mm-416 === Subject: Re: Editors again, New York Journal of Mathematics >Maybe he's not familiar with Jack and Jill or ghlights for Cldren. Both of those are too advanced for s stuff. Remember those Golden Books that had various stories? That's probably the level of s proofs. David === Subject: Solution to sum of a discrete-time decaying complex exponential I am looking for a closed-form solution to the following problem; X(N,w) = SUM( (N^-K/2)*EXP(-j*w*K*ln(N)) ), K = integer from 1 to infinity, N an integer > 1, w an arbitrary angular frequency. Numerical simulations show that the sum X always converges a single complex value. Adams === Subject: Re: Solution to sum of a discrete-time decaying complex exponential >I am looking for a closed-form solution to the following problem; >X(N,w) = SUM( (N^-K/2)*EXP(-j*w*K*ln(N)) ), K = integer from 1 to >infinity, N an integer > 1, w an arbitrary angular frequency. It's just a geometric series. Assuming your j is what mathematicians call i, and your N^-K/2 is N^(-K/2), the series converges and the sum is N^(-1/2) exp(-i w ln(N))/(1-N^(-1/2) exp(-i w ln(N))) = N^(-1/2-iw)/(1-N^(-1/2-iw)) === Subject: Re: Small prime factors and Christian Aebi's 13 problem >> In http://www.oswego.edu/~baloglou/math/trteen.html I >> problem and prepared the trteen web page mostly for fun* >BTW, I haven't visited your website. At the mention of 13 >web pages, warning bells started to ring in my head and the >question, do you really want to subject your eyes to such >an ordeal? was asked (and answered). 'trteen' web page, not trteen web pages -- and that was not the only misunderding on your part, I am afraid... [Hey, if you have trouble with trteen and your eyes are a bit sensitive, how about http://www.oswego.edu/~baloglou/103/seventeen.html? It's more fun!] :-) === Subject: Re: Small prime factors and Christian Aebi's 13 problem charset=iso-8859-1 >> In http://www.oswego.edu/~baloglou/math/trteen.html I >> problem and prepared the trteen web page mostly for fun* >BTW, I haven't visited your website. At the mention of 13 >web pages, warning bells started to ring in my head and the >question, do you really want to subject your eyes to such >an ordeal? was asked (and answered). 'trteen' web page, not trteen web pages -- and that > was not the only misunderding on your part, I am afraid... I bow to your wisdom! :-) > [Hey, if you have trouble with trteen and your eyes are a > bit sensitive, how about > http://www.oswego.edu/~baloglou/103/seventeen.html? It's more > fun!] :-) Indeed! Daniel === Subject: Re: Math rules REVISED Well, remember c is a factor of the *cont* term of the polynomial, > so they're trying to make a new nonsensical math rule. For ince, by such an argument maybe with 2(x^2 + 2x+1) = (x+1)(2x+2) the 2 could in fact somehow, someway be a variable dependent on x. Wch 2? There are two on the left-hand side, two on the > right-hand side. Does it matter? > I could certainly believe that in factoring > x^2 + ax + 12 = (x+r1)(x+r2) over the integers, > that r1 and r2 depend on a, despite the fact that > both must be factors of 12 in the integers. What's so strange about that? - Randy Well, notice that you have r_1 = (-a + sqrt(a^2 -48))/2 arbitrarily picking the positive for r_1, wch you might notice shows the dependency between r_1 and 'a'. The pertinent expression though is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and faced with a mathematically precise argument wch step-by-step shows a way to factor that expression, wch gives a fascinating conclusion, some seem willing to instead try to throw out the math by making a new rule. By their rule factors like f^2 can be decomposed into dependent variables just because they want them to be. So they claim that you have sometng like f^2 = w_1(m) w_2(m) w_3(m) where the w's vary as m varies, so that they can claim that if you divide off f^2, you have a factorization like (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (a_1 x + uf)/w_1(m) (a_2 x + uf)/w_2(m) (a_3 x + uf)/w_3(m) and they want the w's to vary with m because I can prove that in fact they are not dependent on m, as actually you have w_1(m) = f, w_2(m) = f, and w_3 = 1. === Subject: Re: Math rules REVISED > ...and they want the w's to vary with m because I can prove that in fact > they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? -- The only tng that never seems to run of gas is a pompous gasbag. -- -- http://www.crbond.com === Subject: Re: Math rules REVISED > ...and they want the w's to vary with m because I can prove that in fact > they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? -- > The only tng that never seems to run of gas is a pompous gasbag. > -- > But you see, I *do* prove it, and follow rather basic math rules in doing so, wle other people come along, and make up bogus math rules, apparently because they don't like a particular conclusion that follows from the math. Here's the information, where I go step-by-step, and I'm copying from the post that STARTED ts thread. It seems to me that some of you so despise mathematics that even a step-by-step exposition using BASIC math rules is not enough for you to acknowledge a math result you don't want. I find that fascinating, and sad. After all, what are you doing on sci.math then? Lurking to try and destroy the math community by attacking it at its base? The math rules here are so easy I can give them quickly in ts post. For ince, for a factor g of a polynomial P(m), g=r+c exists, where c=g when m=0, and r = g-c. How much simpler can you get? Ok, so you need a ring, and wle my point is it has problems, I'll use them here so that I can say that the other condition is that g is in the ring of algebraic integers for any algebraic integer m. So now I'll go to the following expression: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) If you're a purists you may have problems with me calling that P(m) as you have all those variables running around. I'll try and help out a bit by letting f=5, so that you have P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) and you still may be bothered by a lot of variables, but I need the remaining ones for placement, and you may also wonder what the ring is. Well consider the ring to be algebraic integers, so all of the variables are algebraic integers. Wle I'm looking at the expression as a polynomial P(m), the remaining variables are there as *placeers* so that I can factor it into non-polynomial factors. That is, so that I have 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). That may seem awkward but its purpose provides the need for it. Still by math rules what I've done is quite correct. Now though you can see the purpose of generalizing a factor g, as here I finally have more than a trivial use for g=r+c. Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, because g_1 is a factor of P(m). Letting m=0, with P(m), I get P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), and now you can see another reason for why I need to keep the other variables as placeers. Here that reveals that when m=0, (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), wch means that exactly two of the a's go to 0, when m=0. Well that makes it easy, as my indices are arbitrary, so I can pick a_1 and a_2 to go to 0 when m=0, and now figure out that g_1 = 5u when m=0, wch gives c_1 = 5u, and then I have r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. You may be wondering what's the point, since r and c here map to what would be the r and c if x were the key variable. Well I'm working towards the point as that 25 factor of 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) has just been sitting there, when normally you'd separate that off when factoring, as *normally* it'd make the factorization ambiguous. I'll do that now and consider P(m)/25, so that I have P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3. But what now? Well before I had 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) so I can wonder what happens to (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) when the 25 is divided off. It turns out that focusing on the cont term of P(m) provides an answer as from before I have P(0) = 25u^2(3x + 5u), where intriguingly the 25 is again visible as a factor. So dividing both sides by 25 gives me P(0)/25 = u^2(3x + 5u), and there are no more visible factors of 5, but you might imagine that maybe 'u' or 'x' could have a factor that is 5. Well possibly for specific cases, but definitely not in general, right? After all, I might pick x=7 and u=11, as there's been no constraint on those variables, and I've left them in as placeers. So what's the significance of that result? Well focusing now on g_1 as I go from the general to the specific, as a variable of P(m), it's also affected by the dividing off of 25. I'll imagine a factor w_1 divides off of g_1 when 25 is divided off of P(m). That gives me g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. So using g_1/w_1 = r' + c', I can set use P(0)/25, wch is P(0)/25 = u^2(3x + 5u), to see that c' = u. That forces w_1 = 5, so that I have g_1/w_1 = a_1 x/5 + u wch *should* mean that a_1 has a factor that is 5 in the ring. === Subject: Re: Math rules REVISED I honestly don't get your very first rule, because you define no dependency of P(m) to g (equalling r+c), wch seems typical of your exhortations. if,now, youbeleive that the googolplex has stranded you with an audience that is exclusively composed of hateful morons ... assuming that there really is no-one in these threads who is able to maintain an agreement with your Fine New Math, BUT that it's not just you ... then you should start giving seminars to capable youth, who will then carry-on. heck, if you come to LA, you couldjoin Hulkboy's afterschool curricula; if he should become Governeurateur,then you'd be the Secretary of the New Math -- after you'd convinced m, of course! wait; I realized that I didn't read that, correctly; still, what is the hypothesis that youhave found, relating to the math, other than to mathematicians as a ******* class? more to the point, have you ever proven *any* other mathematical theorem, or geometrical one? > The math rules here are so easy I can give them quickly in ts post. For ince, for a factor g of a polynomial P(m), g=r+c exists, where > c=g when m=0, and r = g-c. How much simpler can you get? Ok, so you need a ring, and wle my point is it has problems, I'll > use them here so that I can say that the other condition is that g is > in the ring of algebraic integers for any algebraic integer m. So now I'll go to the following expression: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) If you're a purists you may have problems with me calling that P(m) as > you have all those variables running around. I'll try and help out a > bit by letting f=5, so that you have P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) and you still may be bothered by a lot of variables, but I need the > remaining ones for placement, and you may also wonder what the ring > is. Well consider the ring to be algebraic integers, so all of the > variables are algebraic integers. Wle I'm looking at the expression as a polynomial P(m), the > remaining variables are there as *placeers* so that I can factor > it into non-polynomial factors. That is, so that I have 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). That may seem awkward but its purpose provides the need for it. Still by math rules what I've done is quite correct. Now though you can see the purpose of generalizing a factor g, as here > I finally have more than a trivial use for g=r+c. Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, > because g_1 is a factor of P(m). Letting m=0, with P(m), I get P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), and now you can see another reason for why I need to keep the other > variables as placeers. Here that reveals that when m=0, (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), wch means that exactly two of the a's go to 0, when m=0. Well that makes it easy, as my indices are arbitrary, so I can pick > a_1 and a_2 to go to 0 when m=0, and now figure out that g_1 = 5u when m=0, wch gives c_1 = 5u, and then I have r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. You may be wondering what's the point, since r and c here map to what > would be the r and c if x were the key variable. Well I'm working towards the point as that 25 factor of 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) has just been sitting there, when normally you'd separate that off > when factoring, as *normally* it'd make the factorization ambiguous. I'll do that now and consider P(m)/25, so that I have P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3. But what now? Well before I had 25((625 m^3 - 75 m^2 + 3m) x^3 - 3(-1+25m )x u^2 + 5u^3) = (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) so I can wonder what happens to (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) when the 25 is divided off. It turns out that focusing on the cont term of P(m) provides an > answer as from before I have P(0) = 25u^2(3x + 5u), where intriguingly the 25 is again visible as a factor. So dividing both sides by 25 gives me P(0)/25 = u^2(3x + 5u), and there are no more visible factors of 5, but you might imagine that > maybe 'u' or 'x' could have a factor that is 5. Well possibly for specific cases, but definitely not in general, > right? After all, I might pick x=7 and u=11, as there's been no constraint on > those variables, and I've left them in as placeers. So what's the significance of that result? Well focusing now on g_1 as I go from the general to the specific, as > a variable of P(m), it's also affected by the dividing off of 25. I'll imagine a factor w_1 divides off of g_1 when 25 is divided off of > P(m). That gives me g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. So using g_1/w_1 = r' + c', I can set use P(0)/25, wch is P(0)/25 = u^2(3x + 5u), to see that c' = u. That forces w_1 = 5, so that I have g_1/w_1 = a_1 x/5 + u wch *should* mean that a_1 has a factor that is 5 in the ring. === Subject: Re: Math rules REVISED ...and they want the w's to vary with m because I can prove that in fact they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? -- > The only tng that never seems to run of gas is a pompous gasbag. > -- > But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. It seems to me that some of you so despise mathematics that even a > step-by-step exposition using BASIC math rules is not enough for you > to acknowledge a math result you don't want. I find that fascinating, > and sad. After all, what are you doing on sci.math then? Lurking to try and destroy the math community by attacking it at its > base? I don't mean to put down the newsgroup or anytng, but if the modern math community is founded on sci.math... then we are in a sad state of affairs. === Subject: Re: Math rules REVISED > >...and they want the w's to vary with m because I can prove that in fact >they are not dependent on m,... >>Then why do keep *asserting* you can prove it, but never actually prove it? >>-- >>The only tng that never seems to run of gas is a pompous gasbag. > But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. > [repost deleted] It would be more helpful if you responded to the critiques of your proof than simply restated it without changes. Or did you make any changes? If so, please repost with the changes ghlighted. Wle you're at it, please explain the flaws in our comments or how they have helped. Ignoring them does not make them go away. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Math rules REVISED ...and they want the w's to vary with m because I can prove that in fact they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? -- > The only tng that never seems to run of gas is a pompous gasbag. > -- > But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. > Oh, good. I posted a response to that two days ago, but you may not have seen it. I have copied it below - > It seems to me that some of you so despise mathematics that even a > step-by-step exposition using BASIC math rules is not enough for you > to acknowledge a math result you don't want. I find that fascinating, > and sad. After all, what are you doing on sci.math then? Lurking to try and destroy the math community by attacking it at its > base? The math rules here are so easy I can give them quickly in ts post. For ince, for a factor g of a polynomial P(m), g=r+c exists, where > c=g when m=0, and r = g-c. How much simpler can you get? It's simpler than that: To be more succinct, define r(x) = g(x) - g(0). That one equation is basically the content of your lemma. >Let's say you were one of the people who accepted that, Few if any of us have not accepted your lemma. After all, it says almost notng. > and yes I know >some of you wish to have the specific ring, but if you know much about >polynomials, then you'll realize that P(x) being a polynomial is >what's important. >Since I know many of you don't underd the details of many math >rules, I'll leave that as a place where you can reply and I'll explain >those rules in more detail, wle for others I'll continue. >So now I'll go to the following expression: > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) >If you're a purists you may have problems with me calling that P(m) as >you have all those variables running around. It does seem odd, given that you are going to factor it as a polynomial in x, not as a polynomial in m. And in your lemma, g and r are also thought of as functions of x, not of m. c = g(0) refers to the value of g when x = 0, not when m = 0. But ts looks like just quibbling over notation. Clearly P is a function of both m and x. You might consider writing it as P(m, x) just to makes tngs clear. (See also below - it's not just quibbling.) >I'll try and help out a >bit by letting f=5, so that you have > P(m) = 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) >and you still may be bothered by a lot of variables, but I need the >remaining ones for placement, and you may also wonder what the ring >is. >Well consider the ring to be algebraic integers, so all of the >variables are algebraic integers. In your applications all the variables m, f, and u are ordinary integers. >Wle I'm looking at the expression as a polynomial P(m), the >remaining variables are there as placeers so that I can factor it >into non-polynomial factors. >That is, so that I have > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) = > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u). Where, presumably, a1, a2, and a3 are algebraic integers - right? >That may seem awkward but its purpose provides the need for it. >Still by math rules what I've done is quite correct. >Now though you can see the purpose of generalizing a factor g, as here >I finally have more than a trivial use for g=r+c. >Letting g_1 = a_1 x + 5u, I want to know what g_1 equals when m=0, >because g_1 is a factor of P(m). >Letting m=0, with P(m), I get > P(0) = 25(3xu^2 + 5u^3) = 25u^2(3x + 5u), >and now you can see another reason for why I need to keep the other >variables as placeers. Here that reveals that when m=0, > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) = 25u^2(3x + 5u), >wch means that exactly two of the a's go to 0, when m=0. >Well that makes it easy, as my indices are arbitrary, so I can pick >a_1 and a_2 to go to 0 when m=0, and now figure out that > g_1 = 5u >when m=0, wch gives c_1 = 5u, and then I have > r_1 = g_1 - c_1 = a_1 x + 5u - 5u = a_1 x. Everytng is OK up to ts point. Of course in general, a1, a2 and a3 are functions of m, and when m <> 0, none of them are zero, in contrast to the m = 0 case, when two of them are zero and P(m) becomes a degree 1 polynomial rather than a degree 3 polynomial. >You may be wondering what's the point, since r and c here map to what >would be the r and c if x were the key variable. >Well I'm working towards the point as that 25 factor of > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) >has just been sitting there, when normally you'd separate that off >when factoring, as *normally* it'd make the factorization ambiguous. >I'll do that now and consider P(m)/25, so that I have > P(m)/25 = (625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3. >But what now? >Well before I had > 25((625 m^3 - 75 m^2 + 3m) x^3 - > 3(-1+25m )x u^2 + 5u^3) = > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) >so I can wonder what happens to > (a_1 x + 5u)(a_2 x + 5u)(a_3 x + 5u) >when the 25 is divided off. >It turns out that focusing on the cont term of P(m) provides an >answer as from before I have > P(0) = 25u^2(3x + 5u), >where intriguingly the 25 is again visible as a factor. >So dividing both sides by 25 gives me > P(0)/25 = u^2(3x + 5u), >and there are no more visible factors of 5, but you might imagine that >maybe 'u' or 'x' could have a factor that is 5. >Well possibly for specific cases, but definitely not in general, >right? R I G H T ! What you have done so far is valid only for m = 0. >After all, I might pick x=7 and u=11, as there's been no constraint on >those variables, and I've left them in as placeers. Worse yet, you might pick m = sometng other than 0. >So what's the significance of that result? >Well focusing now on g_1 as I go from the general to the specific, as >a variable of P(m), it's also affected by the dividing off of 25. >I'll imagine a factor w_1 divides off of g_1 when 25 is divided off of >P(m). >That gives me > g_1/w_1 = r_1/w_1 + c_1/w_1 = a_1 x/w_1 + 5u/w_1 >but as g_1/w_1 is a factor of P(m)/25, I also have an r and c for it. ... noting here that w1 is a function of m, unless you have a proof to the contrary. But you have not provided any proof or evidence of that. >So using g_1/w_1 = r' + c', I can set use P(0)/25, wch is > P(0)/25 = u^2(3x + 5u), >to see that c' = u. Yes. Just fine for m = 0. But c' <> u in general. See below. >That forces w_1 = 5, so that I have > g_1/w_1 = a_1 x/5 + u Yes again: for m = 0, w1 = 5. But not in general. >wch *should* mean that a_1 has a factor that is 5 in the ring. Since all you have considered so far is m = 0, and a1 = 0 when m = 0, a1 = 0 does indeed have a factor of 5, although for m = 0 it also has a factor of q for any other integer q. The factorization in ts case is degenerate. As you have defined it above, c' = c1/w1. But (unless you can prove otherwise), w1 is a function of m. Therefore also c' is a function of m. Ts is true even though you have c = 5*u is NOT a function of m. >You might know that there has been a lot of debate about that result >where several posters have claimed that the result is only true if >m=0. >Well, remember c is a factor of the *cont* term of the polynomial, >so they're trying to make a new nonsensical math rule. Here you are fooling yourself, and it has to do with what I said before about P(m, x). For a fixed value of m, c' = 5*u/w1 is the cont *with respect to x*. But since in general, w1 must be assumed to be a function of m, c' is also a function of m. Ts is a key point wch you seem to have had enormous difficulty underding. Why? >For ince, by such an argument maybe with > 2(x^2 + 2x+1) = (x+1)(2x+2) >the 2 could in fact somehow, someway be a variable dependent on x. You have ALMOST GOT IT. The cont c' here is 2, wch is the cont with respect to x. You do not have an m in your oversimplified example. If you did, you would expect in general that c' would be a function of m. For example: let Q(x) = x^2 + m*x + 2. Suppose Q(x) is factored in the form (x + r1)*(x + r2). Do you actually tnk r1 and r2 are not functions of m ??? >Obviously the posters know that no one would believe such a tng with >a *polynomial* factor, so they rely on the complexity of my using >non-polynomial factors to try and sneak in ts nonsensical rule. You are NOT using non-polynomial factors: g(x) = a1*x + u*f and g'(x) = a1*x/w1 + 5*u/w1 are quite obviously *first-degree polynomials* in x. You do NOT have the complexity of non-polynomial factors. That is NOT the problem. >And it has worked for them, as the sci.math newsgroup has let them go >on for months pusng ts new, nonsensical math rule, wch I've >called voodoo math. >Why would posters try to make up such a new rule? >Why don't you ask them? Absolutely! But let me ask why you keep starting new threads on ts - with ideas that are now becoming *threadbare* - when you have failed utterly to answer questions like the following: You claim to have shown factorization of the form you want in three very special cases: 1. m = 0 2. f = 3 3. f = sqrt(2) What you actually need for your proofs are: 1. f a prime integer larger than 3, and 2. m relatively prime to f. None of the three special cases above cover ts. How can you proceed from m = 0, f = 3 to, for example, m = 1, f = 5 ? Don't you tnk it is odd that you have not succeeded in finding any arguments wch apply for such cases ? It must seem strange to your loyal fans that you claim to have a proof for exactly and *only* the cases that are irrelevant to your main argument. All you have is some circumtial evidence (and in the case of f = 3, I don't agree that you have anytng valid at all). If you had some way to show how m = 0, f = 3 can be used to show what you want for other values, that might work. But you have given no such proof. Again you just have that bit of circumtial evidence in the degenerate case, m = 0. Why don't (or can't) you produce a proof for values like the following ? m = 1, f = 5 m = 4, f = 7 m = 62, f = 107 etc. ... instead of just more irrelevant peripheral 'evidence' ? > === Subject: Re: Math rules REVISED > >...and they want the w's to vary with m because I can prove that in fact >they are not dependent on m,... >>Then why do keep *asserting* you can prove it, but never actually prove it? >>-- >>The only tng that never seems to run of gas is a pompous gasbag. >>-- > But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. > >HERE'S THE SETUP TO THE QUESTIONS<<<<<<<<<<<<<<<<<<<< The problem in the ring of algebraic integers is that you can have algebraic integers 'a', 'b', 'c', and a prime p, such that abc = p but neither 'a', 'b', nor 'c' shares any non-unit factors with p. If you claim to have proven ts, then clearly your methodology is flawed. After all, we have 1. If 'a' is not a unit, then 'a' itself is a non-unit factor of 'a' and 'p'. 2. If 'b' is not a unit, then 'b' itself is a non-unit factor of 'b' and 'p'. 3. If 'c' is not a unit, then 'c' itself is a non-unit factor of 'c' and 'p'. 4. If all are units (the only other possibility), then 'p' is a unit, and thereby not a prime. Note that the fact that p is a prime is only relevant to item 4, so I'll dispense with that in a simpler example: 2*3 = 6 >>HERE ARE THE QUESTIONS<<<<<<<<<<<<<<<<<<<<<<<<<<<<< What if I were to claim that neither 2 nor 3 shared non-unit factors with 6, would you agree? If not, then why not? Is it the fact that 2 *is* a factor of 2 and of 6? Is it the fact that 3 *is* a factor of 3 and of 6? >THOSE WERE THE QUESTIONS<<<<<<<<<<<<<<<<<<<<<<<<<< > Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. > How can I be bothered to care about your silly little games, when you so clearly are busy proving that you don't know what you're talking about? What use is a technique of proof that allows one to deduce falsehoods? > It seems to me that some of you so despise mathematics that even a > step-by-step exposition using BASIC math rules is not enough for you > to acknowledge a math result you don't want. I find that fascinating, > and sad. > Yes, hatred of mathematics includes insisting on correctness. > After all, what are you doing on sci.math then? Lurking to try and destroy the math community by attacking it at its > base? > I thought you were intent on destroying the mathematical community, since you started out with the assumption that it amount to wte-collar welfare? I thought you were acting on a mission because someone once told you that you would aceve greatness, and had learned from your Jehovah's Witness (The Watchtower Bible and Tract Society of New York, Inc) that the world was full of corruption and you had to destroy that corruption. Since you're so intent on attacking the motives of others, perhaps your motives should be questioned. Let's look at the facts: 1. You continually post the same nonsense. 2. You ignore valid refutations of your work. 3. You propose refutations of others' work, mainly by repeating the same nonsense. 4. You have as much as proclaimed your goal of destroying mathematics. 5. Your command of the English language is below subdard (OK, that's irrelevant, but I felt compelled to announce that fact). 6. Your command of mathematics makes your command of the English language seem stellar. Given these objective facts, why should anyone pay any attention? I have two reasons: 1. It's all about the entertainment value. 2. Our allegiance to mathematics and its study make us unwilling to sit back and let someone get away with such crap as you would, if left to your own devices. ... all the rest of that garbage deleted ... Annoyed at having your st pinched off in mid-loaf? Then answer the questions I posed above. Until then, you're just a poor excuse of a contributor, and no worthy debater at all. Dale. === Subject: Re: Math rules REVISED >HERE'S THE SETUP TO THE QUESTIONS<<<<<<<<<<<<<<<<<<<< > The problem in the ring of algebraic integers is > that you can have algebraic integers 'a', 'b', 'c', > and a prime p, such that > abc = p > but neither 'a', 'b', nor 'c' shares any non-unit > factors with p. >If you claim to have proven ts, then clearly your methodology >is flawed. After all, we have > 1. If 'a' is not a unit, then 'a' itself is > a non-unit factor of 'a' and 'p'. > 2. If 'b' is not a unit, then 'b' itself is > a non-unit factor of 'b' and 'p'. > 3. If 'c' is not a unit, then 'c' itself is > a non-unit factor of 'c' and 'p'. Not that I really tnk that James has the faintest clue what he is talking about, but isn't it reasonable to assume that he actually means that neither 'a', 'b', nor 'c' share any non-unit, non-trivial factors with p? I have no idea if ts revised statement is true or not but I'm equally sure that (a) James doesn't either (b) he can't prove it either way and (c) if true, it won't help s argument. Alan -- Defendit numerus === Subject: Re: Math rules REVISED > >HERE'S THE SETUP TO THE QUESTIONS<<<<<<<<<<<<<<<<<<<< >> The problem in the ring of algebraic integers is >> that you can have algebraic integers 'a', 'b', 'c', >> and a prime p, such that >> abc = p >> but neither 'a', 'b', nor 'c' shares any non-unit >> factors with p. >>If you claim to have proven ts, then clearly your methodology >>is flawed. After all, we have >> 1. If 'a' is not a unit, then 'a' itself is >> a non-unit factor of 'a' and 'p'. >> 2. If 'b' is not a unit, then 'b' itself is >> a non-unit factor of 'b' and 'p'. >> 3. If 'c' is not a unit, then 'c' itself is >> a non-unit factor of 'c' and 'p'. > Not that I really tnk that James has the faintest clue what he > is talking about, but isn't it reasonable to assume that he actually > means that neither 'a', 'b', nor 'c' share any non-unit, non-trivial > factors with p? I have no idea if ts revised statement is true or not but I'm equally > sure that (a) James doesn't either (b) he can't prove it either way > and (c) if true, it won't help s argument. Alan JSH is claiming ts to be true in the ring of all algebraic integers; that ring is closed under the extraction of roots of arbitrary order (i.e. for any natural number n, if a is an algebraic integer, so is a^(1/n)). Thus, if by non-trivial factor you mean a factor that is not equal to the given number (i.e., 'a', 'b', or 'c' above), one could just as easily take sqrt(a), sqrt(b), sqrt(c), for ince. Further, for any algebraic integer divisor of any of the three numbers, that divisor is also divisor of p. The only way out is either for the three numbers to be units, or for the numbers to have *no* non-unit divisors. Dale. === Subject: Re: Math rules REVISED ...and they want the w's to vary with m because I can prove that in fact they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. > Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. [snip blah, blah, blah...] [snip blah, blah, blah...] Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you stated it would -- that the w's are independent of m. You merely repeated a tiresome string of calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult Attention Deficit Disorder), since your exposition had notng to do with the original question or your assertion you were addressing it. You might want to consult a qualified physician. -- There are two tngs you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: Math rules REVISED > ...and they want the w's to vary with m because I can prove that in fact they are not dependent on m,... Then why do keep *asserting* you can prove it, but never actually prove it? But you see, I *do* prove it, and follow rather basic math rules in > doing so, wle other people come along, and make up bogus math rules, > apparently because they don't like a particular conclusion that > follows from the math. Here's the information, where I go step-by-step, and I'm copying from > the post that STARTED ts thread. [snip blah, blah, blah...] [snip blah, blah, blah...] Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you > stated it would -- that the w's are independent of m. You merely repeated a tiresome string of > calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult > Attention Deficit Disorder), since your exposition had notng to do with the original > question or your assertion you were addressing it. You might want to consult a qualified > physician. Your claim that I didn't prove that the w's are independent of m merely shows a continuing need to make up mathematics to suit you. I've distilled the argument further to give you less room to dodge the mathematical logic. In the ring of algebraic integers (used for now), given a factor g of a polynomial P(m), r and c exists such that g=r+c, where c=g when m=0, and r=g-c. Consider that c is a factor of the cont term P(0), wch follows as g is a factor of P(m), so g at m=0 is a factor of P(0), but c=g at m=0; therefore, c is a factor of the cont term P(0). Now consider P(m) such that it has f^2 as a factor, and for a factor g, c has f as a factor. Further consider that P(0)/f^2 is coprime to f; therefore, as c is a factor of P(0) a factor of f must divide off of c, when f^2 divides off of P(m). So the objction of the poster C. Bond apparently stems from the bizarre, and wrong belief that though a factor of the cont term P(0), c might actually be dependent on m. But you see, the cont term is found by setting m=0, so how can c be dependent on m? It's the phantom m that some posters seem to believe in, wch gives you another take on the phrase voodoo mathematics. Neat joke, eh? Get it? They believe in phantom m's, and I call it voodoo mathematics. What's funny, or you might say larious, is how many of you LISTENED to people like C. Bond, Nora Baron, Arturo Magidin, and W. Dale Hall!!!!!!! It's so freaking funny. Don't you remember some of the tngs I've said over the years? How about another phrase then: All the world's a stage... === Subject: Re: Math rules REVISED > So the objction of the poster C. Bond apparently stems from the > bizarre, and wrong belief that though a factor of the cont term > P(0), c might actually be dependent on m. Let's consider example P(0)=6 (cont, right?) Now consider (all are integers): c = 2, if m is odd. c = 3, if m is even. So c is a factor of cont 6, but it's actually dependent of m. Do you find my example bizzare? === Subject: Re: Math rules REVISED Visiting Assit Professor at the University of Montana. [.snip.] >Neat joke, eh? Get it? They believe in phantom m's, and I call it >voodoo mathematics. What's funny, or you might say larious, is >how many of you LISTENED to people like C. Bond, Nora Baron, >Arturo Magidin, and W. Dale Hall!!!!!!! Hmmm... Wonder why I did not rate scare quotes? >It's so freaking funny. Don't you remember some of the tngs I've >said over the years? Sure we do: It seems silly to say that obviously my attempt at a proof isn't even close. But I feel guilty enough to do so. Obviously, I was suffering form a bit of delusion wch has amazingly come off and on. I'll only say that I've been under some pressures. It's amazing what the mind will believe if pushed, and I'll continue to believe that Fermat found a simple solution---despite the evidence to the contrary---simply because that's what I chose to believe. In any event, I've found that the mild release making outrageous claims has given me is finally surmounted by a sense of shame; so I'll quit. Since I have notng of value to add to ts newsgroup, I will unsubscribe. My apologies to anyone concerned or in any way interested. I realized just now just how seriously I abused the helpfulness and kindness of people from ts newsgroup and at colleges and It was just too easy to use those to get attention when I was going through a trying time in my life. I do feel honored to have corresponded with some of the top mathematicians in the world, but I'll be forever embarrassed about the subject of those conversations and how I acted. Well, as usual no one actually contacted me to tell me what was wrong with my previous proof. But it doesn't matter because I was having a brain fart anyway. I just realized that all of ts craziness with Fermat's Last Theorem started wle I was wroking as a radiation safety officer. My superior, a cruel and evil major, made me work in the radiation storage area for long periods of time (and he called me nasty names and did other stuff wch made me want to shoot m with an M-16 but then they'd have sent me to Leavenworth or the firing squad and I figured it wasn't worth it but I really wanted to anyway so I got out).. I fear that it affected my mental processes in some way. However, I tnk I'm over it (and I managed to get out of the radiation field). So now that I've returned to sanity (at least temporarily). I can forget about pure math problems. But the comment about Sisyphus did strike a nerve and helped me to decide that succeed or fail I was done at ts point. Anyone who read my BOLD post before probably noticed that the conclusion doesn't follow. I seem to have been seeing an invisible one witn the expression. Frankly, I wanted you guys to do all the work. I figured if I pointed in the right direction, someone would grab it and go. And of course, a famous one: Ts seemed like a good time for a psychology (or I shoudl say, sociology) experiment because I'm almost through with Sci.Math. I've posted my main result on FLT from the beginning. Although I've been told that it's interesting but leads to notng, I've got a couple more angles to check out on it before I give up. Anyway, it's fun as a hobby and it impresses women (yes, honest). Anyway, I'm not a big number theory fan. [...] I'm outa here. Maybe I'll go back to my physics work. ====================================================================== [Gabriele Rossetti] has left a vast body of writings... in wch he has attempted to prove the truth of s unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in wch we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of storical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to m and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied s proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to s own simple soul in a flash of inspiration... In such work as ts... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman ====================================================================== === Subject: Re: Math rules REVISED > -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in > _The Shakespearan Ciphers Examined_, by William F. > Friedman and Elizebeth S. Friedman These quotes of yours are beautiful. Did you find them yourself or is there a web page of these perfect put-downs for net.cooks? How about ts one: [Teilhard de Chardin's book] cannot be read without a feeling of suffocation, a gasping and flailing around for sense ... the greater part of it is nonsense, tricked out by a variety of metaphysical conceits, and its author can be excused of dishonesty only on the grounds that before deceiving others he has taken great pains to deceive mself. P.B. Medawar, Critical Review of 'The Phenomenon of Man', in Mind 70(1961), 99-106, quoted in D. Barrow and Frank J. Tipler, The Anthropic Cosmological Principle. V. -- mail me at lastname at cs utk edu === Subject: Re: Math rules REVISED > Don't you remember some of the tngs I've > said over the years? I believe some of your greatest witticisms can be found on web sites. V. -- mail me at lastname at cs utk edu === Subject: Re: Math rules REVISED ... stuff deleted ... > It's the phantom m that some posters seem to believe in, wch gives > you another take on the phrase voodoo mathematics. Neat joke, eh? Get it? They believe in phantom m's, and I call it > voodoo mathematics. What's funny, or you might say larious, is > how many of you LISTENED to people like C. Bond, Nora Baron, > Arturo Magidin, and W. Dale Hall!!!!!!! > I don't care about what you tnk other posters believe in. I have focused on asking questions that you ought to be able to underd. For ince, your claim that in the factorization of the polynomial x^3 - 12 x^2 + 65 = (x - a1)(x - a2)(x - a3) that any of the a's is coprime to 5. I showed factorizations: 5 = qi ri ai = ri si for i=1,2,3, in the ring of algebraic integers. Further, I showed that ri, the common factor between ai and 5, is not a unit. Somehow, you could not accept the fact that ri serves as a common, non-unit factor shared by ai and 5. In fact, when I proved that the above two equations, *together* with your claim that ai is coprime to 5 proves that 5 must be a unit, you claimed that I had made a circular argument. The failure of your argument became most obvious when you launched into your Dale's a racist diversion. More recently, you have been parading a separate version of your deeply flawed version of algebra, namely ts: The problem in the ring of algebraic integers is that you can have algebraic integers 'a', 'b', 'c', and a prime p, such that abc = p but neither 'a', 'b', nor 'c' shares any non-unit factors with p. If you claim to have proven ts, then clearly your methodology is flawed. After all, we have 1. If 'a' is not a unit, then 'a' itself is a non-unit factor of 'a' and 'p'. 2. If 'b' is not a unit, then 'b' itself is a non-unit factor of 'b' and 'p'. 3. If 'c' is not a unit, then 'c' itself is a non-unit factor of 'c' and 'p'. 4. If all are units (the only other possibility), then 'p' is a unit, and thereby not a prime. Note that the fact that p is a prime is only relevant to item 4, so I'll dispense with that in a simpler example: 2*3 = 6 >>HERE ARE THE QUESTIONS<<<<<<<<<<<<<<<<<<<<<<<<<<<<< What if I were to claim that neither 2 nor 3 shared non-unit factors with 6, would you agree? If not, then why not? Is it the fact that 2 *is* a factor of 2 and of 6? Is it the fact that 3 *is* a factor of 3 and of 6? >THOSE WERE THE QUESTIONS<<<<<<<<<<<<<<<<<<<<<<<<<< How can I be bothered to care about your silly little games, when you so clearly are busy proving that you don't know what you're talking about? What use is a technique of proof that allows one to deduce falsehoods? > It's so freaking funny. Don't you remember some of the tngs I've > said over the years? How about another phrase then: > Do I remember some of the tngs you've said over the years? Yes: Integers are irrational. I'm disgusting. I'm just a pile of st. I should just die like so many of you have said. I hate myself. I despise ts life. I'm notng but a sick joke to be made fun of by those of you who have real educations. People who actually know sometng, when I know notng. I'm just notng. It gets more interesting as the multiplicate inverses of unit algebraic integers can themselves NOT be algebraic integers. The following apropos of ts factorization : 5x^2 + 3x + 3 = (ax + b)(cx + d) (from 03/27/02 on sci.math) Besides, no factorization exists where they're all algebraic integers. And maybe some people might begin to wonder why if you claim some such factorization exists that you haven't given it. The answer is that you can't because it does NOT exist. Why was that date relevant? Well, on (03/25/02), Derek Holt posted ts: then a factorization with integral coefficients for the above is 5x^2 + 3x + 3 = (sqrt(2-r1)x - sqrt(-3)) (sqrt(2-r2)x + sqrt(-3)), where r1, r2 are the roots of y^2 + 3y + 15 = 0 (i.e. 5 times the roots of 5x^2 + 3x + 3 = 0). with ts solution, but, hey. And now ts piece of wisdom: > All the world's a stage... > Apparently that's all you can underd. Often-repeated sound bites. Dale === Subject: Re: Math rules REVISED > Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you > stated it would -- that the w's are independent of m. You merely repeated a tiresome string of > calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult > Attention Deficit Disorder), since your exposition had notng to do with the original > question or your assertion you were addressing it. You might want to consult a qualified > physician. > Your claim that I didn't prove that the w's are independent of m > merely shows a continuing need to make up mathematics to suit you. > I've distilled the argument further to give you less room to dodge the > mathematical logic. 1) My post did *not* reference any made up mathematics. It simply noted that you did not prove that the 'w's are independent of 'm', and asked you to do so.2) Your further distillation below does *not* reference 'w' at all! 3) Your exposition completely dodges the question you claim to be addressing. Tsk, tsk. > In the ring of algebraic integers (used for now), given a factor g of > a polynomial P(m), r and c exists such that g=r+c, where c=g when m=0, > and r=g-c. > Consider that c is a factor of the cont term P(0), wch follows > as g is a factor of P(m), so g at m=0 is a factor of P(0), but c=g at > m=0; therefore, c is a factor of the cont term P(0). > Now consider P(m) such that it has f^2 as a factor, and for a factor > g, c has f as a factor. > Further consider that P(0)/f^2 is coprime to f; therefore, as c is a > factor of P(0) a factor of f must divide off of c, when f^2 divides > off of P(m). > So the objction of the poster C. Bond apparently stems from the > bizarre, and wrong belief that though a factor of the cont term > P(0), c might actually be dependent on m. My objction (sic) stems from the obvservation that you did *not* prove, or even attempt to prove, the very assertion in question: whether the 'w's are independent of 'm'. It has notng to do with beliefs -- right, wrong or indifferent. > But you see, the cont term is found by setting m=0, so how can c > be dependent on m? > It's the phantom m that some posters seem to believe in, wch gives > you another take on the phrase voodoo mathematics. Talk about 'voodoo'. Where did the 'w's go? You claimed to have proven that they were independent of 'm', but you didn't even reference 'w' at all! > Neat joke, eh? Get it? They believe in phantom m's, and I call it > voodoo mathematics. What's funny, or you might say larious, is > how many of you LISTENED to people like C. Bond, Nora Baron, > Arturo Magidin, and W. Dale Hall!!!!!!! If anyone is listening to ts thread, they are hearing requests for you to prove that the 'w's are independent of 'm' and observing that you continually dodge the issue. > It's so freaking funny. Don't you remember some of the tngs I've > said over the years? How about another phrase then: > All the world's a stage... > How about ( instead of posting another off-topic phrase) actually *proving* that the 'w's are indendent of 'm'? Or is your AADD getting in the way? -- There are two tngs you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: Math rules REVISED > Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you stated it would -- that the w's are independent of m. You merely repeated a tiresome string of calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult Attention Deficit Disorder), since your exposition had notng to do with the original question or your assertion you were addressing it. You might want to consult a qualified physician. Your claim that I didn't prove that the w's are independent of m > merely shows a continuing need to make up mathematics to suit you. > I've distilled the argument further to give you less room to dodge the > mathematical logic. 1) My post did *not* reference any made up mathematics. It simply noted that you did not prove that > the 'w's are independent of 'm', and asked you to do so.2) Your further distillation below does > *not* reference 'w' at all! > 3) Your exposition completely dodges the question you claim to be addressing. Tsk, tsk. > Readers of the sci.math newsgroup, notice that C. Bond has continued despite my going into *detail* about the math rules that prove my case. Rather than acquiesce to the mathematics, he makes more bogus claims. That is a disrespect to *you* readers and worse to mathematics as he continues to push the voodoo. > In the ring of algebraic integers (used for now), given a factor g of > a polynomial P(m), r and c exists such that g=r+c, where c=g when m=0, > and r=g-c. Consider that c is a factor of the cont term P(0), wch follows > as g is a factor of P(m), so g at m=0 is a factor of P(0), but c=g at > m=0; therefore, c is a factor of the cont term P(0). Now consider P(m) such that it has f^2 as a factor, and for a factor > g, c has f as a factor. Further consider that P(0)/f^2 is coprime to f; therefore, as c is a > factor of P(0) a factor of f must divide off of c, when f^2 divides > off of P(m). So the objction of the poster C. Bond apparently stems from the > bizarre, and wrong belief that though a factor of the cont term > P(0), c might actually be dependent on m. My objction (sic) stems from the obvservation that you did *not* prove, or even attempt to prove, > the very assertion in question: whether the 'w's are independent of 'm'. It has notng to do with > beliefs -- right, wrong or indifferent. > That's just so bogus!!! Now how in the hell can you not see that if c is NOT dependent on m that it is NOT dependent on m? Your claims come down to trying to snow other readers into tnking sometng about some w's wch you don't show, as you claim I don't prove that the w's are independent on m. Yet your claim comes down to claiming that c, wch is not dependent on m, is dependent on m, and I wonder where is the phantom m? It's voodoo mathematics!!! If you're not lying and trying to piss on other readers, show your w, and explain how what I've pointed out about c doesn't mean what I say it does. > But you see, the cont term is found by setting m=0, so how can c > be dependent on m? It's the phantom m that some posters seem to believe in, wch gives > you another take on the phrase voodoo mathematics. Talk about 'voodoo'. Where did the 'w's go? You claimed to have proven that they were independent of > 'm', but you didn't even reference 'w' at all! Oh. Um, I can throw in some w's. Consider the factorization P(m)/f^2 = (b_1 x + w_1)(b_2 x + w_2)(b_3 x + w_3) and given that P(m) = (a_1 x + fu)(a_2 x + fu)(a_3 x + fu) where I've previously shown that for two of the a's, and f divides off, then you have that the w's are conts where two equal u, and one equals fu. So fu C. Bond. > Neat joke, eh? Get it? They believe in phantom m's, and I call it > voodoo mathematics. What's funny, or you might say larious, is > how many of you LISTENED to people like C. Bond, Nora Baron, > Arturo Magidin, and W. Dale Hall!!!!!!! If anyone is listening to ts thread, they are hearing requests for you to prove that the 'w's are > independent of 'm' and observing that you continually dodge the issue. I'm not dodging. You're either lying or really tnk that the readers of sci.math have such a contempt for mathematics that they'll let you continue to get away with what you've gotten away with before. But I say, fu C. Bond. > It's so freaking funny. Don't you remember some of the tngs I've > said over the years? How about another phrase then: All the world's a stage... How about ( instead of posting another off-topic phrase) actually *proving* that the 'w's are > indendent of 'm'? Or is your AADD getting in the way? === Subject: Re: Math rules REVISED >So fu C. Bond. Oh yeah James, that's real mature. How do you expect people to take you seriously after saying sometng like that? In my eyes, that diminishes what little credibility I thought you had. === Subject: Re: Math rules REVISED Oh yeah James, that's real mature. How do you expect people to take you > seriously after saying sometng like that? In my eyes, that diminishes what > little credibility I thought you had. It was actually larious as that trick took over a year to setup. Now you know more about (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) and what I've been tnking for some of you like Arturo Magidin, and Nora Baron as you made up your voodoo mathematics with its phantom m. It's been there for a wle. And by the way , fu. === Subject: Re: Math rules REVISED > >>Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you >>stated it would -- that the w's are independent of m. You merely repeated a tiresome string of >>calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult >>Attention Deficit Disorder), since your exposition had notng to do with the original >>question or your assertion you were addressing it. You might want to consult a qualified >>physician. >Your claim that I didn't prove that the w's are independent of m >merely shows a continuing need to make up mathematics to suit you. >I've distilled the argument further to give you less room to dodge the >mathematical logic. >>1) My post did *not* reference any made up mathematics. It simply noted that you did not prove that >>the 'w's are independent of 'm', and asked you to do so.2) Your further distillation below does >>*not* reference 'w' at all! >>3) Your exposition completely dodges the question you claim to be addressing. Tsk, tsk. > Readers of the sci.math newsgroup, notice that C. Bond has continued > despite my going into *detail* about the math rules that prove my > case. Rather than acquiesce to the mathematics, he makes more bogus claims. That is a disrespect to *you* readers and worse to mathematics as he > continues to push the voodoo. >In the ring of algebraic integers (used for now), given a factor g of >a polynomial P(m), r and c exists such that g=r+c, where c=g when m=0, >and r=g-c. >Consider that c is a factor of the cont term P(0), wch follows >as g is a factor of P(m), so g at m=0 is a factor of P(0), but c=g at >m=0; therefore, c is a factor of the cont term P(0). >Now consider P(m) such that it has f^2 as a factor, and for a factor >g, c has f as a factor. >Further consider that P(0)/f^2 is coprime to f; therefore, as c is a >factor of P(0) a factor of f must divide off of c, when f^2 divides >off of P(m). >So the objction of the poster C. Bond apparently stems from the >bizarre, and wrong belief that though a factor of the cont term >P(0), c might actually be dependent on m. >>My objction (sic) stems from the obvservation that you did *not* prove, or even attempt to prove, >>the very assertion in question: whether the 'w's are independent of 'm'. It has notng to do with >>beliefs -- right, wrong or indifferent. > That's just so bogus!!! Now how in the hell can you not see that if c > is NOT dependent on m that it is NOT dependent on m? James, why are you talking about c? He asked you to prove sometng about w. Your claims come down to trying to snow other readers into tnking > sometng about some w's wch you don't show, as you claim I don't > prove that the w's are independent on m. What he is claiming is that you must *prove* properties of w that you wish to use, not merely *assert* them. Yet your claim comes down to claiming that c, wch is not dependent > on m, is dependent on m, and I wonder where is the phantom m? How are c and w related? Why must both be dependent or independent on m? It's voodoo mathematics!!! If you're not lying and trying to piss on other readers, show your w, > and explain how what I've pointed out about c doesn't mean what I say > it does. You are the one attempting to convince us. The burden of proof is on you. If we are not convinced, either we are dolts, or you have failed to produce a proof. If we are dolts, why are you bothering to try to educate us? If you have failed to produce a proof, fix it. > >But you see, the cont term is found by setting m=0, so how can c >be dependent on m? >It's the phantom m that some posters seem to believe in, wch gives >you another take on the phrase voodoo mathematics. >>Talk about 'voodoo'. Where did the 'w's go? You claimed to have proven that they were independent of >>'m', but you didn't even reference 'w' at all! > Oh. Um, I can throw in some w's. Consider the factorization P(m)/f^2 = (b_1 x + w_1)(b_2 x + w_2)(b_3 x + w_3) and given that P(m) = (a_1 x + fu)(a_2 x + fu)(a_3 x + fu) where I've previously shown that for two of the a's, and f divides > off, then you have that the w's are conts where two equal u, and > one equals fu. So fu C. Bond. Is ts an attempt to show that fu is a function of a person, or did you just want to appear immature? How does ts relate to the mathematics you so desperately want us to stay focussed on? >Neat joke, eh? Get it? They believe in phantom m's, and I call it >voodoo mathematics. What's funny, or you might say larious, is >how many of you LISTENED to people like C. Bond, Nora Baron, >Arturo Magidin, and W. Dale Hall!!!!!!! >>If anyone is listening to ts thread, they are hearing requests for you to prove that the 'w's are >>independent of 'm' and observing that you continually dodge the issue. > I'm not dodging. You're either lying or really tnk that the readers > of sci.math have such a contempt for mathematics that they'll let you > continue to get away with what you've gotten away with before. But I say, fu C. Bond. If you were not dodging, I would have read the proof. >It's so freaking funny. Don't you remember some of the tngs I've >said over the years? How about another phrase then: >All the world's a stage... >How about ( instead of posting another off-topic phrase) actually *proving* that the 'w's are >>indendent of 'm'? Or is your AADD getting in the way? > fu xy -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Math rules REVISED [snip hysterical diatribe] > fu > Nice proof! Simple, elegant and, as usual, revealing more about you than me. -- There are two tngs you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: Math rules REVISED > >...and they want the w's to vary with m because I can prove that in fact >they are not dependent on m,... >>Then why do keep *asserting* you can prove it, but never actually prove it? >But you see, I *do* prove it, and follow rather basic math rules in >doing so, wle other people come along, and make up bogus math rules, >apparently because they don't like a particular conclusion that >follows from the math. >Here's the information, where I go step-by-step, and I'm copying from >the post that STARTED ts thread. >>[snip blah, blah, blah...] >>[snip blah, blah, blah...] >>Perhaps you failed to notice that your step-by-step exposition still did *not* prove what you >>stated it would -- that the w's are independent of m. You merely repeated a tiresome string of >>calculations based on the condition that m = 0. Possibly you are suffering from AADD (Adult >>Attention Deficit Disorder), since your exposition had notng to do with the original >>question or your assertion you were addressing it. You might want to consult a qualified >>physician. > Your claim that I didn't prove that the w's are independent of m > merely shows a continuing need to make up mathematics to suit you. > I've distilled the argument further to give you less room to dodge the > mathematical logic. In the ring of algebraic integers (used for now), given a factor g of > a polynomial P(m), r and c exists such that g=r+c, where c=g when m=0, > and r=g-c. Consider that c is a factor of the cont term P(0), wch follows > as g is a factor of P(m), so g at m=0 is a factor of P(0), but c=g at > m=0; therefore, c is a factor of the cont term P(0). Now consider P(m) such that it has f^2 as a factor, and for a factor > g, c has f as a factor. Further consider that P(0)/f^2 is coprime to f; therefore, as c is a > factor of P(0) a factor of f must divide off of c, when f^2 divides > off of P(m). So the objction of the poster C. Bond apparently stems from the > bizarre, and wrong belief that though a factor of the cont term > P(0), c might actually be dependent on m. But you see, the cont term is found by setting m=0, so how can c > be dependent on m? It's the phantom m that some posters seem to believe in, wch gives > you another take on the phrase voodoo mathematics. Neat joke, eh? Get it? They believe in phantom m's, and I call it > voodoo mathematics. What's funny, or you might say larious, is > how many of you LISTENED to people like C. Bond, Nora Baron, > Arturo Magidin, and W. Dale Hall!!!!!!! It's so freaking funny. Don't you remember some of the tngs I've > said over the years? How about another phrase then: All the world's a stage... > What do your w's have to do with c? The w's are supposed be factors of the g's, not just the c's. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Math rules REVISED > Well, remember c is a factor of the *cont* term of the polynomial, so they're trying to make a new nonsensical math rule. For ince, by such an argument maybe with 2(x^2 + 2x+1) = (x+1)(2x+2) the 2 could in fact somehow, someway be a variable dependent on x. Wch 2? There are two on the left-hand side, two on the > right-hand side. Does it matter? Uh, yeah. If you are making a statement about the 2 and there are four 2's from, your statement is ambiguous. > I could certainly believe that in factoring > x^2 + ax + 12 = (x+r1)(x+r2) over the integers, > that r1 and r2 depend on a, despite the fact that > both must be factors of 12 in the integers. What's so strange about that? Well, notice that you have r_1 = (-a + sqrt(a^2 -48))/2 arbitrarily picking the positive for r_1, wch you might notice shows > the dependency between r_1 and 'a'. Why did you deem it necessary to tell me that r_1 depends on a, right after I told you that? > The pertinent expression though is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and faced with a mathematically precise argument wch step-by-step > shows a way to factor that expression, wch gives a fascinating > conclusion, some seem willing to instead try to throw out the math by > making a new rule. By their rule factors like f^2 can be decomposed into dependent > variables just because they want them to be. So they claim that you have sometng like f^2 = w_1(m) w_2(m) w_3(m) where the w's vary as m varies, so that they can claim that if you > divide off f^2, you have a factorization like (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (a_1 x + uf)/w_1(m) (a_2 x + uf)/w_2(m) (a_3 x + uf)/w_3(m) With what conditions on the a's and the u's? and they want the w's to vary with m because I can prove that in fact > they are not dependent on m, as actually you have w_1(m) = f, w_2(m) = > f, and w_3 = 1. Well, you can't assert that these choices of w will always give a's and u's with the right properties. Depends on what properties you are asking for. Clearly you could have w_1(m) be any function of m whatsoever and still have a factorization of the form above, so long as w_1*w_2*w_3 = f^2. You've given one choice but not said here what restricts that choice. What stops me from choosing w_1 = log(m), w_2 = f tanh(m), w_2 = f/(tanh(m) log(m))? - Randy === Subject: Re: Factorial/Exponential Identity, Infinity I found a copy of Knuth's Two Notes on Notation. The first note is about Iverson notation, Iverson being the inventor of the programming language APL: A Programming Language, the second note is about Stirling numbers. http://www-cs-staff.ford.edu/~uno/papers/tnn.tex.gz http://pckoenig1.mathematik.tu-muenchen.de/lehre/comb-ws2001/erg/tnn.pdf Knuth notes that Stirling numbers of the first kind were investigated as being the sum of the products of the k-subsets of {1, ..., n} back in the nineteenth century. During the nineteenth century, Stirling's connections with these numbers had been almost entirely forgotten. The numbers themselves were studied, in the role of sums of products of combinations of the numbers {1, 2, ..., n} taken k at a time. Let C_k(n) and Gamma_k(n) denote these sums, when the combinations are respectively without or with repetitions; thus, for example, C_4(4) = 1*2*3 + 1*2*4 + 1*3*4+2*3*4 = 50; Gamma_3(3) = 1*1*1 + 1*1*2 + 1*1*3 + 1*2*2 + 1*2*3 +1*3*3 + 2*2*2 + 2*2*3 + 2*3*3 + 3*3*3 = 90 It turns out that C_k(n) = [ n+1 n+1-k] and Gamma_k(n) = { n+k n }. I want to note that there appears to be an error, C_3(4) is 50, C_4(4) is 24. Then, he explains a rationale for calling Stirling numbers of the first kind Stirling cycle numbers, and Stirling numbers of the second kind Stirling subset numbers, the appelations of first and second kind originate with work in the early twentieth century by a Niels Nielsen, and for bracketing first kind and bracing second kind pairs, where the binomial coefficient uses parentheses. I'm just going to write s(n, m) and S(n, m). I also found a series of posts made to the math-story list circa 1997 that make it clear that a known method of generating Stirling numbers is as the sum of the products of the k-subsets of {1, ..., n}. http://www.polaris.net/~kermit/math/calculus/stirling The methods of calculating s(n+1, n-1), and s(n+1, n-2) were put forth here on ts list, one from a method of calculating s(n, n-2) alternatively to calculating s(n+1, n-1) from ((sum n)^2 - sum (n^2))/2 and another from dividing (sum n)^3 - (sum n)^2 by 6(n+2)/(n-2). A method to calculate s(n+1, n-3) would allow the calculation of a lot of different numbers because of the identity s(n+1, k) = n s(n, k) + s(n, k-1). Let's see, here. s(n+1, k) = n s(n, k) + s(n, k-1) s(n, k-1) = s(n+1, k) - n s(n, k) s(n, k) = (s(n+1, k) - s(n, k-1)) / n Well, maybe it wouldn't. For example if s(n+1, n-3) was known, then s(n+1, n-3) = n s(n, n-3) + s(n, n-4), those values are all easily calculable already by knowing s(n+1, n-2) and s(n+1, n-3). The identity here allows the computation of s(n, x) for n-x < n-k, but there are already formulas for s(n+1, n), s(n+1, n-1), and s(n+1, n-2), as a univariate polynomial of n or a ratio of univariate polynomials of n. Thus, knowing a formula for s(n+1, n-3) wouldn't in itself automatically enable the computation of s(n+1, n-4), but the process of deriving it might. Calculating s(n+1, n-1), the sum of each product of pairs of {1, ..., n}, I was listing the prime and composite factors with their multiple frequencies to then add them together. Then, I noticed that I could just make a square matrix, and have each element be the product of its row and column indices, and remove the diagonal, and remove the upper half, and the sum of the remaining elements was s(n+1, n-1). The sum of the elements of the original matrix was (sum n)^2, removing the diagonal was the subtraction of sum (n^2), and then the simple division by two removed the symmetric upper half. I thought the case for x=3 would be similar, and it is, make a 3d matrix, sum its elements, remove the diagonal, and then divide by sometng, instead of two to halve the remaining elements it is 6(n+2)/(n-2) to divide the remaining elements. I hope that it will be possible in four and more dimensions. Figuring it out would allow the rapid solution of a problem that currently takes quite a few more calculations to compute. Ts wouldn't figure out the sum of the products of k-subsets of an arbitrary set of integers, only those of a set of integers from 1 through n. (Calculating the products of k-subsets of an arbitrary set of integers is about enumerating k-subsets, where enumerating k-subsets of a set {1, ..., n} does immediately apply to enumerating k-subsets of an arbitrary set.) What I'm concerned with are the square, cubic, etcetera k-d Euclidean matrices with equal rank n in each dimension thus that the element at (a, b, c, ..., k) is (a*b*c*... *k). The diagonal of these matrices with a=b=c=... from 1 to n is the sequence (1^k, 2^k, 3^k, ..., n^k). A subset of the elements of ts matrix represent the k-subsets of {1, ..., n}, that is to say for each k-subset of {1, ..., n} that the elements of the k-subset represent coordinates and the element at those coordinates is the product of the elements of the k-subset. The sum of all elements of the matrix is (sum n)^k. The elements of the diagonal do not represent k-subsets, each k-subset has a =/= b =/= c =/= .... Also, for the k-subset {x, y, z, ...} , if a=x, b=y, c=z, ... then there are also the elements a=y, b=z, c=x, ..., etcetera that would be repeated k-subsets. The idea is to sum all the elements of the matrix for each of the k-subsets of {1, ..., n} but with no duplications, i.e., only one element of the matrix per each k-subset is part of the sum. With a square matrix, ts is easy. It's a 2-d matrix, of rank nxn, about the 2-subsets of {1, ..., n}, each element (i, j) for i, j= 1 to n is i * j. We can remove the diagonal and are left with the upper and lower halves of the matrix, they are symmetric about the diagonal, the sum of the elements of the upper half is the same as the sum of the elements of the lower half, dividing the sum of the elements by two yields the result of the sum of the products of the 2-subsets of {1, ..., n}. With gher values of k, an n_1 x n_2 x ... n_k matrix, then besides the diagonal there are subdiagonals with not each coordinate being equal, but at least two coordinates being equal. For example with n=3 for a 3x3x3 cube, there is the diagonal (1, 1, 1), (2, 2, 2), (3, 3, 3) and then there are also the edges of the cube, (1, 1, 2), (1, 1, 3), (2, 2, 1), (2, 2, 3), (3, 3, 1), (3, 3, 2), (1, 2, 1), (1, 3, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3), (3, 2, 3), (2, 1, 1), (3, 1, 1), (1, 2, 2), (3, 2, 2), (1, 3, 3), and (2, 3, 3). None of those points have coordinates that represent a k-subset because at least two of their coordinates are equal. These are about the edges and diagonals of the square sides of the cube of the matrix, and the diagonals between corners of the cube. Here I'm looking at a Rubik's cube. It's jumbled, I can solve a side. Starting from a solved cube I can get the x's, h's or boxcars. Anyways, there are plenty of elements with unequal coordinates, a fraction of these represent unique k-subsets. Consider the cube with 3x3x3 subcubes, the center subcube is not used as it is at (2, 2, 2). Here I'm about to start writing on ts Rubik's cube. OK, I've solved a side. Let's see, the 3-subset of {1, 2, 3} is {1, 2, 3}. Its permutations are (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). In the context of matrices, for an nxnxn matrix, what is 6(n+2)/(n-2) vis-a-vis an nxn matrix and 2? I dig those dual sum integral identities. Ross === Subject: Re: Subgroup vs. Subquotient > By the way, as an exercise, prove that a finite simple group cannot > have a nontrivial cyclic Sylow 2-subgroup, without using transfer > theorems. The first proof I learned for ts did not use the transfer: Consider the Cayley representation of G, wch is the permutation representation of G wch comes from letting G act on itself by right multiplication. Note that the Cayley permutation of any element has all its cycles having the same length and that ts length is the order of the element. Therefore the number of cycles in the Cayley permutation of x is [G:]. Let |G|=2^s*t where s is a nonnegative integer and t is odd. Since a Sylow 2-subgroup of G is nontrivial, s is positive. Then let x be a generator of a Sylow 2-subgroup of G. The Cayley permutation for x has t cycles of length 2^s. Therefore ts Cayley permutation for x can be written as the product of t(2^s-1) transpositions. Since s is positive, 2^s-1 is odd. Since t is also odd, ts means that the Cayley permutation of x is an odd permutation. Therefore the elements of G wch have even Cayley permutations form a subgroup of index 2 in G. Proof complete. ---- David === Subject: Re: Subgroup vs. Subquotient >There is a theorem of finite group theory that is provable using the >transfer that states: If G is a finite group and P is a Sylow p-subgroup of G such that P is >in the center of the normalizer N_G(P), then G contains a normal >subgroup H such that p does not divide |H| and G/H ~= P. I realized recently that ts could answer some of my curiosity about >finite groups. Specifically, I know how to prove: If G is a finite simple group and |G| is congruent to 4 mod 8, then G >has a subquotient isomorpc to the alternating group A_4. Proof. Let K be a Sylow 2-subgroup of G. Then |K|=4 so K is cyclic or >isomorpc to the product of 2 cyclic groups of order 2. If K were >cyclic, G would have a subgroup of index 2 (any finite group with a >cyclic Sylow 2-subgroup has a subgroup of index 2) and ts would be a >proper normal subgroup of G. Since G is simple, we therefore know that >K is isomorpc to the product of 2 cyclic groups of order 2 and >therefore that |Aut(K)|=6. >Now let N be the normalizer of K in G. Since K is abelian, the action >of the quotient group N/K by conjugation on K is well-defined and ts >gives us a homomorpsm p from N/K to Aut(K). >Since K is a Sylow 2-subgroup of G and K is a subgroup of N, K is a >Sylow 2-subgroup of N and therefore |N/K| is odd. Therefore |p(N/K)| >is odd also. We also know p(N/K) is nontrivial since the triviality >of p(N/K) would imply that K is in the center of N and therefore >that G has a proper normal subgroup. >Therefore |p(N/K)|=3. >Then ker(p) is a normal subgroup of N/K having index 3. Let M be the >preimage of ker(p) under the quotient map from N to N/K. By the >definition of M, P is in the center of M. >Therefore M has a normal subgroup B such that |B| is odd and M/B ~= K. >Then |B| is odd and |K|=4, (|B|,|K|)=1 and therefore B and K intersect >trivially by Lagrange's Theorem. Since B and K are both normal in M >and BK=M, ts shows that M is isomorpc to the Cartesian product of >B and K. >Therefore, B is the subgroup consisting of the elements of odd order >in M and therefore B is a characteristic subgroup of M. Since B is >characteristic in M and M is normal in N, B is normal in N. >Let psi be the quotient map sending N to N/B. Then >|N/B|=[N:B]=[N:M][M:B]=3*4=12. >Since B has odd order, K~=psi(K) and since K is normal in N, psi(K) is >normal in N/B. >Finally, let y be an element of N outside M. Then conjugation through >y induces an automorpsm of order 3 on K. Since B centralizes K and K >~= psi(K), ts means that conjugation through psi(y) induces an >automorpsm of order 3 on psi(K). Ts is enough to prove that N/B ~= >A_4, and we are done. Ts does not necessarily mean that G has a subgroup isomorpc to >A_4, since it is easy for a group to have A_4 isomorpc to a >subquotient without being isomorpc to a subgroup: the group >, wch has >order 36, is a good example (or SL_2(Z/(3)), wch is even smaller, >also works). So, as far as ts logic is concerned, it is possible to >have a finite simple group of order congruent to 4 mod 8 with no >subgroup isomorpc to A_4. Does anyone know whether or not ts ever >actually happens? It is true that a finite simple group with Sylow 2-subgroup of order 4 > contains A_4 as a subgroup, but I do not know if there is an easy > or reasonably elementary proof of that. It follows from the classification > of finite (simple) groups with dihedral Sylow 2-subgroups, wch was > published by Gorenstein and Walter in 1962. In particular, if the > Sylow group has order 4, then the group is isomorpc to PSL(2,q) > for q congruent to 3 or 5 mod 8, and all such groups contain A_4. By the way, as an exercise, prove that a finite simple group cannot > have a nontrivial cyclic Sylow 2-subgroup, without using transfer > theorems. The first proof I learned for ts did not use the transfer: Let G be a finite group of even order with a cyclic Sylow 2-subgroup. Look at the Cayley representation of G, wch is the permutation representation of G given by letting G act by right multiplication on itself (I read from left to right). Note that the Cayley permutation of any element of G has all of its cycles having the same length and that ts length is the order of the element. Therefore the number of cycles in the Cayley permutation of x is |G| divided by the order of x. So let x be a generator of a Sylow 2-subgroup. Let |G|=2^s*t where s is a positive integer and t is odd. Then, by the above facts, the Cayley permutation of x has t cycles of length 2^s. Since an n-cycle can be written as the product of n-1 transpositions, the product of t 2^s-cycles can be written as the product of t(2^s-1) transpositions. Since s is a positive integer, 2^s is even so 2^s-1 is odd. Since t is also odd, t(2^s-1) is odd and therefore the Cayley permutation of x is odd. Therefore the elemnts of G wch have even Cayley permutations form a subgroup of index 2 in G. ---- David === Subject: Pronunciation od Godel NNTP-Posting_Host: student-concordia-dorm.student.concordia.net NNTP-P0STING-HOST: 210.117.60.5 everybody, Is the name Godel pronounced Go - Dell or is it, Go - dull? Or am I just way off base? adam === Subject: Re: Pronunciation od Godel X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Griffith X-Treme: C&C,DWS >Is the name Godel pronounced Go - Dell or is it, Go - dull? No. The name G.9adel, or Goedel, however, . . . -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Pronunciation od Godel > Is the name Godel pronounced Go - Dell or is it, Go - dull? Or am > I just way off base? I recorded a .wav with how a German speaker would pronounce the name. Hear: http://www.math.uni-hannover.de/tmp/goedel.wav (Recording quality is not good though, due to my low quality laptop mic). Kai === Subject: Re: Pronunciation od Godel >Is the name Godel pronounced Go - Dell or is it, Go - dull? >Or am I just way off base? The vowel of the first syllable is the same as that of the second syllable of mother. (There is an umlaut over the o.) The vowel in the second syllable is schwa, similar to the vowel in book. The accent is on the first syllable. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Pronunciation od Godel > Is the name Godel pronounced Go - Dell or is it, Go - dull? Perhaps the best advice to an American English speaker is to say girdle but without quite getting to the r. English and Australian speakers of English don't ever get to the r anyway, so they can pronounce it as if it were girdle. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Polynomial algebra (2 questions) Gino Prosapio ha scritto nel messaggio > Thank for your reply. > I tnk (1) is OK. After all, I+x is in B[x]/I, not in B'[X]. === Subject: Group work on Collatz conjecture? You know... I see the Collatz conjecture talked about alot here at sci.math. Perhaps we could really try to proove (or disproove) it here, or contribute sometng to it... just imagine what a tribute to the ability of a group of people that never met, to put their abilities together to solve a problem? Anyway... I was just looking at some sequences and stuff today in reference to the Collatz conjecture, anyway I tnk we all can agree that once a power of two is reached, it will inevitably go to 1. So here's sometng that isn't spectacular in anyway, just may be another tool at looking at when these powers of two come up from the 3x+1 part of the conjecture: (2^(2n) - 1)/3 Where n is the nth number in the sequence of natural numbers that will reveal a power of 2 when multiplied by 3 and added to 1. I might like to note that when looking at the 3x+1 sequence, I noticed no powers of two where the power was odd... I didn't focus on it too much, but perhaps it could be proven? Just a thought, that may not even have anytng to do with the Collatz conjecture === Subject: Re: Group work on Collatz conjecture? > You know... I see the Collatz conjecture talked about alot here at > sci.math. Perhaps we could really try to proove (or disproove) it > here, or contribute sometng to it... just imagine what a tribute to > the ability of a group of people that never met, to put their > abilities together to solve a problem? Anyway... I was just looking at some sequences and stuff today in > reference to the Collatz conjecture, anyway I tnk we all can agree > that once a power of two is reached, it will inevitably go to 1. So > here's sometng that isn't spectacular in anyway, just may be another > tool at looking at when these powers of two come up from the 3x+1 part > of the conjecture: (2^(2n) - 1)/3 Where n is the nth number in the sequence of natural numbers that will > reveal a power of 2 when multiplied by 3 and added to 1. I might like to note that when looking at the 3x+1 sequence, I noticed > no powers of two where the power was odd... I didn't focus on it too > much, but perhaps it could be proven? Just a thought, that may not > even have anytng to do with the Collatz conjecture I couldn't sleep so I was scanning the news. Adding a tid bit. The difference between an odd and the resulting even is always 2(that odd)+y. So for 5 3(5)+1 = 16 16 - 5 = 11 = 2(5)+1 So the difference is 2x+y Also given some odd value such as 59 and 3x+y , y = 1 59,178,89,268 and using A,B,C,D for those values then D - (B+C) = Y .. here 268 - (178+89) = 1 I also can get that relationsp if I project even values such as in the sequence for 57: 57,172,86,43 by expressing 57,172 86,[259] thus projecting an even as 3x+1 259 - (172+86) = 1. When it is a sequence of evens 40 20 10 5 40 [121] 20 [61] B-(C+D) = A. What that is good for who knows. Ernst === Subject: Re: Group work on Collatz conjecture? === >Subject: Group work on Collatz conjecture? >You know... I see the Collatz conjecture talked about alot here at >sci.math. Perhaps we could really try to proove (or disproove) it >here, or contribute sometng to it... just imagine what a tribute to >the ability of a group of people that never met, to put their >abilities together to solve a problem? >Anyway... I was just looking at some sequences and stuff today in >reference to the Collatz conjecture, anyway I tnk we all can agree >that once a power of two is reached, it will inevitably go to 1. So >here's sometng that isn't spectacular in anyway, just may be another >tool at looking at when these powers of two come up from the 3x+1 part >of the conjecture: >(2^(2n) - 1)/3 If you look at those numbers in binary, you will see that they are all alternating bit patterns: 5 = 0101 21 = 010101 85 = 01010101 To multiply a binary number by 3, do a SFT-LEFT and then add it to the original: 01010101 10101010 --------- 11111111 And when you do the +1, you get a power of 2 11111111 1 --------- 100000000 The next number after 01010101 that gives you a power of 2 will be 0101010101 wch is the pervious pattern with another 01 appended. Appending a binary 01 is done by SFT-LEFT SFT-LEFT INCREMENT. Since SFT-LEFT means multiply by 2, ts next number is n*4 + 1, the number after 85 is 85*4+1 or 341. >Where n is the nth number in the sequence of natural numbers that will >reveal a power of 2 when multiplied by 3 and added to 1. >I might like to note that when looking at the 3x+1 sequence, I noticed >no powers of two where the power was odd... I didn't focus on it too >much, but perhaps it could be proven? Couple tngs to note. Since the powers of two come from alternating bit patterns, the resulting binary number will always have an even number of 0s, thus, only even powers of two. Second, the 3x+1 operation always results in a number that is congruent to 1 (mod 3). The powers of two alternate between 1 (mod 3) and 2 (mod 3) with all the even powers being 1 (mod 3). So an odd power of two cannot ever result from 3x+1. >Just a thought, that may not >even have anytng to do with the Collatz conjecture It does and there are many more fascinating tngs like that. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Hypercomplex trigonometry Are trigonometric functions defined for hypercomplex numbers (quarternions, octonions, sedenions)? If so, can anyone suggest any resources or links regarding ts? I found a page wch defines some quaternion functions: http://world.std.com/~sweetser/quaternions/intro/tools/tools.html but I didn't see any references, so I'm unsure how these were derived. Gary === Subject: Re: Hypercomplex trigonometry it'snot hard to comprehend quaternions,if youconsider that Gibbs derived vectors (and the terms, inner & outer product) from amilton. trig.is defined just divvied-up between the X,Y,Z co-ords.of the purely imaginary parts, although it could be generalized (see la boite perfectment .-) I tnk that Niel,below,meant that they can be represented as a *pair* of complex numbers, with the addition of the j operator (as ij=k, ji=-k .-) Are trigonometric functions defined for hypercomplex numbers > (quarternions, octonions, sedenions)? If so, can anyone suggest any --les ducs d'Enron! http://tarpley.net === Subject: Re: Hypercomplex trigonometry >Are trigonometric functions defined for hypercomplex numbers >(quarternions, octonions, sedenions)? If so, can anyone suggest any >resources or links regarding ts? I found a page wch defines some >quaternion functions: >http://world.std.com/~sweetser/quaternions/intro/tools/tools.html In any Banach algebra (including the quaternions), you can define the trig functions by the usual Maclaurin series. I tnk ts is still OK for the octonions and sedenions, since they are power-associative (the subalgebra generated by any element is associative), and the series for sin(x) etc only use powers of x. Of course the real question is not just can you define them?, but what properties do they have?. === Subject: Re: Hypercomplex trigonometry > Of course the real question is not just can you define them?, but > what properties do they have?. Any individual quaternion q can be written in the form a+bu, where a and b are real, and u^2=-1. Fixing that u, the set of all x+yu, x and y real, is isomorpc to the complex numbers. If the complex function sin satisfies sin(a+bi) = c+di, then the quaternion version is: sin(a+bu) = c+du. Similarly for many of the other dard functions. === Subject: Uniformly continuous?? show that f(x) = 1/x (x>0) is not uniformly continuous. ---------------------- in my book, (delta=d , epsilon=e) put d(e,a) = min{a/2 , (1/2)(a^2)e} and ts book showed |f(x)-f(a)| |f(x)-f(y)| > e it is possible that finding the concrete x,a,e?? help me , please~~~