a factor, this is > usually applied to an integer value, i.e. (49 has 7 as a factor, etc). > Are you saying that a_*(x) is integer valued for every x? It seems like you > have only shown this for the case x=0? Maybe you can elaborate here. For any ring R, and elements b and c in R, b is a factor of c if there is an element d in R such that b*d=c. Most of the discussion that James works with occurs in the algebraic integers (roots of monic polynomials with integer coefficients). However, division by 7 is actually defined as multiplying by 1/7, which is not an algebraic integer. So he is working in some larger ring (possibly the algebraic numbers) when does the multiplication by 1/7 and claiming that the result is again in the algebraic integers. How he draws this conclusion is not entirely clear to me. Then again, IÕm not sure he is fully aware of all of the above. >>What can be shown is that in the ring of algebraic integers, if >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) >>is irreducible over Q, with integer coefficients, then the aÕs cannot >>have 7 as a factor in the ring of algebraic integers. > So, are you setting x to be an integer value in order to get integer > coefficients? For which x it is irreducible? That is the key question he has never examined. For x=0, it clearly is reducible. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Point of logic can someone help me out here? I was trying to follow this at home and got stuck out of the gate :( > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). Is there anything special about the polynomial other than it has 49 as a factor. Does it have to be a square factor? > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) Err, where did the 5 and 7 come from? How do we know that the a_*(x) such functions exist, and what can we infer about them - are they polynomials? > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Where did this come from? > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. Err, I didnÕt follow this. Setting x=0 in the above equation yields a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Darren === Subject: Re: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? For each x, let a_1(x), a_2(x), and a_3(x) be the three roots (counting multiplicities) of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Then we have a_1(x) + a_2(x) + a_3(x) = 3(1 - 49x), a_1(x)a_2(x) + a_1(x)a_3(x) + a_2(x)a_3(x) = 0, a_1(x)a_2(x)a_3(x)=49(2401 x^3 - 147 x^2 + 3x), and itÕs not hard to check that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) does indeed equal JamesÕ polynomial. > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Yes, so if you let a_1(0)=0, a_2(0)=0, a_3(0)=3, then 5a_1(0)+7=7, 5a_2(0)+7=7, 5a_3(0)+7=22. These are the constant terms of 5a_1(x)+7, 5a_2(x)+7, 5a_3(x)+7. By the constant term of a function James just means its value when x=0. > Darren === Subject: Re: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? No, it does not have to be a square factor. The polynomial is special in that it can be expanded out so that I can factor it into non-polynomial factors. > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? The aÕs are not polynomials. The factorization comes from P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1+49 x )(5)(7^2) + 7^3 where if you multiply everything out and simplify you will just get P(x) = 14706125 x3 - 900375 x2 - 17640 x + 1078 which you saw above, and thatÕs part of why itÕs special. > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1+49 x )(5)(7^2) + 7^3 with respect to 7 and 5, that is, factoring it as if 7 and 5 were the polynomial variables. ItÕs a neat analysis tool which I discovered from which everything follows. IÕll stick in some variables in key places to show: P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (Y^3) - 3(-1+49 x )(Y)(Z^2) + Z^3 where IÕve used Y and Z. > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Yes, where each root gives one of the aÕs, so if you try that with P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) youÕll notice that it gives the same answer as P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 as they are exactly equal. > Darren YouÕre welcome. James Harris === Subject: Re: Point of logic > ... each root gives one of the aÕs, so if you try that with > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > youÕll notice that it gives the same answer as > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > as they are exactly equal. Yes, and youÕll also notice that P(x) = (245x + 15.013...)(245x - 8.480... )(245x + 8.467...) where the unterminated numbers are 245 times the roots of P(x) = 0. Where are your Ô7Õs and Ô22Õ now? You seem to be unaware that there an unlimited number of ways to express the factors of a polynomial. -- 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: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( >>What I have is a polynomial >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>which has 49 as a multiple, as >>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? The polynomial is related to one that James invented a few years ago in his attempt to find a proof of FermatÕs Last Theorem involving little more than high-school algebra. As an historical aside, James has been beating this particular horse (FLT) for about nine years in this newsgroup. If youÕre *really* interested in the details, check Google groups for JamesÕs posting history. In answer to your second question, part of JamesÕs proof involved showing that in a factorization like the one below, exactly two of the factors were divisible by a prime, so in this context he needs a polynomial that is divisible by the square of a prime. >>I factor the polynomial as >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? The 5 and 7 were more or less picked out of the air. All thatÕs important in terms of JamesÕs magnum opus is that they be primes unequal to 3. Below IÕll show how to get the a_iÕs and show that for any ordinary integer x, a_i(x) will be an algebraic integer. In general the a_iÕs are most emphatically not polynomials. >>when the aÕs are the three roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? In a nutshell, this comes from rewriting P(x) in a form that allows him to perform a somewhat nonstandard factorization. His original polynomial may be written as P(x) = 49(5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x + 22) First, group the terms containing 5^3: P(x) = 49(5^3(7^4*x^3 - 3*7^2*x^2) - 2^3*3^2*5*x + 22) Now add and subtract 5^3*3*x and tidy up: P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 375x - 360x + 22) = 49(5^3(2401x^3 - 147x^2 + 3x) - 735x + 22) = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5*49*x + 22) (HeÕs aiming for a term -1 + 49x here, since that will closely match the simiilar term in his FLT polynomial) Now add and subtract again, this time using 3*5: P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5*49*x + 15 - 15 + 22) and again tidy up: P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5(-1+49*x) + 7) (He probably did these steps in reverse, adjusting things so that he would have a 7 as the last term.) Now write P in another way (which again comes from his FLT proof) P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) and expand the right hand side as (5^3)(a_1 a_2 a_3) + (5^2)(7)(a_1 a_2 + a_1 a_3 + a_2 a_3) + (5)(7^2)(a_1 + a_2 + a_3) + 7^3 Now comes the nonstandard part--in the two expressions for P, equate terms with equal powers of 5 so we have a_1 a_2 a_3 = 49(2401x^3 - 147x^2 + 3x) a_1 a_2 + a_1 a_3 + a_2 a_3 = 0 a_1 + a_2 + a_3 = -3(-1 + 49x) (IÕve supressed the x arguments, but bear in mind that the a_i should be considered to be functions of x.) Now letÕs get a polynomial for which the a_iÕs will be roots Clearly one such is (a - a_1)(a - a_2)(a - a_3) which expands as a^3 - (a_1 + a_2 + a_3)a^2 + (a_1 a_2 + a_1 a_3 + a_2 a_3)a - (a_1 a_2 a_3) Using the values we obtained above, we can then conclude that for any integer x, the a_iÕs will satisfy a^3 + (-1 + 49x)a^2 + 49(2401x^3 - 147x^2 + 3x) = 0 and hence, by the way, will be algebraic integers for any integer x. You can try it yourself, using any old primes (except for 3) in place of the 5 and the 7. >>I determine the constant terms of the three factors by setting x=0, >>which reveals that for two of them the constant term is 7, while for >>one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Sure. a^3 - 3a^2 = (a)(a)(a-3) = 0 so two solutions are 0 and the third is 3. Plug these into (5a_1 + 7)(5a_2 + 7)(5a_3 + 7) to get (5*0 + 7)(5*0 + 7)(5*3 + 7) = (7)(7)(22). I apologize for the length of this, but couldnÕt figure out a shorter way of explaining things. Hope it helps. Rick === Subject: Re: Point of logic I was tearing my hair out trying to see how he got the cubic equation :) Incidently, is equating terms in powers of 5 a mathematically valid thing to do? Darren > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( >>What I have is a polynomial >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>which has 49 as a multiple, as >>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? > The polynomial is related to one that James invented a few years ago > in his attempt to find a proof of FermatÕs Last Theorem involving > little more than high-school algebra. As an historical aside, James > has been beating this particular horse (FLT) for about nine years > in this newsgroup. If youÕre *really* interested in the details, > check Google groups for JamesÕs posting history. > In answer to your second question, part of JamesÕs proof involved > showing that in a factorization like the one below, exactly two > of the factors were divisible by a prime, so in this context he > needs a polynomial that is divisible by the square of a prime. >>I factor the polynomial as >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? > The 5 and 7 were more or less picked out of the air. All thatÕs > important in terms of JamesÕs magnum opus is that they be primes > unequal to 3. Below IÕll show how to get the a_iÕs and show that > for any ordinary integer x, a_i(x) will be an algebraic integer. > In general the a_iÕs are most emphatically not polynomials. >>when the aÕs are the three roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? > In a nutshell, this comes from rewriting P(x) in a form that > allows him to perform a somewhat nonstandard factorization. > His original polynomial may be written as > P(x) = 49(5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x + 22) > First, group the terms containing 5^3: > P(x) = 49(5^3(7^4*x^3 - 3*7^2*x^2) - 2^3*3^2*5*x + 22) > Now add and subtract 5^3*3*x and tidy up: > P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 375x - 360x + 22) > = 49(5^3(2401x^3 - 147x^2 + 3x) - 735x + 22) > = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5*49*x + 22) > (HeÕs aiming for a term -1 + 49x here, since that will closely > match the simiilar term in his FLT polynomial) > Now add and subtract again, this time using 3*5: > P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5*49*x + 15 - 15 + 22) > and again tidy up: > P(x) = 49(5^3(2401x^3 - 147x^2 + 3x) - 3*5(-1+49*x) + 7) > (He probably did these steps in reverse, adjusting things so > that he would have a 7 as the last term.) > Now write P in another way (which again comes from his FLT proof) > P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > and expand the right hand side as > (5^3)(a_1 a_2 a_3) + (5^2)(7)(a_1 a_2 + a_1 a_3 + a_2 a_3) > + (5)(7^2)(a_1 + a_2 + a_3) > + 7^3 > Now comes the nonstandard part--in the two expressions for P, equate > terms with equal powers of 5 so we have > a_1 a_2 a_3 = 49(2401x^3 - 147x^2 + 3x) > a_1 a_2 + a_1 a_3 + a_2 a_3 = 0 > a_1 + a_2 + a_3 = -3(-1 + 49x) > (IÕve supressed the x arguments, but bear in mind that the a_i > should be considered to be functions of x.) > Now letÕs get a polynomial for which the a_iÕs will be roots > Clearly one such is (a - a_1)(a - a_2)(a - a_3) which expands as > a^3 - (a_1 + a_2 + a_3)a^2 > + (a_1 a_2 + a_1 a_3 + a_2 a_3)a > - (a_1 a_2 a_3) > Using the values we obtained above, we can then conclude that > for any integer x, the a_iÕs will satisfy > a^3 + (-1 + 49x)a^2 + 49(2401x^3 - 147x^2 + 3x) = 0 > and hence, by the way, will be algebraic integers for any integer x. > You can try it yourself, using any old primes (except for 3) in > place of the 5 and the 7. >>I determine the constant terms of the three factors by setting x=0, >>which reveals that for two of them the constant term is 7, while for >>one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? > Sure. a^3 - 3a^2 = (a)(a)(a-3) = 0 so two solutions are 0 and the > third is 3. Plug these into (5a_1 + 7)(5a_2 + 7)(5a_3 + 7) to get > (5*0 + 7)(5*0 + 7)(5*3 + 7) = (7)(7)(22). > I apologize for the length of this, but couldnÕt figure out a shorter > way of explaining things. Hope it helps. > Rick === Subject: Re: Point of logic > I was tearing my hair out trying to see how he got the cubic equation :) > Incidently, is equating terms in powers of 5 a mathematically valid thing to > do? Yes. Basically, it would be perfectly valid if one replaced the number 5 with the indeterminate y--all James has done is then formally set y equal to 5. This is why it looks so strange, but thereÕs nothing really wrong with his approach. As Arturo said above, [d]epends on what you do with it. If you were then to deduce a property that depends on what we might call the fiveness of that implicit y, all bets would be off. Rick === Subject: Re: Point of logic days. My association with the Department is that of an alumnus. >I was tearing my hair out trying to see how he got the cubic equation :) >Incidently, is equating terms in powers of 5 a mathematically valid thing to >do? Depends on what you do with it. In the original developement, the roles of 5 and 7 were played by variables, so it was a matter of considering the expression as a polynomial in one of those variables rather than as a polynomial in x. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( James has been working on this as part of a larger project for a while (measured in years). This is a particular example from a far more complicated polynomial in 4 variables. What he has done here is select values for three of them to get a polynomial in x. His idea is to try to support his general conclusion with this example. >>What I have is a polynomial >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>which has 49 as a multiple, as >>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? I believe originally it was t^2 as a factor, so yes. >>I factor the polynomial as >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? The 5 and 7 come from the variables mentioned earlier. The a_i(x) do exist since cubics have closed form solutions. The explicit functions >>when the aÕs are the three roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? Again, prior work that has not been properly referred to. >>I determine the constant terms of the three factors by setting x=0, >>which reveals that for two of them the constant term is 7, while for >>one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? And 5*3+7 = 22, 5*0+7 = 7 Whether it is appropriate to call them constant terms in the sense that we use the phrase when referring to contant terms of polynomials is certainly open to question. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Point of logic >> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? He never answers this type of question. He does not like to provide such details. I thought that this polynomial was a bit contrived and odd too but he seems to think that it is an example that shows some fundamental problem with algebraic integers. He never provides definitions or proofs and he never uses standard mathematical definitions or notations. He implies that he works in one ring, and then squirms out of a debate by retroactively changing the ring he claims he was working in. He seems to like high-school algebra a lot, but seems incapable of understanding anything in abstract algebra. He claims that his work has huge implications in various branches of abstract algebra. But he is never precise in explaining what specific aspcts of abstract algebra are effected by his findings. He claims that Galois theory is destroyed by his results. But he never shows a disproof of any specific theorem in Galois theory. He claims that WilesÕ FLT proof is destroyed by his results. But he also has never pin-pointed in the FLT proof where wiles makes an error. He may have a hatred of Wiles, due to the fact that for years, James has been trying to convince the world that he has a simple FLT proof. Of course, his proof is entirely high school algebra, and is riddled with algebraic errors and invalid logic. And he has experienced an enormous amount of ridicule from almost everyone. He makes lots of silly errors and draws invalid generalizations from special cases, and never seems to understand when his mistakes are pointed out to him. His response to a challenge is no ignore the point, and start a new thread on sci.math repeating the same old crap. At times he has delusions of grandeur, when he thinks that nobody on sci.math understands mathematics as well as he does. In that state, he sees us as playthings, and in time, he believes that his mathematical discoveries will shake the world. At other times, he is contrite, and admits some errors. And at other times he is defiant and claims that we are all out to get him, by censoring his work, in a puddle of self-pity. He gets caught up in futile arguments that have nothing to do with math, like the gender of Nora Baron, as if that mattered to anyone! And then at other times he claims that he is only interested in the math, and emotion and ego have no place in math. It is very sad, James is very sick indeed. === Subject: Re: JSH: Point of logic > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. ItÕs kind of strange to call f(0) the constant term of f when f isnÕt a polynomial. ItÕs called the constant term when f is a polynomial, because f is a sum of finitely many monomial terms, and the constant term is the one of degree 0. Your non-polynomial factors of P(x) have no such expressions. Furthermore, the word constant is really confusing you here. I suggest you stop using it. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). Why should any of the aÕs have 7 as a factor? ThereÕs no reason to assume this division stays inside the ring of algebraic integers for all relevant values of x. > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. If this is true, then it just proves the above division leaves the ring of algebraic integers for some x. ThereÕs nothing that interesting about the fact that not everything in this ring is divisible by 7. > I simply assert that given 7 to get to 1, you must > divide 7 by 7, and that by the distributive property, a factor that > has 7 as a constant term must itself by divided by 7. Then you note that these factors have properties inconsistent with divisibility by 7. When you see that a_1(x), for certain values of x, has a property that rules out its being divisible by 7 in the ring of algebraic integers, then you should conclude that a_1(x)/7 doesnÕt stay in this ring. When I was first exposed to division in elementary school, we used remainders when quotients didnÕt divide evenly. We said, 23 divided by 7 equals 3, remainder 2. Later I learned about fractions, and eventually became used to the idea that 23/7 is a number. Now IÕm used to being able to divide almost any number by 7, but I still need to keep issues of divisibility in mind. It would be a mistake to assume this sort of number shares all the properties of integers. I can only apply the properties of integers to x/7 if I know that x is divisible by 7. === Subject: Re: JSH: Point of logic > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). I factor the polynomial as P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) when the aÕs are the three roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > ItÕs kind of strange to call f(0) the constant term of f when f > isnÕt a polynomial. ItÕs called the constant term when f is a > polynomial, because f is a sum of finitely many monomial terms, > and the constant term is the one of degree 0. Your non-polynomial > factors of P(x) have no such expressions. Furthermore, the word > constant is really confusing you here. I suggest you stop using > it. The terms are constants. And they are 7, 7 and 22, factors of the constant term of the polynomial P(x), which is 1078. So, literally, the terms are constants, so they are constant terms, in that they do not vary. Understand? > By the distributive property for the constant terms of two of (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) to be divided by 7, the full factor has to be divided by 7, which > gives P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). > Why should any of the aÕs have 7 as a factor? ThereÕs no reason > to assume this division stays inside the ring of algebraic > integers for all relevant values of x. The logic is trivial. The factors of P(x) MUST also contain factors of its constant term. You can determine what those factors are, and doing so shows that they are 7, 7 and 22. Now P(x) has 49 as a multiple, and dividing that multiple off divides 49 from the factors of P(x). Understand so far? Looking at the resultant factors of the constant term of P(x)/49, you find that they are 1, 1, and 22, indicating that 7 divides off from two of them. That means that two of the factors of P(x) must have been divided by 7. > What can be shown is that in the ring of algebraic integers, if a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. > If this is true, then it just proves the above division leaves > the ring of algebraic integers for some x. ThereÕs nothing that > interesting about the fact that not everything in this ring is > divisible by 7. Well, the algebra says one thing, but the ring of algebraic integers says another. There is an apparent contradiction which has to be resolved. > I simply assert that given 7 to get to 1, you must > divide 7 by 7, and that by the distributive property, a factor that > has 7 as a constant term must itself by divided by 7. > Then you note that these factors have properties inconsistent with > divisibility by 7. No. They are inconsistent with it IN THE RING OF ALGEBRAIC INTEGERS. So the algebra says one thing, but in the ring of algebraic integers you get something contradictory if the cubic defining the aÕs is irreducible over Q. > When you see that a_1(x), for certain values of x, has a property > that rules out its being divisible by 7 in the ring of algebraic > integers, then you should conclude that a_1(x)/7 doesnÕt stay in > this ring. If the cubic defining the aÕs is irreducible then, yes, itÕs not in the ring, in apparent contradiction with what follows from the algebraic argument. ItÕs neat. A one time thing in math history--apparent contradiction at the heart of whatÕs believed to be mathematically true. Never been seen before like this in all of human history. > When I was first exposed to division in elementary school, we used > remainders when quotients didnÕt divide evenly. We said, 23 > divided by 7 equals 3, remainder 2. Later I learned about > fractions, and eventually became used to the idea that 23/7 is a > number. Now IÕm used to being able to divide almost any number by > 7, but I still need to keep issues of divisibility in mind. It would > be a mistake to assume this sort of number shares all the properties > of integers. I can only apply the properties of integers to x/7 if I > know that x is divisible by 7. You have to follow the logic. If the logic escapes you then youÕll never see the apparent contradiction. If you canÕt see the apparent contradiction, then you have no foothold for understanding the result. Logically, you must believe that the factors of the polynomial of P(x) multiply together to give P(x), including its constant term, 1078. Technically, g_1(x) g_2(x) g_3(x) = P(x) and then you must believe that those factors include factors of the constant term of P(x), and if you do believe then you can accept g_1(0) g_2(0) g_3(0) = P(0) and if those are in fact factors of the constant term of P(x), then you can accept that they themselves are NOT dependent on the value of x, and in fact since they are 7, 7 and 22, that is true. If you believe that constants do not change as x varies, then you can accept that the value of x has nothing to do with the value of 7, 7 and 22. But dividing the polynomial by 49--its multiple--gives constant terms that are 1, 1 and 22, so they changed. Since the polynomial has 49 as a multiple thereÕs no RATIONAL reason to believe that youÕll get some kind of fraction. If you go from 7, 7 and 22 to 1, 1 and 22, then you divided two by 7. ItÕs simple. Now I suggest you try to attack the logic. For instance, do you disagree with the assertion that the factors of P(x) include factors of its constant term? If so, what is the mathematical basis for your disagreement? If not, then how can you not get it? James Harris === Subject: Re: JSH: Point of logic Discussion, linux) > The logic is trivial. The factors of P(x) MUST also contain factors > of its constant term. You can determine what those factors are, and > doing so shows that they are 7, 7 and 22. Now P(x) has 49 as a > multiple, and dividing that multiple off divides 49 from the factors > of P(x). Understand so far? I really donÕt get it. What is the difference between your situation and this one? Let f(x) = x g(x) = x h(x) = x^2 + 2x P(x) = (f(x) + 2)(g(x) + 2)(h(x) + 1) Since P(x) = (x + 2)^2 (x + 1)^2, clearly P(n) is divisible by 4 for every natural number n. 2 divides (f(0) + 2) and (g(0) + 2). Therefore, for every n, 2 divides (f(n) + 2) and (g(n) + 2). But this is clearly false, since 2 does not divide n + 2 for every n. Clearly, itÕs not the distributive law thatÕs relevant, because the distributive law applies equally in my case. So what is it about your case that makes it different than mine? What additional facts (and additional arguments) justify your conclusion but not mine? -- Come on people!!! The US just blew up a lot of people in Iraq, donÕt you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving FermatÕs last theorem === Subject: Re: JSH: Point of logic >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >I factor the polynomial as >P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >when the aÕs are the three roots of >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >I determine the constant terms of the three factors by setting x=0, >which reveals that for two of them the constant term is 7, while for >one it is 22. >>ItÕs kind of strange to call f(0) the constant term of f when f >>isnÕt a polynomial. ItÕs called the constant term when f is a >>polynomial, because f is a sum of finitely many monomial terms, >>and the constant term is the one of degree 0. Your non-polynomial >>factors of P(x) have no such expressions. Furthermore, the word >>constant is really confusing you here. I suggest you stop using >>it. > The terms are constants. And they are 7, 7 and 22, factors of the > constant term of the polynomial P(x), which is 1078. > So, literally, the terms are constants, so they are constant terms, in > that they do not vary. > Understand? Usually people start by defining what a term *is*, such as a section of an expression that is isolated from all other sections by addition (or subtraction). Based on that definition, 22 is not a term at all, simply the value of final factor to explicitly have 22 as a constant term was clearer about this. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Point of logic !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >> >> I factor the polynomial as >> >> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >> >> when the aÕs are the three roots of >> >> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >> >> I determine the constant terms of the three factors by setting x=0, >> which reveals that for two of them the constant term is 7, while for >> one it is 22. >> ItÕs kind of strange to call f(0) the constant term of f when f >> isnÕt a polynomial. ItÕs called the constant term when f is a >> polynomial, because f is a sum of finitely many monomial terms, >> and the constant term is the one of degree 0. Your non-polynomial >> factors of P(x) have no such expressions. Furthermore, the word >> constant is really confusing you here. I suggest you stop using >> it. > The terms are constants. And they are 7, 7 and 22, factors of the > constant term of the polynomial P(x), which is 1078. > So, literally, the terms are constants, so they are constant terms, in > that they do not vary. In that they donÕt vary exactly when _what_ changes? In short: constant with respect to what? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). Excuse me, thatÕs for P(0). > But for P(x)/49, the constant term > is 22. Again, P(0) > So, logically, two of the constant terms were divided by 7. Suppose you noted that (7)(7)(22) = (49)(2)(11)? Now *one* of the constant terms is divided by 7 -- twice. You only get (7)(7)(22) with your factorization when x=0. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. Not true for the 22. The numbers between parentheses *are* functions of ÔxÕ and they will change with ÔxÕ, even though the 7s donÕt. But the parenthetic value (22) is computed from (5*a_3(x) + 7) = (5*a_3(0) + 7) = (5*3 + 7) = (22). IsnÕt the Ôconstant termÕ = 7 here, as it is in the other groups? Why not? You have called 22 a constant term, but clearly the value 22 is computed from the sum of the constant 7 and a function of ÔxÕ , 5a_3(x). None of the 7Õs change with ÔxÕ, BUT THE 22 DOES. > Dividing P(x) by 49 changes the constant terms. ...again, P(0)... > They go from being 7, 7 and 22 to being 1, 1 and 22. Remember, the 22 is *non* a constant. It is the result of evaluating an expression involving ÔxÕ, e.g. (5a_3(x) + 7) = 22 *only* when x = 0. It is not 22 otherwise. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. Yes, but if 7/w(x) = 1 when x = 0, then w(0) = 7. You donÕt know anything about w(x) or 7/w(x) for other values of ÔxÕ. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. > So is there a problem with the logic in my argument above? Major problems with your whole exposition. > How can there be? I simply assert that given 7 to get to 1, you must > divide 7 by 7, and that by the distributive property, a factor that > has 7 as a constant term must itself by divided by 7. ThatÕs not all youÕre asserting. You are arguing a general case from a specific case. DoesnÕt work. > The logic is trivial. For the simple special cases you offer as a substitute for proof, yes. But there is no rigorous argument proving any of your main points and there are counter examples which clearly disprove them. > If thatÕs not true, the the entire logical framework of mathematics > itself is crap. No, the enormous leap you have made from an oversimplified situation to a complex situation without proof is crap. > Notice, I depend on the distributive property and 7/w = 1, meaning > w=7. > So why all the arguing? Because thatÕs not all you are doing. You have erected a straw man consisting of simple unambiguous values and relations as substitutes for the structures in your main argument. They donÕt apply. You cannot take simple symbols, such as ÔwÕ, in a simple expression and solve for them unambiguously and hope that this sheds light on a complex expressions involving functions of some variable. > Well, my work shows that some people didnÕt properly understand > algebraic integers, and they built a lot of proofs on their ßawed > understanding. Algebraic integers have nothing to do with it. Your claim that P(x) is a multiple of 49 is only true if ÔxÕ is an ordinary integer. There is no appeal to algebraic integers in any of the above exposition. [snip nasty and unnecessary parting shot] -- 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: JSH: Point of logic Discussion, linux) [...] > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, How does this follow from the distributive property? Give us a few more steps, please. If I told you that x is divisible by w (in some fixed ring R) and x + y is divisible by w (in R) and therefore, by the distributive property, y is divisible by w, you might ask how the distributive property applies. I could then say this: By assumption, there is a z (in R) such that x = wz and also a u (in R) such that x + y = wu. Therefore, y = (x + y) - x = wu - wz = w(u - z) (By the distributive law). Since u is in our ring and z is too, so is u - z and thus y is a multiple of w. Think you could spell things out so clearly for the above argument? -- Jesse F. Hughes I have written many words to sci.math, some of them are not even meaningless. --Ross Finlayson === Subject: Re: JSH: Point of logic > [...] > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, > How does this follow from the distributive property? > Give us a few more steps, please. > If I told you that x is divisible by w (in some fixed ring R) and > x + y is divisible by w (in R) and therefore, by the distributive > property, y is divisible by w, you might ask how the distributive > property applies. I could then say this: > By assumption, there is a z (in R) such that x = wz > and also a u (in R) such that x + y = wu. Therefore, > y = (x + y) - x > = wu - wz > = w(u - z) (By the distributive law). > Since u is in our ring and z is too, so is u - z and thus y is a > multiple of w. > Think you could spell things out so clearly for the above argument? P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) Notice that if you let x=0, you have that the factors give the appropriate constants, as 5 a_1(0)/7 + 1 = 1 5 a_2(0)/7 + 1 = 1 5 a_3(0) + 7 = 22. The algebra is basic. James Harris === Subject: Re: JSH: Point of logic > Notice that if you let x=0, you have that the factors give the > appropriate constants, as > 5 a_1(0)/7 + 1 = 1 > 5 a_2(0)/7 + 1 = 1 > 5 a_3(0) + 7 = 22. The value Ô22Õ was arrived at by evaluating a_3(x) at x = 0. Therefore it is a function of ÔxÕ and will change when ÔxÕ changes. On the other hand the values 5 and 7 above are constants and will *not* change with ÔxÕ. Please clarify what you mean by ÔconstantÕ. If you mean a value that is fixed and does not change with ÔxÕ, the value 22 is disqualified. > The algebra is basic. Please apply it here. -- 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: JSH: Point of logic > Notice that if you let x=0, you have that the factors give the > appropriate constants, as > 5 a_1(0)/7 + 1 = 1 > 5 a_2(0)/7 + 1 = 1 > 5 a_3(0) + 7 = 22. > The value Ô22Õ was arrived at by evaluating a_3(x) at x = 0. Therefore it is a >function of ÔxÕ and Nope. 22 is NOT a function of x. ItÕs the number 22. HereÕs how it works. You have the polynomial P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) and I factor it into three factors that here IÕll just call g_1(x), g_2(x), and g_3(x), so you have g_1(x) g_2(x) g_3(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) and I assert that the factors of P(x) have factors of the constant term of P(x), and that constant term is 49(22). Well, I can *test* that assertion by setting x=0, and I find that g_1(x) contributes 7, g_2(x) contributes 7, and g_3(x) contributes 22. Technically, g_1(0) g_2(0) g_3(0) = 49(22) so the gÕs MUST have factors of the constant term. You see, 22 is not a function of x. It is a constant, and it is a factor of the constant term of P(x). > will change when ÔxÕ changes. On the other hand the values 5 and 7 above are constants and will > *not* change with ÔxÕ. Please clarify what you mean by ÔconstantÕ. If you mean a value that is > fixed and does not change with ÔxÕ, the value 22 is disqualified. > The algebra is basic. > Please apply it here. Locked inside the factors g_1(x), g_2(x), and g_3(x) are factors of the constant term. You see, the factors of the polynomial multiply together to give that polynomial, so they must have pieces of the polynomial to multiply together. The polynomial P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) has one piece that is constant, as it is 49(22), and factors of the polynomial have to produce that piece. If it is not produced by the polynomial factors, then where does it come from? So they have constants within them, and you can find what those are by clearing out x, by setting it to 0. ItÕs easy. Basic math. James Harris === Subject: Re: JSH: Point of logic > Notice that if you let x=0, you have that the factors give the > appropriate constants, as 5 a_1(0)/7 + 1 = 1 5 a_2(0)/7 + 1 = 1 5 a_3(0) + 7 = 22. > The value Ô22Õ was arrived at by evaluating a_3(x) at x = 0. Therefore it is a >function of ÔxÕ and > Nope. 22 is NOT a function of x. ItÕs the number 22. Now your dissembling. You presented a factorization which contained the term(5*a_3(x) + 7) and demonstrated that (5*a_3(0) + 7) = (22). You also then claimed that the value (22) was *not* a function of ÔxÕ. Of course it is a number, but let me be more precise: the number 22 appears between those parentheses when Ôx = 0Õ. A different number will appear in that position when ÔxÕ has a different value. ThatÕs what is meant by a number or value being a function of a variable. Stop kidding around. Everyone knows whatÕs going on in this discussion. You have made a serious and unrecoverable blunder. -- 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: JSH: Point of logic it could already in a book of world records, or known as JamesÕ Last Recovered Fumble. > ThatÕs what is meant by > a number or value being a function of a variable. --Advice 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: JSH: Point of logic > Locked inside the factors g_1(x), g_2(x), and g_3(x) are factors of > the constant term. Locked inside the factors? Locked? What the hell does locked mean? Talk mathemtaics, idiot! > So they have constants within them, and you can find what those are by > clearing out x, by setting it to 0. But that only tells you about the situation when x=0. You cannot generalize what you find in that case with the general case where x=/=0. Can you see that? > ItÕs easy. Basic math. It is not easy. It is meaningless. Do you wonder why so many people tell you are full of crap, if what you say is both easy and basic? Freaking idiot! === Subject: Re: JSH: Point of logic > Nope. 22 is NOT a function of x. ItÕs the number 22. But it is 22 when x=0. It is other values when x =/= 0. So rather than being a constant, 22 is just a specific case of a variable that can take on values other than 22 as x varies. The problem is that you go on to genearlize from the case where the variable is 22 and x = 0 to other cases where the variable is not 22 for x =/= 0. That is your blunder... you make a generalization that is not valid for all values of x. === Subject: Re: JSH: Point of logic >> How does this follow from the distributive property? >> Give us a few more steps, please. ... >> Think you could spell things out so clearly for the above argument? > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) > Notice that if you let x=0, you have that the factors give the > appropriate constants, as > 5 a_1(0)/7 + 1 = 1 > 5 a_2(0)/7 + 1 = 1 > 5 a_3(0) + 7 = 22. > The algebra is basic. Answer the question. -- --Tim Smith === Subject: Re: Point of logic > Well, my work shows that some people didnÕt properly understand > algebraic integers, and they built a lot of proofs on their ßawed > understanding. Phhhhhhht! Your work shows that some people didnÕt properly understand algebraic integers? What a -up you are... I cannot follow any of your posts, James. They all seem to be vague hand waving arguments with disconnected and wrong pseudo-arithmetic ramblings. They are not clear or precise. Your arithmetic is at a pre-university level with loads of errors. Can you be more professional in your arguments? For example, what ring are you actually working in? Can you define it? What results in Galois theory are killed by your work? What part of WilesÕ FLT proof is in error, according to your results? Indeed, what are your results? What does properly unit mean. I see little in the way of actual algebra in your writings. To make your case, you need to follow conventions, and stick to established definitions. Where are your definitions, lemmas, theorems, proofs, corollaries, etc. What are you actually trying to prove? It would be nice to know what that is up front. I think that you avoid all this hard stuff because you are not clear on what you are trying to say. Is that to leave you wiggle-room? You say that your paper is accepted for review at Annals of Mathematics (at Princeton?). Well, I can tell you that they have never published any paper written in the silly style that you write in. It is too sloppy, too vague, and too full of errors. And it is just not interesting. It just will not get published that way. I guarantee it. And how do you figure that all the thousands of mathematicians over the last several hundred years missed the supposedly simple and trivial problem with algebraic numbers that you believe you have discovered. Seems unlikely? And after your FLT and APF fiascos, why do you have any confidence in this latest nonsense of yours? And of course, we would all like to know. what the is The Hammer? === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as Do you object to standard mathematical terminology? Would you say that 15 has 5 as a multiple? Or rather, would you say that 15 has 5 as a FACTOR? === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. WhatÕs that supposed to mean? Of course you can have a factorization P(x)/49 = [(5 a_1(x) + 7)/7][(5 a_2(x) + 7)/7][(5 a_3(x) + 7)] where a_1(0)=0, a_2(0)=0, and a_3(0)=3. But thereÕs no guarantee that (5 a_1(x) + 7)/7 and (5 a_2(x) + 7)/7 will be algebraic-integer-valued functions of x. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. Okay. So in fact the functions (5 a_1(x) + 7)/7 and (5 a_2(x) + 7)/7 arenÕt algebraic-integer-valued functions of x. WhatÕs the problem then? > So is there a problem with the logic in my argument above? What were you arguing? === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. As has been pointed out many times before this is only one of an infinite number of possible factorizations of P(x). In all cases the product of the constant terms is 1078. But since (if w1(0) = 1) (5 a_1(x) + 7) and (5 a_1(x)/w1(x) + 7/w1(x)) both have a constant term of 7 this does not tell us much. - William Hughes === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). LetÕs parallel your example with a simpler one and see where you went wrong. What I have is a polynomial Q(x) = 7(25x^2 + 30x + 2) [1] = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2 I factor the polynomial as Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2] where the aÕs are the two roots of a^2 - (x - 1)a + 7(x^2 + x) [3] > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. I set x = 0, which reveals that for one of the terms in [2] the constant term is 7 and the other is 2. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). That corresponds with the constant terms of Q(x) being 14, which is (7)(2) > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. Now I note that dividing both sides by 7 gives me Q(x)/7 = 25x^2 + 30x + 2 which factors as Q(x)/7 = (5 a_1(x) + 7)(5 a_2(x)+ 7)/7 and I note that the constant term is now 2. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. So, logically, one of my constant terms was divided by 7 > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. Dividing Q(x) by 7 changes the constant terms. They go from being 7, 2 to being 1 and 2. Up to here, thereÕs no real problem with what youÕve written. Now, however, you start veering off the tracks, as weÕll see. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. > So is there a problem with the logic in my argument above? Yes. You distribute 49 across the factors as 7, 7, 1 which is certainly okay in case x = 0, just as I did by distributing 7 across the factors as 7, 1. However, youÕre incorrect in your assertion that the pattern that obtains when x = 0 WILL GENERALLY HOLD when x is not zero. In my example, you would have us write each factorization as Q(x)/7 = (5(a_1(x)/7) + 1)(5a_2(x) + 7) [4] regardless of the value of x. The problem is that your restriction in [4] depends strongly on the polynomial used to define the terms a_1(x) and a_2(x). For x = 0, the factorization in [4] worked because of the way the a polynomial splits into linear factors, but consider the case when x = 1. Now we have Q(1) = 399 = (7)(57) where the aÕs satisfy a^2 + 14 = 0 which gives us a_1(1) = sqrt(-14), a_2(1) = -sqrt(-14) and so we have the factorization Q(1) = (5 sqrt(-14) + 7)(-5 sqrt(-14) + 7) which is indeed equal to 399, as expected. However, itÕs immediately obvious that sqrt(-14) isnÕt divisible by 7, so the factorization in [4] does not hold in general. Instead of distributing the value 7 as 7 in the first factor and 1 in the second, we have the distribution sqrt(7) in both factors, so we have the factorization Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] * [(-5 sqrt(-14) + 7)/sqrt(7)] [BTW, you might argue that 7 might divide sqrt(-14) in some ring other than the algebraic integers, and indeed it might. However, that would result in 7 being a unit in that ring, which is something youÕve repeatedly said you donÕt want.] Rick === Subject: Re: JSH: Point of logic > [BTW, you might argue that 7 might divide sqrt(-14) in some > ring other than the algebraic integers, and indeed it might. > However, that would result in 7 being a unit in that ring, > which is something youÕve repeatedly said you donÕt want.] Provided the ring contains all the algebraic integers. > Rick === Subject: Re: JSH: Point of logic [BTW, you might argue that 7 might divide sqrt(-14) in some > ring other than the algebraic integers, and indeed it might. > However, that would result in 7 being a unit in that ring, > which is something youÕve repeatedly said you donÕt want.] Provided the ring contains all the algebraic integers. No, you only need sqrt(-14) and sqrt(-14)/7. Consider a ring R with unit which contains sqrt(-14) and sqrt(-14)/7. Then R also contains (sqrt(-14)/7)^2 = -14/49 = -2/7 and 1-3*(-2/7) = 1/7 (R contains the integers) So 7 is a unit in R -William Hughes === Subject: Re: JSH: Point of logic >A mathematical proof must be logical. What I aim to do here is >elaborate on the logic, which proves IÕm correct, and letÕs see what >happens. >What I have is a polynomial >P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >which has 49 as a multiple, as >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >I factor the polynomial as >P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >when the aÕs are the three roots of >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >I determine the constant terms of the three factors by setting x=0, >which reveals that for two of them the constant term is 7, while for >one it is 22. >That corresponds with the constant term of P(x) being 1078, which is >7(7)(22). >Now I note that dividing both sides by 49 gives me >P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, >which factors as >P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 >and I note that the constant term is now 22. >Now what do I know about the constant terms? >Well they are constant, and are specific numbers, specifically 7, 7 >and 22, for the factors of P(x). But for P(x)/49, the constant term >is 22. Exactly up to this point, everything is OK. >So, logically, two of the constant terms were divided by 7. No, NOT logically. All you require when you divide by 49 is that the resulting quotients are algebraic integers. You can split 49 up into three pieces, each of which is a function of x, such that when you divide the three factors (5 a_1(x) + 7), (5 a_2(x) + 7), and (5 a_3(x) + 7) by these three factors of 49, each of the results is an algebraic integer. We have shown you REPEATEDLY how this can be done. Even seeing it, you STILL keep insisting that, logically, it CANNOT be done. >That is the essential point on which everything hinges. >Notice how simple it is, the constant terms are actual numbers. There is huge irony in this statement. It is totally correct! And you proceed almost immediately to ignore it! You define the constant term of a function G(x) to be G(0). Then you start considering something like g_3(x) = r_3(x) + 22, where r_3(0) = 0. Of course by your definition, the constant term of g_3(x) is g_3(0) = r_3(0) + 22 = 0 + 22 = 22. That is completely correct and proper use of YOUR DEFINITION of constant term. But then you start considering what might happen if you divide g_3(x) by another function, say, w_3(x): the result is h_3(x) = g_3(x)/w_3(x) = (r_3(x) + 22)/w_3(x) = r_3(x)/w_3(x) + 22/w_3(x). And you ask: what is the constant term of h_3(x) ? And YOUR answer is: 22/w_3(x). What your answer SHOULD be, USING YOUR OWN DEFINITION, is 22/w_3(0). Why canÕt you get this? JUST USE YOUR OWN DEFINITION !!! If you do, you will NOT get 22/w_3(x) as the constant term. Instead you throw your own definition right out the window and you fall back to your old high-school habits of thinking: the constant term, that has to be the thing on the end, i.e., it MUST be 22/w_3(x). See that x in there? ThatÕs a VARIABLE. w_3(x) is a VARIABLE function of the VARIABLE x. ItÕs NOT CONSTANT. You have departed right here from the use of your OWN PERFECTLY GOOD DEFINITION of constant term. >Numbers like 7 and 22 are NOT variables, and they remain the same >without regard to the value of x. Who says otherwise ???? >Dividing P(x) by 49 changes the constant terms. >They go from being 7, 7 and 22 to being 1, 1 and 22. Only if you split up 49 in the WRONG WAY, the way which is guaranteed NOT TO WORK, if you want (as is your goal) to end up with algebraic integer factors. >Therefore, exactly two of them were divided by 7. The rest of what you say, after this point, is wrong because you have jumped to assuming what you want to be true. >There is no other way to get from 7 to 1, and you can write it out >algebraically. >7/w = 1, giving w = 7. But if w is a FUNCTION of x - a VARIABLE function, as you know it must be - then 7/w is NOT the constant term of (5 a_1(x) + 7)/w. >By the distributive property for the constant terms of two of >(5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) >to be divided by 7, the full factor has to be divided by 7, which >gives >P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) >indicating that two of the aÕs have 7 as a factor, and IÕve >arbitrarily selected a_1(x) and a_2(x). >What can be shown is that in the ring of algebraic integers, if >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) >is irreducible over Q, with integer coefficients, then the aÕs cannot >have 7 as a factor in the ring of algebraic integers. >So is there a problem with the logic in my argument above? YES The fact that you donÕt understand what your error in logic is, does not mean it isnÕt there. >How can there be? I simply assert that given 7 to get to 1, you must >divide 7 by 7, and that by the distributive property, a factor that >has 7 as a constant term must itself by divided by 7. >The logic is trivial. ItÕs trivially wrong. I actually despair of getting through to you on this. We have shown you everything that you need to know to understand the error. It appears to me that you actually DO understand the little individual bits. What you donÕt seem to be capable of getting is putting it all together to reach the conclusion: you reasoning is wrong because you donÕt adhere to your own definition in a rigorous manner. >If thatÕs not true, the the entire logical framework of mathematics >itself is crap. No, sorry, whatÕs crap is your own logic. >Notice, I depend on the distributive property and 7/w = 1, meaning >w=7. Here is your reasoning. The product of the constant terms must equal the constant term of the product. That is actually a true statement! Then you look at the three expressions above. You think: the constant term of (r_1(x) + 7)/w1 must be 7/w1. Since this is the constant term, IT MUST BE CONSTANT. Since w1 = 7 when x = 0, that must be true in general. So whereÕs the error ? The error is in saying 7/w1 is the constant term. w1 cannot be assumed to be constant. We have defined *nonconstant* functions w1 = w_1(x), w2 = w_2(x), and w3 = w_3(x), such that 1. w_1(x) * w_2(x) * w_3(x) = 49 2. w_1(x) is a divisor of 7 (similary for w_2(x), w_3(x) 3. w_1(x) is a divisor of r_1(x), etc. (All divisors within the ring of algebraic integers). So why isnÕt 7/w_1(x) the constant term of (r_1(x) + 7)/w_1(x) ??? Two reasons: 1. ItÕs not constant. 2. ItÕs not YOUR OWN DEFINITION of constant term. Your own definition of constant term tells you exactly what the *real* constant term is. It is: (r_1(0) + 7)/w_1(0) = 7/w_1(0). See? There is no VARIABLE x in this (correct) constant term. You keep saying that we claim that the constant term is not constant. Nothing could be further from the truth. In fact, when you insist that 7/w_1(x) is the constant term, it is YOU who is claiming that the constant term is not constant! >So why all the arguing? HereÕs why. If you accept what is clearly, obviously true - that is, that the constant terms are 7/w_1(0), 7/w_2(0), and 22/w_3(0), then your whole proof just collapses. You know this. You are terrified of it. You have erected a Berlin Wall within your mind to keep from acknowledging it. I want to show you just how blatant your misunderstanding of this actually is. Below is a direct quote from a post by you on November 14. There you defined b_1(x) = a_1(x)/w_1(x) and v_1(x) = 7/w_1(x), etc., and v_3(x) = 22/w_1(x): Here is the quote. >P(x)/49 = > (5 b_1(x) + v_1(x))(5 b_2(x) + v_2(x))(5 b_3(x) + v_3(x)) >and if you allow that the factors are each factors of the constant >term as before, then you have > v_1(0) v_2(0)(15 + v_3(0)) = 22 >at x=0, and when x does not equal 0, you have ******************************************************* AND HERE IS THE BIG FAT ERROR ******************************************************* > v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 >where I introduce u_3(x) to handle any further weirdness with how >a_3(x) behaves, where u_3(0) = 3, to agree with previous results See that? ItÕs unbelievable! First you write out the constant term CORRECTLY, adhering perfectly well to your own definition: v_1(0) v_2(0)(15 + v_3(0)) = 22 And THEN IMMEDIATELY you throw that definition RIGHT OUT THE WINDOW, and say v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 where everywhere you have replaced the CONSTANT 0 (which is correct) by the VARIABLE x (which is wrong) !!! How much more obvious can it get? This is a truly low-level, rank-beginner-nonmathematician mistake. A SUBSTITUTION mistake. No high-powered abstract math is involved. No group theory, no Galois, no Gauss, no Hilbert, no calculus, no high-school math, even! As you once described many of us here in sci.math this is truly a LOW SKILL LEVEL mistake. ItÕs pathetic that you keep defending it, not getting it. Big fat hint: trying sticking with your own definition. >Well, my work shows that some people didnÕt properly understand >algebraic integers, and they built a lot of proofs on their ßawed >understanding. >They did and people long dead did as well. >That means that a lot of prestige is at stake, and egos and other >boring things that help make the world a very nasty place. >So these people fight the logic and the conclusion in order to try and >keep the status quo. They are trying to maintain homeostasis in their >environment. >Trouble is theyÕre living a lie, and continuing to teach it to fresh >students who deserve to be taught true mathematical ideas. The shoe is utterly, completely on the other foot. You are fighting tooth and nail to keep your own ego from collapsing. We are just saying what is obviously and rigorously correct. You have an enormous emotional investment in this: more than a yearÕs work, two or three journal submissions, rejections, counterexamples, interminable hostile arguments with people that you have called liars and cheats and incompetents a thousand times. You know that if you dare let any of this in, past your internal censor, all of your work for the past 20 months or so is reduced to zero. Count on it. Nora B. >James Harris >http:// mathforprofit.blogspot.com / === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. This happens: > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. Yes. Nobody has a problem with that. > That is the essential point on which everything hinges. It is *not*. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, And *this* is what is wrong. Consider the three function w_1(x), w_2(x) and w_3(x) such that w_1(0) = w_2(0) = 7 and w_3(0) = 1. Furthermore, w_1(x).w_2(x).w_3(x) = 49 for *all* x. In the factorisation: [(5 a_1(x) + 7)/w_1(x)][(5 a_2(x) + 7)/w_2(x)][(5 a_3(x) + 7)/w_3(x)], the constant terms are now 1, 1 and 22, *regardless of other properties of the wÕs*. I do not know why the distributive property dictates that the wÕs should be constant, because there is no need for that. For the constant terms, by your own definition of constant term, only the values of the wÕs for x = 0 are important. > So is there a problem with the logic in my argument above? Yup. > How can there be? I simply assert that given 7 to get to 1, you must > divide 7 by 7, and that by the distributive property, a factor that > has 7 as a constant term must itself by divided by 7. This is where your error is. It must be divided by a function which has the value 7 when x = 0, because by your own definition, the constant term of a function is the value of the function when x = 0. It is not necessarily a constant function, and that is what you claim. > Well, my work shows that some people didnÕt properly understand > algebraic integers, and they built a lot of proofs on their ßawed > understanding. Oh, yes. You may start with Dedekind as the first wrong-doer. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Limit approaching infinity. The problem is to evaluate the limit as x approaches pos. infinity, for (x)/[sqrt {(x^2) +1000}]. Naturally, I turn the denominator into 1 + (1000)/(x^2), which allows me to turn it into 1 + 0 = 1. This leaves me with x / 1. However, the solutions book turns the numerator x into 1 in the first step, for a final answer of 1/1 and I donÕt see how they turned the x into a 1 (unless thereÕs some property that allows x/1 as x-->inf. to become 1, but IÕd imagine that inf./1 would just be inf.) === Subject: Re: Limit approaching infinity. > The problem is to evaluate the limit as x approaches pos. infinity, for > (x)/[sqrt {(x^2) +1000}]. > Naturally, I turn the denominator into 1 + (1000)/(x^2), which allows me > to turn it into 1 + 0 = 1. This leaves me with x / 1. > However, the solutions book turns the numerator x into 1 in the first > step, for a final answer of 1/1 and I donÕt see how they turned the x into > a 1 (unless thereÕs some property that allows x/1 as > but IÕd imagine that inf./1 would just be inf.) I never saw anything ÔnaturalÕ about complicated, unnecessary algebraic manipulations to evaluate limits at infinity. These are among the easiest limits to evaluate (for the most part) and most can be done in a second or two in your head, without ever touching pencil to paper. sqrt(x^2) is |x| but since x-->+oo you can remove the abs. value bars. What you have in essence is a limit of the form x/(x+1000). The limit is just the ratio of leading coefficients (disregard any lower powered terms and constant terms, they matter not as x gets big.) By the exact same reasoning, you can say the following limit is 5pi/6: (5pi)x^24 + (36pi/17)x^23 + 77x^7 + 23x^2 ------------------------------------------ 6x^24 + 252x^11 + 97x^9 + 887x^5 + 16x Now talk your teacher into giving you 100 of these for your next test. -- Darrell === Subject: Re: Limit approaching infinity. >I never saw anything ÔnaturalÕ about complicated, unnecessary algebraic >manipulations to evaluate limits at infinity. These are among the easiest >limits to evaluate (for the most part) and most can be done in a second or >two in your head, without ever touching pencil to paper. Of course, one with experience can see what the answer is by comparing the relative orders. But the unnecessary manipulations are in fact quite necessary for a student to work through to gain a real understanding of the subject. After doing enough such problems, a beginner will eventually see, and understand, the heuristic reasoning. >sqrt(x^2) is |x| but since x-->+oo you can remove the abs. value bars. What >you have in essence is a limit of the form x/(x+1000). Actually, itÕs not of that form, itÕs more essentially of the form x/x. When you write it as: x / {x sqrt( 1 + 1000/x)} that sqrt factor is a factor, not a term, and it is more like 1, not 1000, especially as x is large. > The limit is just >the ratio of leading coefficients (disregard any lower powered terms and >constant terms, they matter not as x gets big.) >By the exact same reasoning, you can say the following limit is 5pi/6: >(5pi)x^24 + (36pi/17)x^23 + 77x^7 + 23x^2 >------------------------------------------ >6x^24 + 252x^11 + 97x^9 + 887x^5 + 16x Of course you can, and on a multiple choice test you could even get away with it. Personally, I would expect students in my class to show the work. Otherwise they could get by with just memorizing the shortcut with no understanding. --Lynn === Subject: Re: Limit approaching infinity. >>I never saw anything ÔnaturalÕ about complicated, unnecessary algebraic >>manipulations to evaluate limits at infinity. These are among the easiest >>limits to evaluate (for the most part) and most can be done in a second or >>two in your head, without ever touching pencil to paper. > Of course, one with experience can see what the answer is by comparing > the relative orders. But the unnecessary manipulations are in fact > quite necessary for a student to work through to gain a real > understanding of the subject. After doing enough such problems, a > beginner will eventually see, and understand, the heuristic reasoning. I suppose thatÕs one reasonable way of teaching it, but my teacher felt differently, and I had a perfectly clear understanding of it at the time (as I was a beginner taking calculus). It may not have been the most rigorous way of teaching it, but it got the job done quite well. It is a very intuitive concept. As x increases without bound, what difference does adding 1000 make? None, in the limit, so donÕt even worry about it. Throw it out. The same can be said for anything other than the highest powered terms. And of those highest powered terms (the highest in the numerator and denominator) it is also just as intuitively seen that if the degree of the numerator is greater than that of the denominstor, the limnit does not exist (blows up) or if the degree of the denominstor is the higher one, the limit is 0. A number, divided by a MUCH BIGGER number, is a real SMALL number. A number divided by a MUCH SMALLEr number, is a really BIG number. If the numbers are about the same size (ie the numerator and denominator are of equal degree) then those variables will CANCEL and what you have left is just the ratio of leading coefficients. Speak for yourself, but I understand that reasonably well. The readerÕs mileage may vary. >>sqrt(x^2) is |x| but since x-->+oo you can remove the abs. value bars. >>What >>you have in essence is a limit of the form x/(x+1000). > Actually, itÕs not of that form, itÕs more essentially of the form > x/x. When you write it as: > x / {x sqrt( 1 + 1000/x)} > that sqrt factor is a factor, not a term, and it is more like 1, not > 1000, especially as x is large. >> The limit is just >>the ratio of leading coefficients (disregard any lower powered terms and >>constant terms, they matter not as x gets big.) >>By the exact same reasoning, you can say the following limit is 5pi/6: >>(5pi)x^24 + (36pi/17)x^23 + 77x^7 + 23x^2 >>------------------------------------------ >>6x^24 + 252x^11 + 97x^9 + 887x^5 + 16x > Of course you can, and on a multiple choice test you could even get > away with it. I got away with it just fine on OPEN ENDED tests. One does not need to see a list of choices in order to tell just by LOOKING that the limit is 5pi/6. > Personally, I would expect students in my class to show > the work. Otherwise they could get by with just memorizing the > shortcut with no understanding. You incorrectly suggest that not showing the work implies misunderstanding. Wrong. I, for one, understand quite well, as do others. Not everyone will, but then again there is no one way of teaching something that everyone understands. To the OP: Do whatever works for YOU, not for Lynn, or for me! Some people simply do not like others presenting methods of solution they do not personally approve of, and will try to discredit it. If you want to write a half page of algebraic manipulation (as a practical exercise in algebraic manipulation, I guess) for every one of these limits, then go right ahead! If you want to evaluate these limits quickly and easily, I just showed you a much more efficient method. Some people seem to equate quick and easy with not good enough. There is also the issue of the real world. In the real world you want to FIND that damned thing and apply it for whatever you need it for, not waste time doing it the long, hard way. But then again, in the real world the average person will never NEED to find a limit, but thatÕs the topic of a different discussion entirely. -- Darrell === Subject: Re: Limit approaching infinity. >> Personally, I would expect students in my class to show >> the work. Otherwise they could get by with just memorizing the >> shortcut with no understanding. >You incorrectly suggest that not showing the work implies >misunderstanding. Wrong. I donÕt care to get into a big debate here, but no, I do not suggest that not showing the work implies misunderstanding. What not showing the work does is leave uncertain in the teacherÕs mind whether the student understands what is going on or has just memorized a short-cut. > I, for one, understand quite well, as do others. >Not everyone will, but then again there is no one way of teaching something >that everyone understands. >To the OP: Do whatever works for YOU, not for Lynn, or for me! Some people >simply do not like others presenting methods of solution they do not >personally approve of, and will try to discredit it. If you want to write a >half page of algebraic manipulation (as a practical exercise in algebraic >manipulation, I guess) for every one of these limits, then go right ahead! >If you want to evaluate these limits quickly and easily, I just showed you a >much more efficient method. Some people seem to equate quick and easy with >not good enough. That last paragraph leaves me thinking you should read my post again, given the way you misrepresent it. --Lynn === Subject: Re: Limit approaching infinity. > Personally, I would expect students in my class to show > the work. Otherwise they could get by with just memorizing the > shortcut with no understanding. >>You incorrectly suggest that not showing the work implies >>misunderstanding. Wrong. > I donÕt care to get into a big debate here, Nothing wrong with some healty debate! > but no, I do not suggest > that not showing the work implies misunderstanding. Then why the fuss about misunderstanding? > What not showing > the work does is leave uncertain in the teacherÕs mind whether the > student understands what is going on or has just memorized a > short-cut. So you insist justification must accompany any ÔshortcutÕ to be convinced the student knows what he is doing? I donÕt think you really mean that. Should I suppose after performing the algebraic manipulation you desire on a limit at infinity, you also insist on them proving, by the definition of limit, that the limit really is what that process (ie shortcut) _says_ it is. For every application of the product rule (a shortcut), do you insist on verification by the definition of derivative which is of course defined in terms of a limit (bring out the limit definition again.)> Would you ever get ANYWHERE with them if shortcuts were not allowed? How do you know the student ÔunderstandsÕ what he is doing by way of using ANY shortcut? Good sense and judgment, thatÕs how. And, sometimes, you DONÕT know. Literacy is sometimes all that is required (and you have time for) at the cost of deeper understanding of the reasons why. Most of everything the student does, that you teach them, is in some way, shape, or form, is a ÔshortcutÕ for some other longer, more rigorous process. ItÕs a judgement call where to draw the line for how much justification of any given process is needed to demonstrate competency of that process, but yours is not the only judgment on the matter. You seem to be more concerned with the demonstration of certain tedious algebraic manipulation vs. a demonstration of an intuitive and elegant approach to the same problem. ThatÕs your judgment to make of course, but realize not everyone is necessarily evaluating oneÕs ability to perform tedious algebra by hand on these types of problems. >> I, for one, understand quite well, as do others. >>Not everyone will, but then again there is no one way of teaching >>something >>that everyone understands. >>To the OP: Do whatever works for YOU, not for Lynn, or for me! Some >>people >>simply do not like others presenting methods of solution they do not >>personally approve of, and will try to discredit it. If you want to write >>half page of algebraic manipulation (as a practical exercise in algebraic >>manipulation, I guess) for every one of these limits, then go right ahead! >>If you want to evaluate these limits quickly and easily, I just showed you >>much more efficient method. Some people seem to equate quick and easy >>with >>not good enough. > That last paragraph leaves me thinking you should read my post again, > given the way you misrepresent it. Well, being that you tried to discredit a simpler, more elegant, and perfectly valid method of solution is very representative of itself! And yes, you obviously do equate quick and easy with not good enough. You just said so yourself that doing so does not satisfy your requirement. You insist on seeing Ôthe work.Õ May I humbly suggest, although there is obvious need to show Ôthe workÕ at times, there are also times when seeing a light turn on demonstrates much more than doing a bunch of tedious algebraic manipulation. Call me crazy, but I just donÕt see how doing a half page of algebra lights a brighter bulb than elegantly realizing, just by looking at it using the process mentioned, what the limit is. And, I stand 110% behind the suggestion that the OP use whichever method works best FOR HIM. He has been given no less than two valid ways of approaching the problem. He is the sole judge which method, if either, to use. You simply do not know the specific circumstances. For all you know, he may approach his teacher and state the reasons why the ÔshortcutÕ is valid, so that the teacher is reasonably convinced he ÔknowsÕ what heÕs doing, and allow him to simply write the answer without showing the Ôwork.Õ Then again, the teacher may be more like you, and demand each and every thing be accompanied by rigorous justification. -- Darrell === Subject: Re: Limit approaching infinity. >Well, being that you tried to discredit a simpler, more elegant, and >perfectly valid method of solution is very representative of itself! And >yes, you obviously do equate quick and easy with not good enough. You just >said so yourself that doing so does not satisfy your requirement. You >insist on seeing Ôthe work.Õ You certainly are master of the straw man tactic of debate. The paragraph in question is: Of course, one with experience can see what the answer is by comparing the relative orders. But the unnecessary manipulations are in fact quite necessary for a student to work through to gain a real understanding of the subject. After doing enough such problems, a beginner will eventually see, and understand, the heuristic reasoning. The length of your replies notwithstanding, I will stand by what I actually said in that paragraph. I also note you snipped without comment the only non-subjective assertion I made, namely, in response to your statement: >What you have in essence is a limit of the form x/(x+1000). Actually, itÕs not of that form, itÕs more essentially of the form x/x. When you write it as: x / {x sqrt( 1 + 1000/x^2)} that sqrt factor is a factor, not a term, and it is more like 1, not 1000, especially as x is large. Do you have difficulty acknowledging objective errors? --Lynn === Subject: Re: Limit approaching infinity. >>Well, being that you tried to discredit a simpler, more elegant, and >>perfectly valid method of solution is very representative of itself! And >>yes, you obviously do equate quick and easy with not good enough. You >>just >>said so yourself that doing so does not satisfy your requirement. You >>insist on seeing Ôthe work.Õ > You certainly are master of the straw man tactic of debate. The > paragraph in question is: > Of course, one with experience can see what the answer is by > comparing the relative orders. I donÕt think youÕre listening to what IÕm saying. Even the INEXPERIENCED can see what the answer is by comparring relative orders. > But the unnecessary manipulations are > in fact quite necessary for a student to work through to gain a real > understanding of the subject. ThatÕs your opinion. Not everyone shares it. > After doing enough such problems, a > beginner will eventually see, and understand, the heuristic > reasoning. Or, they can see and understand it from the get-go. A number divided by a much smaller number, is a very big number (oo). A number divided by a much larger number, is a very small number (0). These are not difficult things to understand to grasp. > The length of your replies notwithstanding, I will stand by what I > actually said in that paragraph. > I also note you snipped without comment the only non-subjective > assertion I made, namely, in response to your statement: >>What you have in essence is a limit of the form x/(x+1000). > Actually, itÕs not of that form, itÕs more essentially of the form > x/x. When you write it as: > x / {x sqrt( 1 + 1000/x^2)} > that sqrt factor is a factor, not a term, and it is more like 1, not > 1000, especially as x is large. > Do you have difficulty acknowledging objective errors? Lynn, I didnÕt comment on it because frankly I felt it was not WORTH commenting on. Why are you searching for an argument where one does not exist? It seems like such a nitpick and IMO we were discussing much more interesting things. Since you are being so nitpicky to the point of mentioning this yet again, one could just as easily say that x/x is really of the form 1. This is really a silly sidebar to the much more legitimate discussion we were having. As much as you apparently donÕt like it, it is fact that lim x-->oo x/(x+1000) is the same as lim x-->oo x/[sqrt(x^2)+1000). I simply chose to first address that sqrt(x^2) for large x is the same as x BEFORE I adressed the negligibility of 1000. It is simply the order I chose to address it. It is NOT errornious. ItÕs a judgment call, Lynn, as to how much ÔworkÕ needs to be shown in order to convince someone that a student is ÔgettingÕ it. As much as you apparently do not like it, yours is definitely NOT the only judgment that matters. Some feel there is beauty and elegance in such a short and sweet method of solution. Perhaps what bothers you more than the brevity of the method is the fact that I (and in fact, just about ANY student first learning limits at infinity) am/is quite capable of not only applying the method but UNDERSTANDING it without having to do a page gull of algebra, despite your insistence that only the ÔexperiencedÕ can. Lynn, even my 6th grade son understands that a small number divided by a large number is a small number, a large number divided by a small number is a large number, and that two (nonzero) numbers that are the same, divide to 1. Good grief, itÕs simply ONE way of approaching these types of problems. If you hate seeing alternate methods so much, then Usenet is not the place for you. In YOUR CLASS you may decide which is acceptable. Here, The READER decides which method, if any, he will use. The fact is both methods are OK so get your panties out of a wad. EVEN IF the OP is not allowed to use the method I suggested, he can still use it as very fast and efficient CHECK to see if he got the right answer. -- Darrell === Subject: Re: Limit approaching infinity. Typo alert: >x / {x sqrt( 1 + 1000/x)} should be x / {x sqrt( 1 + 1000/x^2)} Lynn === Subject: Re: Limit approaching infinity. >The problem is to evaluate the limit as x approaches pos. infinity, for >(x)/[sqrt {(x^2) +1000}]. >Naturally, I turn the denominator into 1 + (1000)/(x^2), which allows me to >turn it into 1 + 0 = 1. This leaves me with x / 1. It appears that you divided the argument of sqrt by x^2. That isnÕt necessarily a bad thing to do, even shows promise. But if you remember algebra you canÕt just multiply or divide some part of an expression by some arbitrary value without changing the value of that expression. However you can multiply or divide by 1, or 1 in various disguises, without changing the value of the expression. So, the hint is, what can you do to not change the value of the expression? Or how can you multiply or divide by something that is equivalent to 1 and get where you need to go? If you read this a few times this should be enough of a hint for you to figure this out, and hints teach a lot more than someone just blurting out the answer. === Subject: Re: Limit approaching infinity. >The problem is to evaluate the limit as x approaches pos. infinity, for >(x)/[sqrt {(x^2) +1000}]. >Naturally, I turn the denominator into 1 + (1000)/(x^2), I hope itÕs not very natural, because itÕs wrong. The denominator is |x| * sqrt(1 + 1000/x^2) or at least I think thatÕs the closest correct value to what you typed. Since x is approaching _positive_ infinity, |x| = x. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Limit approaching infinity. > The problem is to evaluate the limit as x approaches pos. infinity, for > (x)/[sqrt {(x^2) +1000}]. > Naturally, I turn the denominator into 1 + (1000)/(x^2), How? What became of the sqrt? As x gets big, sqrt(x^2 + 1000) approaches x. So the limit is 1. > which allows me to > turn it into 1 + 0 = 1. This leaves me with x / 1. > However, the solutions book turns the numerator x into 1 in the first step, > for a final answer of 1/1 and I donÕt see how they turned the x into a 1 > (unless thereÕs some property that allows x/1 as x-->inf. to become 1, but > IÕd imagine that inf./1 would just be inf.) === Subject: Re: Limit approaching infinity. >The problem is to evaluate the limit as x approaches pos. infinity, for >(x)/[sqrt {(x^2) +1000}]. As x --> oo, the 1000 in the denom becomes negligible. Drop it, and see what happens. bob >Naturally, I turn the denominator into 1 + (1000)/(x^2), which allows me to >turn it into 1 + 0 = 1. This leaves me with x / 1. >However, the solutions book turns the numerator x into 1 in the first step, >for a final answer of 1/1 and I donÕt see how they turned the x into a 1 >(unless thereÕs some property that allows x/1 as x-->inf. to become 1, but >IÕd imagine that inf./1 would just be inf.) === Subject: Re: Limit approaching infinity. > The problem is to evaluate the limit as x approaches pos. infinity, for > (x)/[sqrt {(x^2) +1000}]. > Naturally, I turn the denominator into 1 + (1000)/(x^2), which allows me to > turn it into 1 + 0 = 1. This leaves me with x / 1. How do turn sqrt(x^2 + 1000) into 1 + 1000/x^2 ?? ThatÕs wrong right from the start! sqrt(x^2 + 1000) = sqrt[x^2 (1 + 1000/x^2)] = x sqrt(1 + 1000/x^2) (for x>0, which weÕre talking about) So now you have x -------------------- x sqrt(1 + 1000/x^2) 1/sqrt(1 + 1000/x^2) which is obviously 1 in the limit as x->+oo -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Algebraic question. Does (x-1) / abs (x-1) = -1? Context: lim x--> 1-, in the equation (x-1)/ abs((x^2) -1) = x-1 / abs(x+1)abs(x-1) = -1 / x+1 = -1/2 ThatÕs from the solution book. === Subject: Re: Algebraic question. > Does (x-1) / abs (x-1) = -1? Substitute in two values of x, one which makes x - 1 positive and one which makes it positive. Thus get the answer no. > Context: lim x--> 1-, in the equation (x-1)/ abs((x^2) -1) > = x-1 / abs(x+1)abs(x-1) > = -1 / x+1 > = -1/2 > ThatÕs from the solution book. === Subject: Re: Algebraic question. > Does (x-1) / abs (x-1) = -1? > Substitute in two values of x, one which makes x - 1 positive and one > which makes it positive. Thus get the answer no. Idiot me. One of those positives should be negative. === Subject: Re: Algebraic question. days. My association with the Department is that of an alumnus. >Does (x-1) / abs (x-1) = -1? Depends on the value of x. It is true when x<1. It is false when x>=1. >Context: lim x--> 1-, in the equation (x-1)/ abs((x^2) -1) >= x-1 / abs(x+1)abs(x-1) >= -1 / x+1 Because, since x-->1-, it is in particular smaller than 1, so therefore abs(x-1) = 1-x. So (x-1)/abs(x-1) = (x-1)/(1-x) = -1. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Can someone check my proof please? (real analysis) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB16EpE25938; IÕm trying to prove or disprove the following: Suppose {f_n} is a sequence of real valued functions that converge to f(x) on D such that each f_n(x) is bounded. Then prove or disprove f:D->R is bounded. --- Notice that the problem does not state what type of convergence is occurring (i.e. pointwise vs. uniform). IÕved proved the assertion for uniform convergence as follows: Since each {f_n} is bounded, we know that for each natural number n, there exists some real number M_n such that |f_n(x)| < M_n. We also know that if {f_n:D->R} converges uniformly to f:D->R, we can choose some epsilon > 0 such that |f(x) - f_n(x)| < epsilon, for all n > N and all x in D. Using the above information, observe that |f(x)| < |f(x) - f_N+1(x) + f_N+1(x)| < |f(x) - f_N+1(x)| + |f_N+1(x)| < epsilon + M_N+1. Thus we have shown that f(x) is bounded under the given circumstance. --- Notice that this proof relies upon uniform convergence... which is NOT necessarily the type of convergence that is happening (the problem was vague!). Does this proof fail for pointwise convergence? I seem to think so, since in that case each bound M will depend on the value of x. Any help would be greatly appreciated (THE SOONER THE BETTER!!!). === Subject: Re: Can someone check my proof please? (real analysis) >IÕm trying to prove or disprove the following: >Suppose {f_n} is a sequence of real valued functions that converge to >f(x) on D such that each f_n(x) is bounded. Then prove or disprove >f:D->R is bounded. >--- >Notice that the problem does not state what type of convergence is >occurring (i.e. pointwise vs. uniform). IÕved proved the assertion >for uniform convergence as follows: Oops. If the statement is exactly as above, and if they havenÕt said somewhere else very explicitly that convergence means uniform convergence, then the problem means just that f_n -> f pointwise. >Since each {f_n} is bounded, we know that for each natural number n, >there exists some real number M_n such that >|f_n(x)| < M_n. >We also know that if {f_n:D->R} converges uniformly to f:D->R, we can >choose some epsilon > 0 such that >|f(x) - f_n(x)| < epsilon, for all n > N and all x in D. >Using the above information, observe that >|f(x)| < |f(x) - f_N+1(x) + f_N+1(x)| > < |f(x) - f_N+1(x)| + |f_N+1(x)| > < epsilon + M_N+1. >Thus we have shown that f(x) is bounded under the given circumstance. >--- >Notice that this proof relies upon uniform convergence... which is NOT >necessarily the type of convergence that is happening (the problem was >vague!). Does this proof fail for pointwise convergence? Yes. >I seem to >think so, since in that case each bound M will depend on the value of >x. Any help would be greatly appreciated (THE SOONER THE BETTER!!!). ************************ David C. Ullrich === Subject: Re: Can someone check my proof please? (real analysis) === Subject: Can someone check my proof please? (real analysis) > Suppose {f_n} is a sequence of real valued functions that > converge to f(x) on D such that each f_n(x) is bounded. > Then prove or disprove f:D->R is bounded. f_n(x) = x if x in [-n,n] = -n if x < n = n if x > n > IÕved proved the assertion for uniform convergence as follows: > Since each {f_n} is bounded, we know that for each > natural number n, there exists some real number M_n such that > |f_n(x)| < M_n. > We also know that if {f_n:D->R} converges uniformly to f:D->R, > we can choose some epsilon > 0 such that > |f(x) - f_n(x)| < epsilon, for all n > N and all x in D. > Using the above information, observe that > |f(x)| < |f(x) - f_N+1(x) + f_N+1(x)| > < |f(x) - f_N+1(x)| + |f_N+1(x)| > < epsilon + M_N+1. Write f_(N+1) and M_(N+1) for ascii clarity. > Thus we have shown that f(x) is bounded under the given > circumstance. Well done. > Does this proof fail for pointwise convergence? Yes. See counterexample at top with uniformly continuous f_nÕs. ---- === Subject: probability q of the Society of Actuaries A 35-year old can buy a one-year term $100000 insurance policy for premium amount P. The insurance company assumes (i) probability of death in the next year for a 35-year old is 0.001, (ii) insurance premium is paid at beginning of year, and invested at an annual interest rate of 5%, (iii) if death occurs, the payment of $100000 is made to the beneficiary at the end of the one-year term. 1. Let L be the loss random variable from the point of view of the insurance company. a.. What are the possible values of the loss and their corresponding probabilities?a What is the expected loss as a function of P? There is a loss in event of death, and a gain=negative loss in event of no death. b.. What should the premium P be in order that the expected value of the loss is 0? 2. Now suppose 1000 different 35-year olds buy the one-year term $100000 insurance policy for premium P. Let the losses to the company be L1,...,L1000. What is the premium P to charge in order that the probability of a positive loss over the 1000 policies is less than or equal to 0.05 ? === Subject: Re: probability q of the Society of Actuaries posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9 > A 35-year old can buy a one-year term $100000 insurance policy for premium > amount P. The insurance company assumes (i) probability of death in the next > year for a 35-year old is 0.001, (ii) insurance premium is paid at beginning > of year, and invested at an annual interest rate of 5%, (iii) if death > occurs, the payment of $100000 is made to the beneficiary at the end of the > one-year term. > 1. Let L be the loss random variable from the point of view of the insurance > company. > a.. What are the possible values of the loss and their corresponding > probabilities? There are only two scenarios: policyholder lives, or policyholder dies. The probabilities are stated in the problem, so all you need to do is work out the P&L (profit/loss) in each case. Policyholder lives: insurer pockets the premium and pays nothing. Policyholder dies: insurer pockets the premium but has to make the $100,000 payout. The complication is that you must be clear about the VALUATION DATE (the as of date for which you calculate P&L). The premium is received today, yet the payout (if any) is made in one year, so the amounts canÕt be directly added. To calculate the P&L now (i.e. the date the policy is taken out), you need to discount the payout back to today, but leave the premium alone. You need to choose an appropriate discount rate. The problem suggests that the appropriate discount rate is 5% but that is not explicitly stated. Alternatively, to calculate the P&L in one year, you need to grow the premium by the stated investment rate but leave the payout alone. > What is the expected loss as a function of P? There is a > loss in event of death, and a gain=negative loss in event of no death. Suppose that a variable, x, can take n discrete values x_i, with respective probabilities p_i, where i ranges from 1 through n. The expected value of x, denoted E(x), is given by E(x) = sum{i = 1 to n: x_i * p_i}. In this case, the variable x is the P&L and n = 2 (only two possibilities). You know both values of x_i and p_i from the previous answer, so just plug these into the formula. Assuming discount rate = investment rate, youÕll see that E(P&L) as of today is equal to E(P&L) in one year, discounted back to today. Or, looking at it the other way around, E(P&L) in one year is equal to E(P&L) now invested for one year at the investment rate. > b.. What should the premium P be in order that the expected value of the > loss is 0? Set E(P&L) = 0 and solve for P. (Terminology point. In my view itÕs much easier if you use the term P&L. E.g. Expected P&L = 0 rather than Expected value of the loss is 0. Positive P&L means profit, negative P&L means loss, P&L = 0 means break-even. Then you donÕt get confused about the signs, negative losses becoming profits, use of ugly terms like positive loss etc.) > 2. Now suppose 1000 different 35-year olds buy the one-year term $100000 > insurance policy for premium P. Let the losses to the company be > L1,...,L1000. What is the premium P to charge in order that the probability > of a positive loss over the 1000 policies is less than or equal to 0.05 ? Do you mean the LOWEST premium? (E.g. you could charge $1,000,000 and be guaranteed to never lose whatever happened. You probably wouldnÕt sell too many policies though!). Assuming you do mean the lowest premium, IÕd probably work it out like this... Find the smallest number n, such that the probability of n or more deaths is less than or equal to 0.05. Then set P so that at you break even at n-1 deaths. Then you will lose only if there are n or more deaths (with probability <= 0.05). HTH === Subject: JSH: Nothing new so far If you think that my showing non-polynomial factorization with an actual polynomial P(x) is new, think again. If you think my focusing on actual constant terms 7, 7 and 22 is new, think again. If you think my pointing out that 7 and 22 are constants and not functions of x is new, think again. If you think my explaining that the result I have follows from the fact that to get from 7 to 1 you need to divide 7 by 7, and the distributive property, is new, think again. I talked about all of that over a year ago, in discussions on the sci.math newsgroup. Like I said, the readership of sci.math IS stupid. So then, how can I end up arguing with people on Usenet about a very basic argument that proves a fundamental problem in the area of algebraic number theory, when I submitted a paper for publication, and actually had reason to believe it was published for a while until some of the people IÕd argued with on sci.math, inexplicably succeeded in getting it withdrawn by ganging together to send some emails? IÕve mentioned that an early draft of a paper in this area went to Barry Mazur who commented on it (encouragingly, actually). And I sent a paper Andrew Granville to submit it to the New York Journal of Mathematics, so he at least looked at it, as well. IÕve explained the result over a period of months to a Cornell math grad student who contacted *me* claiming that he would help me if I could explain my work to him. I went to Vanderbilt University in person and talked to a professor McKenzie, explaining it all in detail, answering his various questions. So how is this possible in the field of mathematics? Well, people claiming to be mathematicians often say that mathematics is different from other disciplines in that it is immune from upheaval. They claim that the base of mathematics--its core--is perfect, having been worked over hundreds or even thousands of years so that there is no concern about error. They live in a fantasy world of the fieldÕs perfection. Also, quite a few men are heroes to the world now, especially the math world because of results that IÕve shown are specious. While itÕs relatively quiet like this, they can continue to feel proud of their accomplishments. Sure as time passes and the information circulates they may more and more be like the walking dead in the eyes of their colleagues and students like yourselves, but denial is a special human characteristic, which is quite powerful. So time passes. Usenet is one of the few areas where I can talk mathematics, so itÕs convenient to me. And talking out my ideas is fun, and helps my own understanding. But donÕt think that what IÕm saying now is new. IÕve explained it all before. James Harris === Subject: Re: JSH: Nothing new so far Discussion, linux) > If you think my explaining that the result I have follows from the > fact that to get from 7 to 1 you need to divide 7 by 7, and the > distributive property, is new, think again. I notice that you still havenÕt really explained the relevance of the distributive property. I havenÕt seen any point at which you use an equation of the form: a(b + c) = ab + ac I havenÕt seen any point at which you used the related bit of reasoning: x divides y and x divides (y + z), so x divides z. What IÕve seen is that you use something like: 49 divides (f(x) + 7)(g(x) + 7)(h(x) + 22) and f(0) = g(0) = h(0) = 0. Therefore, 7 divides f(x) and g(x) (for all x, mind you!). I donÕt see how the distributive property applies here. Help me out with this basic math. How do you get this from the distributive property? -- And yes, for those who think that just maybe I did find a short proof of FermatÕs Last Theorem, and THE prime counting function, if I succeed at what IÕm working on now world economy as you know it will be gone. -- James Harris branches out. === Subject: Re: JSH: Nothing new so far >If you think that my showing non-polynomial factorization with an >actual polynomial P(x) is new, think again. If you think my focusing >on actual constant terms 7, 7 and 22 is new, think again. >If you think my pointing out that 7 and 22 are constants and not >functions of x is new, think again. >If you think my explaining that the result I have follows from the >fact that to get from 7 to 1 you need to divide 7 by 7, and the >distributive property, is new, think again. >I talked about all of that over a year ago, in discussions on the >sci.math newsgroup. >Like I said, the readership of sci.math IS stupid. >So then, how can I end up arguing with people on Usenet about a very >basic argument that proves a fundamental problem in the area of >algebraic number theory, when I submitted a paper for publication, and >actually had reason to believe it was published for a while until some >of the people IÕd argued with on sci.math, inexplicably succeeded in >getting it withdrawn by ganging together to send some emails? >IÕve mentioned that an early draft of a paper in this area went to >Barry Mazur who commented on it (encouragingly, actually). Exactly what did he say? >And I sent >a paper Andrew Granville to submit it to the New York Journal of >Mathematics, so he at least looked at it, as well. >IÕve explained the result over a period of months to a Cornell math >grad student who contacted *me* claiming that he would help me if I >could explain my work to him. >I went to Vanderbilt University in person and talked to a professor >McKenzie, explaining it all in detail, answering his various >questions. >So how is this possible in the field of mathematics? How is what possible? YouÕre not even _claiming_ that any of those people you mention agreed your results were actually correct. >Well, people claiming to be mathematicians often say that mathematics >is different from other disciplines in that it is immune from >upheaval. >They claim that the base of mathematics--its core--is perfect, having >been worked over hundreds or even thousands of years so that there is >no concern about error. >They live in a fantasy world of the fieldÕs perfection. >Also, quite a few men are heroes to the world now, especially the math >world because of results that IÕve shown are specious. While itÕs >relatively quiet like this, they can continue to feel proud of their >accomplishments. >Sure as time passes and the information circulates they may more and >more be like the walking dead in the eyes of their colleagues and >students like yourselves, but denial is a special human >characteristic, which is quite powerful. >So time passes. Usenet is one of the few areas where I can talk >mathematics, so itÕs convenient to me. And talking out my ideas is >fun, and helps my own understanding. >But donÕt think that what IÕm saying now is new. >IÕve explained it all before. And youÕll explain it all again in the future. ThatÕs not going to make it correct. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Nothing new so far Discussion, linux) > Well, people claiming to be mathematicians often say that mathematics > is different from other disciplines in that it is immune from > upheaval. > They claim that the base of mathematics--its core--is perfect, having > been worked over hundreds or even thousands of years so that there is > no concern about error. Who would think that? Only someone that knows just as bloody little about history of mathematics as you do, I guess. It wasnÕt too long ago that mathematics suffered a crisis (at least for the foundations folk, but it probably didnÕt stir much interest among the working mathematicians). Anyone with a smattering of interest in history of mathematics knows about the problems RussellÕs paradox caused. Other crises arenÕt too hard to find. If you had found a real problem in algebraic number theory, it would raise questions about how mathematics is done, but it wouldnÕt shake the foundations of mathematical theory. -- Jesse F. Hughes I often told you of the dangers of hubris, and most importantly of all, I TOLD you that I wanted to change the institution of mathematics worldwide. -- James Harris, on the evils of pride === Subject: Re: JSH: Nothing new so far > If you think that my showing non-polynomial factorization with an actual > polynomial P(x) is new, think again. If you think my focusing on actual > constant terms 7, 7 and 22 is new, think again. > If you think my pointing out that 7 and 22 are constants and not functions > of x is new, think again. > If you think my explaining that the result I have follows from the fact > that to get from 7 to 1 you need to divide 7 by 7, and the distributive > property, is new, think again. And if you think JSH ignoring the numerous people who have provided simple, clear, examples of the various errors in his work is new, think again. James, are you ever going to learn that simply reposting the same wrong arguments in a new thread wonÕt make the refutation go away? -- --Tim Smith === Subject: Solving Solvable Sextics Using Polynomial Decomposition Hello all, For those who like solvable equations and groups, hereÕs something that might be interesting: Solving Solvable Sextics Using Polynomial Decomposition ABSTRACT: Using basic results established by Etienne Bezout (1730-1783) and Niels Henrik Abel (1802-1829), we devise a general method to solve the solvable sextic in radicals by deriving two kinds of resolvents, one kind of the 15th degree and the other of the 10th degree, that factors when the sextic is solvable and thus enabling us to decompose the solvable sextic either as: a) three quadratics whose coefficients are determined by a cubic or b) two cubics whose coefficients are determined by a quadratic. Mathematics Subject Classification: Primary: 12E12. http://www.geocities.com/titus_piezas/sextics.html Just click at the link to the .pdf file. --Titus === Subject: Re: Solving Solvable Sextics Using Polynomial Decomposition by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB1L7Il08353; >Hello all, >For those who like solvable equations and groups, hereÕs something that might >be interesting: >Solving Solvable Sextics Using Polynomial Decomposition >ABSTRACT: Using basic results established by Etienne Bezout (1730-1783) >and Niels Henrik Abel (1802-1829), we devise a general method to solve >the solvable sextic in radicals by deriving two kinds of resolvents, >one kind of the 15th degree and the other of the 10th degree, that factors >when the sextic is solvable and thus enabling us to decompose the solvable >sextic either as: a) three quadratics whose coefficients are determined by >a cubic or b) two cubics whose coefficients are determined by a quadratic. >Mathematics Subject Classification: Primary: 12E12. href=http://www.geocities.com/titus_piezas/sextics.html>http: //www.geoci ties.com/titus_piezas/sextics.htmlJust click at the link to the .pdf file. > The paper is good. Readability needs improvement. IÕd suggest > typesetting the paper using LaTeX, and try to cut down on the > paragraph breaks. Interesting work, nonetheless. --Titus === Subject: Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB1L7I508357; HereÕs a magic-square puzzle in which the unbranded eight heads of of beef must be banded with a prime number (numbers can only be divided evenly by one and itself) in such a way that each horizontal, vertical, and diagonal row of three totals 111. The lowest and only non-prime number brand (1: an even more elite digit) in the puzzle has been placed to give a head start. What prime numbers, when added diagionally, vertical, and horizontal in a square of nine blocks will give you 111? The number 1 was given === Subject: solving recurrence relations I hope this isnÕt too long of an example. Here goes... I am studying a math book called Concrete Mathematics: A Foundation of Computer Science, Gramham, Knuth & Patashnik. Not for any classes, just on my own. On page 25, section 2.2 - Sums and Recurrences, is the following example, as best I can do using a text editor. The sum Sn = sigma notation where k=0 to n, ak is equivalent to the recurrence S0 = a0; Sn = Sn-1 + an, for n > 0; (2.6) Therefore we can evaluate sums in closed form by using the methods we learned in Chapter 1 to solve recurrences in closed form. For example, if an is equal to a constant plus a multiple of n, the sum-recurrence (2.6) takes the following general form: Ro = alpha; Rn = Rn-1 + beta + gama x n, for n > 0; //x = multiplication So, in general the solution can be written in the form Rn = A(n) x alpha + B(n) x beta + C(n) x gama, where A(n), B(n) and C(n) are the coefficients of dependence on the general parameters alph, beta and gama. The repertoire method tells us to try plugging in simple functions of n for Rn, hoping to find constant parameters alpha, beta and gama where the solution is especially simple. Setting Rn = 1 implies that alpha = 1, beta = 0 and gama = 0; hence A(n) = 1. Setting Rn = n implies that alpha = 0, beta = 1 and gama = 0; hence B(n) = n. Setting Rn = n^2 implies that alpha = 0, beta = -1 and gama = 2; hence 2C(n) - B(n) = n^2 and we have C(n) = (n^2 + n)/2. Easy as pie. Right. I follow until the last sentence where Rn = n^2. How in the world would a person just guess the general parameters of alpha = 0, beta = -1 and gama = 2? It seems to me that this is a bit of a contrived example where the writer knows the solution and fits the parameters to that solution. In most cases I wonÕt know the solution, so I am a bit confused by the example. Another thing that confuses me is that C(n) = (n^2 + n)/2. Without knowing that this is the solution, how in the world would I ever realize that. http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- === Subject: Re: solving recurrence relations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Wu31365; >I hope this isnÕt too long of an example. Here goes... >I am studying a math book called Concrete Mathematics: A Foundation of >Computer Science, Gramham, Knuth & Patashnik. Not for any classes, just on >my own. >On page 25, section 2.2 - Sums and Recurrences, is the following example, as >best I can do using a text editor. Please read the following suggestions for writing in ascii. >The sum >Sn = sigma notation where k=0 to n, ak Sn = Sum_{k = 0 to n} ak >is equivalent to the recurrence >S0 = a0; >Sn = Sn-1 + an, for n > 0; (2.6) >Therefore we can evaluate sums in closed form by using the methods we >learned in Chapter 1 to solve recurrences in closed form. >For example, if an is equal to a constant plus a multiple of n, the >sum-recurrence (2.6) takes the following general form: >Ro = alpha; >Rn = Rn-1 + beta + gama x n, for n > 0; //x = multiplication Please, please donÕt write x for multiplication -- it looks too much like a variable x, which is *very* distracting. Rn = R_{n-1} + beta + (gamma)n >So, in general the solution can be written in the form >Rn = A(n) x alpha + B(n) x beta + C(n) x gama, Rn = A(n)alpha + B(n)beta + C(n)gamma >where A(n), B(n) and C(n) are the coefficients of dependence on the general >parameters alph, beta and gama. >The repertoire method tells us to try plugging in simple functions of n for >Rn, hoping to find constant parameters alpha, beta and gama where the >solution is especially simple. >Setting Rn = 1 implies that alpha = 1, beta = 0 and gama = 0; hence >A(n) = 1. >Setting Rn = n implies that alpha = 0, beta = 1 and gama = 0; hence >B(n) = n. >Setting Rn = n^2 implies that alpha = 0, beta = -1 and gama = 2; hence >2C(n) - B(n) = n^2 >and we have C(n) = (n^2 + n)/2. Easy as pie. >Right. I follow until the last sentence where Rn = n^2. How in the world >would a person just guess the general parameters of alpha = 0, beta = -1 and >gama = 2? S/he wouldnÕt guess -- s/he would use the recurrence R_n = R_{n-1} + beta + (gamma)n n^2 = (n-1)^2 + beta + (gamma)n n^2 = n^2 - 2n + 1 + beta + (gamma)n to see beta = -1, gamma = 2. And alpha = R_0 = 0, as you say above. Then, from this example, R_n = n^2 = A(n)alpha + B(n)beta + C(n)gamma = 1(0) + n(-1) + C(n)2 because we know B(n) from before; hence 2C(n) - n = n^2, and this explains the thing which confused you below [C(n) = (1/2)(n^2 + n)]. It seems to me that this is a bit of a contrived example where the >writer knows the solution and fits the parameters to that solution. In most >cases I wonÕt know the solution, so I am a bit confused by the example. >Another thing that confuses me is that C(n) = (n^2 + n)/2. Without knowing >that this is the solution, how in the world would I ever realize that. Hope this helps, Todd Trimble === Subject: Re: solving recurrence relations in alt.math.undergrad: > I hope this isnÕt too long of an example. Here goes... > I am studying a math book called Concrete Mathematics: A Foundation of > Computer Science, Gramham, Knuth & Patashnik. Not for any classes, just on > my own. > On page 25, section 2.2 - Sums and Recurrences, is the following example, as > best I can do using a text editor. > The sum > Sn = sigma notation where k=0 to n, ak > is equivalent to the recurrence > S0 = a0; > Sn = Sn-1 + an, for n > 0; (2.6) Subscripts are usually denoted by underscores. IÕm going to rewrite some of these to make them easier for me to read. S_n = Sum[k=0 to n; a_k] S_0 = a_0; S_n = S_{n-1} + a_n, for n > 0. [...] > For example, if an is equal to a constant plus a multiple of n, the > sum-recurrence (2.6) takes the following general form: > Ro = alpha; > Rn = Rn-1 + beta + gama x n, for n > 0; //x = multiplication R_0 = alpha, R_n = R_{n-1} + beta + gamma*n, for n > 0. > So, in general the solution can be written in the form > Rn = A(n) x alpha + B(n) x beta + C(n) x gama, R_n = A(n)*alpha + B(n)*beta + C(n)*gamma > where A(n), B(n) and C(n) are the coefficients of dependence on the general > parameters alph, beta and gama. > The repertoire method tells us to try plugging in simple functions of n for > Rn, hoping to find constant parameters alpha, beta and gama where the > solution is especially simple. > Setting Rn = 1 implies that alpha = 1, beta = 0 and gama = 0; hence > A(n) = 1. > Setting Rn = n implies that alpha = 0, beta = 1 and gama = 0; hence > B(n) = n. > Setting Rn = n^2 implies that alpha = 0, beta = -1 and gama = 2; hence > 2C(n) - B(n) = n^2 > and we have C(n) = (n^2 + n)/2. Easy as pie. > Right. I follow until the last sentence where Rn = n^2. How in the world > would a person just guess the general parameters of alpha = 0, beta = -1 and > gama = 2? You donÕt guess them: you solve for them. You assume that R_n = n^2 is a solution to the recurrence R_0 = alpha, R_n = R_{n-1} + beta + gamma*n, for n > 0. Since in that case R_0 = 0^2 = 0, that forces alpha to be 0. Similarly, substituting R_n = n^2 and R_{n-1} = (n-1)^2 into the recurrence yields n^2 = (n - 1)^2 + beta + gamma*n (for n > 0), n^2 = n^2 - 2n + 1 + beta + gamma*n (for n > 0), and after a little simplification (gamma - 2)*n + (beta + 1) = 0 for all n > 0. Let c = gamma - 2 and d = beta + 1. You have cn + d = 0 for all positive integers n. ItÕs easy to see that this can happen only if c = d = 0, i.e., gamma - 2 = beta + 1 = 0, or, finally, gamma = 2 and beta = -1. Since you have the general relationship R_n = A(n)*alpha + B(n)*beta + C(n)*gamma, you can write n^2 = 0*A(n) + (-1)*B(n) + 2*C(n), n^2 = 2C(n) - B(n). [...] > Another thing that confuses me is that C(n) = (n^2 + n)/2. Without knowing > that this is the solution, how in the world would I ever realize that. You already know from considering the solution R_n = n that the coefficient function B is the identity function: B(n) = n. Substitute into my last line above: n^2 = 2C(n) - n, n^2 + n = 2C(n), C(n) = (n^2 + n)/2. The basic procedure here is the following. Pick as many test solutions as you have coefficient functions; here thatÕs three, and the test solutions are R_n = 1, R_n = n, and R_n = n^2. If your test solutions really are solutions, each one will correspond to a specific choice of the parameters alpha, beta, etc. After you find the parameters corresponding to a solution, you can use them and the relationship R_n = A(n)*alpha + B(n)*beta + C(n)*gamma to get an equation involving the coefficient functions. If youÕve picked independent test solutions, the resulting equations will be independent. In this simple example they were A(n) = 1, B(n) = n, and n^2 = 2C(n) - B(n). You then solve this system for the coefficient functions, and you have your general solution. Here itÕs R_n = alpha + beta*n + gamma*(n^2 + n)/2. [...] Brian === Subject: Re: solving recurrence relations > in alt.math.undergrad: > The basic procedure here is the following. Pick as many > test solutions as you have coefficient functions; here > thatÕs three, and the test solutions are R_n = 1, R_n = n, > and R_n = n^2. If your test solutions really are solutions, > each one will correspond to a specific choice of the > parameters alpha, beta, etc. After you find the parameters > corresponding to a solution, you can use them and the > relationship Would you elaborate on If your test solutions really are solutions, ...? What if I had selected other values for R_n to solve for, other than R_n = 1, R_n = n and R_n = n^2? Mike http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- === Subject: Re: solving recurrence relations in alt.math.undergrad: >> in alt.math.undergrad: >> The basic procedure here is the following. Pick as many >> test solutions as you have coefficient functions; here >> thatÕs three, and the test solutions are R_n = 1, R_n = n, >> and R_n = n^2. If your test solutions really are solutions, >> each one will correspond to a specific choice of the >> parameters alpha, beta, etc. After you find the parameters >> corresponding to a solution, you can use them and the >> relationship > Would you elaborate on If your test solutions really are solutions, ...? > What if I had selected other values for R_n to solve for, other than R_n = > 1, R_n = n and R_n = n^2? You might have picked functions that donÕt satisfy the recurrence. Recall that it was R_0 = alpha, R_n = R_{n-1} + beta + gamma*n, for n > 0. Suppose that you try R_n = n^3. Then R_0 = 0^3 = 0, so alpha = 0, and you also have n^3 = (n - 1)^3 + beta + gamma*n, for n > 0. Expand and simplify: n^3 = n^3 - 3n^2 + 3n - 1 + beta + gamma*n, 3n^2 - 3n + 1 - beta - gamma*n = 0, and finally 3n^2 - (gamma + 3)*n + (1 - beta) = 0 for all n > 0. matter what beta and gamma are, the function of n on the left is a quadratic and has at most two zeroes. Remember that the general solution turns out to be R_n = A(n)*alpha + B(n)*beta + C(n)*gamma, where A(n), B(n), and C(n) are three *particular* functions. In linear algebraic terms, the solutions of the recurrence, as alpha, beta, and gamma run over all possible triples of real numbers, are just the linear combinations of the functions A(n), B(n), and C(n). Obviously there must be many functions that simply arenÕt linear combinations of these three, and if you try one of them, you wonÕt be able to solve successfully for alpha, beta, and gamma. Brian === Subject: Re: solving recurrence relations > in alt.math.undergrad: > [...] > Brian Brian, I apologize for the notation. Next time I will know. I will study your response and get back with you if I have questions. I wish the mathematicians who write math textbooks would assume that the reader is a novice. I love math but sometimes I get so stuck because I have not yet learned to read between-the-lines. Mike http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- === Subject: Re: solving recurrence relations alt.math.undergrad: >> in alt.math.undergrad: > I apologize for the notation. Next time I will know. No need to apologize: itÕs not something youÕd be likely to guess if you hadnÕt seen ASCII-fied mathematics before. > I will study your response and get back with you if I have questions. Feel free: I occasionally teach a course from Graham, Knuth, & Patashnik, so IÕm pretty familiar with quite a bit of it. > I wish the mathematicians who write math textbooks would > assume that the reader is a novice. GK&P is mostly pretty clear, but the discussion of the repertoire method does leave quite a bit to be desired. (I believe that thereÕs even a marginal note to that effect.) ItÕs too bad that this expository low point comes so early in the book. [...] Brian === Subject: Re: solving recurrence relations >I hope this isnÕt too long of an example. Here goes... > I am studying a math book called Concrete Mathematics: A Foundation of > Computer Science, Gramham, Knuth & Patashnik. Not for any classes, just > on my own. > On page 25, section 2.2 - Sums and Recurrences, is the following example, > as best I can do using a text editor. > The sum > Sn = sigma notation where k=0 to n, ak > is equivalent to the recurrence > S0 = a0; > Sn = Sn-1 + an, for n > 0; (2.6) > Therefore we can evaluate sums in closed form by using the methods we > learned in Chapter 1 to solve recurrences in closed form. > For example, if an is equal to a constant plus a multiple of n, the > sum-recurrence (2.6) takes the following general form: > Ro = alpha; > Rn = Rn-1 + beta + gama x n, for n > 0; //x = multiplication > So, in general the solution can be written in the form > Rn = A(n) x alpha + B(n) x beta + C(n) x gama, > where A(n), B(n) and C(n) are the coefficients of dependence on the > general parameters alph, beta and gama. > The repertoire method tells us to try plugging in simple functions of n > for Rn, hoping to find constant parameters alpha, beta and gama where the > solution is especially simple. > Setting Rn = 1 implies that alpha = 1, beta = 0 and gama = 0; hence > A(n) = 1. > Setting Rn = n implies that alpha = 0, beta = 1 and gama = 0; hence > B(n) = n. > Setting Rn = n^2 implies that alpha = 0, beta = -1 and gama = 2; hence > 2C(n) - B(n) = n^2 > and we have C(n) = (n^2 + n)/2. Easy as pie. > Right. I follow until the last sentence where Rn = n^2. How in the world > would a person just guess the general parameters of alpha = 0, beta = -1 > and gama = 2? It seems to me that this is a bit of a contrived example > where the writer knows the solution and fits the parameters to that > solution. In most cases I wonÕt know the solution, so I am a bit confused > by the example. What if I had guessed alpha = 1, beta = -2 and gama = 3? Where would that leave me? Mike http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- === Subject: JSH: Rest of story Some of you may wonder where I get the polynomial P(x) that I factor so creatively, or how I figured out that factorization. Well, for a lot of years I tried to find a short and easy proof of FermatÕs Last Theorem. What IÕm doing is pulling from *some* of that research. Actually IÕm pulling a few sentences from a certain proof. Those sentences I expanded out into a paper, which is that paper that went to Southwest Journal of Pure and Applied Mathematics: The editors at that journal yanked my paper after some sci.math posters emailed them claiming it was wrong, and did so, within less than 24 hours of the emailings as I noticed when the sci.mathÕers talked about emailing the journal, and my paper was yanked the NEXT DAY. The journal had the paper for over nine months, and claimed it peer reviewed it, and the editors told me that it was well reviewed and thanked me for the paper, and then the sci.mathÕers emailed. IÕve re-written the argument, have a new paper, and a new title, and itÕs currently at the Annals in Princeton. IÕve been arguing about it for a couple of years on sci.math and other newsgroups, and that comes from a couple of sentences from the full proof. Technically, I use x^{2p} - x^p y^p + y^{2p} = z^{2p} to get the polynomials that you see posted with which I explain out how the math works. ThatÕs from a few sentences that I pulled out. So why that equation and not x^p + y^p = z^p? Well, IÕm not proving FLT, now am I? IÕm explaining an error in algebraic number theory, so I donÕt need the FLT equation, and found it convenient to use a variation on it. Basically the mathematics that is right that you are being taught is primitive. And as for whatÕs wrong, itÕs just criminal that youÕre still being taught it. The correct mathematics is far more powerful. Small minds will continue to attack it, as they usually do. People threatened by the truth will fight it, and one of the ways to do that is dragging of the heels. ItÕs passive-aggressive b.s. but thatÕs basically what math society is doing now. They try to sit on their hands, drag things out, and I guess hope that IÕll just go away. I can imagine them praying every night for the bad man to please stop. Except IÕm the good guy, so who are they really praying to, for what? We got here from people like me pushing the envelope, making the discoveries, and discovering the truth. And people like me have to deal with people who fight the truth for selfish reasons. And people like me, win. What I challenge you to do is pay careful attention and check for yourselves. Put not your trust in some professor or some book or some Usenet poster making big claims. Put your trust where it belongs--in mathematical truth. Trust no one. Check for yourself. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Rest of story > The editors at that journal yanked my paper after some sci.math > posters emailed them claiming it was wrong, and did so, within less > than 24 hours of the emailings as I noticed when the sci.mathÕers > talked about emailing the journal, and my paper was yanked the NEXT > DAY. Evidently the editing of that periodical is a little careless, and someone let JSHÕs paper in. After being informed by many sci.math readers of their error, they quickly yanked the paper. This shows that journal editors are human, not that the entire math establishment is a conspiracy to suppress the results of JSH. This incident does not reßect well on anyone, and unfortunately it gives JSH fuel for his delusional fire. === Subject: Re: Rest of story > Some of you may wonder where I get the polynomial P(x) that I factor > so creatively, or how I figured out that factorization. Well, considering that it is pretty much useless, I figure that it just came from your mindless mental meanderings. > Well, for a lot of years I tried to find a short and easy proof of > FermatÕs Last Theorem. And failed. > What IÕm doing is pulling from *some* of that research. Oooooooh... research... right... Idiot! > Actually IÕm pulling a few sentences from a certain proof. You proved something? I must have missed that. > sentences I expanded out into a paper, which is that paper that went > to Southwest Journal of Pure and Applied Mathematics: Where it was accidentally published by an editor asleep at the switch. > The editors at that journal yanked my paper after some sci.math > posters emailed them claiming it was wrong, and did so, within less > than 24 hours ... Because the editor realized his mistake. > The journal had the paper for over nine months, and claimed it peer > reviewed it, and the editors told me that it was well reviewed and > thanked me for the paper, and then the sci.mathÕers emailed. How do you know it was actually peer reviewed? Did you get anything back from a reviewer? And if it was peer reviewed, does that prove that your paper was correct? > IÕve re-written the argument, have a new paper, and a new title, and > itÕs currently at the Annals in Princeton. It will not get published there. I guarantee it. I would bet my Golden RetrieverÕs life on it. > Well, IÕm not proving FLT, now am I? Nope. You tried that for long enough and got nowhere. > IÕm explaining an error in > algebraic number theory, so I donÕt need the FLT equation, and found > it convenient to use a variation on it. There is no error in algebraic number theory. The error is in your feeble efforts in algebra, not a problem with the discipline. > And as for whatÕs wrong, itÕs just criminal that youÕre still being > taught it. Criminal? You have shown that you do not understand much about copyright law, and now you show that you do not understand much about criminal law. What criminal law are you talking about, and in what jurisdiction? > The correct mathematics is far more powerful. What is correct mathematics? How is it more powerful. > Except IÕm the good guy, so who are they really praying to, for what? > Trust no one. Check for yourself. I did... and I believe you are full of it. === Subject: Softmath? Alebrator http://www.softmath.com Does anyone know anything about that program? I want to know if itÕs worth the money. I have an online math class, and the tutoring you get via the Internet is not enough. CP === Subject: looking for f(x), no Lagrange Polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Rr31087; finding the function f(x) such that when x=1 f(1)=5; x=2 f(2)=1; x=3 f(3)=2; x=4 f(4)=3; x=5 f(5)=4. no Lagrange Polynomials === Subject: Re: looking for f(x), no Lagrange Polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2G66l11666; >finding the function f(x) such that when x=1 f(1)=5; x=2 f(2)=1; x=3 >f(3)=2; x=4 f(4)=3; x=5 f(5)=4. >no Lagrange Polynomials We might think about cyclic maps ; here combine f(f(5)=3... => f^[5](5)=5,f^[5](4)=4... cycle=5 . Or notice f(x)=x-1 except for 1 ,f(1)=5 ;let us add a correcting term ,nul when x=2,3,4,5 for instance:Int((6-x)/5) Int or ßoor() and adjust to the needed value f(1)=5: f(x)= x-1+ 5*Int((6-x)/5) , There are many other solutions... good luck,Alain. === Subject: Re: looking for f(x), no Lagrange Polynomials > finding the function f(x) such that when x=1 f(1)=5; x=2 f(2)=1; x=3 > f(3)=2; x=4 f(4)=3; x=5 f(5)=4. > no Lagrange Polynomials How about... f(x) = x-1 for x != 1 f(x) = 5 for x = 1 Of course there are many other alternatives that work. Mike. === Subject: Re: looking for f(x), no Lagrange Polynomials > finding the function f(x) such that when x=1 f(1)=5; x=2 f(2)=1; x=3 > f(3)=2; x=4 f(4)=3; x=5 f(5)=4. > no Lagrange Polynomials f(x) = 5 if x=1 1 if x=2 2 if x=3 3 if x=4 4 if x=5 0 for all other values of x There you go -- itÕs a function and it doesnÕt use Lagrange polynomials. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Help on Uniform Convergence topic please? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Xe31410; I was asked to prove the following assertion: Prove that {f_n:D->R} converges uniformly to f:D->R if and only iff lim [ sup |f_n(x) - f(x)| ] = 0. n->oo[ x in D ] --- To be honest this isnÕt intuitively obvious to me just yet. Perhaps I am understanding the supremum statement incorrectly. Define S_n = sup |f_n(x) - f(x)|. x in D Then, does lim S_n = 0 n->oo imply that for all x in D and epsilon > 0, there exists some natural number N such that |f_n(x) - f(x)| < sup |f_n(x) - f(x)| = |f_N(x) - f(x)| < epsilon, for all n > N ??? I donÕt think I am making any sense, so help would be greatly appreciated! === Subject: Re: Help on Uniform Convergence topic please? in alt.math.undergrad: > I was asked to prove the following assertion: > Prove that {f_n:D->R} converges uniformly to f:D->R if and only iff > lim [ sup |f_n(x) - f(x)| ] = 0. > n->oo[ x in D ] > --- > To be honest this isnÕt intuitively obvious to me just yet. Perhaps I > am understanding the supremum statement incorrectly. Define > S_n = sup |f_n(x) - f(x)|. > x in D > Then, does > lim S_n = 0 > n->oo > imply that for all x in D and epsilon > 0, there exists some natural > number N such that >|f_n(x) - f(x)| < sup |f_n(x) - f(x)| = |f_N(x) - f(x)| < epsilon, > for all n > N ??? It means that for any e > 0 there is a natural number N such that |S_n| < e. (Since itÕs clear that S_n >= 0, this amounts to saying that S_n < e.) This is equivalent to the statement that for any e > 0 thereÕs a nat. number N such that |S_n| <= e (why?), which is slightly easier to work with here. Now ask yourself what it means for S_n to be less than or equal to e: sup{|f_n(x) - f(x)| : x in D} <= e. This is exactly equivalent to the statement that |f_n(x) - f(x)| <= e for each x in D. Thus, the statement that the limit of the S_n is 0 means that for any e > 0 there is a nat. number N such that for all x in D and all n >= N, |f_n(x) - f(x)| <= e. [...] Brian === Subject: Re: Help on Uniform Convergence topic please? >in alt.math.undergrad: >> I was asked to prove the following assertion: >> Prove that {f_n:D->R} converges uniformly to f:D->R if and only iff >> lim [ sup |f_n(x) - f(x)| ] = 0. >> n->oo[ x in D ] >> --- >> To be honest this isnÕt intuitively obvious to me just yet. Perhaps I >> am understanding the supremum statement incorrectly. Define >> S_n = sup |f_n(x) - f(x)|. >> x in D >> Then, does >> lim S_n = 0 >> n->oo >> imply that for all x in D and epsilon > 0, there exists some natural >> number N such that >>|f_n(x) - f(x)| < sup |f_n(x) - f(x)| = |f_N(x) - f(x)| < epsilon, >> for all n > N ??? >It means that for any e > 0 there is a natural number N such >that |S_n| < e for all n > N. >. (Since itÕs clear that S_n >= 0, this >amounts to saying that S_n < e.) This is equivalent to the >statement that for any e > 0 thereÕs a nat. number N such >that |S_n| <= e (why?), which is slightly easier to work >with here. >Now ask yourself what it means for S_n to be less than or >equal to e: > sup{|f_n(x) - f(x)| : x in D} <= e. > >This is exactly equivalent to the statement that > |f_n(x) - f(x)| <= e for each x in D. >Thus, the statement that the limit of the S_n is 0 means >that for any e > 0 there is a nat. number N such that > for all x in D and all n >= N, |f_n(x) - f(x)| <= e. >[...] >Brian ************************ David C. Ullrich === Subject: Re: Maths, english, science and geography by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2IUlH25856; === Subject: Break Even analysis and TFC by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2JDY930052; can anyone help on this problem? Given the following information: Selling price per unit(p) = $10 Variable cost per unit(v) = $7 Plant capacity = 125,000 units Sales demand = 80,000 units Profit = $90,000 Total fixed costs (TFC) are equal to:??? this is givin me trouble can anyone help?? Chrisothoulos === Subject: Re: Break Even analysis and TFC > Given the following information: > Selling price per unit(p) = $10 > Variable cost per unit(v) = $7 > Plant capacity = 125,000 units > Sales demand = 80,000 units > Profit = $90,000 > Total fixed costs (TFC) are equal to:??? Start from the top... Profit = Revenue - Total Cost You are given profit and are told enough to determine revenue, so you can solve for the total cost. Then, also by definition... Total Cost = Variable Cost + Total Fixed Costs You are given enough info to figure out the variable cost, you know total cost from the previous step, so you can find TFC. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Break Even analysis and TFC > can anyone help on this problem? > Given the following information: > Selling price per unit(p) = $10 > Variable cost per unit(v) = $7 > Plant capacity = 125,000 units > Sales demand = 80,000 units > Profit = $90,000 > Total fixed costs (TFC) are equal to:??? > this is givin me trouble can anyone help?? > Chrisothoulos Setting aside fixed costs for the moment, you make $3/unit. Right? So if you sold 80,000 you make in $s how much? And if your profit was $90,000 how much was your fixed cost? Bill === Subject: Statistic and probability problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2JbVD32379; I really need help on this problem, can anyone help? problem: According to the U.S. National Center for Health Statistics, 21.3% of all adults will develop a heart condition. If you randomly select 6 adults, what is the probability that 2 of them will develop a heart condition? === Subject: Re: Statistic and probability problem >I really need help on this problem, can anyone help? > problem: > According to the U.S. National Center for Health Statistics, 21.3% of > all adults will develop a heart condition. If you randomly select 6 > adults, what is the probability that 2 of them will develop a heart > condition? This sound like a binomial distribution. http://stat.tamu.edu/stat30x/notes/node66.html Bill And you need to be clear on whether you mean exactly 2 or 2 or more. === Subject: Re: Break Even analysis and TFC by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2Kx1E08188; ok so 80,000 X 3 is 240,000 and 90,000 profit you would just subtract 90,000 from 240,000 which would be 150,000 so 150,000 is fixed cost..... that sounds about right.... chrisothoulos === Subject: JSH: Polynomial factors versus non-polynomial ones Essentially debates with me about non-polynomial factorization boil down to some insistent belief that non-polynomial factors follow different rules from polynomial factors. Like consider P(x) = (ax + 2)(x+1) = 2(x^2 + 2x + 1) and the question, what is a? I doubt many of you need to actually solve for ÔaÕ to figure out that itÕs 2. But how do you know? Here you can *see* that itÕs 2, but can you trace out the logic? If you let g_1(x) = ax + 2, and g_2(x) = x+1, then at x=0, you have g_1(0) g_2(0) = 2 with constant terms that are 2 and 1. If you divide 2 from both sides then you have P(x)/2 = x^2 + 2x + 1 and at x=0, then you have a constant term that is 1. So a 2 divided off from one of the constant terms, and from the distibutive property, you know g_1(x) has 2 as factor, therefore ÔaÕ has it as a factor, and a = 2, is in fact the solution. Now thatÕs trivial with polynomials, but what about non-polynomial factors? CanÕt it all change? Must it not change? Well, logically, you still have factors of the full polynomial. No matter how weird they may be, the factors of a polynomial are the factors of the polynomial. Those factors multiply to give the polynomial in all its parts, including its constant term. The factors of the constant term are determinable by clearing out the variable i.e. setting the polynomial variable to 0. But what if you have this nagging feeling that maybe thereÕs no such thing as a constant term for certain non-polynomial variables? Well, your nagging feeling is a non-separability feeling. That is, you feel that possibly the factors of the constant term are not separable within non-polynomial functions from the factors of the rest of the polynomial. Like, with P(x) = x^2 + 2x + 1, you believe that if I have non-polynomial factors g_1(x), and g_2(x) where g_1(x) g_2(x) = x^2 + 2x + 1 that the factors of 1 within g_1(x) and g_2(x) are not separable from the factors of x^2 + 2x. Now then, if you donÕt realize thatÕs what you must believe then no progress can be made. After all, if you believe they ARE separable, then thereÕs little room to argue with my results! So why would you believe that they are not separable? Well, for one thing, you canÕt SEE them, like you can with the simpler polynomial factors. However, logically you can prove separability. But my problem is that no matter how many times I step through the logic, and no matter how many times I show the result follows mathematically, I run into many of you who believe something different, which boils down to the deep belief that separability doesnÕt exist with non-polynomial factors. My guess is that you learn it intuitively, like by looking a something like f(x) = sqrt(x+1) where YOU canÕt separate things out and see. So I keep running into a wall of intuition thatÕs just wrong, and unfortunately few of you seem to be taught to trust logic over intuition, so the arguments continue. James Harris === Subject: Re: JSH: Polynomial factors versus non-polynomial ones > Essentially debates with me about non-polynomial factorization boil > down to some insistent belief that non-polynomial factors follow > different rules from polynomial factors. > Like consider > P(x) = (ax + 2)(x+1) = 2(x^2 + 2x + 1) > and the question, what is a? > I doubt many of you need to actually solve for ÔaÕ to figure out that > itÕs 2. > But how do you know? > Here you can *see* that itÕs 2, but can you trace out the logic? Sure. Given: (ax + 2)(x+1) = 2(x^2 + 2x + 1) Simplifying each side (by the distributive property, and associative and commutative properties of addition) gives: ax^2 + (2+a)x + 2 = 2x^2 + 4x + 1 For these to be equal, the coefficients must be equal, so we get: a=2 and 2+a=4. By the addition property of equations I can add -2 to both sides of the second equation and get: a=2 and a=2. Therefor, a=2. > If you let g_1(x) = ax + 2, and g_2(x) = x+1, then at x=0, you have > g_1(0) g_2(0) = 2 > with constant terms that are 2 and 1. > If you divide 2 from both sides then you have > P(x)/2 = x^2 + 2x + 1 > and at x=0, then you have a constant term that is 1. At this point you are doing nothing close to what I would do. You may think I am being deliberately contrary here, but what I put above is 10000% more likely to be my approach than what you are doing. > So a 2 divided off from one of the constant terms, and from the > distibutive property, you know g_1(x) has 2 as factor, therefore ÔaÕ > has it as a factor, and a = 2, is in fact the solution. This is about as clear as mud. g_1(x) = ax+2. HOW do you know it has 2 as a factor? It would be nice to mention that polynomials are a Unique Factorization Domain, at least. Then you could justify that 2 is a factor of P(x) and is not a factor of g_2(x), therefor it must be a factor of g_1(x). Therefor a must have 2 as a factor. You havenÕt stated why a cannot be 4, however. I know a is not 4, but you havenÕt shown that. > Now thatÕs trivial with polynomials, but what about non-polynomial > factors? ThatÕs a strange way to do it with polynomials. Perhaps if you backed up some of your steps with algebraic properties it would become clearer. > CanÕt it all change? Must it not change? I would recommend completely changing your approach to solving such problems. > Well, logically, you still have factors of the full polynomial. > No matter how weird they may be, the factors of a polynomial are the > factors of the polynomial. > Those factors multiply to give the polynomial in all its parts, > including its constant term. > The factors of the constant term are determinable by clearing out the > variable i.e. setting the polynomial variable to 0. Be careful here. Constant term is a defined concept when discussing polynomials. It is not a defined concept when discussing functions in general. Term is not necessarily a defined or applicable concept when discussing functions in general. For example: f(x)=sqrt(x+1) does not have a constant term as a radical function. However, you are claiming that it has a constant term of 1 since f(0) = 1. Your use of these concepts does not agree with standard definitions IÕm familiar with (at least in intermediat/college algebra textbooks). > But what if you have this nagging feeling that maybe thereÕs no such > thing as a constant term for certain non-polynomial variables? > Well, your nagging feeling is a non-separability feeling. That is, > you feel that possibly the factors of the constant term are not > separable within non-polynomial functions from the factors of the rest > of the polynomial. So youÕre saying that f(x)=sqrt(x+1) should have been written as f(x)=[sqrt(x+1) -1] + 1? You can write it that way but the notion of terms is normally applied to the simplified form, which the seperated version clearly is not. > Like, with P(x) = x^2 + 2x + 1, you believe that if I have > non-polynomial factors g_1(x), and g_2(x) where g_1(x) g_2(x) = x^2 + > 2x + 1 that the factors of 1 within g_1(x) and g_2(x) are not > separable from the factors of x^2 + 2x. > Now then, if you donÕt realize thatÕs what you must believe then no > progress can be made. Perhaps using different terminology that doesnÕt clash with standard definitions would help as well. > After all, if you believe they ARE separable, then thereÕs little room > to argue with my results! Sorry, your results do not hang on this concept. > So why would you believe that they are not separable? > Well, for one thing, you canÕt SEE them, like you can with the simpler > polynomial factors. IÕm glad you know what I can and canÕt see. > However, logically you can prove separability. Start by clearly defining it. I have a notion of what you mean based on the rewrite above, but it could be wrong. > But my problem is that no matter how many times I step through the > logic, and no matter how many times I show the result follows > mathematically, I run into many of you who believe something > different, which boils down to the deep belief that separability > doesnÕt exist with non-polynomial factors. > My guess is that you learn it intuitively, like by looking a something > like > f(x) = sqrt(x+1) > where YOU canÕt separate things out and see. What do you want seperated out? I already demonstrated how to seperate out a +1 as a seperate term. > So I keep running into a wall of intuition thatÕs just wrong, and > unfortunately few of you seem to be taught to trust logic over > intuition, so the arguments continue. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Polynomial factors versus non-polynomial ones days. My association with the Department is that of an alumnus. >Sure. Given: (ax + 2)(x+1) = 2(x^2 + 2x + 1) >Simplifying each side (by the distributive property, and associative and >commutative properties of addition) gives: >ax^2 + (2+a)x + 2 = 2x^2 + 4x + 1 >For these to be equal, the coefficients must be equal, so we get: >a=2 and 2+a=4. Careful: this is true if we assume the two expressions are equal ->for all x<-, and even then, we must be careful (as to what the xÕs can be; e.g., elements of a ring, finite field, what?). -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Polynomial factors versus non-polynomial ones >>Sure. Given: (ax + 2)(x+1) = 2(x^2 + 2x + 1) >>Simplifying each side (by the distributive property, and associative and >>commutative properties of addition) gives: >>ax^2 + (2+a)x + 2 = 2x^2 + 4x + 1 >>For these to be equal, the coefficients must be equal, so we get: >>a=2 and 2+a=4. > Careful: this is true if we assume the two expressions are equal ->for > all x<-, and even then, we must be careful (as to what the xÕs can be; > e.g., elements of a ring, finite field, what?). True. I was interpretting it as occurring in Z[x] or R[x] and looking for a value of a in Z or R. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Polynomial factors versus non-polynomial ones Any news back from the Annals yet? === Subject: Acceleration Problem Assume that a fully loaded plane starting from rest has a constant acceleration while moving down a runway. The plane requires .8 mile of runway and a speed of 160 mph in order to lift off. What is the planeÕs acceleration? === Subject: Re: Acceleration Problem >Assume that a fully loaded plane starting from rest has a constant >acceleration while moving down a runway. The plane requires .8 mile >of runway and a speed of 160 mph in order to lift off. What is the >planeÕs acceleration? You know that v = at and s = (1/2)at^2. You have s and v. Eliminate t between the two equations and solve for a. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Acceleration Problem Use V^2 = U^2 + 2as V=160 mph U=0 mph s=0.8 miles solve for a. (The result will be measured in mile per hour per hour. Multiply by appropriate factor to get different units). > Assume that a fully loaded plane starting from rest has a constant > acceleration while moving down a runway. The plane requires .8 mile > of runway and a speed of 160 mph in order to lift off. What is the > planeÕs acceleration? === Subject: JSH: Did you ever finish your FLT proof? I was wondering... you never talk about FLT these days... Did you give up? Did you admit failure? Did you try sending that to the Annals for publication? And your attempts at RSA factoring... did you ever send in a solution for the money prize? And you recent work on APF, which you sent to the Annals of Mathematics............ any word back yet? And another question... could you tell us exactly what Barry Mazur said to you about your ideas? In other words, have you ever succeeded anywhere? You see yourself as a great mathematician... but have you convinced anyone other than yourself of that? === Subject: JSH: Look at it backwards I usually start with one polynomial and then talk about dividing 49 off from it, but here IÕll start *after* 49 has been divided off: S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 and consider the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) and the assertion that the dÕs must have factors w_1(x), w_2(x), and w_3(x) that multiply to give w_1(x) w_2(x) w_3(x) = 7. But thereÕs no factors of 7 anywhere. So why should the dÕs have functions that are factors of 7? Ok, maybe that seems unfair, as there could be LOTS of different functions you can plug in for the cÕs and dÕs, but why should ANY of them have functions of x that are factors of 7? The equation has no memory. You have a memory, so if I start with the polynomial multiplied by 49, then you can say to yourself that thereÕs some dependency on 7. But itÕs a mirage. Mathematically a constant multiple is not a big deal. ItÕs just a constant multiple that you can divide off, leaving a result that--you guessed it--has no memory of the multiple! One of the weirder things about the discussions here, which I assume escapes most of you is some fascinating belief that the equation has a memory. You see something like P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 and your brain apparently HOLLERS at you that 49 is still there, but no, itÕs gone. When I show something like P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) you brain insists that 7 is still there. I mean just LOOK! You can see 7 dividing two of the aÕs and for GodÕs sake!!! ThereÕs a 7 in that last factor, you see it, donÕt you? In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of course 7 is still there!!! So some poster comes at you claiming that 49 divides off from P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) in a way that varies dependent on the value of x, and your brain tells you, OK!!! You have a memory. To you 7 is still there, even though the factor is divided off. Follwing the sci.mathÕers insistent raving you get P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) where the wÕs are factors of 7, so that youÕre left with functions of 7. But they would STILL be there with the factorization of S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 inexplicably still there despite the polynomial not having any factors of 7. Your eyes fool you. Your memory betrays you. The math doesnÕt bother with such nonsense. A multiple of a polynomial can be divided off and itÕs gone. It doesnÕt leave a trace. In a way whatÕs happening now is an instructive lesson in the limitations of the human brain. Your brains SEE something, and insistently tell you that 7 is STILL THERE, and so the arguments go on for years. After all, you can SEE the 7Õs. Come on, whoÕs fooling who, right? Dammit. You can see the 7Õs in there, canÕt you? James Harris === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here IÕll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) > and the assertion that the dÕs must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3). Then the w_i are factors of the d_i in the ring of algebraic numbers. (Are they factors in any other ring? Who knows? Who cares? No one except you has made such a claim.) > But thereÕs no factors of 7 anywhere. So why should the dÕs have > functions that are factors of 7? the dÕs have functions that are factors of 7 is gibberish. -William Hughes === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here IÕll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the dÕs must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But thereÕs no factors of 7 anywhere. So why should the dÕs have >functions that are factors of 7? You are getting maximally confused. In your original approach, you are factoring P(x) as a polynomial in 5, not as a polynomial in x. The constant term with respect to 5 and the constant term with respect to x are different. It was a stupid mistake to think you were simplifying by treating 5 as if it were the polynomial variable in the first place. And that is what is causing your confusion here. >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the cÕs and dÕs, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. The memory problem here is, you have forgotten what you started with. To see this, you need to go back to the way you were doing things originally: for example, P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) To put this in terms of what you are doing now: replace x in this expression by 5, and replace m by x, and replace f by 7. Also replace u by 1. This gives P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), and you can clearly see that, as a polynomial in 5, the constant term is 49 * 7 = 7^3, not 49 * 22. Go back to the expression above for P(m). The constant term with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to 49*22 when u = 1, f = 7, and x = 5. But the constant term with respect to x is u^3 f^3. You have forgotten this and gotten mixed up, thatÕs all. >You have a memory, so if I start with the polynomial multiplied by 49, >then you can say to yourself that thereÕs some dependency on 7. But >itÕs a mirage. >Mathematically a constant multiple is not a big deal. ItÕs just a >constant multiple that you can divide off, leaving a result that--you >guessed it--has no memory of the multiple! >One of the weirder things about the discussions here, which I assume >escapes most of you is some fascinating belief that the equation has a >memory. >You see something like >P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 >and your brain apparently HOLLERS at you that 49 is still there, but >no, itÕs gone. No - my brain does no such hollering. >When I show something like >P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) Key thing to note here: as I said, the polynomial variable with respect to which you are factoring is 5, not x. See? >you brain insists that 7 is still there. >I mean just LOOK! You can see 7 dividing two of the aÕs and for GodÕs >sake!!! >ThereÕs a 7 in that last factor, you see it, donÕt you? >In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of course 7 is >still there!!! >So some poster comes at you claiming that 49 divides off from >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) >in a way that varies dependent on the value of x, and your brain tells >you, OK!!! >You have a memory. To you 7 is still there, even though the factor is >divided off. As a polynomial in 5, itÕs still there. It is not a coincidence that 22 = 3 * 5 + 7. >Follwing the sci.mathÕers insistent raving you get >P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + >7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) >where the wÕs are factors of 7, so that youÕre left with functions of >But they would STILL be there with the factorization of >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >inexplicably still there despite the polynomial not having any factors >of 7. >Your eyes fool you. Your memory betrays you. No, itÕs YOUR memory that is the problem. You have forgotten how you were factoring: not with respect to x, but with respect to 5. >The math doesnÕt bother with such nonsense. A multiple of a >polynomial can be divided off and itÕs gone. >It doesnÕt leave a trace. >In a way whatÕs happening now is an instructive lesson in the >limitations of the human brain. Your brains SEE something, and >insistently tell you that 7 is STILL THERE, and so the arguments go on >for years. >After all, you can SEE the 7Õs. Come on, whoÕs fooling who, right? You are hopelessly tangled in your own dimwitted oversimplification. You have forgotten what you were doing. >Dammit. >You can see the 7Õs in there, canÕt you? Sure. On *both* sides. Just look at: P(x)/49 = (x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7, and note that the constant term is ... 7, just like it should be, if you regard 5 as the polynomial variable! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards >I usually start with one polynomial and then talk about dividing 49 >off from it, but here IÕll start *after* 49 has been divided off: >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >and consider the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >and the assertion that the dÕs must have factors >w_1(x), w_2(x), and w_3(x) >that multiply to give w_1(x) w_2(x) w_3(x) = 7. >But thereÕs no factors of 7 anywhere. So why should the dÕs have >functions that are factors of 7? > You are getting maximally confused. In your original > approach, you are factoring P(x) as a polynomial in 5, > not as a polynomial in x. The constant term with > respect to 5 and the constant term with respect to > x are different. It was a stupid mistake to think you > were simplifying by treating 5 as if it were the > polynomial variable in the first place. And that is what > is causing your confusion here. There is only one variable Nora Baron, so what other variable would you have listed with S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) considering that S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? So whoÕs confused? >Ok, maybe that seems unfair, as there could be LOTS of different >functions you can plug in for the cÕs and dÕs, but why should ANY of >them have functions of x that are factors of 7? >The equation has no memory. > The memory problem here is, you have forgotten what you > started with. To see this, you need to go back to the way you > were doing things originally: for example, Now note, the polynomial S(x) doesnÕt have 7 as a factor, as this time IÕm starting with the result of dividing off the multiple. So the poster Nora Baron (actually a guy as revealed in another post when he ended with a male name) is trying to show how 7 gets back into the expression. That is, heÕs trying to prove that the factorization DOES have a memory of the multiple 49. > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > To put this in terms of what you are doing now: replace x in this > expression by 5, and replace m by x, and replace f by 7. Also > replace u by 1. This gives > P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), > and you can clearly see that, as a polynomial in 5, the constant > term is 49 * 7 = 7^3, not 49 * 22. No. It can be factored with respect to 5 and 7, but 5 is not a variable. The polynomial variable is x. Now, IÕve explained lots of times to the Nora Baron poster, and what I want you all to consider is that the poster isnÕt really that dense, but instead knows you better than you know yourselves. Essentially the basic strategy is just to disagree. Time after time, and even in answer to surveys that IÕve done, readers who normally lurk will admit that they primarily rely on the fact that people argue with me, assuming that if I were right, then others wouldnÕt disagree! So, for sci.mathÕers like Nora Baron, the strategy is clear--just disagree. They often donÕt even TRY to actually make mathematical senze. > Go back to the expression above for P(m). The constant term > with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to > 49*22 when u = 1, f = 7, and x = 5. But the constant term with > respect to x is u^3 f^3. You have forgotten this and gotten mixed > up, thatÕs all. Now the expressions actually is S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 where the factorization S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) is being considered. Now how do you get 7 into that? Well, thatÕs not the issue for the poster! All he has to do is disagree. For most readers thatÕs what works. Well, that enough here. James Harris === Subject: Re: JSH: Look at it backwards by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB42XUG02024; >>I usually start with one polynomial and then talk about dividing 49 >>off from it, but here IÕll start *after* 49 has been divided off: >>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >>and consider the factorization >>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >>and the assertion that the dÕs must have factors >>w_1(x), w_2(x), and w_3(x) >>that multiply to give w_1(x) w_2(x) w_3(x) = 7. >>But thereÕs no factors of 7 anywhere. So why should the dÕs have >>functions that are factors of 7? >> You are getting maximally confused. In your original >> approach, you are factoring P(x) as a polynomial in 5, >> not as a polynomial in x. The constant term with >> respect to 5 and the constant term with respect to >> x are different. It was a stupid mistake to think you >> were simplifying by treating 5 as if it were the >> polynomial variable in the first place. And that is what >> is causing your confusion here. >There is only one variable Nora Baron, I didnÕt say there were other variables in your present polynomial. I was describing its origin, when there were 4 variables, which explains why you are now confused about the whole thing. > so what other variable would >you have listed with >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >considering that >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22? >So whoÕs confused? No question about that. You are. >>Ok, maybe that seems unfair, as there could be LOTS of different >>functions you can plug in for the cÕs and dÕs, but why should ANY of >>them have functions of x that are factors of 7? >>The equation has no memory. >> The memory problem here is, you have forgotten what you >> started with. To see this, you need to go back to the way you >> were doing things originally: for example, >Now note, the polynomial S(x) doesnÕt have 7 as a factor, I DIDNÕT SAY IT DID, ASSHOLE > as this time >IÕm starting with the result of dividing off the multiple. >So the poster Nora Baron (actually a guy as revealed in another post >when he ended with a male name) is trying to show how 7 gets back into >the expression. It gets back in in the constant term, because you have been factoring this function as if it were a polynomial in 5. >That is, heÕs trying to prove that the factorization DOES have a >memory of the multiple 49. Not at all. The 49 is obviously gone. But the constant term at the end, when P(x)/49 is viewed as a polynomial in 5 [your idea, not mine!], is still there, and it is 7, not 22. >> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) >> To put this in terms of what you are doing now: replace x in this >> expression by 5, and replace m by x, and replace f by 7. Also >> replace u by 1. This gives >> P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7), >> and you can clearly see that, as a polynomial in 5, the constant >> term is 49 * 7 = 7^3, not 49 * 22. >No. It can be factored with respect to 5 and 7, but 5 is not a >variable. You have treated it exactly as such. When you factor P(x) in the form P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7), you are NOT factoring it as a polynomial in x. If you were, the terms a_1(x), etc., would themselves be polynomials in x. As you yourself have pointed out MANY TIMES, this is not a polynomial factorization when considered as a function of x. (In fact you have claimed that such Ônonpolynomial factorizationÕ was one of your great conceptual breakthroughs. Now you want to deny it ???). However it CLEARLY has the FORM of a factorization in 5, where 5 is treated as a polynomial variable. >The polynomial variable is x. It most certainly is not [yes, I know you are going to howl about this, but I stand by what I say]. When you factor a b x^2 + (a + b) x + 1 as (a x + 1)(b x + 1), THEN you are factoring in terms of the polynomial variable x. The coefficients of x in the factors are functions of a and b, NOT functions of x, as in your present factorization of P(x). Have you really no recollection of how you got started on this track? Originally, x was m, and 5 was x. At that time you were considering factors like a_1(m) x + u f, etc. This was a factorization in terms of the polynomial variable x, with coefficients which were algebraic integer functions of m. Just plug in x = 5, u = 1, f = 7, and m = x, and you have your present factorization - as a polynomial in 5, not in x !! ItÕs easy, actually, to see how you have gotten confused, especially if (unlike real mathematicians) you have a poor memory. >Now, IÕve explained lots of times to the Nora Baron poster, and what >I want you all to consider is that the poster isnÕt really that dense, >but instead knows you better than you know yourselves. >Essentially the basic strategy is just to disagree. I only disagree when you are wrong. >Time after time, and even in answer to surveys that IÕve done, readers >who normally lurk will admit that they primarily rely on the fact that >people argue with me, assuming that if I were right, then others >wouldnÕt disagree! I donÕt recall people saying that. What lurkers do seem to think, when they speak up, is that YOUR math is vague, hand-waving and illogical. ThatÕs not my fault. Plus they seem to think that your habit of making nasty personal attacks damages your credibility. ThatÕs not my fault either. >So, for sci.mathÕers like Nora Baron, the strategy is clear--just >disagree. Again, I only disagree when you are wrong. Of course, that happens to be virtually all the time. >They often donÕt even TRY to actually make mathematical senze. This is gratuitous and false in my case, as you well know. Also in the cases of Dik Winter, Rick Decker, Arturo Magidin, ÔRupertÕ, Will Twentyman, Dale Hall, and others. >> Go back to the expression above for P(m). The constant term >> with respect to m is f^2 (3 xu^2 + u^3 f). That happens to reduce to >> 49*22 when u = 1, f = 7, and x = 5. But the constant term with >> respect to x is u^3 f^3. You have forgotten this and gotten mixed >> up, thatÕs all. >Now the expressions actually is >S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 >where the factorization >S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) >is being considered. >Now how do you get 7 into that? Well, thatÕs not the issue for the >poster! It was explained in what I posted. You need to consider how you arrive at S(x), and for that you need to consider the form of the original function P(x): as a polynomial in 5, it has constant term 7^3. I thought you were simply confused on this point, having forgotten how you got here. I am now inclined to think you are trying to deny the obvious correct explanation, though I am not sure why. Plus throughout your entire reply you seem to be playing to the grandstand. You whine, exaggerate, and attempt to discredit. The actual math you have hardly addressed at all. You seem to think that you can win mathematical arguments by an appeal to some great silent majority of people out there who you can cajole into thinking that we critics are evil, cheating liars who are jealous of your discoveries. (If that were true, one of us evil cheating liars would have published it long ago, of course giving no credit to James S.Harris. Funny, that hasnÕt happened. Wonder why? Why isnÕt anyone stealing your ideas?) You seem to have the impression that math is a popularity contest. ItÕs not. The only way to win is through rigorous proof. You donÕt have one. ThatÕs the main reason you are not winning. >All he has to do is disagree. >For most readers thatÕs what works. Right. Go ahead and ask your faithful admiring grandstand if thatÕs what works. >Well, that enough here. Yup - Duh - that enough ! Nora B. >James Harris === Subject: Re: JSH: Look at it backwards > So the poster Nora Baron (actually a guy as revealed in another post > when he ended with a male name) is trying to show how 7 gets back into > the expression. What difference does it make what NoraÕs gender is? What does it have to do with math? Has anybody but you made note of the question of NoraÕs gender? No, because it is not relevant. Your obsession with her (or his) gender just screams mental illness. It is (probably) just a pseudonym, nothing more. So what? My pseudonym is o[CapitalYAcute]in, so do you claim that I am pretending to be a Norse god? Well, nobody cares!!! Idiot!!! === Subject: Re: JSH: gametes So... James Harris got me wondering... which type of gametes do you actually produce, Nora Baron? === Subject: Re: JSH: gametes > So... James Harris got me wondering... which type of gametes do you actually > produce, Nora Baron? See how that works? You start by mocking JSH, and after a while you start to think like him. HeÕs addictive. HeÕs actually making some very good points about the social aspect of proof. Philosophers of math make the same arguments ... that a proof is whatever convinces a majority of working mathematicians; that whatÕs considered a proof in one era is regarded as utterly lacking in rigor in another. And now heÕs got you caring about the gender of a palindromic poster. === Subject: Re: JSH: gametes > See how that works? You start by mocking JSH, and after a while you > start to think like him. HeÕs addictive. HeÕs actually making some very > good points about the social aspect of proof. Philosophers of math make > the same arguments ... that a proof is whatever convinces a majority of > working mathematicians; that whatÕs considered a proof in one era is > regarded as utterly lacking in rigor in another. > And now heÕs got you caring about the gender of a palindromic poster. Yes. You are 100 correct in all that... but now I really do want to know... Damn it!!! === Subject: Re: JSH: gametes Nora B. Baron comes from a long line of palindromic forearmed folks; if you can make that stick for two generations, thatÕs farther than IÕve gotten! > And now heÕs got you caring about the gender of a palindromic poster. --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: JSH: Look at it backwards > I usually start with one polynomial and then talk about dividing 49 > off from it, but here IÕll start *after* 49 has been divided off: > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > and consider the factorization > S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x)) Ok, at this point there is no indication as to what these c_i(x) and d_i(x) might be, aside from the obvious necessary properties to generate the coefficients of S(x) when the factors are multiplied together. Are they meant to be continuous functions? Continuous almost everywhere? What are their domain/range? > and the assertion that the dÕs must have factors > w_1(x), w_2(x), and w_3(x) > that multiply to give w_1(x) w_2(x) w_3(x) = 7. If they have such factors, then the range must not be in the algebraic integers. Note: I mention that since you usually say something about everything is in the algebraic integers. > But thereÕs no factors of 7 anywhere. So why should the dÕs have > functions that are factors of 7? They can quite easily in the appropriate ring. For example, 22 has 7 as a factor in the rational numbers, algebraic numbers, reals, etc. What makes you say thereÕs no factors of 7 anywhere? > Ok, maybe that seems unfair, as there could be LOTS of different > functions you can plug in for the cÕs and dÕs, but why should ANY of > them have functions of x that are factors of 7? *IN WHAT RING?* Do you not see that you are making some sort of assumption that you have not shared with us? > The equation has no memory. Huh? Why would you even speak of such a thing? > You have a memory, so if I start with the polynomial multiplied by 49, > then you can say to yourself that thereÕs some dependency on 7. But > itÕs a mirage. Actually, you are the one who keeps talking about 7 with such a polynomial. > Mathematically a constant multiple is not a big deal. ItÕs just a > constant multiple that you can divide off, leaving a result that--you > guessed it--has no memory of the multiple! Clue: you are think of things as being processes. Try thinking of them as being equivalent statements. You can talk about two different equations being equivalent, but they are different equations. The notion of a process, or memory of prior steps in the process, is an artifact of the work a person does to work towards a goal. The result in something like the above is simply a set of equivalent statements and some justification for them being equivalent. > One of the weirder things about the discussions here, which I assume > escapes most of you is some fascinating belief that the equation has a > memory. Then why are you the only one who talks about it? > You see something like > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > and your brain apparently HOLLERS at you that 49 is still there, but > no, itÕs gone. I see it quite clearly. On the left side of the equation. Where it was in the previous equation (if at all) is irrelevent. > When I show something like > P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) > you brain insists that 7 is still there. > I mean just LOOK! You can see 7 dividing two of the aÕs and for GodÕs > sake!!! I see three 7s, actually. > ThereÕs a 7 in that last factor, you see it, donÕt you? > In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of course 7 is > still there!!! Precisely. > So some poster comes at you claiming that 49 divides off from > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) > in a way that varies dependent on the value of x, and your brain tells > you, OK!!! Wrong. Simple examples make it clear that, in general, this is how things work when viewed from the perspective of working on *factors* of a polynomial with a goal of keeping the terms of each factor in a particular ring. If you focus on terms of the factors or donÕt care about staying within a ring at any particular level, then it doesnÕt matter how you write it. > You have a memory. To you 7 is still there, even though the factor is > divided off. I see only /7, /7, and +7. > Follwing the sci.mathÕers insistent raving you get > P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) + > 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x)) > where the wÕs are factors of 7, so that youÕre left with functions of > 7. Functions of 7? Huh? > But they would STILL be there with the factorization of > S(x) = 300125x^3 - 18375 x^2 - 360 x + 22 > inexplicably still there despite the polynomial not having any factors > of 7. So, it would appear that an appropriate definition for your functions at the top of your post, to tie it in with whatÕs down here would be: given the a_i(x)s and w_i(x)s, then: c_i(x) = 5a_i(x)/w_i(x) and d_i(x)=7/w_i(x). However, that would mean that d_i(x)*w_i(x)=7, which does not suggest anything about d_i(x) having a factor of 7 as a factor. So, assuming that the d_i(x) was arrived at as above, and that the w_i(x) are the same in both cases, where is the notion of w_i(x) being a factor of d_i(x) coming from? > Your eyes fool you. Your memory betrays you. > The math doesnÕt bother with such nonsense. A multiple of a > polynomial can be divided off and itÕs gone. > It doesnÕt leave a trace. > In a way whatÕs happening now is an instructive lesson in the > limitations of the human brain. Your brains SEE something, and > insistently tell you that 7 is STILL THERE, and so the arguments go on > for years. Apparently you did not understand any of the arguments, or you wouldnÕt be saying any of this. > After all, you can SEE the 7Õs. Come on, whoÕs fooling who, right? > Dammit. > You can see the 7Õs in there, canÕt you? I see two division by 7Õs and a +7. Now, which 7s are supposed to still be there? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Look at it backwards > The equation has no memory. This is new! What the are you on about now? > You have a memory, so if I start with the polynomial multiplied by 49, > then you can say to yourself that thereÕs some dependency on 7. But > itÕs a mirage. Mirage? Ummmmm..... Holy moly... I have no words for this... Ummmm.... OK, go on.... > Mathematically a constant multiple is not a big deal. Can you prove that assertion, and can you first define mathematically what a big deal is. Just joking, I know you are talking figuratively... but really James, do you have to add in all this goofy non-math talk in every post? > ItÕs just a > constant multiple that you can divide off, leaving a result that--you > guessed it--has no memory of the multiple! What does divide off mean? I think I know what you mean, but can you not try to use standard terminology? > Your eyes fool you. Your memory betrays you. And Wiles was tricked in this way in his FLT proof, right? === Subject: Re: Linear Algebra Text Suggestion by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB30maT28993; how do you do it === Subject: Graph Theory Problem I canÕt figure out where to start with this question. Any hints? Suppose that M is a matching of graph G that is not contained in any larger matching, and that MÕ is a maximum matching of G. Show that |MÕ|<=2|M|. I canÕt find any ways to relate the two. === Subject: Re: Graph Theory Problem alt.math.undergrad: > I canÕt figure out where to start with this question. Any hints? > Suppose that M is a matching of graph G that is not > contained in any larger matching, and that MÕ is a > maximum matching of G. Show that |MÕ|<=2|M|. > I canÕt find any ways to relate the two. Suppose that |MÕ| > 2|M| and show that M must be contained in a larger matching. Hint: M involves only 2|M| of the vertices of G, and each of these vertices is adjacent to at most one edge of MÕ. Brian === Subject: Re: Graph Theory Problem > alt.math.undergrad: >>I canÕt figure out where to start with this question. Any hints? >>Suppose that M is a matching of graph G that is not >>contained in any larger matching, and that MÕ is a >>maximum matching of G. Show that |MÕ|<=2|M|. >>I canÕt find any ways to relate the two. > Suppose that |MÕ| > 2|M| and show that M must be contained > in a larger matching. Hint: M involves only 2|M| of the > vertices of G, and each of these vertices is adjacent to at > most one edge of MÕ. > Brian === Subject: Vectors question 2 I am working through a question for a university assignment. I am not wanting to be given the answer, but am after some pointers. The question is: A train is travelling in a straight line at 24km/h. A movie stunt-man moves at 4km/h straight across its roof, at right angles to the direction of its motion. Draw a clear sketch of his resultant movement relative to the ground, and find the speed and direction of that movement. My take on this is as follows (way below is a very bad ASCII representation of the type of graph that is in my head...definetly not to any sort of scale) I first create a graph with x-axis distance travelled, and y-axis time. I can then plot u = 24i + j to show the speed that the train is travelling at. The line perpendicular to to u is the speed and direction of the stunt man, the length of that line will be 4. The line that forms the hyptonuse (v), represents the speed (length), and direction (x,y) that the stuntman has travelled. Am I on the right track (no pun intended)? If not, could you offer some suggestions. ^ | | 0 | 0 0 | 0 0 | v 0 0 (24,1) | 0 0 | 0 0 | 0 0 u | 0 0 |0____________________________> Please note, this is an assignment, so I am not wanting anyone to solve it for me. I do not want any reason to be accused of wrongdoings. Cassandra === Subject: THANKS!!! (calculus) has turned out to be much easier then the first half. I think tonight is the last night I can afford to give to my assignment (other essays beckon). All in all I had a thoroughly enjoyable weekend and am looking forward to the next assignment. double checking my work, and most importantly, patiently explaining concepts to me. Cassandra Thompson === Subject: Lines, points and planes Aah, too tired to explain too much. I have answered the following question but fear I am very incorrect. Q) Find the equation of the plane that passes through the origin and is perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) A) Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real number. v = 3i + j - k Q1 = (0, 0, 0) a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 3x + y [CapitalEth] z = 0 Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. I was quite confident that it was correct. Except now I am attempting a question in which we are given a point and a plane, and asked to find the equation of the line perpendicular to the plane. I thought I would just work backwards using the above question, but I feel like my answers are all wrong. Any suggestions? Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 > 3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 > 3x + y ? z = 0 > Therefore the equation of the plane is 3x + y ? z = 0. > I was quite confident that it was correct. Except ... Omigod. Google turned some of your minus signs into ?s when I went into preview mode. Why, I donÕt know. (You didnÕt use some weird character like a dash instead of an ASCII hyphen did you?) Apart from the Google trauma, your answer looks fine to me. If you are unsure whether your answers are right, itÕs a great idea to check them with a numerical example (if you havenÕt already done so). In this example I would check the answer as follows... 1. Find the intersection of the line and the plane - call it point A. 2. Pick any point on the line (other than the point of intersection) - call it point B. 3. Pick any point on the plane (other than the point of intersection) - call it point C. Check that ABC is a right-angled triangle with right angle at A (use Pythagoras). If it is, then that gives you great confidence that you got the answer right. You have to be REALLY unlucky for the check to work but the answer to still be wrong. > ... now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? > Cassandra. The equation of the plane will immediately give you the normal vector (i.e. the vector of the required perpendicular line). Specifically, if the equation of the plane is given in Cartesian form as ax + by + cz + d = 0, then the normal vector is (a,b,c) (or any non-zero scalar multiple thereof). But you knew that, right? Now that you have the line vector and the coordinates of one point on the line, it is straightforward to find the (parametric) equation of too, so I think you should be OK... You can check your answer using a similar right-angled triangle construction. === Subject: Re: Lines, points and planes >>Aah, too tired to explain too much. I have answered the following >>question but fear I am very incorrect. >>Q) Find the equation of the plane that passes through the origin and is >>perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) >>A) >>Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real >>number. >>v = 3i + j - k >>Q1 = (0, 0, 0) >>a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 >>3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 >>3x + y ? z = 0 >>Therefore the equation of the plane is 3x + y ? z = 0. >>I was quite confident that it was correct. Except ... > Omigod. Google turned some of your minus signs into ?s when I went > into preview mode. Why, I donÕt know. (You didnÕt use some weird > character like a dash instead of an ASCII hyphen did you?) I am typing my assignment in MS Word, I copied the above workings straight from word. Yes, it would have been dashes not minus signs as I notice that MS Word does alot of auto-correcting with regard to minus-signs/dashes. Sorry about that. Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 > 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 > 3x + y [CapitalEth] z = 0 > Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. Starting from the end... The normal vector of the plane Ax + By + Cz + D = 0 is (A, B, C), so the normal of your plane is (3, 1, -1). The direction vector of the line you were given is also (3, 1, -1), so the line is indeed perpendicular to the plane. 3 * 0 + 0 - 0 = 0, so your plane contains the origin. Thus, the plane you got is the right answer. > I was quite confident that it was correct. Except now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? If you have point P = (Xp, Yp, Zp) and plane Ax + By + Cz + D = 0, then the normal to the plane is n = (A, B, C) and the line through P in the direction of n is simply l = (Xp, Yp, Zp) + t * (A, B, C). Based on the knowledge you demonstrated in the rest of your posts here, my hunch is that thatÕs exactly what you did, that your answers are probably right, and that you just need to convince yourself of that :-) but if you really are wrong or unsure, post a specific example and weÕll try to find specific errors. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Proving planes are not parallel The definition that I have is as follows: Two planes are parallel if their normal vectors are parallel, that is, if the cross product of their normal vectors is zero. However my understanding is that a cross product yields another vector, not a scalar. So I am confused. My testbook has one so-called worked example, but it borders more along the lines of weÕll leave the proof as an exercise, ie it leaves out the middle bits. *************************************** Two parallel planes. P_1: 2x + 3y - z = 3 P_2: -4x - 6y + 2z = 8 so n_1 = 2i + 3j - k, n_2 = -4i - 6j + 2k The planes are parallel because: n_2 = -2n_1 and n_1 x n_2 = 0 *************************************** I am trying to prove two planes are not parallel. Should I write them as vectors, then show that the vectors are not multiples of each other? ie show n_1 <> x(n_2) How does n_1 x n_2 = 0 in the above example? I thought n_1 x n_2 = -1i + 0j + 4k Cassandra === Subject: Re: Proving planes are not parallel > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. There is a zero vector in every vector space, such that if 0 is the zero scalar and Z is the zero vector and . indicates product of scalar with vector then 0.Z = Z Since it is usually obvious from the context when a zero object is to be a scalar and when a vector, the same word is used ambiguously for both, as in this case. To say that the cross product of two vectors is zero clearly must mean the zero vector, not the zero scalar. Does this clear things up a bit? >... > Cassandra === Subject: Re: Proving planes are not parallel Cassandra Thompson a .8ecrit : > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. In those formulations, zero stands for the number 0 or for the null vector > My testbook has one so-called worked example, but it borders more along > the lines of weÕll leave the proof as an exercise, ie it leaves out > the middle bits. > *************************************** > Two parallel planes. > P_1: 2x + 3y - z = 3 > P_2: -4x - 6y + 2z = 8 > so > n_1 = 2i + 3j - k, > n_2 = -4i - 6j + 2k > The planes are parallel because: > n_2 = -2n_1 > and n_1 x n_2 = 0 > *************************************** > I am trying to prove two planes are not parallel. Should I write them as > vectors, then show that the vectors are not multiples of each other? > ie show > n_1 <> x(n_2) It is one possibility (better to write n_1 <> k(n_2) ) ; another one is to calculate the cross-product of n_1 and n_2 and to show it is not null > How does n_1 x n_2 = 0 in the above example? I thought > n_1 x n_2 = -1i + 0j + 4k You must have made a calculation mistake. First, n_1 is indeed equal to -2 n_2 (check it). Second, the i component of the cross-product n_1 x n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... > Cassandra === Subject: Re: Proving planes are not parallel > Cassandra Thompson a .8ecrit : >> The definition that I have is as follows: >> Two planes are parallel if their normal vectors are parallel, that is, >> if the cross product of their normal vectors is zero. >> However my understanding is that a cross product yields another >> vector, not a scalar. So I am confused. > In those formulations, zero stands for the number 0 or for the null vector >> My testbook has one so-called worked example, but it borders more >> along the lines of weÕll leave the proof as an exercise, ie it >> leaves out the middle bits. >> *************************************** >> Two parallel planes. >> P_1: 2x + 3y - z = 3 >> P_2: -4x - 6y + 2z = 8 >> so >> n_1 = 2i + 3j - k, >> n_2 = -4i - 6j + 2k >> The planes are parallel because: >> n_2 = -2n_1 >> and n_1 x n_2 = 0 >> *************************************** >> I am trying to prove two planes are not parallel. Should I write them >> as vectors, then show that the vectors are not multiples of each other? >> ie show >> n_1 <> x(n_2) > It is one possibility (better to write n_1 <> k(n_2) ) ; another one is > to calculate the cross-product of n_1 and n_2 and to show it is not null >> How does n_1 x n_2 = 0 in the above example? I thought >> n_1 x n_2 = -1i + 0j + 4k > You must have made a calculation mistake. First, n_1 is indeed equal to > -2 n_2 (check it). Second, the i component of the cross-product n_1 x > n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... >> Cassandra simple algebra errors that I have posted to the newsgroup. Although I have to admit I find the nature of cross-products begs for mistakes. Perhaps I should learn to calculate the cross-product using a matrix, or better still get a graphing calculator! I have answered the assignment equation using the n_1 <> k(n_2) method, product and null vector. Cassie === Subject: Re: Proving planes are not parallel >simple algebra errors that I have posted to the newsgroup. Although I >have to admit I find the nature of cross-products begs for mistakes. >Perhaps I should learn to calculate the cross-product using a matrix, or >better still get a graphing calculator! >I have answered the assignment equation using the n_1 <> k(n_2) method, >product and null vector. >Cassie DonÕt feel ashamed Cassie. We all make mistakes, from inadvertent typoÕs to just ßat being wrong. In my opinion it is always better to check whether v2 = c * v1 than v1 cross v2 = zero vector if testing for parallelness exactly because it is simpler and a less error - prone method. The cross product does have many uses, though, and you will need to be able to calculate them with a minimum of errors. There are several methods in use. My favorite, which requires a little more writing but, at least for me, keeps the errors to a minimum is the following. To cross with I first make a helper where I write the vectors starting and ending with the middle component: b c a b e f d e Now the components of the cross product are gotten by computing the left, middle, and right 2x2 determinants in the helper: In particular, this gets the sign of the middle component correct without giving it any additional thought. --Lynn === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: > Talk about typos...., of course the first one is bf - ec :-) --Lynn === Subject: Re: Proving planes are not parallel >>b c a b >>e f d e >>Now the components of the cross product are gotten by computing the >>left, middle, and right 2x2 determinants in the helper: >> :-) > --Lynn Yes, that is a much sinpler way of doing it. I have been writing the equations down, then writing down the general formula, then putting small labels on the equation (a_1, b_1 etc) then trying to look in three places at once to be sure I am putting the right number inthe right place. Of course this could easily be done on a calculator, scilab, or even using MS Excel, but I need to master it before I turn to technology. Cassandra === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: > Talk about typos...., of course the first one is bf - ec >> :-) >> --Lynn >Yes, that is a much sinpler way of doing it. >I have been writing the equations down, then writing down the general >formula, then putting small labels on the equation (a_1, b_1 etc) then >trying to look in three places at once to be sure I am putting the right >number inthe right place. Ugh!!! Then IÕm glad I posted that for you. I always told my classes to not even bother to memorize the formula. Just use the helper. --Lynn === Subject: Unit vectors... I am nearing the end of my bunch of vector questions (...hopefully...) I have found a unit vector in the direction of b = (3, 1, 0) below. I have also taken a crack at finding a vector four units long in the opposite direction. I am not sure about the math for the latter. |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) To find a vector four units long in the opposite direction, I imagine I would only need multiple the unit vector by the scalar -4? ie -4u = (-12/sqrt(10), -4/sqrt(10), 0). Is it this simple? cassandra === Subject: Re: Unit vectors... have also taken a crack at finding a vector four units long in the > opposite direction. I am not sure about the math for the latter. > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) Indeed, when |v| /= 0, |v/|v|| = (1/|v|)|v| = 1 > To find a vector four units long in the opposite direction, I imagine I > would only need multiple the unit vector by the scalar -4? > ie > -4u = (-12/sqrt(10), -4/sqrt(10), 0). > Is it this simple? Yup. Surprise, not everything is hard. Exercise: whatÕs the length of (1,1,1,1)? Show (1/2, 1/2, 1/2, 1/2) is a unit vector. === Subject: Re: Unit vectors... >>I have found a unit vector in the direction of b = (3, 1, 0) below. I >>have also taken a crack at finding a vector four units long in the >>opposite direction. I am not sure about the math for the latter. >>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >>u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) > Indeed, when |v| /= 0, |v/|v|| = (1/|v|)|v| = 1 >>To find a vector four units long in the opposite direction, I imagine I >>would only need multiple the unit vector by the scalar -4? >>ie >>-4u = (-12/sqrt(10), -4/sqrt(10), 0). >>Is it this simple? > Yup. Surprise, not everything is hard. > Exercise: whatÕs the length of (1,1,1,1)? > Show (1/2, 1/2, 1/2, 1/2) is a unit vector. Yes, when doing questions that take me hours, a 2 minute question makes me wonder what I am doing wrong. u = (1,1,1,1) = sqrt(4) = 2 v = (1/2, 1/2, 1/2, 1/2 = sqrt(1) = 1 Ahhh, very nice. This has been a great weekend, hard but satisfying, I really enjoy learning this stuff. I also really appreciated the help because there is something satisfying about understanding it at the end. Just lovely. === Subject: Re: Unit vectors... Cassandra: What text are you using? -- Casey === Subject: Re: Unit vectors...(Textbook) > Cassandra: > What text are you using? > -- > Casey It is a study guide (I am an external student so we are sent a 300 page study guide and told to buy two textbooks). The textbooks for the course (Algebra and Calculus) are: Calculus: single and multivariable Hughes-Hallet, Gleason & McCallum Elementary Linear Algebra Larson & Edwards The study guide has a number of very large textbook excerpts. For the topic of vectors we have been working straight from one of these excerpts: Elementary Linear Algebra Grossman, pp154-207 Why do you ask? Cassie === Subject: Re: Unit vectors... >I am nearing the end of my bunch of vector questions (...hopefully...) > I have found a unit vector in the direction of b = (3, 1, 0) below. I > have also taken a crack at finding a vector four units long in the > opposite direction. I am not sure about the math for the latter. > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) > To find a vector four units long in the opposite direction, I imagine I > would only need multiple the unit vector by the scalar -4? > ie > -4u = (-12/sqrt(10), -4/sqrt(10), 0). > Is it this simple? yes. === Subject: Finding the equation of a plane. Okay me again..I should explain that I am an external student. Which means I donÕt have any lectures, nor do I even have contact with any other students in my class. I hope this explains why I am asking so many questions. If I had study group I would be very involved in it. plane for vector a and b. I think I have done it correctly (hopefully..). a = (1, -2, 1) b = (3, 1, 0) w = a x b = (-1, 3, 7) Find the plane passing through the point A = (1, -2, 1) having the normal vector n = [CapitalEth]1i + 3j + 7k. a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 -1(x [CapitalEth] 1) + 3(y + 2) + 7(z [CapitalEth] 1) = 0 [CapitalEth]x + 1 + 3y + 6 + 7z [CapitalEth] 7 = 0 [CapitalEth]x + 3y + 7z = 0 Confirm this by finding the plane passing through the point B = (3, 1, 0) having the normal vector n = [CapitalEth]1i + 3j + 7k. a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 [CapitalEth]1(x [CapitalEth] 3) + 3(y [CapitalEth] 1) + 7(z [CapitalEth] 0) = 0 [CapitalEth]x + 3 + 3y [CapitalEth] 3 + 7z [CapitalEth] 0 = 0 [CapitalEth]x + 3y + 7z = 0 Therefore, the equation of the plain containing a and b is [CapitalEth]x + 3y + 7z = 0 Cassandra === Subject: Re: Finding the equation of a plane. >Okay me again.. Cassandra, many of us read several math newsgroups. I think you will find that you donÕt really need to post your questions to more than one newsgroup. When you cross-post lots of questions we find ourselves re-reading the same questions, not sure whether you are posting something new. So I would suggest you pick the newsgroup you like the best and try posting to just it, or at least donÕt post the same questions in different groups. --Lynn === Subject: Re: Finding the equation of a plane. >>Okay me again.. > Cassandra, many of us read several math newsgroups. I think you will > find that you donÕt really need to post your questions to more than > one newsgroup. When you cross-post lots of questions we find ourselves > re-reading the same questions, not sure whether you are posting > something new. So I would suggest you pick the newsgroup you like the > best and try posting to just it, or at least donÕt post the same > questions in different groups. > --Lynn Cassandra. === Subject: Re: Finding the equation of a plane. Cassandra, Lynn: Cross-posting is actually quite *acceptable* practice in Usenet, if done correctly and in moderation. After looking at CassandraÕs original post in this thread, what she did as far as the cross-posting aspect is concerned, is EXACTLY CORRECT and conforms to standard. Lynn may have the best of intentions, but she needs to get the facts straight. Posting a message to more than one relevant newsgroup exposes the question to a larger audience. There will be some people that subscribe to all of thse newsgroups, but there will also be some that may only subscribe to one or two, so cross-posting usually results in more (hopefully useful) responses. Some have a thing for cross-posting, as evidenced here already. ThereÕs absolutely NOTHING wrong with what Cassandra did in the original post. She is preceisely what she is supposed to do, if she indeed intended to address both groups. If readers have their client software configured appropriately (ie donÕt show messages that have already been read) then they only see the post *one* time, even if they subscribe to all the newsgroups being cross-posted to. There is a valid argument some make about posting the same message to more than one newsgroup. Unlike Cassandra, many people do not cross-post but rather send the same message, individually, to multiple newsgroups. This results in inconvience for many and will get you ßamed. A reader may spend time writing a response only to find out, right after they sent it, the question has already been answered by another who replied to the one of the other messages sent, individually, to another newsgroup that is not subscribed to by that person. Not to mention the original poster has to send multiple messages that say the same thing. There is also the issue of system resources. The end user may not care, but it is more efficient for a server to carry a single copy of a message that is addressed to multiple groups, than to have several individual messages all saying the same thing, but each adressed to a single newsgroup. The proper way to accomplish getting your message to multiple newsgroups is to cross-post, which is exactly what Cassandra did. It (cross-posting) means to write a single message but address it to multiple newsgroups on the Newsgroup: line of whatever program is being used to compose the message the message only once, allows only a single copy of the message to reside on each server that carries the newsgroups, and is better for those that read the message, too. By default when anyone replies to a cross-posted message, the reply is sent to *all* the newsgroups you originally addressed, even if the person responding only subscribes to one of them. Everyone reading *any* of the addressed newsgroups sees *all* the replies from *everyone* who replied, and the OP only need follow up with any *one* of these newsgroups, to see *all* replies. Cross-posting is your friend, just be sure to do it correctly. All you have to remember is to address all the newsgroups in a single post, not send multiple posts each to a single newsgroup. Oh, and be very conservative with the number of newsgroups you cross-post to. Software resides on many many newsgroups. Usually, there will only be a very few newsgroups anyone would want to cross-post to anyway (just the ones where themessage will be on-topic.) All that said, I would not have addressed CassandraÕs question about equations of planes to sci.math, NOT because it was cross-posted there, but because the topic is probably a little too elementary for that group. Lynn, if youÕre still skeptical about cross-posting, just go back and look at your response, suggesting the OP only post to a single newsgroup. Your reply was actually addressed to BOTH of the newsgroups origianlly addressed by the OP. This is by DESIGN, for the aforementioned reasons, and IÕm sure you didnÕt realize it at the time, else why would one be so hypocritical as to suggest addressing only one newsgroup when the very suggestion to do so was itself addressed to multiple newsgroups! -- Darrell === Subject: Re: Finding the equation of a plane. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB6DWSe30415; cross-posting rules, but you did a very good job of laying them out. Joseph A. >Cassandra, Lynn: >Cross-posting is actually quite *acceptable* practice in Usenet, if done >correctly and in moderation. After looking at CassandraÕs original post in >this thread, what she did as far as the cross-posting aspect is concerned, >is EXACTLY CORRECT and conforms to standard. Lynn may have the best of >intentions, but she needs to get the facts straight. Posting a message to >more than one relevant newsgroup exposes the question to a larger audience. >There will be some people that subscribe to all of thse newsgroups, but >there will also be some that may only subscribe to one or two, so >cross-posting usually results in more (hopefully useful) responses. >Some have a thing for cross-posting, as evidenced here already. ThereÕs >absolutely NOTHING wrong with what Cassandra did in the original post. She which >is preceisely what she is supposed to do, if she indeed intended to address >both groups. >If readers have their client software configured appropriately (ie donÕt >show messages that have already been read) then they only see the post *one* >time, even if they subscribe to all the newsgroups being cross-posted to. >There is a valid argument some make about posting the same message to more >than one newsgroup. Unlike Cassandra, many people do not cross-post but >rather send the same message, individually, to multiple newsgroups. This >results in inconvience for many and will get you ßamed. A reader may spend >time writing a response only to find out, right after they sent it, the >question has already been answered by another who replied to the one of the >other messages sent, individually, to another newsgroup that is not >subscribed to by that person. Not to mention the original poster has to >send multiple messages that say the same thing. There is also the issue of >system resources. The end user may not care, but it is more efficient for a >server to carry a single copy of a message that is addressed to multiple >groups, than to have several individual messages all saying the same thing, >but each adressed to a single newsgroup. >The proper way to accomplish getting your message to multiple newsgroups is >to cross-post, which is exactly what Cassandra did. It (cross-posting) >means to write a single message but address it to multiple newsgroups on the >Newsgroup: line of whatever program is being used to compose the message to write >the message only once, allows only a single copy of the message to reside on >each server that carries the newsgroups, and is better for those that read >the message, too. By default when anyone replies to a cross-posted message, >the reply is sent to *all* the newsgroups you originally addressed, even if >the person responding only subscribes to one of them. Everyone reading >*any* of the addressed newsgroups sees *all* the replies from *everyone* who >replied, and the OP only need follow up with any *one* of these newsgroups, >to see *all* replies. >Cross-posting is your friend, just be sure to do it correctly. All you have >to remember is to address all the newsgroups in a single post, not send >multiple posts each to a single newsgroup. Oh, and be very conservative >with the number of newsgroups you cross-post to. Software resides on many *too* many >newsgroups. Usually, there will only be a very few newsgroups anyone would >want to cross-post to anyway (just the ones where themessage will be >on-topic.) >All that said, I would not have addressed CassandraÕs question about >equations of planes to sci.math, NOT because it was cross-posted there, but >because the topic is probably a little too elementary for that group. >Lynn, if youÕre still skeptical about cross-posting, just go back and look >at your response, suggesting the OP only post to a single newsgroup. Your >reply was actually addressed to BOTH of the newsgroups origianlly addressed >by the OP. This is by DESIGN, for the aforementioned reasons, and IÕm sure >you didnÕt realize it at the time, else why would one be so hypocritical as >to suggest addressing only one newsgroup when the very suggestion to do so >was itself addressed to multiple newsgroups! >-- >Darrell === Subject: Re: Finding the equation of a plane. Darrell: My suggestion to Cassandra doesnÕt require an essay from you explaining how you think I donÕt understand how usenet works and implying that I donÕt know what I am doing or she needs to get the facts straight. And if you wish to refer to me correctly, use Lynn, Mr., Professor, or Dr., but not she. Get lost. --Lynn === Subject: Re: Finding the equation of a plane. > explaining how you think I donÕt understand how usenet works and > implying that I donÕt know what I am doing or she needs to get the > facts straight. And if you wish to refer to me correctly, use Lynn, > Mr., Professor, or Dr., but not she. Get lost. Lynn is a common female name. I apologize for the incorrect assumption. As for the rest, I see... You donÕt require an essay on how Usenet works, you just want to lecture others on how Usenet works, when they have done absolutely nothing that suggests they donÕt know how Usenet works. Good Grief, itÕs bad enough we canÕt even make valid mathematical contribution without the third degree from you, now we canÕt cross-post anymore lest it meet with your disapproval as well. I think you just hate being wrong. Merry Xmas! === Subject: Re: Finding the equation of a plane. > a(x - x0) + b(y - y0) + c(z - z0) = 0 > -1(x - 1) + 3(y + 2) + 7(z - 1) = 0 > -x + 1 + 3y + 6 + 7z - 7 = 0 > -x + 3y + 7z = 0 ThatÕs correct. In general, if you are given N, the normal to the plane, and P, the point the plane is to pass through, the equation of the plane is (where . is the dot product and r=(x, y, z)). N.(r - P) = 0 If you consider the geometrical meaning of the dot product of two vectors being zero, it should be obvious why thatÕs the equation of the plane normal to N passing through P. Anyways, you can manipulate that into N.r = N.P or (-1, 3, 7).(x, y, z) = (-1, 3, 7).(1, -2, 1) -x + 3y + 7z = -1(1) + 3(-2) + 7(1) -x + 3y + 7z = 0 -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Finding the equation of a plane. > In general, if you are given N, the normal to the plane, and P, > the point the plane is to pass through, the equation of the > plane is (where . is the dot product and r=(x, y, z)). > N.(r - P) = 0 Absolutely. IMHO, this is by far the best way to remember both the geometry and vector algebra needed to represent a plane in R^{n}. Just as identifying a line in R^{n} requires a point x= (x1,x2,...,xn) and a direction v=(v1,v2,...,vn). Then the line is simply r(t) = x+vt, where r(t) is a function from the real numbers to R^{n}. DM === Subject: Magnitude and area of a parallelogram Firstly, I am not sure if it annoying that I am using the same thread to ask so many question, however I thought it might be equally annoying to be starting lots of new threads. At least if it is in the one thread you have a good idea of my background. This question is not part of the assignment, however working through the assignment has shed light on a couple a few equations, which in turn have confused me. I am hoping to ask some questions. We have two vectors: a = (1, -2, 1) b = (3, 1, 0) The cross-product of these two vectors is: a x b = (-1, 3, 7) let w = (-1, 3, 7) Okay all good so far I think. Now if I want to find the magnitude of w, I treat it exactly the same as I would if I wanted to find the magnitude of a and b. |w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) However I also know that the following formulae is true: |a x b| = |a||b| sin& Is the |a x b| in this formulae the same as |w|? Does |w| = |a||b| sin & hold true? Does |a x b| = sqrt(17)? I guess I am just confused by the to different formulaes. ie |w| = sqrt(w_1^2 + w_2^2 + w_3^2) |a x b| = |a||b| sin& Cassandra. === Subject: Re: Magnitude and area of a parallelogram > Firstly, I am not sure if it annoying that I am using the same thread to > ask so many question, however I thought it might be equally annoying to > be starting lots of new threads. At least if it is in the one thread you > have a good idea of my background. > This question is not part of the assignment, however working through the > assignment has shed light on a couple a few equations, which in turn > have confused me. I am hoping to ask some questions. > We have two vectors: > a = (1, -2, 1) > b = (3, 1, 0) > The cross-product of these two vectors is: > a x b = (-1, 3, 7) > let w = (-1, 3, 7) > Okay all good so far I think. Now if I want to find the magnitude of w, > I treat it exactly the same as I would if I wanted to find the magnitude > of a and b. > |w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) > However I also know that the following formulae is true: > |a x b| = |a||b| sin& > Is the |a x b| in this formulae the same as |w|? > Does |w| = |a||b| sin & hold true? > Does |a x b| = sqrt(17)? > I guess I am just confused by the to different formulaes. > ie > |w| = sqrt(w 1^2 + w 2^2 + w 3^2) > |a x b| = |a||b| sin& ThatÕs all correct. This is one way you can determine & (within -pi/2 to pi/2). Another way is via the dot products (within 0 and pi). And the parallelogram? Tomasso. === Subject: Re: Magnitude and area of a parallelogram - angle from cos and sin >>Firstly, I am not sure if it annoying that I am using the same thread to >>ask so many question, however I thought it might be equally annoying to >>be starting lots of new threads. At least if it is in the one thread you >>have a good idea of my background. >>This question is not part of the assignment, however working through the >>assignment has shed light on a couple a few equations, which in turn >>have confused me. I am hoping to ask some questions. >>We have two vectors: >>a = (1, -2, 1) >>b = (3, 1, 0) >>The cross-product of these two vectors is: >>a x b = (-1, 3, 7) >>let w = (-1, 3, 7) >>Okay all good so far I think. Now if I want to find the magnitude of w, >>I treat it exactly the same as I would if I wanted to find the magnitude >>of a and b. >>|w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) >>However I also know that the following formulae is true: >>|a x b| = |a||b| sin& >>Is the |a x b| in this formulae the same as |w|? >>Does |w| = |a||b| sin & hold true? >>Does |a x b| = sqrt(17)? >>I guess I am just confused by the to different formulaes. >>ie >>|w| = sqrt(w_1^2 + w_2^2 + w_3^2) >>|a x b| = |a||b| sin& >> >ThatÕs all correct. This is one way you can determine & (within -pi/2 to pi/2). >Another way is via the dot products (within 0 and pi). >And the parallelogram? >Tomasso. The best way to determine & is from =both= its cosine and sine. If one uses only sin(&) to determine &, then one gets & with a severe loss of precision if sin(&) happens to be close to 1; in general, the absolute error in f(x) = |f Ô(x)| times the absolute error in x. And |d(arcsin x)/dx| = 1/sqrt (1 - xx). Whenever one knows both sin(&) and cos(&) (which is not always the case) one does not suffer loss of precision in &. Johan E. Mebius === Subject: Re: Magnitude and area of a parallelogram > ThatÕs all correct. This is one way you can determine & (within -pi/2 to pi/2). > Another way is via the dot products (within 0 and pi). > And the parallelogram? > Tomasso. I am very confused right now. Last night I went to bed thinking I understood this stuff, now I realise I have errors everywhere and non of it makes sense. If I were to give your two vectors, how would you go about finding the parallelogram formed by them? Last night I thought that that was what the cross-product formed. Now I have no idea why I came to that conclusion. The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w = (-1, 3, 7). Which clearly does not form a parallelogram. I have create a graph, this graph shows that the parallelogram formed by a and b has the fourth point x = (4, -1, 1). so how did I get this point? I got it straight off the graph. How do I get it mathematically? I could just add the points ie x = (3 + 1, -2 + 1, 1 + 0). But I donÕt recall seeing this formula. So is it the right thing to do. And what of the cross-product. Is it just some point in space that doesnÕt even lie on the parallelogram? If so is its sole relationship to the parallelogram the fact that its magnitude (length) is equal to the area of the parallelogram (in square units)? If this is all true, now I am going to attempt to use the cross-product of a and b (ie w) to find the equation of the plane containing a and b. I am assuming this plane doesnÕt contain w, but the vector w just has these wonderful properties that assist us in finding the plane? Time for breakfast I think. cassandra === Subject: Re: Magnitude and area of a parallelogram - >> ThatÕs all correct. This is one way you can determine & (within -pi/2 >> to pi/2). Another way is via the dot products (within 0 and pi). >> And the parallelogram? >> Tomasso. > I am very confused right now. Last night I went to bed thinking I > understood this stuff, now I realise I have errors everywhere and non > of it makes sense. > If I were to give your two vectors, how would you go about finding the > parallelogram formed by them? > Last night I thought that that was what the cross-product formed. Now > I have no idea why I came to that conclusion. > The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is > w = (-1, 3, 7). Which clearly does not form a parallelogram. > I have create a graph, this graph shows that the parallelogram formed > by a and b has the fourth point x = (4, -1, 1). > so how did I get this point? I got it straight off the graph. How do I > get it mathematically? I could just add the points ie x = (3 + 1, -2 + > 1, 1 + 0). But I donÕt recall seeing this formula. So is it the right > thing to do. > And what of the cross-product. Is it just some point in space that > doesnÕt even lie on the parallelogram? If so is its sole relationship > to the parallelogram the fact that its magnitude (length) is equal to > the area of the parallelogram (in square units)? > If this is all true, now I am going to attempt to use the > cross-product of a and b (ie w) to find the equation of the plane > containing a and b. I am assuming this plane doesnÕt contain w, but > the vector w just has these wonderful properties that assist us in > finding the plane? > Time for breakfast I think. > cassandra Or perhaps time for a nightrest... I guess that you will learn about vectors and bivectors and about polar and axial vectors at a later time, but let me give you a foretaste. In 3D vector algebra and analysis it is common practice to identify a vector V of a certain length L with a pancake P of =area= L in a plane perpendicular to V. The exact positions of the vector and the pancake in space and the exact shape of the pancake do not matter. What does matter is the sense of orientation in the plane of the pancake; by convention the direction of V and the rotation sense in P fit to each other like the translation and rotation of a =right-handed= screw. Now look at the projections of V onto the axes of a Cartesian coordinate system (components of V), and also at the projections of P onto the coordinate planes of the same system (components of P). They are pairwise equal: Vx = Pyz, Vy = Pzx, Vz = Pxy. And now the main reason why V and P can be identified: their components change in identical ways when V and P are rotated together, maintaining their perpendicular relation. Finally the vector product: two vectors A and B with a common origin O span a parallelogram (O, A, A+B, B). Its area is |A|.|B|.|sin (A, B)|. This is also the length of the vector product A x B. It is a property of 3D space that all vectors that are perpendicalar to both A and B are scalar multiples of A x B. Try and find out for yourself what happens when one does not allow only rotations, but also reßections. Enjoy your breakfast! Johan E. Mebius === Subject: Re: Magnitude and area of a parallelogram Cassandra Thompson dixit: >> ThatÕs all correct. This is one way you can determine & (within -pi/2 to pi/2). >> Another way is via the dot products (within 0 and pi). >> And the parallelogram? >> Tomasso. >I am very confused right now. Last night I went to bed thinking I >understood this stuff, now I realise I have errors everywhere and non of >it makes sense. >If I were to give your two vectors, how would you go about finding the >parallelogram formed by them? >Last night I thought that that was what the cross-product formed. Now I >have no idea why I came to that conclusion. >The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w >= (-1, 3, 7). Which clearly does not form a parallelogram. >I have create a graph, this graph shows that the parallelogram formed by >a and b has the fourth point x = (4, -1, 1). >so how did I get this point? I got it straight off the graph. How do I >get it mathematically? I could just add the points ie x = (3 + 1, -2 + >1, 1 + 0). But I donÕt recall seeing this formula. So is it the right >thing to do. This is simple vector addition, if you consider vector ÔaÕ starts at the origin then you have a+b = x, so yes this formula is correct to get the other point of the parallelogram. The vector (-1,3,7) which has magnitude sqrt(59) and not sqrt(17) as you said in your 1st message, is perpendicular to the plane that contains vectors a and b (and also x, if you consider it a vector). See this: http://members.tripod.com/~Paul_Kirby/vector/ Vcrossproduct.html toward the bottom of the page there is a diagram explaining this. See the rest of the site too: http://members.tripod.com/~Paul_Kirby/vector/VectorLand.html >And what of the cross-product. Is it just some point in space that >doesnÕt even lie on the parallelogram? If so is its sole relationship to >the parallelogram the fact that its magnitude (length) is equal to the >area of the parallelogram (in square units)? >If this is all true, now I am going to attempt to use the cross-product >of a and b (ie w) to find the equation of the plane containing a and b. >I am assuming this plane doesnÕt contain w, but the vector w just has >these wonderful properties that assist us in finding the plane? For the plane equation see http://www.netcomuk.co.uk/~jenolive/vect11.html http://www.netcomuk.co.uk/~jenolive/vect12.html http://www.netcomuk.co.uk/~jenolive/vect15.html and the rest of the site also. >Time for breakfast I think. >cassandra === Subject: Re: Magnitude and area of a parallelogram > Cassandra Thompson dixit: >ThatÕs all correct. This is one way you can determine & (within -pi/2 to pi/2). >Another way is via the dot products (within 0 and pi). >And the parallelogram? >Tomasso. >>I am very confused right now. Last night I went to bed thinking I >>understood this stuff, now I realise I have errors everywhere and non of >>it makes sense. >>If I were to give your two vectors, how would you go about finding the >>parallelogram formed by them? >>Last night I thought that that was what the cross-product formed. Now I >>have no idea why I came to that conclusion. >>The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w >>= (-1, 3, 7). Which clearly does not form a parallelogram. >>I have create a graph, this graph shows that the parallelogram formed by >>a and b has the fourth point x = (4, -1, 1). >>so how did I get this point? I got it straight off the graph. How do I >>get it mathematically? I could just add the points ie x = (3 + 1, -2 + >>1, 1 + 0). But I donÕt recall seeing this formula. So is it the right >>thing to do. > This is simple vector addition, if you consider vector ÔaÕ starts at > the origin then you have a+b = x, so yes this formula is correct to > get the other point of the parallelogram. The vector (-1,3,7) which > has magnitude sqrt(59) and not sqrt(17) as you said in your 1st > message, is perpendicular to the plane that contains vectors a and b > (and also x, if you consider it a vector). See this: > http://members.tripod.com/~Paul_Kirby/vector/ Vcrossproduct.html > toward the bottom of the page there is a diagram explaining this. > See the rest of the site too: > http://members.tripod.com/~Paul_Kirby/vector/VectorLand.html >>And what of the cross-product. Is it just some point in space that >>doesnÕt even lie on the parallelogram? If so is its sole relationship to >>the parallelogram the fact that its magnitude (length) is equal to the >>area of the parallelogram (in square units)? >>If this is all true, now I am going to attempt to use the cross-product >>of a and b (ie w) to find the equation of the plane containing a and b. >>I am assuming this plane doesnÕt contain w, but the vector w just has >>these wonderful properties that assist us in finding the plane? > For the plane equation see > http://www.netcomuk.co.uk/~jenolive/vect11.html > http://www.netcomuk.co.uk/~jenolive/vect12.html > http://www.netcomuk.co.uk/~jenolive/vect15.html > and the rest of the site also. >>Time for breakfast I think. >>cassandra I found this link http://www.phy.syr.edu/courses/java-suite/crosspro.html off one of the above pages. The applet on that page, and the explanation below was extremely simple and did a wonderful job of explaining it all. I now have 2 vectors and the normal to those two vectors. I am now trying to find the equation of the plane. Not sure how to go about it. I see lots of equations for finding the equation of a plane, but none for the situation that I have two vectors and the normal. I am guessing I am going to need to do some manipulating? Cassandra === Subject: Re: Magnitude and area of a parallelogram > If I were to give your two vectors, how would you go about finding the > parallelogram formed by them? > Last night I thought that that was what the cross-product formed. Now I > have no idea why I came to that conclusion. >... > And what of the cross-product. Is it just some point in space that > doesnÕt even lie on the parallelogram? If so is its sole relationship to > the parallelogram the fact that its magnitude (length) is equal to the > area of the parallelogram (in square units)? Some questions to get your brain re-focussed: (1) What is the area of a triangle? Given the base and the height? (2) What is the area of a triangle? Given two sides and the included angle? (3) Can you construct a parallelogram from, say, two triangles? (4) What was your formula for ||w||? (5) Does the direction of w have anything to do with the area of the parallelogram? ItÕs plane? etc. Tomasso. === Subject: Re: Magnitude and area of a parallelogram >>If I were to give your two vectors, how would you go about finding the >>parallelogram formed by them? >>Last night I thought that that was what the cross-product formed. Now I >>have no idea why I came to that conclusion. >>... >>And what of the cross-product. Is it just some point in space that >>doesnÕt even lie on the parallelogram? If so is its sole relationship to >>the parallelogram the fact that its magnitude (length) is equal to the >>area of the parallelogram (in square units)? > Some questions to get your brain re-focussed: > (1) What is the area of a triangle? Given the base and the height? > (2) What is the area of a triangle? Given two sides and the included angle? > (3) Can you construct a parallelogram from, say, two triangles? > (4) What was your formula for ||w||? > (5) Does the direction of w have anything to do with the area of the parallelogram? ItÕs plane? etc. > Tomasso. Showing on a 3d graph that the formulae |u||v|sin& is the area of the paralleagram. But your questions 4 & 5 are where I am stuck. My formula for ||w|| is |w| = a_1*b_1i + a_2*b_2j + a_3*b_3k = -1j + 3j + 7k Does the direction of w have anything to do with the area of the parallelogram, and its plane? I have no idea. I am assuming by your question that it does. So I will go look. But all incite would be appreciated, because I am really clueless on this one. Cassandra === Subject: Re: Magnitude and area of a parallelogram > But your questions 4 & 5 are where I am stuck. > My formula for ||w|| is > |w| = a 1*b 1i + a 2*b 2j + a 3*b 3k > = -1j + 3j + 7k Nearly. That is w (vector). ||w|| = area of parallelogram = sqrt (1+9+49) = your answer. w is normal to the plan of the parallelogram. Tomasso. === Subject: Re: Magnitude and area of a parallelogram >>But your questions 4 & 5 are where I am stuck. >>My formula for ||w|| is >>|w| = a_1*b_1i + a_2*b_2j + a_3*b_3k >> = -1j + 3j + 7k > Nearly. That is w (vector). > ||w|| = area of parallelogram = sqrt (1+9+49) = your answer. > w is normal to the plan of the parallelogram. > Tomasso. had thought that area of the parallelogram to be sqrt(1 + 9 + 7^2). (But wasnÕt sure, and I appreciate the confirmation). doing the following: sqrt((-1)^2 + 3^2 + 7^2) = sqrt(1 + 9 + 7) = sqrt(17). I so much appreciated being able to show my results. Just to be able to check that I havenÕt done any silly typos like the above. Cassandra === Subject: Proving the cross product is orthogonal I hope I donÕt seem like one of those louts that only turn up at assignment time!! However I am struggling with a proof. My answer is not coming out as expected, and I think I am making a silly mistake somewhere. If anyone could have a glance over it, that would be appreciated. Q) a =(1, -2, 1) b =(3, 1, 0) Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b. A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product] a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7k Let w = -3i + 2j + 7k Prove vector a is perpendicular to w a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (a.w)/(|a||w|) = 0 Prove vector a is perpendicular to w b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) I know that is wrong, can anyone see what my mistkae is?? Cassandra === Subject: Re: Proving the cross product is orthogonal > I hope I donÕt seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is not > coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular to > each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] Do you know about the dot product? === Subject: Re: Proving the cross product is orthogonal >>I hope I donÕt seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] > Do you know about the dot product? Yes, I like the dot product...much easier then the cross product I think! It turned out my main mistake was silly mistakes in my workings (ie 2*0 = 2 etc) I did this two places, I think it was only ßuke that my workings for a being perpendicular to a x b, after the silly errors where fixed both proofs worked. Cassandra === Subject: Re: Proving the cross product is orthogonal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMN503869; >I hope I donÕt seem like one of those louts that only turn up at >assignment time!! However I am struggling with a proof. My answer is not >coming out as expected, and I think I am making a silly mistake >somewhere. If anyone could have a glance over it, that would be appreciated. >Q) a =(1, -2, 1) b =(3, 1, 0) >Find a x b and prove that your cross-product vector is perpendicular to >each of the vectors a and b. >A) [I have worked out the cross-product, and I have proven that a is >perpendicular to the cross-product, however I am struggling to prove >that b is perpendicular to the cross-product] >a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k Right hereÕs your mistake. You seem to be thinking -2*0 = -2 and 1*0 = 1. Oops! >Let w = -3i + 2j + 7k >Prove vector a is perpendicular to w >a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >cos&= (a.w)/(|a||w|) = 0 >Prove vector a is perpendicular to w >b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >I know that is wrong, can anyone see what my mistkae is?? Todd Trimble === Subject: Re: Proving the cross product is orthogonal >>I hope I donÕt seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is > not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be > appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular > to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] >>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k > Right hereÕs your mistake. You seem to be thinking -2*0 = -2 > and 1*0 = 1. Oops! >>Let w = -3i + 2j + 7k >>Prove vector a is perpendicular to w >>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >>cos&= (a.w)/(|a||w|) = 0 >>Prove vector a is perpendicular to w >>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >>I know that is wrong, can anyone see what my mistkae is?? > Todd Trimble Big ÔoopsÕ alright, I checked that about 5 times and still didnÕt see it...*sigh* === Subject: Re: Proving the cross product is orthogonal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMNR03877; >I hope I donÕt seem like one of those louts that only turn up at >assignment time!! However I am struggling with a proof. My answer is not >coming out as expected, and I think I am making a silly mistake >somewhere. If anyone could have a glance over it, that would be appreciated. >Q) a =(1, -2, 1) b =(3, 1, 0) >Find a x b and prove that your cross-product vector is perpendicular to >each of the vectors a and b. >A) [I have worked out the cross-product, and I have proven that a is >perpendicular to the cross-product, however I am struggling to prove >that b is perpendicular to the cross-product] >a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k >Let w = -3i + 2j + 7k >Prove vector a is perpendicular to w >a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >cos&= (a.w)/(|a||w|) = 0 >Prove vector a is perpendicular to w >b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >I know that is wrong, can anyone see what my mistkae is?? >Cassandra Yes, cosine= (b.w)/(|b||w|)= 0 but did you notice that that only requires that b.w be 0? It is not necessary to find |b| or |w|. As to what you did wrong, I hate to say it but: Arithmetic! you have: a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7k Now what, exactly, is -2*0-1*1? 1*3- 1*0? You seem to be under the impression that a*0= a. Most people say a*0= 0. === Subject: Re: Proving the cross product is orthogonal >>I hope I donÕt seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is > not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be > appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular > to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] >>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k >>Let w = -3i + 2j + 7k >>Prove vector a is perpendicular to w >>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >>cos&= (a.w)/(|a||w|) = 0 >>Prove vector a is perpendicular to w >>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >>I know that is wrong, can anyone see what my mistkae is?? >>Cassandra > Yes, cosine= (b.w)/(|b||w|)= 0 but did you notice that that only > requires that b.w be 0? It is not necessary to find |b| or |w|. > As to what you did wrong, I hate to say it but: Arithmetic! > you have: > a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k > Now what, exactly, is -2*0-1*1? 1*3- 1*0? > You seem to be under the impression that a*0= a. > Most people say a*0= 0. *grin* you are right, most people would say a*0=0. Normally I say the same....not sure what was up last night! Cassie === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess that you were writing too small print too tightly in the left upper corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); etc., etc. IHTH - Johan E. Mebius > I hope I donÕt seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is > not coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be > appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular > to each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] > a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k > Let w = -3i + 2j + 7k > Prove vector a is perpendicular to w > a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 > |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (a.w)/(|a||w|) = 0 > Prove vector a is perpendicular to w > b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) > I know that is wrong, can anyone see what my mistkae is?? > Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess > that you were writing too small print too tightly in the left upper > corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); > etc., etc. > IHTH - Johan E. Mebius >> I hope I donÕt seem like one of those louts that only turn up at >> assignment time!! However I am struggling with a proof. My answer is >> not coming out as expected, and I think I am making a silly mistake >> somewhere. If anyone could have a glance over it, that would be >> appreciated. >> Q) a =(1, -2, 1) b =(3, 1, 0) >> Find a x b and prove that your cross-product vector is perpendicular >> to each of the vectors a and b. >> A) [I have worked out the cross-product, and I have proven that a is >> perpendicular to the cross-product, however I am struggling to prove >> that b is perpendicular to the cross-product] >> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k >> Let w = -3i + 2j + 7k >> Prove vector a is perpendicular to w >> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (a.w)/(|a||w|) = 0 >> Prove vector a is perpendicular to w >> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >> I know that is wrong, can anyone see what my mistkae is?? >> Cassandra Much appreciated. I have been writing my equations in MS Word using the equation editor. Perhaps I should write them on paper Ôbig and largeÕ first!! Cassie === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > Much appreciated. I have been writing my equations in MS Word using the > equation editor. Perhaps I should write them on paper Ôbig and largeÕ > first!! With a pencil. The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose and sometimes because they donÕt check their work. === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE >>Much appreciated. I have been writing my equations in MS Word using the >>equation editor. Perhaps I should write them on paper Ôbig and largeÕ >>first!! > With a pencil. > The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose > and sometimes because they donÕt check their work. Yes, one of my biggest downfalls in life is that I make lots of mistakes. I once had to do a speed & accuracy test as part of a pysch test. I scored full marks (not sure how they calculated it), but I remember the pysch saying something about me getting alot further through the test then most, but making more mistakes then most. I used to have a turtle on my desk to remind myself to slow down and double check things. I find it so difficult, but it is something I should really work on. I have made two very elementary mistakes in the course of this thread...I do need to improve. === Subject: Just checking line slope I hope this doesnÕt sound dumb, but is the following an acceptable formula for a straight line. I am sure it is, but just wanted to check. y = x/6 === Subject: Re: Just checking line slope > I hope this doesnÕt sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Add or subtract any number to your x,or y or 6 in this equation and it is still a straight line. I first learnt it by visual detection: If the plot or graph is straight then it is linear, but if bent it is not so.There is only one way keeping it straight, so many ways to bend it. === Subject: Re: Just checking line slope <3Mcsd.58178$K7.34752@news-server.bigpond.net.au I hope this doesnÕt sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Yes. Checking your work is never dumb. In general the equation of a straight line in the plane is y = mx + b or x = a Are you familiar with this general equation? Your equation fits that general form for what values of m and b? === Subject: Re: Just checking line slope >>I hope this doesnÕt sound dumb, but is the following an acceptable >>formula for a straight line. I am sure it is, but just wanted to check. >>y = x/6 > Yes. Checking your work is never dumb. > In general the equation of a straight line in the plane is > y = mx + b or x = a > Are you familiar with this general equation? > Your equation fits that general form for what values of m and b? x/6 part. I have been struggling with some of those finer rules for what constitutes a linear equation. My textbook says this... The following *are* linear equations: 1) 3x + 2y = 7 2) x1-2x2 +10x3 + x4 = 0 3) 1/2x + y - (pi)z = sqrt(2) 4) (sin((pi)/2))x1 - 4x2 = e^2 The following *are not* linear equations: 5) xy + z = 2 6) e^x - 2y = 4 7) sin(x1) + 2x2 - 3x3 = 0 8) 1/x + 1/y = 4 Firstly my apologies for being unsure of ASCII syntax. Where I have written a number after the variable I am trying to indicate the number is a subscript, ie x1 is x subscript 1. What is the correct syntax? While I can understand why most of these are linear/not linear. Some have me confused. Why is 4 okay, but not 7? What is wrong with 8? Cassie === Subject: Re: Just checking line slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMN803873; >I hope this doesnÕt sound dumb, but is the following an acceptable >formula for a straight line. I am sure it is, but just wanted to check. >y = x/6 >> Yes. Checking your work is never dumb. >> In general the equation of a straight line in the plane is >> y = mx + b or x = a >> Are you familiar with this general equation? >> Your equation fits that general form for what values of m and b? >x/6 part. I have been struggling with some of those finer rules for what >constitutes a linear equation. >My textbook says this... >The following *are* linear equations: >1) 3x + 2y = 7 >2) x1-2x2 +10x3 + x4 = 0 >3) 1/2x + y - (pi)z = sqrt(2) >4) (sin((pi)/2))x1 - 4x2 = e^2 >The following *are not* linear equations: >5) xy + z = 2 >6) e^x - 2y = 4 >7) sin(x1) + 2x2 - 3x3 = 0 >8) 1/x + 1/y = 4 >Firstly my apologies for being unsure of ASCII syntax. Where I have >written a number after the variable I am trying to indicate the number >is a subscript, ie x1 is x subscript 1. What is the correct syntax? >While I can understand why most of these are linear/not linear. Some >have me confused. >Why is 4 okay, but not 7? >What is wrong with 8? >Cassie 4 has (sin((pi)/2)) which is a number, a constant. 4 is exactly the same as Ax_1- 4x_2= B 7, on the other hand, has sin(x1) and x1 is a VARIABLE, not a constant. It is functions of VARIABLES that count. 8 has 1/x and 1/y, which are quite different from just x and y. A linear function has the variables themselves, no powers or functions, multiplied by numbers (not other variables) then added or subtracted. === Subject: Re: Just checking line slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB4GMPG03973; >I hope this doesnÕt sound dumb, but is the following an acceptable >formula for a straight line. I am sure it is, but just wanted to check. >y = x/6 >> Yes. Checking your work is never dumb. >> In general the equation of a straight line in the plane is >> y = mx + b or x = a >> Are you familiar with this general equation? >> Your equation fits that general form for what values of m and b? >x/6 part. I have been struggling with some of those finer rules for what >constitutes a linear equation. >My textbook says this... >The following *are* linear equations: >1) 3x + 2y = 7 >2) x1-2x2 +10x3 + x4 = 0 >3) 1/2x + y - (pi)z = sqrt(2) >4) (sin((pi)/2))x1 - 4x2 = e^2 >The following *are not* linear equations: >5) xy + z = 2 >6) e^x - 2y = 4 >7) sin(x1) + 2x2 - 3x3 = 0 >8) 1/x + 1/y = 4 >Firstly my apologies for being unsure of ASCII syntax. Where I have >written a number after the variable I am trying to indicate the number >is a subscript, ie x1 is x subscript 1. What is the correct syntax? Many people use what you use (x1, etc.), but perhaps more common is x_1, or x_(1) or x_{1} when clarity is needed. The main thing required is that the notation be uncluttered and unambiguous; youÕre doing well on both counts. >While I can understand why most of these are linear/not linear. Some >have me confused. >Why is 4 okay, but not 7? >What is wrong with 8? (4) is ok because itÕs in the form ax + by = c where a, b, c are *constants* [sin(pi/2) and e^2 are constants; remember that a function applied to a constant yields another constant]. On the other hand, (7) is not linear because sin(x1) is not linear in x1 [is not a constant times x1]. In (8), neither 1/x nor 1/y is linear (neither is a constant times the appropriate variable). Todd Trimble === Subject: Re: Just checking line slope The following *are* linear equations: > 1) 3x + 2y = 7 > 2) x1-2x2 +10x3 + x4 = 0 > 3) 1/2x + y - (pi)z = sqrt(2) > 4) (sin((pi)/2))x1 - 4x2 = e^2 > The following *are not* linear equations: > 5) xy + z = 2 > 6) e^x - 2y = 4 > 7) sin(x1) + 2x2 - 3x3 = 0 > 8) 1/x + 1/y = 4 > Firstly my apologies for being unsure of ASCII syntax. Where I have > written a number after the variable I am trying to indicate the number > is a subscript, ie x1 is x subscript 1. What is the correct syntax? x_1 and x^n for n-th power of x. Best readable ascii style is to persist like youÕre doing for the most part with adequate spaces like x1 - 2x2 + 10x3 + x4 = 0 > While I can understand why most of these are linear/not linear. Some > have me confused. > Why is 4 okay, but not 7? sin pi/2 is a constant. sin x_1 isnÕt ax_1 for some a in R. > What is wrong with 8? It doesnÕt have the form ax + by = c, the most general equation for a line, hence the expression linear, line like. Indeed, y + x = 4xy Were you to graph the equations, what is linear and what is not would likely jump out at you for only lines are linear. === Subject: Re: Just checking line slope > Were you to graph the equations, what is linear and what is not > would likely jump out at you for only lines are linear. an equation looks linear on a graph, then it is linear. Aah, I enjoy maths so much...but it is also very hard at times...posibly why I like it so much! cassandra === Subject: Re: Vectors question 2 > I am working through a question for a university assignment. I am not > wanting to be given the answer, but am after some pointers. > The question is: > A train is travelling in a straight line at 24km/h. A movie stunt-man > moves at 4km/h straight across its roof, at right angles to the > direction of its motion. Draw a clear sketch of his resultant movement > relative to the ground, and find the speed and direction of that movement. > My take on this is as follows (way below is a very bad ASCII > representation of the type of graph that is in my head...definetly not > to any sort of scale) > I first create a graph with x-axis distance travelled, and y-axis time. I would give up on attempting to depict time in this picture, unless you had some real attachment to drawing a 3-dimensional picture. Why? Well, the trainÕs motion takes up one dimension, time a second one, and the stuntmanÕs motion (which is not collinear with the trainÕs, if I read the problem correctly) a third one. Rather, imagine the displacements traversed over one unit of time. That unit can be of your choosing: one second, one hour, one millenium, ... you get to name it (I prefer a unit that makes calculations trivial, and that would suggest one hour; the problem is that you would then need to imagine a train that is 4 km wide. While the mental image that emerges is difficult to believe in, the mental image is absolutely irrelevant; if the mental image is important, imagine the time interval to be 1 second or 1 millisecond). You can do this based on an assumption of constant velocities, and the simple (vector) relation s = vt, where s represents the (vector) displacment, v the (vector) velocity, and t the (scalar) time, which (by assumption) is a constant. If you have a fixed time interval, the equation s = vt takes a velocity v, multiplies it by a fixed time interval t, and obtains a vector that points from the starting position, to the position reached at the end, at time t. The problem is to deal with the vector addition (which is what is meant by the resultant) of velocities, and what you need to do is to represent that. So, let the train be travelling along the +x axis, and the stuntman travelling along the +y direction. The resultant, which is the sum of the two vectors v_train and v_stuntman, represents the velocity vector of the stuntman with respect to the ground. (It would alternatively represent the velocity of the train with respect to the ground, if it were the stuntman running along the ground carrying a moving train. IÕll ignore that interpretation.) > I can then plot u = 24i + j to show the speed that the train is > travelling at. > The line perpendicular to to u is the speed and direction of the stunt > man, the length of that line will be 4. Note that you are devoting the vector i to displacement along a single (spatial) direction, and j to displacement along the time direction. This is incorrect. The vector u is fine. However, your vector perpendicular to u will be (proportional to) either -i + 24 j or i - 24 j, since those are along the (only) direction perpendicular to u in the plane containing the directions i and j. In the first case, the stuntman is moving along the -i direction. That canÕt be right, since he must be moving perpendicular to the trainÕs direction. In the second case, time is moving backwards. That might be nice for a sci-fi ßick, but it doesnÕt really happen that way in real life. You need to notice that the stuntmanÕs motion is perpendicular in spatial terms: if the train moves east, he must be going north or south. ItÕs the spatial content that you must model. Everyone and everything moves through time in exactly the same way (ignoring the subtleties that relativity forces us to deal with). Thus, the description of the motions of the train and the stuntman must show their motions to be perpendicular *in space*. Their behavior in time can be reßected in dealing with the magnitudes of their velocity vectors, and those can be made spatial in character by using the formula s = vt that I mentioned earlier. The problem is that the i: space and j: time coordinate system makes for a single spatial dimension, and that doesnÕt allow for two separate spatial directions. ThatÕs why I said earlier that this approach would require three dimensions to depict everything, and that it would be best to ignore devoting a dimension to the time variable. > The line that forms the hyptonuse (v), represents the speed (length), > and direction (x,y) that the stuntman has travelled. > Am I on the right track (no pun intended)? If not, could you offer some > suggestions. Sorry, you have missed the train entirely (that pun was irresistable). Another will be along shortly, provided you follow the instructions I gave earlier. > | 0 > | 0 0 > | 0 0 > | v 0 0 (24,1) > | 0 0 > | 0 0 | 0 0 u > | 0 0 > |0____________________________ Please note, this is an assignment, so I am not wanting anyone to solve > it for me. I do not want any reason to be accused of wrongdoings. > Cassandra I hope the information I gave was both sufficient to get you on your way, and yet not too much. Dale. === Subject: Re: Vectors question 2 >> I am working through a question for a university assignment. I am not >> wanting to be given the answer, but am after some pointers. >> The question is: >> A train is travelling in a straight line at 24km/h. A movie stunt-man >> moves at 4km/h straight across its roof, at right angles to the >> direction of its motion. Draw a clear sketch of his resultant movement >> relative to the ground, and find the speed and direction of that >> movement. > I hope the information I gave was both sufficient to get you on your > way, and yet not too much. > Dale. Dale, written a few more times I think before I understand it all. So after reading your response I began to think. Ignoring vectors and i & j etc. If I wanted to figure this out a few weeks ago (prior to learning this stuff) my method would have been as follows. Put the direction of the train on the x-axis, the direction of the stuntman on the y-axis. (this is I think what you are trying to say). So now for the first part of the question --Sketch the resultant movement of the stuntman relative to the ground-- The sketch would be the line y = x/6, for x >= 0. The second part of the question --find the speed--- is answered by finding the length of the line. speed = sqrt(4^2 + 24^2) = 4*sqrt(37) km/h The final part of the question then --find the...direction of that movement-- This would be where the vector math comes in. tan& = b / a = 4 / 24 = 1 / 6 & = tan^-1(1/6) = 0.165 Phew, this seems alot nicer. got me started at any rate. Any further comments? Now for question number three... Cassandra === Subject: Re: Vectors question 2 > So after reading your response I began to think. Ignoring vectors and i > & j etc. If I wanted to figure this out a few weeks ago (prior to > learning this stuff) my method would have been as follows. > Put the direction of the train on the x-axis, the direction of the > stuntman on the y-axis. (this is I think what you are trying to say). That method is correct, and produces the correct answers... so now my question is: what is it that you learned over the last few weeks that confused you into trying to plot time as well? meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Vectors question 2 >>So after reading your response I began to think. Ignoring vectors and i >>& j etc. If I wanted to figure this out a few weeks ago (prior to >>learning this stuff) my method would have been as follows. >>Put the direction of the train on the x-axis, the direction of the >>stuntman on the y-axis. (this is I think what you are trying to say). > That method is correct, and produces the correct answers... so now my question > is: what is it that you learned over the last few weeks that confused you into > trying to plot time as well? > meeroh To answer your question, I overthought the question. We are studying vectors, including 3-dimensional vectors and projections. Perhaps it was because I had only just finished studying projections. Or because I thought that question must be more complex then it looked etc. I really donÕt know why, however I am sort of glad because it has allowed me to gain alot better understanding into using vectors, but then making mistakes does usually have that effect. cassandra === Subject: Re: Vectors question 2 >>So after reading your response I began to think. Ignoring vectors and i & j >>etc. If I wanted to figure this out a few weeks ago (prior to learning this >>stuff) my method would have been as follows. >>Put the direction of the train on the x-axis, the direction of the stuntman >>on the y-axis. (this is I think what you are trying to say). > That method is correct, and produces the correct answers... so now my > question is: what is it that you learned over the last few weeks that > confused you into trying to plot time as well? To answer your question, I overthought the question. We are studying > vectors, including 3-dimensional vectors and projections. Perhaps it was > because I had only just finished studying projections. Or because I thought > that question must be more complex then it looked etc. > I really donÕt know why, however I am sort of glad because it has allowed me > to gain alot better understanding into using vectors, but then making > mistakes does usually have that effect. Well, if you want to consider this and include time as one of your dimensions, then you have to realize that you have to project the space-time coordinate system into a space-only coordinate system before trying to describe it in terms of visible effects. For example, consider a single point traveling along a straight line. Let the line of travel be the X axis and let time T be perpendicular to it. Then, in the XT plane, the point will be describing some curve (continuous, unless your universe includes teleportation). For example, if the point is oscillating about the origin, then the space-time curve describing its motion would be a sine wave on the T axis. But a description of that motion would probably not refer to a sine wave about the T axis, it would refer to an oscillatory motion about the origin. This seems like a trivial point until you start considering more dimensions: consider an XY plane and a time T axis perpendicular to it, and two points traveling in the plane -- one at a fixed velocity along the X axis, starting at the origin, and the other at the same velocity along the Y axis, also starting at the origin. If you look at them in the XY plane, then their trajectories (namely, the X and Y axes) are perpendicular. However, if you consider them to be points in XYT space, then the first oneÕs trajectory is a straight line in the XT plane, and the second oneÕs trajectory is a straight line in the YT plane, and those trajectories are not perpendicular! It is only their projections into the XY place (the space-only coordinate system) that are perpendicular. Which brings me to your question; your mistake in considering this with distance on the X axis and the time on the T axis was that when the description of physical motion of the stunt man was described as being perpendicular to the motion of the train, you drew the perpendicular in the XT plane, which doesnÕt work. What you need to do here is count the number of dimensions of space (clearly, you need at least two, as the man the the train are moving at right angles to each other. Then you add one for time, and you see that you need a three-dimensional system. In that system, the trajectory of the train will be a straight line; if the train travels along the X axis, as it does in your original graph, then its XYT trajectory will be a straight line in the XT plane, as it is in your original graph. However, the motion of the man is, according to the problem statement, perpendicular to that of the train -- but the problem statement refers to them in the space-only coordinate system, which means that you have to have the man moving perpendicular not to the XYT time-space trajectory of the train, but to the XY spatial trajectory of the train. If you do that, then everything comes out right, but itÕs somewhat complicated to draw that. Now, there is no really good reason to involve time in this, as the movement of the train (relative to the ground) and the man (relative to the train) are constant with time, so what you are asked for (movement of the man relative to the ground) is also constant and you can draw it without involving the T axis at all. However, I think you should know why your initial solution didnÕt work and how you could have made it work. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Proving Trig Identities? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3Frlo09977; hello. i need some help proving some identities. #1. cos40=1-8sin^2thetacos^2theta #2. sin(pi/6-s)+cos(pi/3-s)=cos s === Subject: Re: Proving Trig Identities? >hello. i need some help proving some identities. >#1. cos40=1-8sin^2thetacos^2theta I have no idea what this means. Please post again in some notation that we can understand. >#2. sin(pi/6-s)+cos(pi/3-s)=cos s There may be a more clever way, but you could use the formulas for sin(A-B) and cos(A-B) and simplify. What have you tried? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Proving Trig Identities? >hello. i need some help proving some identities. >#1. cos40=1-8sin^2thetacos^2theta Assuming thatÕs 4theta on the left side. Think of cos(4theta) as cos(2*2theta) and use the double angle for cosine that involves only sines. That should get you started. >#2. sin(pi/6-s)+cos(pi/3-s)=cos s Use the addition formula for sin(a - b) and cos(a - b) and the values for the functions of pi/6 and pi/3. --Lynn === Subject: TFC and contribution by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3HG1V18324; can you help me with this question guys? The Selling price per unit (p) plus the Total Variable Cost(xv) less the Total Fixed Costs (TFC) per unit is equal to: a. Contribution (C) to Total Fixed Costs (TFC) b. Contribution (C) to loss, before the B.E.P. c. Contribution (C) profit, after the B.E.P. d. Both a and c e. none of the above im just confused on this === Subject: Re: TFC and contribution > can you help me with this question guys? > The Selling price per unit (p) plus the Total Variable Cost(xv) less > the Total Fixed Costs (TFC) per unit is equal to: > a. Contribution (C) to Total Fixed Costs (TFC) > b. Contribution (C) to loss, before the B.E.P. > c. Contribution (C) profit, after the B.E.P. > d. Both a and c > e. none of the above > im just confused on this I have no idea. WhatÕs a B.E.P.? Also, it would be helpful for you to show any work/insights you have gained so far in solving the problem. Bill === Subject: Re: JSH: Two sides [snip delusion] What does properly a unit mean? Can you answer that simple question James Harris? === Subject: Re: JSH: Two sides > [snip delusion] > What does properly a unit mean? > Can you answer that simple question James Harris? JSH does not answer questions. === Subject: Animation on Mathematica or Maple by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3LRdo09446; Using Mathematica or Maple, I am trying to illustrate a sphere S1 (stationary) with a much smaller sphere S2 (animated) orbiting S1. I would like to have the distance between S1 and S2 as close to zero as possible. One last thing IÕd like to do is be able to view the animation from different angles (I know that Mathematica allows you to spin 3D objects in real time). Any suggestions on how to make this happen. So far IÕve only got a stationary preview of S1. Help would be greatly appreciated!! Joseph A. === Subject: Re: Animation on Mathematica or Maple >Using Mathematica or Maple, I am trying to illustrate a sphere S1 >(stationary) with a much smaller sphere S2 (animated) orbiting S1. I >would like to have the distance between S1 and S2 as close to zero as >possible. One last thing IÕd like to do is be able to view the >animation from different angles (I know that Mathematica allows you to >spin 3D objects in real time). >Any suggestions on how to make this happen. So far IÕve only got a >stationary preview of S1. Help would be greatly appreciated!! >Joseph A. Ask and ye shall receive :-). I had a little time to kill this evening, so here is some maple code -- it might not display to well here but, hey, itÕs free and it works. I have put line spaces between each line of Maple instructions. R is the big sphere radius, r the small one, and d the separation between them. The number 80 in the display statement can be raised for better resolution. Enjoy. --Lynn >restart; >R:=10;r:=1;d:=0.2; >big:=plot3d([R*sin(phi)*cos(theta),R*sin(phi)*sin(theta),R* cos(phi)],theta= 0..2*Pi,phi=0..Pi,color=cyan): >small:=t->plot3d([(R+r+d)*cos(t),(R+r+d)*sin(t),0]+[r*sin( phi)*cos(theta),r *sin(phi)*sin(theta),r*cos(phi)], theta=0..2*Pi,phi=0..Pi,style=patchnogrid,color=brown): > orbit:=display(seq(small(k*2*Pi/80),k=0..80),insequence=true): > display({orbit,big},scaling=constrained); === Subject: Re: Animation on Mathematica or Maple by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB42XU702002; >Using Mathematica or Maple, I am trying to illustrate a sphere S1 >(stationary) with a much smaller sphere S2 (animated) orbiting S1. I >would like to have the distance between S1 and S2 as close to zero as >possible. One last thing IÕd like to do is be able to view the >animation from different angles (I know that Mathematica allows you to >spin 3D objects in real time). >Any suggestions on how to make this happen. So far IÕve only got a >stationary preview of S1. Help would be greatly appreciated!! >Joseph A. Here is the stationary view of the two spheres. I want S2 to revolve around S1. How can I do this using mathematica? http://www.mindcraftstudios.com/mathematics/orbit.jpg === Subject: Re: Animation on Mathematica or Maple by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB42XTU01997; Here is a stationary view of S1 (large sphere) and S2 (small sphere). What I want is S2 to revolve around S1. Also, I want to be able to see the rotation from any view. Joseph A. === Subject: Re: Animation on Mathematica or Maple >Here is a stationary view of S1 (large sphere) and S2 (small sphere). >What I want is S2 to revolve around S1. Also, I want to be able to >see the rotation from any view. << Graphics`Shapes` s = Table[Show[ TranslateShape[Graphics3D[Sphere[0.1,12,12]],{1.5Cos[t], 1.5Sin[t],0}], Graphics3D[Sphere[1, 12, 12]], Boxed -> False], {t, 0, 2Pi, Pi/24}]; Export[spheres.gif, s] There are gif-specific options that Export will accept to modify some of the presentation. There is also an option to Show or maybe Graphics3D, which I have forgotten and have spent ten minutes unsuccessfully searching for, that will keep the spheres from wobbling back and forth as the animation progresses. But maybe this will give you an idea how to get started with this. If you get in a bind and this isnÕt enough to let you figure it out then throw mail at me and IÕll fiddle with this a little more. There are some bugs in Exported animated gifÕs, IÕve reported the ones that IÕve found and havenÕt heard whether they have been fixed in the latest version or not. I just tried rendering this with the Windows XP SP2 Picture and Fax Viewer and it fails to display this. But IrFanView is happy with the result. You will likely find that if you push the edges of the implementations of animated gif viewers that you will turn up bugs and features that you didnÕt expect. === Subject: JSH: James, you be the judge... Here is how James Harris operates: 1. Start with some goofy polynomial that was a leftover from one of his many failed FLT proofs. No explanation of motivation or reasoning provided for choosing that particular polynomial or explanation of why it might be meaningful in his new context. 2. Avoid being specific about such important details as which ring he intends to work in. 3. Dream up a bunch of weird non-standard terminology like properly unit, dividing off, etc. Use terms incorrectly, such as referring to multiple when he should use the term factor. Also, make use of standard terminology like distributive property and constant term without actually understanding it or using it properly. Also, add lots of vague distractions such as the equation has no memory, you can see the 7Õs in there, canÕt you? to act as a smoke screen. 4. Apply a bunch of algebraic manipulations, some of which are just wrong. Make inappropriate generalizations from specific cases to general cases, etc. 5. Get a result that seems contradictory, and assume that the error was with the core of algebra rather than the much more likely explanation that he himself might have made an error. 6. Through out a bunch of paranoid ad hominem attacks on critics. Then hypocritically claim that all his critics resort to social crap and personal attacks rather than sticking to mathematics, logic, and reasoning. We are supposed to believe that his arbitrarily chosen goofy polynomial just happens to one that, when probed by the genius of James Harris, give results that shake mathematics to its core! James, you be the judge. Are you a pile of crap? === Subject: JSH: Simply fascinating The math here is so readily understandable that itÕs actually almost as interesting watching how people react to it, as anything else. For instance, IÕve given a polynomial P(x) repeatedly where I factor it into three factors. I point out that the factors must include factors of the constant term of the polynomial P(x). I note that the factors of the constant term are independent of x, as they are, in fact, constant. Mathematically itÕs easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the polynomial results from multiplying together the three factors which IÕve called g_1(x), g_2(x), and g_3(x), then you can get the factors of the constant term, by setting x=0. ItÕs that simple. Notice (1) you have the factors of P(x), (2) the constant term of P(x) is determinable by setting x=0, (3) you also get the factors of the constant term. Mathematically, itÕs simple to the point of trivial. Now then, if you have *constants* which are factors of another constant then why would anyone try to argue that they are actually variables? You have x, as the variable. Besides x there are just these numbers. If you clear out x, then whatÕs left are constants. Letting x=0, clears it out, leaving the constants visible. Some may say, yeah, sure, at x=0, but what about when x doesnÕt equal 0? Um, if the numbers are constant, and so are independent of x, then, duh, why should it matter what value x has? The logic is inescapable. In terms of difficulty, my proof is about as easy as it gets in algebraic number theory, in terms of the actual mathematics. But the concepts are where there is a problem, and the social hang-up is in accepting that thereÕs this simple technique that shows a BIG problem, which can invalidate claims of proof for, well, over a hundred plus years. So the mathemtics is EASY for a trained mathematician to follow. The social implications are hard, if social stuff is important to you, and clearly from what IÕve seen it is to many of you. For instance, at this point IÕve removed all objections raised in detail. Like I can explain supposed counter examples to my work. I can give an actual example where you can see the factorization play out--just as the theory shows it must. And you probably know that my research is the work that can be said to have gone to a journal which at least claims it does formal peer review. They thanked me for the paper said the reviewers liked it, and then some sci.mathÕers got together--actually literally conspiring online in posts on sci.math--sent them emails and the editors yanked my papers THE NEXT DAY. They had it for nine months. IÕd corresponded with them for a while, even corrected them when they called me Dr. Harris as I donÕt have a Ph.d, and I told them I was an independent researcher with concerns about how my work would be handled. They kept saying no problem, ok, all that matters is whatÕs correct. Then they yanked my paper after sci.mathÕers emailed them: All the pertinent facts are in my favor. So whatÕs the hold-up? My research shows that some mistakes were made over a hundred years ago, and a lot of people missed them, and gave proofs which were not, and are not proofs. Mathematics is unforgiving. It doesnÕt care about the social implications of the truth. So it doesnÕt matter mathematically that a LOT of people out there are terribly dependent on the false beliefs and incorrect results, but it DOES matter a lot to those people! I call their behavior passive-aggessive, as by dragging their feet, taking as long as possible before acknowledging my research, or worse, hoping to NEVER acknowledge it at all, they are passively hoping to escape mathematical truth, in what amounts to a very aggressive way. I liken their behavior to judges at a race, who watch a runner break a world record, and then lie about his time, refuse to admit he even finished the race, and some even call him names!!! TheyÕre turning the way itÕs supposed to work with a major discovery, upside down. And itÕs silly behavior as eventually the truth will come out, and you know what IÕll do then? Probably go to the beach. IÕll also hang out in some bars. Yup, IÕll definitely hang out in some bars, preferably near a beach. Yup, you guessed it, IÕll do my best to forget about them, as why bother worrying about silly people who do silly things. LifeÕs too short. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Simply fascinating > IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). In general thatÕs not true at all. For example, consider: (x+1)*(x+2)*(x+3) If you expand it out to get a polynomial in the usual form, you see that the constant term is 6. But 6 is not a factor of any of the three polymomial factors, in fact neither 2 nor 3 is a factor of any of them say what you really meant to say? Or maybe you were totally mistaken and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. ItÕs true that some mistakes were made long ago, the worst in my opinion being the first developers of set theory who concocted a set of all sets, which was subsequently proven not to exist. But IÕve seen no evidence that *you* have ever done any research that showed any such past mistakes of others. Instead all I see is *you* making mistakes yourself and not admitting them. Do you know how to factor 7 in the ring of integers with sqrt(7) adjoined? ThatÕs easy, of course you do, right? But do you know how to factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can figure out that simple arithmetic problem, hereÕs something more interesting: Find all primes p (other than 2 and 7 which IÕve already shown you) such that 7 can be nontrivially factored in the ring of integers with sqrt(p) adjoined. Finally, find all finite sets of two or more primes p1,p2,...,pn such that 7 is composite in the ring you get when you adjoin all the sqrts of those primes but itÕs prime in the ring you get when you adjoin all but one of those sqrts. Same question if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a prime. (Note: When I refer to a prime p or primes p1,p2..., I mean prime in the ring of integers. When I refer to 7 being prime or composite, I mean prime or composite in the ring of integers adjoined with the various sqrts.) === Subject: Re: JSH: Simply fascinating > IÕve given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > In general thatÕs not true at all. For example, consider: > (x+1)*(x+2)*(x+3) But he said, e.g. in your example, having factored your polynomial into (x+1), (x+2), and (x+3), 1, 2, and 3 must be factors of 6, the constant term of your expanded polynomial. > If you expand it out to get a polynomial in the usual form, you see > that the constant term is 6. But 6 is not a factor of any of the three > polymomial factors, in fact neither 2 nor 3 is a factor of any of them He didnÕt say, with respect to your example, that 6 must be a factor of 1, 2, and 3. Rather that 1, 2, and 3 must be factors of 6. KeithK > say what you really meant to say? Or maybe you were totally mistaken > and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. > ItÕs true that some mistakes were made long ago, the worst in my > opinion being the first developers of set theory who concocted a set > of all sets, which was subsequently proven not to exist. But IÕve seen > no evidence that *you* have ever done any research that showed any such > past mistakes of others. Instead all I see is *you* making mistakes > yourself and not admitting them. > Do you know how to factor 7 in the ring of integers with sqrt(7) > adjoined? ThatÕs easy, of course you do, right? But do you know how to > factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can > figure out that simple arithmetic problem, hereÕs something more > interesting: Find all primes p (other than 2 and 7 which IÕve already > shown you) such that 7 can be nontrivially factored in the ring of > integers with sqrt(p) adjoined. Finally, find all finite sets of two or > more primes p1,p2,...,pn such that 7 is composite in the ring you get > when you adjoin all the sqrts of those primes but itÕs prime in the > ring you get when you adjoin all but one of those sqrts. Same question > if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a > prime. (Note: When I refer to a prime p or primes p1,p2..., I mean > prime in the ring of integers. When I refer to 7 being prime or > composite, I mean prime or composite in the ring of integers adjoined > with the various sqrts.) === Subject: Re: JSH: Simply fascinating