mm-2989 === Subject: Re: Classification error, algebraic integer issue > It's fresh ground. > OOOPS again! It's fresh ... what ? Fresh ground coffee. Hey, sometimes people just forget a when they type. After all, didn't someone just say JSH is doing marketing research? He's selling _coffee_ fergoodnesssake. No wonder we didn't recognize it as _mathematics_! dave === Subject: Re: Complex numbers and 2x2 matrices (was Re: Complex numbers and 2x2 matrixes) >Also, can we have things such as e^(quat) or ln(quat) where quat is a >quaternion? Has anybody ever figured out how to do this? Exponentials, and logarithms of invertible elements, exist in any Banach algebra. Look up holomorphic functional calculus. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: sci.math competition > There have been several of these posts lately, all coming from the > Math Forum, which allows people to post follow-up messages to > ancient and obviously expired threads. > Those who post from Google Groups can also post > follow-ups to ancient messages (going back to > early 80's !!) === Subject: Re: Sum [n in Set] (1/n) >I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) >where Set is a subset of the positive integers. I seem to recall it >brought in measure theory. Does anyone know a reference? Or at least the >statement? (I'm actually only interested in the statement, not the >proof.) Well, there are infinitely many such statements, hard to know which one you're referring to. Hmm. Say S is the set, and d(n) is card(S intersect {1,2,...n}) / n. Then I think it's clear that the sum_{n in S} 1/n converges if and only if sum(d(2^n)) converges. >Tia, >msh210@math.wustl.edu Of a reply, then, if you have been cheated, >http://math.wustl.edu/~msh210/ Likely your mail's by mistake been deleted. === Subject: Re: Marketing shift, core issues >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. You know, in case you're curious, this sounds really really stupid. If your results were correct you'd be able to convince people of them by explaining the proofs carefully. But in fact they're wrong, people continually explain what the errors are, and tactics from marketing books are not going to change that. >I just wanted to warn readers that I may be employing various tactics >from modern research on human psychology to see if I can't break >through the logjam. >It seemed to me that it'd be a good idea to warn you ahead of time, as >I'd prefer it that the math by itself would be enough, but that's not >the way it works. I guess. >At least now I won't have to feel guilty if I decide to use tactics, >as I've given fair warning. >James Harris >http://mathforprofit.blogspot.com/ === Subject: Re: partial deriviatives! > Kinda basic question... > if you have a function F(q(t),t) > and you have the derivative > d/dt( d/dq[F(q(t),t) ) > then it should be > d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] > right? > Or am i mistaken...?? I think mistaken. The notation is a mess. Start with F(q,r) as a function of two variables. Then G(q,r) = [dF/dq](q,r) is also function of two variables. You want dG(q(t),t)/dt, which equals [dG/dq](q(t),t)*dq/dt + [dG/dr](q(t),t)*1. Recalling what G is, the answer is [d^2F/dq^2](q(t),t)*dq/dt + [d^2F/drdq](q(t),t). === Subject: Re: Largest number ever written down? > John Tapper scribbled the following: > If you allow finite numbers constructed from a series of mathematical > formulae, I think Graham's number takes the cake, by far. This might be considered cheating, but I believe N=B(B(B(B(9)))) (where B denotes the busy-beaver function) beats all the suggestions so far. The trouble is that the busy-beaver function is rather hard to calculate. (The busy-beaver function B(n) is the maximum number of 1s that can be printed by an n-state 2-colour Turing machine which halts. To show that N is bigger than an of the entry M posted so far, it suffices to come up with a Turing machine with at most B(B(B(9))) states which prints more than M 1s. Note that B(6) is at least 10^865, so this shouldn't be too hard to do.) -- Simon Nickerson === Subject: Call for Participation Hello All, NFSNET is a distributed computing project devoted to pushing factorization limits. It allows users to run client software which sends data back to a central site. The data is collected and eventually used to produce a factorization. However, the data is sent via an HTTP connection, which means that clients must be on-line. I am working on trying to do some slightly smaller numbers, but are on the 'MOST-WANTED' list for factorizations. I could use some help, as I have very few machines. I will provide executables (even source if people want to look at it) and data via email. People will then run the code OFF-LINE then email the results back. The current project is to factor 2^653+1, which is the smallest unfactored number of the form 2^n+1 and is in fact THE 'Most-Wanted' factorization. We will then proceed with other numbers of the form 2^n+1. I solicit your help. Please email me at rsilverman@draper.com. I am posting from my AOL account because I do not have newsgroup access at work. The method for doing this work is the way NFSNET used to work, before it was automated. But this older method allows off-line processing. Bob Silverman You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Semdirect product of categories? |Semidirect product is a common construction in group theory. But it is |obvious that if one relaxes the required assumptions, it can be |generalized to monoids. | |Now it occurs to me that another simple generalization step can bring |the concept to a categorical formulation: let A,B be categories and |let (A,A) (the monoid of functors A->A) be regarded as a category |itself. Then choose a functor sigma:B->(A,A) and define | |(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) | |for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1). | |It is straightforward to check that the definition is well-posed. But |now the question is: is it of any utility? Is it taken into account |somewhere? actually i'm not sure i understand your notation well enough to see whether what you're describing actually works, but i'm a bit skeptical about it because you don't seem to be giving the most straightforward generalization of the semi-direct product construction to a context involving arbitrary categories. the most straightforward such generalization is some version of the homotopy colimit construction. you start with a category c and a functor (or in some versions a pseudo-functor) f from c to the category of categories. the homotopy colimit of f is the category where an object is a pair (x,y) with x an object in c and y an object in f(x), and a morphism from (x,y) to (z,w) is a pair (m,n) with m:x->z in c and n:f(m)(y)->w in f(z), with composition of morphisms defined in a semi-obvious way. semi-direct product of monoids is then the special case where c has a unique object x and f(x) has a unique object y. one version of the homotopy colimit construction is called the grothendieck construction and has some pretty important uses in mathematics. the name homotopy colimit that i used above is perhaps not very standard terminology, but considering the many different version of the construction which have been studied, i'm not sure whether standard terminology really exists in this case. some names for other versions (or in some cases maybe the same version) of the construction that you might want to look up include 2-colimit, pseudo-colimit, lax colimit, and so forth. by the way the most general version of the construction is general enough to encompass not just semi-direct products of groups but in fact arbitrary group extensions. the extra twisting data needed to describe an arbitrary group extension can be considered as part of the data forming a pseudo-functor. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: partial deriviatives! >> if you have a function F(q(t),t) >> and you have the derivative >> d/dt( d/dq[F(q(t),t) ) >> then it should be >> d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] >> right? >> Or am i mistaken...?? >> /M >Assuming that you are asking for the total time derivative (not the partial >time derivative), then the answer is [shnip] but the real question is... is the d/dq a total q derivative? So (with the notation for partial derivative as pF/px): let q(t) be invertable d/dq[ F(q,t(q)) ] = (pF/pq) + (pF/pt)(dt/dq) let pF/pq == G(q(t),t) let pF/pt == H(q(t),t) d/dt[ G(q(t),t) + H(q(t),t) (dt/dq) ] = pG/pt + (pG/pq)(dq/dt) + pH/pt (dt/dq) + (pH/pq)(dq/dt)(dt/dq) = pG/pt + pH/pq + (pG/pq)(dq/dt) +(pH/pt)(dt/dq) == (p^2F/ptpq) + (p^2F/pqpt) + (p^2F/pq^2)(dq/dt) + (p^F/pt^2)(dt/dq) = ((p^2F/pq^2)(dq/dt) + (p^2F/pqpt)) + (p^F/pt^2)(dt/dq) + (p^2F/ptpq) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ where this is the answer if only the OP d/dq means partial adam === Subject: Re: Technical Name Of Equatorially Concentric Rings? >> it's impossible to see the labels on your picture. >> >> I'm not saying geodesics travel along these rings (anymore than they >> travel along latitudinal rings). >> Hmmm....maybe a picture will put it in perspective: >> >> http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=62326&fname=ringna m e.gif >They seem readable at this end--are your browser settings off? On the outside chance the OP might not have noticed, recent Mozilla and IE browsers will shrink large images to fit the screen, by default. While inside the image, Mozilla will change the cursor to a magnifying glass with a + sign. Click to enlarge to readable. While large, cursor will contain a - sign to shrink. IE will make the image small, then you have to hover your cursor inside the lower right corner until the pillow appears. Click it to enlarge, reverse to shrink. HTH. === Subject: Re: Cardinality of 2^n numbers? > However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. > No. 2^K is the power set of K, which is the set of all subsets of K. > For example, one such subset is K_p = { 2^p : p is prime }. This is a > set, not a number, and therefore 2^K is not a set of numbers and > certainly is not equal to N_0. I never said 2^K WAS the set of all natural numbers. I said 2^K was BIJECTIVE with the set of all natural numbers. (...Starblade Riven Darksquall...) === Subject: Re: Roots of a simple third-degree equation As several posters have pointed out, if you are just looking for integer (or even rational) roots, it is not too hard. In general, though, it's no easier to solve a cubic with no quadratic term than it is to solve the general case. The reason is as follows: ax^3 + bx^2 + cx + d = 0 Substitute x = y - (b/3a). If you multiply it all out, you'll find that the new polynomial in y has no quadratic term. So if you can find a root y=r of this new equation, then x = r-(b/3a) is a root of the original equation. The method for solving Ay^3 + By + C = 0 was discovered (invented?) in Italy a very long time ago. It's usually called Cardano's formula, although it appears that Cardano actually stole it from someone else. You can find the formula (and it's derivation) at http://mathworld.wolfram.com/CubicEquation.html JoeS -------- > I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: > -r^3 + 3r + 2 = 0 > How does one do it? > Fran.8dois === Subject: Re: Cardinality of 2^n numbers? > >Suppose we had the set of all integers which resembled an integer 2^n >where n is any integer. This would start from 1 in the case where n = >0 to indefinately high. Let's call this set Q. >No, let's call it K. Q is taken already by the set of rational numbers. > ... stuff deleted ... >>Since the alephs are defined as the successive cardinals, it is *this* >>that is a definition (if I understand the definitions correctly). >>However, 2^K is surely not the smallest infinite set. The existence >>of the alephs has nothing to do with 2^K, and it is certain that they >>do *not* show that 2^K is the smallest infinite set. > > However, 2^K is the set of all natural numbers. And the set of all > natural numbers is of cardinality N_0. However, K is also of > cardinality N_0. If the power set of any set is strictly larger (IE > has a greater cardinality) then how is it possible? This would require > that N_0 > N_0, which is the contradiction I'm talking about. > > This is simply not so. If K is the set of powers of 2, it is just not > the case that 2^K is the set of all natural numbers. Since you're the > one making this assertion, I'll ask you to tell me how one associates > a natural number with an arbitrary subset of K, so that every natural > number is matched with a subset, and so that no two different subsets > are matched to the same natural number. According to Cantor's theorem > it cannot be done. Given any set X, its collection of all subsets has > more elements than X. You're misusing cantor's theorum. I never said that K wasn't greater than K. I said N was equal in size to the power set of K. Cantor's theorum does not involve this kidn of thing. The power set of K is related to the natural numbers if you do this: For each element in the power set of K, add all the numbers together. You will always get a unique natural number. For instance, take K = (1, 2, 4). Take the power set of K = (Null, 1, 2, 1&2, 4, 1&4, 2&4, 1&2&4). Now this relates to the natural numbers Num = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the power set of K. However, the power set of K, as the size of the power set approaches infinity, also approaches the set of the natural numbers (Even if 0 is not a natural number, you can just add 1 to each member of the set so that it goes from 1 to 8 rather than 0 to 7.) and is therefore bijective. (...Starblade Riven Darksquall...) === Subject: Re: What constitutes an implicit use of a function? (Was: a confusing group theory conundrum) at 06:37 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >1. If two statements are the same, then they both make use of the >same things What do you mean by the same? If you really mean that they are the same, then your next statement is false. >2. The statements x < x^2 and x < f(x) are the same (when >f(x)=x^2) Only if you are using a different meaning of the same from the one in 1. above. >Now I am not trying to be stubborn, it's just that I am the type of >person who urgently needs to understand things. Try reading the responses you already got. >Clearly a definition like h(x)=h(x) is circular. Don't confuse a definition with a theorem. >the replacement axiom Applies to theorems. >This is very weird, Yes. The Devil is in the details, and you need to start paying attention to the details or you will keep getting wierd results. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: The need of geodesics in physics <3fad7a93$1$fuzhry+tra$mr2ice@news.patriot.net> not specific to operator algebras. Correct, although I don't know whether the more general meaning is ever used in QM. I'm certainly not aware of the field[1] being other than R, C or H. [1] If you use the definition that requires commutativity, then omit H. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: What constitutes an implicit use of a function? (Was: a confusing group theory conundrum) >if one has a given property, What do you mean by has a given property? Your usage seems to transcend the values of the propositional functions, in which case two propositional functions with the same values need not have the same properties. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Cardinality of 2^n numbers? > Starblade Darksquall says... >What I'm saying is that K is the set of all numbers which are a power >of 2... >So K is of cardinality N_0, from what you said, but so is its power >set, since its power set is bijective with the set of natural numbers. > No, the power set of K is not bijective with the naturals. > What is true is the following: > 1. There is a bijection between the power set of K and the > set of *infinite* sequences of 1s and 0s. For each subset > K' of K, associate the infinite sequence s as follows: > s[i] = 1 if 2^i is in K', and 0 otherwise. > 2. There is a bijection between the naturals and the set of > *finite* sequences of 1s and 0s (namely, you can write every > natural number in binary notation). > 3. But there is no bijection between the naturals and the set of > infinite sequences of 1s and 0s. > The proof of 3 is exactly Cantor's theorem: assume that there is > a bijection f between N and the infinite sequences of 1s and 0s, > and show that leads to a contradiction. The only way I see there being a contradiction is if you require that K and N both be of cardinality N_0. However, I don't see why there can't be a non-N_0 cardinality of K such that when its power set is taken, its power set is of cardinality N_0. That will also have to be explained to me. (...Starblade Riven Darksquall...) === Subject: Re: I can't stand it anymore > amanda replied: > [...] >as they have been told by the 19th century European racists >>Is there an Asian country which is not racist? Whya re you talking abput cpuntries. Why not individuals? >>Cause it's not just individuals, it's a cultural thing. >>Why don't you talk about individual Americans? You are the one who accused all Asian countries as racists. > It was an observation. I stand by it. All? Even Thialand? > You do not understand how Western culture (as they see from the > movies) scare the parents in those countries. These governments feel > obligated to deter the bad part of western culture. > I'd say that it often goes much farther than this. Just like with every other group. > In doing so, > sometimes they go extreme. When the movie Flash Dance came out, any > kid who twist their body would be taken away by the police, at least > for a moment. My cousin, about 13, was really good at it. his father > never knew that he could do that. If the father knew, he would > probably scold the son. >Just look at the nice >>things high Japanese govt officials have said about Americans. >who believed that they were superior to all other humans. >>As do all Asian races currently. Currently is the keyword but you are wrong to generalize all Asians. >>So you say. Nice that you are using what Mlaysia said as the ruler. Never mind > that Malaysian government a Muslim government. Show me any Muslims > who thinks western culture is suitable to their children. > Actually, Malaysia is not your typical Muslim country, it's as good as > it gets. But Malaysia is about a quarter chinese. > Chinese represent nearly a quarter of the population, and > Indians are some 8%, but provide 20% of the lawyers and 15% > of doctors. Any racially inflammatory news tends to be self > censored in the major media and there is a conscious > effort among all ethnic groups to work together. Free elections > have been held since 1969 and the standard of living, in 1960 > on the level of Haiti, now surpasses much of Eastern Europe > and Latin America. Education represents over 20% of the > national budget and tens of thousands of students are sent > abroad. Well, there was a time where there is a limit on how many properties Chinese can own. > How about Japan? I never mentioned which particular Asian group are racist in my opinion and which one are not. Only that I do not agree that all Asians are racists. By the way Japan has a lof of foreigners from other Asian countries over staying there and work. If Japanese goovernment do not want *them* stay there illegally (to work), why don't they punish the employers who hire them? It's double standard. > http://www.japantoday.com/e/?content=news&cat=9&id=278105 > Kanagawa governor apologizes for calling foreigners 'thieves' > for calling foreigners sneaky thieves while campaigning over > the weekend for a candidate running in Sunday's House of > Representatives general election. > My strong desire to maintain security led me to make the false, > inappropriate remarks. I apologize and take them back, the > governor said. There are some foreigners who come to Japan > with student visas but end up overstaying illegally and commit > serious crimes. I want to control such crimes. > The Kanagawa prefectural government received more than 100 > emails after the speech both criticizing and supporting the > remarks, Matsuzawa said. (Kyodo News) > Or China? Chinese in general is one group that I now *claim* who think they are superior to all races. > http://tibet.dharmakara.net/ictracism.html > Racism: China's Secret Scourge > ICT Report Refutes Beijing's Denial of Racism in China > Washington, D. C. - International Campaign for Tibet (ICT) > will release a report at the UN World Conference Against > Racism (WCAR) documenting the origin and nature of racism > against Tibetans and how the Chinese government perpetuates > racist attitudes and policies. > The 60- page report, entitled Jampa: The Story of Racism > in Tibet, describes how racist language and concepts permeate > China's constitution, laws and policy and how this has > contributed to the racism and discrimination Tibetans face > today. It is the first comprehensive analysis of this > phenomenon, a subject that has not been widely addressed by > scholars, human rights groups and others who generally focus > on more conventional human rights violations in Tibet. > While highlighting racism in the west, China has effectively > suppressed racism as a domestic issue. This is their shameful > secret, said Tsering Jampa, Director of International Campaign > for Tibet- Europe. > In the months leading up to the World Conference on Racism, > China has portrayed racism as a Western phenomenon that does > not exist in China. In a February 2001 submission to the UN, > China stated that all ethnic groups are living in harmony > in China. > The Chinese government's denial that racism is a significant > problem in China is a policy which prevents Tibetans and others > from addressing racism in meaningful, constructive ways, said > John Ackerly, President of ICT. > The title of the report, Jampa, refers to the protagonist > of a ubiquitous 1963 Communist Party propaganda film depicting > Tibetans as a backward people who can only be uplifted by the > civilizing force of the Chinese. > All Tibetans live under the shadow of this film, said Tsering > Jampa. The Chinese government has used it to denigrate Tibetan > culture and justify its occupation of Tibet. > At the conference ICT will urge the government of China to > acknowledge the extent of the problem and to remove derogatory, > chauvinist or paternalistic language from laws and policy > statements. ICT is also urging Chinese non-governmental > organizations based in the west to work with Tibetan groups > on educational programs and initiatives to help combat this > long-standing problem. > Although China tried to block the accreditation of Tibetan > human rights groups to the World Conference against Racism > a vote by UN member counties approved accreditation for > ICT and one other Tibetan organization. > ICT has invited also Xiao Qiang, Director of Human Rights > in China, whose organization was not accredited, to join > its delegation to the conference. > But then, you probably don't care. Your problem is that you assume too much. But you use probably and so I will give you a break. >Asian racism is probably not a > problem you face. Again, you assume TOO much. >> http://news.bbc.co.uk/1/hi/entertainment/showbiz/2582833.stm >> 'Anti-Asian' Brad Pitt advert banned >> A car advert featuring Brad Pitt has been banned in Malaysia >> because it is an insult to Asians, according to reports >> in the country. > [...] >My saying that Asian do not need a test invented by someone else >>See? You're into 'us' and 'them'. Obviously for you 'us' is Asians, an > d >>'them' is everyone else. Again, another silly accustaions that I am into us and them. > It's what you said. You took it out of context, i.e I didn't not say Us were superior to them. It was merely as reference to groups > Beside, we both know who invented IQ tests. > Tell me who. C'mmon. > So don't make it look like I meant everyone else when Is aid them. > That's what them means in this context. The context was when the person from that group accused me of not doing well on the IQ test and so my using them* refers to the group he belong to and not everyone else. >>And you find 'them' wanting in the same fashion >>as Zainuddin Maidin above. Pretty bold accusation. You think you know but you do not know how > Asians see western culture very well. > I think that they see it much as Ghandi did. Do you prefer that they say with enthusiasm, we should adopt it? > Reporter to Mahatma Ghandi: Mr. Ghandi, what do you think > of Western Civilization? > Ghandi: I think it would be a good idea. >If you can't understand that, that's your problem. >>That's the problem. I see what you actually write and object to, and >>it's no different from the stuff others post. You find things wanting >>just because they are not Asian in origin. You can't see it, I underst > and >>that. Whatever? I do not know how you concluded that I belive the Asians are > superior to anyone else. You seem to be upset that I implied that > Asians are superior to you. May eb you had have some experience with > soem Asians acting like that. > Like it's a rare and unexpected thing? So you did have some experience. And I got picked on because of that? But it depends on which Asians you met and where. In US, most Asians would act snobbish. > Heck..I have met Chinese who claimed to be superior to anyone else. > You talk like there are Chinese who don't believe this. Yes there are. I mean there are truly Intelligent Chinese. > In fact, one time, when this friend of mine said that America is doing > well because of its Chinese population. Immediately after I smiled > and replied I don't think so, I realized that that moment was the > beginning of the end of our friendship. > Interesting. > And she wasn't the first Chinese I met who put down on Amercans > (talking about white Americans). Everytime, I have defended from the > Americans' side. For what? To be accused by people like you that I > think white American are inferior to Asians. Very nice! > Oddly, I've not seen anything from you that looks like a defense of > white Americans. Why should I be defending for them when I didn't say that they are inferior. Oh .. I see. To you defending white Americans would be to say that they are superior. >>Why don't you also complain about Asian racism? Why should I complain about it here? >>Because I'm asking you to, just to show that you care about racism, no > t >>just Asians. But I wasn't talking about racism. I was talking about the miuse of term Asian. If I didn't care about racism, I would not have started this thread. > As I said, there is a difference between objecting to racism against > *anyone* and racism against the group you happen to belong to. Few will > actually do the first, no matter what they claim. So you think because I am Asian, I won't point out racism against among Asian group? Why are you insistent on that? Like I said, I was after the term Asian. I didn't initiate this thread to talk about Asian racism. > Why are you clueless about that? > Just said it again. Maybe you'll catch it this time. Maybe not. Again ... I was after the term Asian. I didn't initiate this thread to talk about Asian racism. >I have complained many times about Indian caste system somewhere else > . >>Elsewhere? What are your complaints? Long time ago - it was soon after 9/11 - I happened to go into an > egpytian group thinking I would learn more about the ancient egyptian > culture. Instead, some Indian Hindu hate mongers were trashing the > Muslims (mainly Pakistanis) pickign on their religion and treatment on > women. So I borught up suttee, caste system, and treatment of women > in their society. Caste systen has been abolished but has it really > gone? I told them to clean up their hosue before picking on others. I > didn't use the same user name. > I found it. > > It's always someone else's fault, isn't it? > Just shows how ill-informed persone you really are. I suggest you > read history of your religion of 'peace' aka Islam, What made you say that my religion is Islam? >and what it did > when when it ransacked flourishing cultures and civilization. > Islamic doctrine is like Nazisms, which it more than matches in > fanaticism and fascism, especially when it come to lying and > distorting history. Not surprised. There are many variations (of books) on that subject. It's up to the personw which view to hold. I know which one you chose. By the way, tell me which Muslim armies went to what is now Malaysia and Indenoesia and ransacked and forced conversion? > In other few hundred years Nazis will also say that Hitler had a > profoundly > positive and spiritual influence on barbaric jews. Just as shameless > retards > are denying Islamic butchery and barbarity of over the past millenia. > [...] > > Rare..amusing. Forogt suttee too? > Suttee was common among Rajput wives of warriors who did not want to > be captured by invading Muslims. Yeah..right. You are so misinformed. Ever watched the movie Far Pavillion? Based on a true story. Was she being chased byt he Muslims she wanted suttee? > The number of women who commit suttee > among 0.8 billion Hindus is around one every few decade. The fact is that it existed. >It voluntary > and even then against the law. Namely, it was voluntary. >Now compare this with brutal, but > legally enforced murder of thousands of Muslim women for committing > adultery. The latter is sanctioned by Allah, No it was not. It's human who did that. > and Pakistan, > Afghanistan, Iran, Saudis implement it. Pakistan, Afhganistan, And Saudi Arabia are not the place where most Muslims are. You seem to be very ill-informed about Muslim and Islam. > Limb chopping being other favorite pastime of pedophile prophets' > retarded followers. Oh..completely misinformed about him too. I suggest you get better sources for you info. > [...] > > What percentage of Muslims women are covered in black tent? > > In Islamic paradises like Afghanistan, Saudi Arabia, Iran it is close > 100%. > If the women dont obey they are beaten mercilessly. > > married at like cattle at any age > > How old were the Hindu girls when sold by their parents to rapists? > > You are thinking of many cases of Indian Muslims who sell their > preteen daughters to dirty old Arabs, especially in Hyderabad. The keyword is Indian Muslims. They have carried on their pre-Islamic culture into their current lives. > They > are rescued by Hindu police. Where were that Hindu police when the Hiundus were setting fires on Muslims, including babies and pregnanat women, using gasoline. It happend just not too long ago. Were you out of this planet at the time? > [...] > > An ignorant hatemonger calling one of the greatest human being in > > history a pedophile doesn't count. Get that into your tick head. > Just like Hitler is called the greatest human being in history by > Nazis. Pol Pot by his own follwer. Just because he butchered people > and had temporal lobe epilepsy with paranoid psychosis and delusions, > does not mean he was great. Today he would be in lunatic asylum, where > he belonged. There would have been no islamic mass murderers like Bin > Ladens, Gaznavis and Timmurs. > Regarding his pedophilia; truth always hurts. Were you raised as Baptist. If so, it is very understanding that you ahve been given misinformation. > Did he not marry Ayesha > when she was 6 year old and he 54 yr old, and had sex with her when > she was only 9 yr old? He married here to make peace with that group. He did not consummate the marraige until she was 13, which was normal for the culture of the time. >Did he not marry young wife of his own adopted > son? Adopted son is not his son. >Did he not marry a Jewish women he captured after slaughtering > her entire family and 700 plus surrendered jewish men? No he did not. Find books not written by the Baptists and Christaain propagandists, will you? > He even gave instruction on how to divide booty from loot. Step in the time and see things form the perspective of the time, will you? ALso, do not forget how he pardoned the enemies and made peace with them. > BTW, it looks like bride burning is today's problem. What say you about > this? Oh did you just get back form outer space? > No one knows how suttee started, or for that matter where or when. > It's not unique to India--widow suicide is known to have occurred > among the Egyptians, Chinese, Vikings, and others. Some say its > origin on the subcontinent dates back 5,500 years, while others > believe it arrived much later, around 1 AD. I've heard Indians > deny there's anything specifically Hindu about it, in that it > doesn't figure in Hinduism's core texts. Today it's most closely > associated with remote villages dominated by the Rajputs. It was the British who outlawed the suttee. India was the only country who practised it as long as they did. > Suttee is different from bride burning, in which a newly married > Indian woman is burned to death by her in-laws for failing to meet > demands for a larger dowry, the traditional gift given to the > couple by the bride's parents. Thousands of such murders have been > reported. Why are you telling me this as if I don't know? > In contrast, suttee is a voluntary act, theoretically at > least, Yes. theorectical ONLY. >meant to atone for the couple's sins and ensure their > reunion in the afterlife. Failed to consider the politics behind it. > But horrified Indian feminists say > that in practice the suttee victim often had little choice. They were not horrified faminists. They are humane human. > Sometimes family members, including other women, browbeat her > into it; sometimes she was bound or hopped up on drugs. Much > of the time even that wasn't enough. It's said music was > played at high volume during suttee so no one could hear the > widow's screams. Seems you're as down on Hindu culture as you > are on Muslim culture. No..I am down on the abusive practises of people (human). > I'm having a difficult time reconciling the claims about the frequency of > suttee with the last accounts of how it occurred in practice. The last official one was when the British outlawed it but it continued to happen .. > Nontheless, > Hindu culture hardly seems blemish free and I disagree that you have to > be perfect to point out faults you see in others. > BTW, I agree with you about islam. No surprise there. You are missing out a lot on Islam. Now, do not get confused thinking that I am praising the Arabs. >As a minority in the country I was born (my parent were born and >raised thare too) who suffered discrimination and were not considered >as full-citizens, you must be joking accusing me as a racist. >>What does that have to do with whether you are racist or not? > Because it was due to race that we were discriminated. > The issue is whether *you* are racist, not others. The issue which I came to understand PERFECTLY now is that you want to me to say Asians (including I am) are racists. >It is very clear > that you think you are saying that since you were discriminated against > you cannot be racist. You twisted what I said. I replied that it was due to race that we were discriminted. Now you are using it to attack me. >>I don't think you really understand the concept. You were the one who didn't understand what I said. > I do understand it, you still don't. >>It's like saying you are black, you can't be racist. >>I got news for ya, if you can say that, you are. > I got news for you. You underestimated me in assuming that I think > black cannot be racists. > But you think that since *you* were discriminated against, that you > cannot be racist. What's the difference? I repeat..You twisted what I said. I replied that it was due to race that we were discriminted. Now you are using it to attack me. >Beside, as a person whose blood covers basically almost all races of >the world (not through Caucasoid and Negroid per se but through >Indo-Aryan which you probably have no clue about), including semite >(not jewish), you gotta be kidding to accuse me as a racist. >>You've gotta be kidding if you think this denies that you can be >>a racist. >>It's clear that you will profit from any pro-Asian racism. What are you talking about? >>Think Affirmative Racism. Think 'diversity'. >I was/am going with the second one, WHICH WAS MY WHOLE POINT OF >criticizing the use of the term Asian in IQ test, duh... >>That is was not invented by an Asian. That was your objection, how >>dare a non-Asian even imply that they can say anything about an Asian. My objection was that I as an Asian did not need to run and take the > IQ test to find out my intelligence level when, I repeat WHEN *that* > guy told me that I must not have done well blah..blah..blah and ehnce > against the IQ tests. > I disagree. Your objection was, and I quote... > Don't you realize that us Asians don't need the tests invented > by non-Asians to find out whether we are intellgent or not? > Your objection was that the teste were invented by non-Asians. Presumab > ly > had the tests been invented by Asians this would be a non-issue, as the > tests would be fine. Presumably is the keyword. >>It's almost certain that you have. The most stupid thing I have ever heard. What I have had was getting >pulled over by the red-neck police >>Racism. Red neck is red neck. > A racist term is a racist term. Simple as that. Oh..now you want me to be politcally correct? > Simple as that. Here in CA, I approached a > police car once to ask for some direction. The first things he said, > with a very pleasant manner was what can I do for you?. In > Houston, it would be like What's your problem?. Another time, as > the police blocked the route to my place, I asked the police standing > there what I should do and he was really nice. In Houston, it would be > like I am busy; get lost. > Seems you are comparing your experiences in CA with your prejudices > about Houston. I am comparing with my experiences with the red necks from Houston. > I even told that (that most of them are red neck) to a police officer > once. It was 11 pm at night, not on the highway but he happened to be > really closed by and I didn't see his car because I wanted to get home > which was not too far and did not slow down as he expected. I was the > only car around there. > He told you I am busy; get lost? No he did not. One did give that kind of impression once. (In words, it would be I am busy; get lost. ) When he saw my disappointment on my face, he immediately said I am sorry. By then, I already got my car window rolling up. > I told him to just give me the ticket and spare the lecture. He > asked me why I talked like that. I told him that I haven't met a > police officer who did not give me a ticket. I told him that most are > red necks. He just laughed. May be he didn't get mad because I was not > far from the campus and he knew that I was a student. (I did tell him > that I was tired and didn't see him when he asked why I didn't slow > down after seeing him). Or may be I looked cute to him (he wasn't very > old) when i said to him that most of them are red necks. Who knows? As > he gave me the ticket, he said that he would not have given me the > ticket if I didn't act the way I did. I replied - silently - yeah, > right! > So you have no issues with stereotyping Americans. No... not Americans. Only red necks, i.e red necks are (many) individuals. >Odd that you have > issues with stereotyping Asians. Why is this? I do not stereotype any group based on race or ethnicity. >in Houston for driving a beat-up >car in my student days because I look a bit Hispanic or South >Americans. >>Odd, I was always pulled over by the cops when I had my old 1965 >>Mustang convertible. Racism obviously, right? In my case, there were other cars speeding but they didn't get > stopped. > I'm not asking about your case. When you got pulled over, it was for your old Buick? When I got pulled over, it was for speeding but he didn't stop the SUV next to me. Why? because it was drven by a white? >>Rich Note: I will not be wasting any more time defending groundless >accusations. >>You've already proven that they are not groundless. Whatever. > Ahh, the superior Asian IQ rears it's ugly head. I will not be wasting any more time defending groundless accusations. > Rich >>Rich > Good to know that I was right that you had some experience with some Asians acting superior to you. I guess it hurts since you expected that you would be exempt form it for being white, huh? Sorry for that experience though. You may be surprised that I almost got that experience too (I may have gotten it somewhere without knowing it) until that Chinese girl realized that way back in my ethnic group, there was Mongoloid mix. Her comment was this: You are smart because of the Mongolid blood. Well..I just wanted to remind you that initiated this thread to express my frustration about the term used in IQ tests in grouping people. So, here is the list of my suggestions: - Get over your hurt witht he Chinese and stop expecting me to say whites are superior to Asians (inclduign Chinese) . - Stop accusing me that I think Asians are superior to white. - Stop expecting me to say that all Asians are racists. - Stop assuming that my being Asian has benefitted me. And - Do not twist my sentences. === Subject: Re: Cardinality of 2^n numbers? (*Snip*) You are very unhelpful. Are you sure you're not just here to be as completely useless as possible? (...Starblade Riven Darksquall...) === Subject: Re: Sum [n in Set] (1/n) > I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) > where Set is a subset of the positive integers. I seem to recall it > brought in measure theory. Does anyone know a reference? Or at least the > statement? (I'm actually only interested in the statement, not the > proof.) Perhaps that the span of {x^s : s in S} is dense in C[0,1] iff sum 1/s is infinite? I forget whose theorem it is. The proof is in Rudin's Real and Complex Analysis. (And I may have some details wrong. Perhaps it's the ALGEBRA generated by these which is dense iff...) --Ron Bruck === Subject: Re: Four Color Graphs > Every complete 4-partite graph is four-colorable. The Mathworld > descriptions of Complete Graph, Complete k-Partite Graph, > k-Partite Graph are essentially appropriate to the following > discussion. > The complete graph Kn is of course n-partite. In Mathworld, the > complete k-partite graph (Ck) is denoted Kp,q, ... ,r; where p+q+ ... > +r = n. Here, a slightly different nomenclature is adopted; ie, Ck = > (P1,P2,P3, ... Pk). > To minimize confusion, let Pi represent partition 'i' and pi represent > the nummer of vertices in partition 'i'. So we can also write that > Ck = {p1, p2, p3, ... , pk} > Then, C4 = {p1, p2, p3, p4}. The number of edges (Ec4) in C4 is, > Ec4 = p1*(p2+p3+p4) + P2*(p3+p4) + p3*p4 > > For example, let p1 = p2 = p3 = p4 = n/4; then > Ec4 = n/4*(n/4+n/4+n/4) + n/4*(n/4+n/4) + n/4*n/4 = > [6*(n/4*n/4)} > Ec4 = .375*(n^2) > Let n = 16, then pi = 4; and Ec4 = .375*(16*16) = 96. To check > Ec4 = 4*(4+4+4) + 4*(4+4) + 4*4 = 48 + 32 + 16 = 96. > If we let Ec4 = INT(0.375*(n^2)), then the formula works for all > values of n when p1 ~ p2 ~ p3 ~ p4. For example; for n = 14, C4 = {4, > 4, 3, 3}. Without wrting out the calculations in detail; > Ec4 = 4*10 + 4*6 + 9 = 40 + 24 + 9 = 73 > Ec4 = INT(.375*(14^14)) = INT(.375*(196)) = 73 > At the other end of the spectrum, let p1= (n-3), p2 = p3 = p4 = 1. > Then > Ec4 = (n-3)*(1+1+1) + 1*(1+1) + 1*1 = 3*n -6 > We propose that 3*n-6 <= Ec4 <= 0.375*(n^2) for all n and all possible > partitionings thereof. Let n =8. The following partitionings are possible; {2,2,2,2), {3.2.2,1}, {3,3,1,1}. {4,2,1,1} & {5,1,1,1}. Howwever, no planar graph may have more than INT(n/2) vertices in any one partition. So the partitioning {5,1,1,1} is not applicable. Then Partitioning max edges max diagonals {2,2,2,2} 24 16 {3,2,2,1} 23 15 {3,3,1,1} 22 14 {4,2,1,1} 21 13 Note that any planar graph may be depicted as an n-sided polygon with up to (2*n-6) diagonals. If n = 8, then 2n-6 = 10. Therefore, the partitioning {2,2,2,2) can be reduced to Binomial(16,10) = 8,008 different 4-color graphs with (3n-6)= 18 edges. But there are only 2,772 planar graphs with 8 vertices and 18 edges. (see HYPOTHESIS below) The total number of configurations due to all four partitionings is 12,298. It is impossible to determine the number of duplicate configurations. Further, there are Binomial(20,16) = 4,845 complete 4-partite graphs with n=8 and {2,2,2,2} partitioning. So it is conservatively safe to say that there are > 8,008 4-colorable graphs with 8 vertices and 18 edges. Of which, only 2,772 may be planar. HYPOTHESIS, The number of maximal planar graphs is [C_(n)*C_(n-1)]/2; where C_(n) and C_(n-1) are the Catalan numbers for (n) and (n-1). Specifically; C_(8) = 132, C_(7) = 42. Afterthought. A possible counter example to the FCT would be a complete 4-partite graph with less than 3n-6 edges; if one existed. As n increases, the ratio of all 4-color graphs to 4-color planar graphs increases rapidly. === Subject: Re: Playing Doctor >So,...... you wanna play doctor with me now, after you have failed >miserably in politics, sociology, psychology, philosophy & economics? >..........AHAHAH.....ahahahaha...... Babe, listen, you are not very >successful, except when it comes to nagging and whining >....about YOUR own problems.......ahahahahaha.... >ahahahah....ahahahahah...ahahahaha.....ahahahanson >>Having a woman volunteer to play doctor is bad? Howzzat???? > Ilsa: SS Doctor Newsflash: Ilsa does not qualify as a woman. Female, perhaps. === Subject: Re: Cardinality of 2^n numbers? ... stuff deleted ... >However, 2^K is the set of all natural numbers. And the set of all >natural numbers is of cardinality N_0. However, K is also of >cardinality N_0. If the power set of any set is strictly larger (IE >has a greater cardinality) then how is it possible? This would require >that N_0 > N_0, which is the contradiction I'm talking about. >>This is simply not so. If K is the set of powers of 2, it is just not >>the case that 2^K is the set of all natural numbers. Since you're the >>one making this assertion, I'll ask you to tell me how one associates >>a natural number with an arbitrary subset of K, so that every natural >>number is matched with a subset, and so that no two different subsets >>are matched to the same natural number. According to Cantor's theorem >>it cannot be done. Given any set X, its collection of all subsets has >>more elements than X. > You're misusing cantor's theorum. I never said that K wasn't greater > than K. I said N was equal in size to the power set of K. Cantor's > theorum does not involve this kidn of thing. I am certainly *not* misusing Cantor's theorEm. K is, by construction, of the same cardinality as N. After all, it is produced by taking the image of N via an invertible function. Just because I stepped from N to a set of the same cardinality, that does not invalidate the application of Cantor's theorem. Your statement that N is equal in size to the power set of K is not true. Recall the mapping I provided (I've corrected a typo: originally mapping): The standard mapping from 2^{1, ..., N} with the set of numbers {0, 1, ..., 2^(N-1)} is achieved by taking the function X |----> sum(x(k)*2^(k-1); k =1, ..., N-1 ), where X is any subset of {1, ..., N}, and x(k) is the function from {1, ..., N} that is 1 when x is in X and 0 otherwise. This is the mapping you're using here: > The power set of K is related to the natural numbers if you do this: > For each element in the power set of K, add all the numbers together. > You will always get a unique natural number. Do those sums exist for *every* subset of K? What about the set of all elements of K? Every other element of K? You should realize that the only subsets you can map to N via this mapping are the FINITE ones. It is true that the collection of FINITE subsets of N has the same cardinality as N itself. However, unless you contest the existence of infinite sets of integers (such as the set of even integers, or the set of all integers greater than 5, for instance), you must admit that those sums do NOT all converge, and thus your alleged mapping fails to be defined everywhere on the power set of K. > For instance, take K = (1, 2, 4). Take the power set of K = (Null, 1, > 2, 1&2, 4, 1&4, 2&4, 1&2&4). Now this relates to the natural numbers > Num = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the power > set of K. However, the power set of K, as the size of the power set > approaches infinity, also approaches the set of the natural numbers > (Even if 0 is not a natural number, you can just add 1 to each member > of the set so that it goes from 1 to 8 rather than 0 to 7.) and is > therefore bijective. > (...Starblade Riven Darksquall...) Ignoring the bulk of subsets of 2^N doesn't buy you any credibility, BTW. Dale. === Subject: Re: Complex numbers and 2x2 matrices (was Re: Complex numbers and 2x2 matrixes) >>Also, can we have things such as e^(quat) or ln(quat) where quat is a >>quaternion? Has anybody ever figured out how to do this? >Exponentials, and logarithms of invertible elements, exist in any Banach >algebra. Look up holomorphic functional calculus. I should qualify that. The holomorphic functional calculus exists in complex Banach algebras, while the usual quaternions are an algebra over the reals. Of course there's no trouble complexifying to get an algebra of quaternions over the complex numbers (with the slight notational annoyance that the quaternion i and the complex number i are not the same). The exp of a (real) quaternion is a (real) quaternion; ln (using the principal branch) of a (real) quaternion whose spectrum does not intersect the nonpositive reals is a (real) quaternion. This is because exp and this branch of ln can be uniformly approximated on a neighbourhood of the spectrum by polynomials with real coefficients. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Playing Doctor >>hanson [ retorted to a woman, not you, Bill, >>[ sheesh, you're too old and ugly.....ahahaha..]: >So,...... you wanna play doctor with me now, after you have failed >miserably in politics, sociology, psychology, philosophy & economics? >..........AHAHAH.....ahahahaha...... Babe, listen, you are not very >successful, except when it comes to nagging and whining >....about YOUR own problems.......ahahahahaha.... >ahahahah....ahahahahah...ahahahaha.....ahahahanson > [Bill] >>Having a woman volunteer to play doctor is bad? Howzzat???? > .....with all her nagging and bitching? .....ahahahahhaa.......... > I doubt that even a superb dictation, nor much less a profound > rectification will keep her quiet and make her happy. And in case > of her having keen oral interests, even such a notorious machination > will not keep her occupied in silence and solace for long neither. > Do you have any advanced ideas, Bill, to bring inner peace to that > poor and agitated woman? We must be good Samaritans, don't we? > ahahahaha........ahahahahanson...... Pacification is just around the corner in Amanda's case. In fact I think it an easily achievable goal. In general, I think passion a good thing. If I'm wrong in everything else in this exchange, I do know she's given some very good replies. Keeping true to your persona, don't fret over it. There's room for all of this in sci.physics. === Subject: Re: I can't stand it anymore > I found it. I took a look and you got the posts written by I also noticed that you copied the stuff written by the guy whom I referred to as Indian Hindu hate monger. All he did was coming up with crap. I am sorry that you took him seriously. You should have got some books written by unbiased historians before you decided to trash Islam and its Prophet, will you? There is not much difference in Islam and Christianity when it comes to crap, crpas which were made by humans. For the story of Muhammed, I strongly suggest that you get a few books written by real historians. > BTW, I agree with you about islam. Which part did you agree with? Are you sure that what you read was written by me? By the way, here is something you may want to read about Muhammed. He was Caesar and Pope in one; but he was Pope without Pope's pretensions, Caesar without the legions of Caesar: without a standing army, without a bodyguard, without a palace, without a fixed revenue; if ever any man had the right to say that he ruled by the right divine, it was Mohammed, for he had all the power without its instruments and without its supports. ---------------------------------------------------------------------------- ----- MAHATMA GANDHI, speaking on the character of Muhammad, (pbuh) says in (YOUNG INDIA): I wanted to know the best of one who holds today's undisputed sway over the hearts of millions of mankind....I became more than convinced that it was not the sword that won a place for Islam in those days in the scheme of life. It was the rigid simplicity, the utter self-effacement of the Prophet, the scrupulous regard for his pledges, his intense devotion to this friends and followers, his intrepidity, his fearlessness, his absolute trust in God and in his own mission. These and not the sword carried everything before them and surmounted every obstacle. When I closed the 2nd volume (of the Prophet's biography), I was sorry there was not more for me to read of the great life. -------------------------------------------------------- My choice of Muhammad to lead the list of the world's most influential persons may surprise some readers and may be questioned by others, but he was the only man in history who was supremely successful on both the religious and secular level. Michael H. Hart The 100: A Ranking of the Most Influential Persons in History. New York: Hart Publishing Company, Inc. 1978, p. 33. ---------------------------------------------------------------------------- ------- If greatness of purpose, smallness of means, and astounding results are the three criteria of human genius, who could dare to compare any great man in modern history with Muhammad? The most famous men created arms, laws and empires only. They founded, if anything at all, no more than material powers which often crumbled away before their eyes. This man moved not only armies, legislations, empires, peoples and dynasties, but millions of men in one-third of the then inhabited world; and more than that, he moved the altars, the gods, the religions, the ideas, the beliefs and souls. . . his forbearance in victory, his ambition, which was entirely devoted to one idea and in no manner striving for an empire; his endless prayers, his mystic conversations with God, his death and his triumph after death; all these attest not to an imposture but to a firm conviction which gave him the power to restore a dogma. This dogma was twofold, the unity of God and the immateriality of God; the former telling what God is, the latter telling what God is not; the one overthrowing false gods with the sword, the other starting an idea with words. Philosopher, orator, apostle, legislator, warrior, conqueror of ideas, restorer of rational dogmas, of a cult without images; the founder of twenty terrestrial empires and of one spiritual empire, that is Muhammad. As well ask, is there any man greater than he? Lamartine, HISTOIRE DE LA TURQUIE, Paris, 1854, Vol. II, pp. 276-277. ---------------------------------------------------------------------------- ------- What nonMuslim say about Muhhamed http://cyberistan.org/islamic/quote1.html Tolerance in Islan by Pickthall, an English convert http://cyberistan.org/islamic/toleran1.html I apologized those at sci.math,sci.physics,sci.chem; this is my last post on this thread. === Subject: Re: differentiable...problem... >> if f is differentiable on (0, infinite) >> and lim [f(x) + f'(x)] = L (x->infinite) >> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) >| f e^x (f + f') e^x >| lim f + f' = L => lim f = lim ----- = lim ------------ = L >| x->oo x->oo x->oo e^x x->oo e^x > Provided f e^x -> +-oo, in which case > 0 = L - L = lim f+f' - lim f = lim f' No proviso is needed. This form of L'Hopital's rule needs only that the denominator is infinite, not also the numerator [1][2]. So there's no need to treat this case specially as you do below. -Bill Dubuque [1] A. E. Taylor, L'Hospital's Rule Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24. http://links.jstor.org/sici?sici=0002-9890(195201)59:3%3C20%3E [2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital Rule Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242. http://links.jstor.org/sici?sici=0002-9890(197604)83:3%3C239%3E > However when f e^x -> k; then > f = fe^x e^-x -> 0 > f e^-x = f e^x e^-2x -> 0 > L = lim f+f' - lim 2f = lim f'-f > 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L > 0 = lim f+f' - lim f = lim f' > So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f' > k = lim(x->oo) f = k + lim f'; lim f' = 0 > If lim f' doesn't exist, then for n >= 2 > f_n(x) = sin(x^n)/x -> 0 > (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation === Subject: Re: Ordering a Power Set > This question arises from economics, but I'll give you the question > first, the context later: > Suppose we have a set of n objects, and a binary relation S that is a > linear ordering, ie irreflexive, asymmetric, complete, transitive. >> Is > there any way to use this relation to induce some soft of ordering on > the power set of our set of objects? >> This arises from an examination of equilibrium in the jungle: we >> have > N agents, and S is strength(1 is stronger than 2, etc.) To see if > this equilibirum is coalition-proof we need some way to order > coalitions of agents. Which I don't know how to do. > What do you mean coalition proof? Can't the top dogs always beat up on > the bottom dogs. Even in US democracy, we see how well the rich top have > coaluded to succefully control, own and extort the poorer bottom. Well, it's a well defined concept in general equilibrium theory. >> Subsets of an n-element set correspond to strings of length n made up >> of the symbols 0 and 1... 1 means this object is in the set 0 means >> this object is not in the set. But strings of 0s and 1s of length n >> correspond to binary representations of the integers from 0 to 2^n-1. >> There is a natural way to order THIS set, right? >> but not one that in any way uses S. The idea is to extend the idea of >> strength of individuals to strength of coalitions. > Use weights. Give each a weight or strength. Make the weights linearly > independent so that no sum of weights of one subset equals the sum of > weights of another subset. For example square roots of different primes. > Then order subsets by sum of weights of members. The order of the > subsets > will depend upon the weights assigned each individual. > Why insist upon a complete order? Because if I don't have one I can say whether or not a particular coalition would be able to appropriate some allocation from another person. -- Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/ === Subject: Re: Usenet Posting Guide? > With all respect, sir, I can guess your politics right across the > board. Just for openers, you're a Sierra Club, save-the-whales type. > Gotcha, huh? > You're about as far off the mark as it is possible to get. Not that I see > the relevance, but I'm a very conservative, Religious Right type -- not > far removed from what most people would call a fundamentalist (though > I don't use that label myself). I'm irrevocably opposed to both abortion > and gun control, voted for Bush (both of them) and supported the Iraq war > (both of them), and certainly do not support the Sierra Club, Greenpeace, > PETA, or any other environmentalist or animal-rights group. And no, > I don't care to discuss or defend any of those positions in sci.math. You're right as right can be. Taking what you say at face value, which I do, I was indeed off the mark in that assessment. My apologies. You also sound like my kind of guy, for that describes me absent the religious overtones. I despise political liberals, and I'll let it go at that as well in that most writers are political liberals, enviro-whackos and PETA types. Ursula === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. Just so that we all understand you this once, do you mean that you've decided to give up on math and become a bookseller or do you mean that you've given up on the math and are hoping to use modern marketing techniques to push your self-described proof on a gullible public? Either way, best of luck. Rick p.s. I guess you missed my earlier post, so I'm reprinting it here. Let's take a simpler example and see how your argument works. Let Q(x) = 7(25x^2 + 30x + 2) [1] = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2 and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2] Now we mirror your construction and observe that we may take a_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3] We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots a_1(0) = 0, a_2(0) = -1. We see that this assignment is consistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4] and we see that we can rewrite [4] as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2) or, emphasizing the constant term in [1] we may write, letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2) Now, however, you would have us write each factorization over the algebraic integers as Q(x)/7 = (5(a_1(x)/7) + 1)(5b_2(x) + 2) [5] regardless of the (rational integer) value of x. This constraint in [5] does indeed work in some cases. For example, if we take x = -1 we see that Q(-1) = -21 = (7)(-3) r(-1) = a^2 + 2a and we see that we may take a_1(-1) = 0, a_2(-1) = -2 and that doing so gives us the factorization Q(-1) = (5(0) + 7)(5(-2) + 7) and we see that Q(-1)/7 = (5(a_1(-1)/7) + 1)(5a_2(-1) + 7) = (1)(-3) The problem is that your restriction in [5] depends strongly on the polynomial r(x) used to define the terms a_1(x) and a_2(x). For x = 0, -1 the factorization in [5] worked because of the way r(x) splits into particular linear factors, but consider the case when x = 1. Now we have Q(1) = 399 = (7)(57) = (7)(3 * 19) r(1) = a^2 + 14 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, it's immediately obvious that sqrt(-14) isn't divisible by 7 in the algebraic integers, so the factorization in [5] does not hold. 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 in algebraic integers Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] * [(-5 sqrt(-14) + 7)/sqrt(7)] In fact, your constraint on the way 7 (or 49, in the example you have been using) is split among the terms in your nonpolynomial factorization is valid only in a small number of cases for x. R. === Subject: Re: Ordering a Power Set Right. So strength will be a function of some sort, that representive relative power, and lets us say how much stronger 1 is than 2, etc. The properties of the equilibirum will then depend pretty strongly on how we specify this function, won't it? >> This question arises from economics, but I'll give you the question >> first, the context later: >> Suppose we have a set of n objects, and a binary relation S that is a >> linear ordering, ie irreflexive, asymmetric, complete, transitive. Is >> there any way to use this relation to induce some soft of ordering on >> the power set of our set of objects? >> This arises from an examination of equilibrium in the jungle: we have >> N agents, and S is strength(1 is stronger than 2, etc.) To see if >> this equilibirum is coalition-proof we need some way to order >> coalitions of agents. Which I don't know how to do. > In order to transfer unambiguously the order on the underlying > ordered set to an order on (finite) subsets or coalitions, the > underlying order must do something like allowing comparisons of > intervals, or differences in order, so that, say, given w, x, y and > z in the underlying set, one may compare w-x to y-z ( find whether w > as much greater than x as y is greater than z). > Since this requires more than simple ordering, there cannot be any > unique extention to coalitions based only on simple ordering. > The question is just how much more than simple ordering is required > on the underlying set in order to make such comparisons of > coalitions meaningful or useful. -- Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/ === Subject: Re: Cardinality of 2^n numbers? Starblade Darksquall says... >The only way I see there being a contradiction is if you require that >K and N both be of cardinality N_0. It's not a requirement, it is a theorem. There is a bijection between N and K, namely f(i) = 2^{i}. By the definition of having the same cardinality, N and K have the same cardinality. -- Daryl McCullough Ithaca, NY === Subject: JSH: Deprogramming needed? (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math discoveries? After all, I've been at this since April 1995 having spent a lot of time and effort, with literally thousands of posts along with all kinds of other activities, websites, and email to mathematicians all over the world. But, what if I'm wrong? What if all the time and energy I've invested in my work has made it difficult for me to see, along with the *harsh* and unforgiving hostility from several people who seem to have made it their mission to make me miserable and find pleasure in mocking or trying to humiliate me, have made it extremely difficult for me to see the truth? Think about the crushing sense of shame and misery if indeed I find out that the logical connections I so carefully and impatiently discovered over the years are simply not really there, but are a need induced delusion. It seems to me that marketing ploy though these statements may be, dealing with people who've made it their business to try and make me miserable, only to at times claim they're trying to help me is just too much. Aren't there any *other* people who suppose they are rational, who can follow a logical argument, who might comment? Why is it always the same people? Or people imitating them in hostile and mocking displays of animosity or anger? Aren't there any rational people who can trace out the steps in the work I've presented, who might step forward at this time, and demonstrate an ability to just be objective? I don't want any replies of people offering, but if you might consider it, I just want you to think about it. Sure I've been reading this marketing book, but it seems to me that still *someone* out there might be the right person, as of course, I don't believe I'm wrong. I've traced out every step in the arguments I have, and I think that irrational people have been dominating the discussion using group effects to hide the truth. Think about it. I'll come back to the subject later. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Basic factorization ideas >If you saw > >(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > >with the c's algebraic integers, I think few of you would have a >problem realizing that only two of the c's have 7 as a factor. > Personally I have a problem with the fact that I can think of no > values for c_1, c_2 and c_3 such that c_1, c_2 and c_3 are algebraic > integers and the left and the right side of the equation are equal. James has tricked you. In this version the roots of the cubic are units. Part of his marketing strategy, I think. -- 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: Deprogramming needed? James Harris grava .88 la saucisse et au marteau: > Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? I don't know the french equivalent for algebraic integers. If someone tells me what set it is (is it the positive and negative integers?), I may have the courage to take a look. But anyhow, your proof is hard to read. You should at least follow the indications the others have given you (clearly define your functions, explicitly when you can and so on). > Why is it always the same people? Or people imitating them in hostile > and mocking displays of animosity or anger? Because I suppose very few people have had the courage to read your proof. And the one who didn't saw the former(s) commenting it and pointing out some problems you didn't solve (you keep claiming they're wrong, but your explications are more confuse than their). I've known better motivations. -- Nicolas === Subject: Re: On Reading Textbooks of points have been raised, and some questions asked, so I shall address those now. Firstly, there is the matter of my mathematical background before I began my independent study. In Britain, people typically study 3 or 4 subjects at A ( for advanced) Level, between the ages of 16 and 18. In my case these were Maths, Further Maths, English Literature and History. I got A grades ( the highest) in the two maths qualifications. The syllabus included material like Pure Maths: differentiation including product, quotient and chain rules; integration including substitution, parts and reduction formulae; arc-lengths, areas under curves, and volume and surface area of surfaces of revolution; differential equations, going up to 2nd order linear equations with constant coefficients; partial differentiation and maxima/minima/saddle points of functions of 2 variables; arithmetic and geometric progressions; complex numbers, including complex loci and De Moivre's Theorem; proof by induction; hyperbolic functions; Cartesian coordinates, covering lines, circles, ellipses, parabolas and hyperbolas; polar coordinates; vectors, including scalar and vector products and the equations of lines and planes. Probability: basic combinations and permutations, conditional probability, discrete and continuous random variables, expectation and variance, probability generating functions. We did briefly cover moment generating functions and maximum likelihood estimators, but I couldn't do those now. Mechanics: Newton's Laws, conservation of energy and momentum, acceleration in polar and intrinsic coordinates, impulse and restitution, using calculus to solve problems with variable forces, moment of inertia and angular momentum. I did buy a couple of A Level textbooks to remind myself of all this stuff before starting the current books. As for my current situation, no I am no longer at university. I decided that seven years was enough, and that it was time to start earning some money. I have the textbooks, my native wits, and whatever help I can get here. A few people asked what I was aiming to achieve with my study. I think my epiphany came when I read Simon Singh's book on Fermat's Last Theorem and watched the associated documentary. But it was not the positive reaction you might expect. I asked myself what I now knew about the theorem and its proof, and was forced to answer Practically nothing. I know what a Diophantine Equation is and what Fermat's conjecture says. I gathered that the theorem is not intrinsically important, but that attempts to solve it led to the development of some very important mathematics. I know that Wiles's proof involved the work of people called Taniyama, Shimura, Flach, Kolyvagin and Iwasawa ( amongst others). But what are the main steps in the proof? I have no idea. I know that modular forms are extremely important, but have not the least idea what one is. What is Iwasawa theory? What sort of situations does it describe? How does it relate to FLT? Sorry, I haven't a clue. I do not mean to imply that the point of my study is to be able to understand the proof of FLT, for I understand that this is extremely advanced subject-matter. But I enjoyed my mathematical studies at school, and I would like to study something a bit more advanced. I can solve a quadratic equation or find the intersection of a line and a plane, but there is a whole body of knowledge, on which a lot of extremely intelligent people spend a lot of time, that I know absolutely nothing about. I want to understand some real maths, as that term is understood by a mathematician. In short, I want to study mathematics for the same reason I studied history: it seems to me that it is interesting and worth knowing. Finally, I asked about how many requests for help I can reasonably make. Certainly, I have no worries about appearing stupid, by asking silly questions, or repeating questions until I understand the answers. I want to do this properly ( and anyway, I am most unlikely to meet any of you face-to-face). My question related more to the reaction of you, the expert readers of the forum. I do not want to reach the stage where people become so bored by my constant questions that they simply scroll past my posts and I receive no ( useful) replies. John Harrison === Subject: Re: JSH: Deprogramming needed? > James Harris grava .88 la saucisse et au marteau: > Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? > I don't know the french equivalent for algebraic integers. .82 Nombres alg.8ebriques é. Ciao, Renaud Dreyer === Subject: Re: Basic factorization ideas > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. A bit of tricking? Your marketing? Let r1, r2 and r3 be the roots of the cubic in some order, than we have: c1 = 7/r1, c2 = 7/r2, c3 = 1/r3 all algebraic integers. So indeed, c1 and c2 are divisible by 7 in the algebraic integers. > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 22. What has 7 being a factor or not of 22 to do with it? This is also true when 7 is a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Oh, you mean that when 7 is a factor of 22 than there are other ways to do the division, as in that case 7 is a unit in the ring. Moreover, there may be values of x such that (f3(x) + 1) is divisible by 7, I think. If this is the case there are other ways to distribute the 7's. ... > Now then, in my advanced factorization work, I just use functions of x > that are a lot more complicated than f_1(x) = c_1 x, and unfortunately > there are people who can use an unfamiliar leap in complexity to > confuse others. Unfortunately, the functions here are trivially divisible by 7 in the ring of algebraic integers. In your advance factorisation work they are *not*. But you think that is a flaw... But indeed, that your functions are much more complicated *makes* a difference. c1 x *is* divisible by 7 because c1 is divisible by 7, regardless of the value of x (c1 is a constant), so (c1 x + 7) is trivially divisible by 7. (c3 x + 1) is *not* divisible by 7, unless x has a specific value as I mentioned above, which might or might not be possible, but that is irrelevant, because the first two factors are trivially divisible by 7. In your (more complicated case) you *know* that a1(x) and a2(x) are *not* divisible by 7 for all x (you say that is a flaw). It has been shown that (b3(x) + 22) is divisible by factors of 7 for almost all x. And those factors compensate for the lack of factors in a1(x) and a2(x). -- 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: Deprogramming needed? > James Harris grava .88 la saucisse et au marteau: > > Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? > > I don't know the french equivalent for algebraic integers. > .82 Nombres alg.8ebriques é. Ciao, D.8esol.8e, c'est .82 entiers alg.8ebriques é, of course. Ciao, Renaud Dreyer === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) How long does it take your Marketing ploy warning to become a threat? -- 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: Deprogramming needed? > (Marketing ploy warning) > What if indeed I *am* wrong, and I don't have these great math > discoveries? > After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. > But, what if I'm wrong? > What if all the time and energy I've invested in my work has made it > difficult for me to see, along with the *harsh* and unforgiving > hostility from several people who seem to have made it their mission > to make me miserable and find pleasure in mocking or trying to > humiliate me, have made it extremely difficult for me to see the > truth? > Think about the crushing sense of shame and misery if indeed I find > out that the logical connections I so carefully and impatiently > discovered over the years are simply not really there, but are a need > induced delusion. > It seems to me that marketing ploy though these statements may be, > dealing with people who've made it their business to try and make me > miserable, only to at times claim they're trying to help me is just > too much. > Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? > Why is it always the same people? Or people imitating them in hostile > and mocking displays of animosity or anger? > Aren't there any rational people who can trace out the steps in the > work I've presented, who might step forward at this time, and > demonstrate an ability to just be objective? > I don't want any replies of people offering, but if you might consider > it, I just want you to think about it. Sure I've been reading this > marketing book, but it seems to me that still *someone* out there > might be the right person, as of course, I don't believe I'm wrong. > I've traced out every step in the arguments I have, and I think that > irrational people have been dominating the discussion using group > effects to hide the truth. > Think about it. I'll come back to the subject later. > James Harris > http://mathforprofit.blogspot.com/ Does your definition of objective mean: agreeing with you? I think these people are being objective. Lurch === Subject: Re: Deprogramming needed? : Does your definition of objective mean: agreeing with you? No, unless it is about matters of objective fact. : I think these : people are being objective. Well, they aren't. There is no objective answer to the question of whether anybody does or doesn't deserve the sort of abuse that JSH has been getting from some of these people, and more to the point, given that everybody participating in the discussion is doing so from mindsets that SHARE a certain kind of academic- Americanness, it IS objectively knowable that that kind of abuse, EVEN it is deserved, is counter-productive. JSH arrives here in 2 kinds of error. One involves a series of small errors in the factorizations, in the computations, in the alleged possibilities around the reasoning steps. The other involves idiotic over- generalizations about the character of mathematicians generally. Tragically, some of the people in the discussion are reacting to him in ways that tend to make them individual supporting examples of the insupportable generalization. If everybody else could emulate Dik Winter and confine their criticisms to the math, this would all go away very quickly. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America === Subject: Re: Factorization dispute, again > who are these *others* that you refer to?... perhaps, > you have an audience that is not known to the rest of us, > the desingated Peanut Gallery of would-be critics & helpmeets. > or is it just the entire, 'virtual crowd of all > of the folks who are in the googolplex, > that *could* lurk on your bifurcating threads -- or > maybe they should? > mea culpa, dood. I have this sense that you, BQH, are more artist than, say, mathematician - and sometimes incredibly funny, and highly literate. I don't always read what you post, and when I do I don't always understand it, but there is *something*. So I hope you won't completely give up this addiction - Nora B. Les ducks d'endrun! Les quacks d'bon-bon! Ils douches de Maman! > keep talking as if convincing *others* changes mathematical truth. > --ils dcues d'Enron! > http://larouchepub.com/ === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. > I just wanted to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. You are more liable to break the logjam by playing Russian roulette solitaire, as the logjam is all in your imagination. > It seemed to me that it'd be a good idea to warn you ahead of time, as > I'd prefer it that the math by itself would be enough, but that's not > the way it works. I guess. > At least now I won't have to feel guilty if I decide to use tactics, > as I've given fair warning. As JSH's past performances lead one to believe that he would not feel guilty even for using an Abomb on the mathematical world..... === Subject: Re: Vedic Mathematics --- Myth and Reality > > While one does not know will happen after 2000 years, one thinks one > knows what happened 2000 years ago, when one is guided by one's trust > in the people he considers worthy of respect, > > So claims don't need to be verified by evidence? > So, just as one takes as genuine evidence the word of a witness > who cannot be proved false, one takes as genuine evidence the > statements of people in positions of respect. Values that are based on respect/reverance for personalities fall within the boundaries of temples/mosques/churches. Historical research doesn't depend on such beliefs. > Arindam Banerjee. === Subject: Re: JSH: Deprogramming needed? Renaud Dreyer grava .88 la saucisse et au marteau: > D.8esol.8e, c'est .82 entiers alg.8ebriques é, of course. Ciao, Le probl.8fme, c'est que j'ai jamais entendu l'expression entier alg.8ebrique en frn.8dais, c'est bien le probl.8fme. Je connais les entiers naturels, les entiers relatifs et les nombres alg.8ebriqu.8es qui sont les nombres r.8eels non transcendants (donc pas forc.8ement tr.8fs entiers). Quid? -- Nicolas === Subject: Re: Probability > You need a triple integral, integrating over all three variables A, B, > and > C. The domain of integration is determined by the condition B^2 >= 4 A > C. > The integrand is just a unit function, and hence the integral can be > interpreted geometrically as the volume of the domain of integration. > The > latter is some section of a hyperboloid, but of course bounded by the > additional constraints that all three variables are between 0 and 1. > > Forgetting about the geometrics and just doing the integration is pretty > straightforward, and I get the final value (5+6*ln 2) / 36 = 0.254413... > Well, back to the geometry, what did you use for your limits of > integration on each of the variables A, B, and C? > Well, I threw away my calculations, but I'll try to reconstruct. The > integration limits are 0 restriction that B^2>4*A*C, i.e. B > 2 * sqrt(A*C). I first did the integral > over B. Holding A and C fixed there are two cases to consider: A*C < 1/4 and > A*C > 1/4. In the former case, the integral is over 2 * sqrt(A*C) < B < 1, > and hence has the value 1 - 2 * sqrt(A*C). In the latter case, the integral > is over the interval from 0 to 1, and hence has the value 1. Next I > eliminated the C integral, and - holding A fixed - once again had to > consider two cases: A < 1/4 and A > 1/4. Are you able to do the rest from > here? > Also, does anyone have handy the discriminant of a cubic? Maybe we > can apply the same idea to this? > Er, what for? Do you have some application in mind? I mean, why uniform in > the interval [0,1]? Why not standard deviation, say, or some other > distribution? > -Michael. No particular reason, I just thought it would be interesting to see if there was some noticeable pattern in the probabilities. I suppose as n tends to infinity the probability of an arbitrary n-th degree polynomial having n real roots tends to 0, but it would still require proof. Anyway it basically boils down to just being curious. === Subject: [JSH] Crank Net!: Re: Basic factorization ideas > I need those of you interested in mathematics to focus on the basics, > so that you can understand the advanced. http://www.crank.net/harris.html === Subject: [JSH] Crank Net!: Re: Deprogramming needed? http://www.crank.net/harris.html > (Marketing ploy warning) http://www.crank.net/harris.html > What if indeed I *am* wrong, and I don't have these great math > discoveries? http://www.crank.net/harris.html > Think about it. I'll come back to the subject later. http://www.crank.net/harris.html > James Harris http://www.crank.net/harris.html === Subject: Re: Sum [n in Set] (1/n) >I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) >where Set is a subset of the positive integers. I seem to recall it >brought in measure theory. > I believe the set of polynomials { t^n, n in Set } spans a dense > subset of the vector space of continuous functions on [0,1] iff > Sum [n in Set] (1/n) diverges. Could that (or something like that) > be what you meant? Yes! > There is also a conjecture of Erdos which I think states that > the sum diverges iff Set contains arbitrarily long arithmetic > progressions. Ah, that I'd never heard. In the future, you can post or mail; doing both is unnecessary. msh210@math.wustl.edu Of a reply, then, if you have been cheated, http://math.wustl.edu/~msh210/ Likely your mail's by mistake been deleted. === Subject: Re: Cube (the movie) > Sometimes, you just have to let art flow over you, whether it's Pi or > JSH. > Cube, that movie where the lady (who is supposed to be a math graduate) > spends half a minute trying to determine whether an even number is prime or > composite? (eheh) I'll have to get that! Sounds like it's a classic along with the Wizard of Oz and This Island Earth. === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) > What if indeed I *am* wrong, and I don't have these great math > math discoveries? Then you would be empirically sane. No chance (note your inserted if despite an extraordinarily large body of trivial disproof submitted by more than a score of posters), http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: question about canonical commutation relations [P,Q]=i*hbar*I I am a college student who heard this problem from one of my linear algebra professors. Question: Find matrices of infinite size Q and P such that [P,Q]=i*hbar*I, hbar is a constant,I is the identity matrix, i^2=-1 I've tried to prove that there do not exist nxn complex matrices Q and P such that [P,Q]=i*hbar*I. Proof. Suppose there exist nxn complex matrices Q and P such that [P,Q]=i*hbar*I. [P,Q]=i*hbar*I => trace([P,Q])=trace(i*hbar*I) trace([P,Q])=trace(P*Q-Q*P)=trace(P*Q)-trace(Q*P)=0 trace(i*hbar*I)=i*n*hbar<>0 => trace([P,Q])<>trace(i*hbar*I). It's a contradiction. Q.E.D. I've also tried to find the matrices of infinite size. Let P,Q be matrices of infinite size. p[i,1]=1 p[i+1,2]=1 ... p[i+n,n+1]=1 ... q[1,i]= -i*hbar q[2,i+1]= -i*hbar ... q[n+1,i+n]= -i*hbar ... when i->infinity [P,Q]=P*Q-Q*P=i*hbar*I. Can anybody explain where the error (if any!) comes into my derivation? Any comments would be appreciated. === Subject: Re: question about canonical commutation relations [P,Q]=i*hbar*I John grava .88 la saucisse et au marteau: > Find matrices of infinite size Q and P such that What is a matrice of infinite size? IMO, this is a complete nonsense since results on matrices specifically use that we are in finite dimension. -- Nicolas === Subject: *OT* BBS's Remeber these?? Remember BBS's?? I was looking through some old BBS programs last night then decided to check out the internet to see if anyone still uses BBS's. I came along http://Synchro.net (Synchronet) last night (some of you may remember it). They went open source and are now offering it free - it uses Telnet... so, I downloaded and installed it on my Compaq Proliant 2500 Server. tried it out, woah - the memories... am going to leave it up as its pretty neat... so come check it out... (don't mind the BBS name, I was tired). You'll need to have outgoing access to port 23 (telnet). I even went as far to get it working with DOVE-Net!!! Maybe we can get a SCI.Math BBS going. lol Anyways here is the Link: telnet://cyberhood.servehttp.com Mike Curry === Subject: Re: Deprogramming needed? > (Marketing ploy warning) > What if indeed I *am* wrong, and I don't have these great math > discoveries? > After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. > But, what if I'm wrong? > What if all the time and energy I've invested in my work has made it > difficult for me to see, along with the *harsh* and unforgiving > hostility from several people who seem to have made it their mission > to make me miserable and find pleasure in mocking or trying to > humiliate me, have made it extremely difficult for me to see the > truth? > Think about the crushing sense of shame and misery if indeed I find > out that the logical connections I so carefully and impatiently > discovered over the years are simply not really there, but are a need > induced delusion. So if you don't get your way, you're going to cry? === Subject: Re: Largest number ever written down? >> John Tapper scribbled the following: >> If you allow finite numbers constructed from a series of mathematical >> formulae, I think Graham's number takes the cake, by far. > This might be considered cheating, but I believe N=B(B(B(B(9)))) > (where B denotes the busy-beaver function) beats all the suggestions > so far. The trouble is that the busy-beaver function is rather hard to > calculate. > (The busy-beaver function B(n) is the maximum number of 1s that can be > printed by an n-state 2-colour Turing machine which halts. To show > that N is bigger than an of the entry M posted so far, it suffices to > come up with a Turing machine with at most B(B(B(9))) states which > prints more than M 1s. Note that B(6) is at least 10^865, so this > shouldn't be too hard to do.) In the same spirit, if one wants to think of B as a not-very-fast-growing function ... http://www.cs.berkeley.edu/~aaronson/bignumbers.html === Subject: Re: Image of a straight line in a convex cylindrical mirror > In 3D, y+b=0 is a plane. Right? Are you looking for its reflection, or > reflection of a line? > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? It's probably safe to assume, since this is a line on the floor of the mall, that the equation is y+b = 0 and z = 0. Now that it's been a few days, can I give a hint? The symmetry of the problem -- what coordinate system should you be using? === Subject: Re: Greek Alphebet >>Do the Greeks use Latin and Greek letters in a similar way to us Latin >>alphabet users? How about the Russians, do they spurn their own >>alphabet and use Latin and Greek letters? >> Others have answered your question about the Russians, and I can assure you >> that Greeks use Greek letters the same way everybody else does: they are not >> Greek to them :-) >I did guess that the Greeks would know the Greek alphabet. But >although the use of the Greek alphabet is not as fixed as the OP >seemed to expect, the Greek letters do seem to be used by us in a more >restricted way than the Latin letters. Do Greeks use pi almost >exclusively for 3.14159 . . . and capital sigma for sum etc. To put >it another way, if I extracted a series of formulae from English and >Greek text books, would you easily tell which had come from which >book? The answer to the first question is yes, and the answer to the second question is no. (But make obvious exceptions for sine or cosine abbreviations (im for imitonon/semi-tone and syn for synimitonon/co-semi-tone), vertices denoted by Capital Greek rather than capital Latin letters, etc.) There is a third question (going against your intuition*): is there *heavier* mathematical use of Greek letters in papers or books published in Greece? I tend to think that the answer is no. *I assume that you would expect the sudden appearance of Greek letters (of mathematical meaning) in a Greek text to cause confusion: that's not the case. baloglouAToswego.edu === Subject: Practical Statistics (Help) Hello All, Background Info: I'm looking for help on what I precieve as real world statistics problem. I'm an electrical engineer currently working on a problem involving the production testing of a wireless network device. We need to come up with a test that can be performed on every device built. The test must verify that the recieve sensitivity is within specification. The specification is that the device must operate up to 100 m away from a certain type of transmitter in a certain environment. Performing this test directly is very impractical in a mass production environment. Therefore, I would like to perform a test such that the Device Under Test (DUT) is only inches from a lower power transmitter in a controlled environment (a shielded box). To do this, I must determine what power level (P) the lower power tranxmitter must be set to in order to simulate a normal transmitter 100 m away. The customer has rejected all proposals of theoretical calculations of P and wants us to perform a study to determin P. My job is to come up with a plan to do this. I would like to take a sample of known good devices and test them by lowering transmitter power (P) until they fail. After I collect this data, I would like to use it to determine what P should be set to for testing of the 1000's of new devices that will be built every day until the end of the product's life. The Math Questions: What method should I use to determine the sample size for the study? I have a pretty good educated guess at the range of values that I will get for P. dev. with high confidence. I can then use this information to find a good P for my future testing. I know this probably sounds like a problem from a text book, but it's not. I've been reading through an old statistics book and I think that I'm starting to talk the talk a little, but I still can't walk the walk. Any suggestions would be appreaciated. Warm Wishes, -Andy === Subject: Re: Basic factorization ideas > If you saw > > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > > 49(x^3 + 5x^2 + 3x + 1) > > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. ... > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > Oh, you mean that when 7 is a factor of 22 than there are other ways > to do the division, as in that case 7 is a unit in the ring. > Moreover, there may be values of x such that (f3(x) + 1) is divisible by > 7, I think. If this is the case there are other ways to distribute the > 7's. Lo and behold, there are indeed (not 7, but factors of 7)! When x = 7y + 2 for some integer y, (c3 x + 1) is *not* coprime to 7. This example is too easy to crack. Define: w3(x) = gcd(c3 x + 1, 7) (this one is not always 1 (*)), w2(x) = 7/w3(x). We now could divide the first factor by 7, the second by w2(x) and the third by w3(x) and remain in the algebraic integers. So, again, the possible ways to divide 49 through may very well depend on the value of x. ---- (*) Proof: We have (c3 x + 1) where c3 is 1/r, with r the negative of one of the roots of x^3 + 5 x^2 + 3 x + 1 Now (x/r + 1) = (r + x)/r, so the first is not coprime to 7 when (r + x) is not coprime to 7. Set x = 7y + 2. (r + x) is not coprime to 7 when (r + 7y + 2) is not coprime to 7, and that is when r + 2 is not coprime to 7. You may verify that (r + 2) is a root of y^3 - 11 y^2 + 35 y - 35. So it is an algebraic integer and not coprime to 7. -- 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: question about canonical commutation relations [P,Q]=i*hbar*I >Question: >Find matrices of infinite size Q and P such that >[P,Q]=i*hbar*I, hbar is a constant,I is the identity matrix, i^2=-1 >I've tried to prove that there do not exist nxn complex matrices Q and >P such that [P,Q]=i*hbar*I. ... Your proof is correct. >I've also tried to find the matrices of infinite size. >Let P,Q be matrices of infinite size. >p[i,1]=1 It's best not to use i here because it will be confused with sqrt(-1). Note that the product of two infinite matrices is well defined if they have only finitely many nonzero entries in each row (there are more general conditions, but this one is enough for your case). Hint: I think you'll want to define P and Q so each has only one nonzero entry in each row: one of them only above the main diagonal, the other only below the main diagonal. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Cube (the movie) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2wVZ22788; >Saw it yesterday. >In German. >Uhm, did the synchro botch everything up or is it >a lot of senseless mathbabble already in the original? :-) >-- >Hauke Reddmann <:-EX8 >Private email:fc3a501@math.uni-hamburg.de >For our chemistry workgroup,remove math from the address >For spamming, remove anything else Saw it too. The math is not completely senseless: the idea to interpret the numbers as coordinates is quite obvious, isn't it? Later on they used the same numbers to describe the path a subcube follows over time. It's not so clear how to do that, although it's possible. Then suddenly the property of being a prime power played a role in identifying those rooms, that are traps. At that point I guess it was just nonsense. It seems difficult to me to encode two types of information into the number triplets shown in the movie. H === Subject: Prove: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2wUQ22782; If g.c.d (x,y)=1 and x divides z and y divides z then xy divides z === Subject: Re: if x has normally distributed digits, does 1/x? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2wmw22889; > Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part of some >(irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? > > I seriously doubt it. Here's an indication of why: > > Let's talk about base 2. Start with x = .01010101... . > Of course x is not normal, but it does contain the right > number of 0's and 1's; it's weakly normal in base 2, in > a sense. But x = 1/3, so 1/x = three = 11.000... , which is > far from normal. >>You're in the land of rationals. >>Rationals and normality don't mix. >>(Take that out of context, man!) >Well, that's amusing enough to make it seem possible that you >were just trying to be funny. In case you were also trying to >make a serious point: >I certainly didn't mean that this was a counterexample, or >something that would lead to a counterexample with no new >ideas required. But what I had in mind was something >_vaguely_ like this: Suppose we could find x_n such that >each n-bit sequence in the binary expansion of x_n occurs >with the right frequency, but 1/x_n has a terminating >binary expansion. Then let x consist of a long stretch >of the bits of x_1, followed by a much longer stretch >of the bits of x_2, etc. For suitable values of long >and longer x will be normal, but it doesn't seem so >unlikely that 1/x would turn out to be abnormal. >>Phil >>-- >>Unpatched IE vulnerability: window.open search injection >>Description: cross-domain scripting, cookie/data/identity >> theft, command execution >>Reference: http: //safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTMExploit: http:/ /safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htmIt tutns out there's a quick and basic argument which conclusively >proves that there are numbers left out by the definition of algebraic >integers. >Consider >(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) >where you might wonder what the c's are, and that's what is covered in >this post. >Now then, you have as a zero of the factorization x = -7/c_1, so let >x= -7/c_1, so you have >49(-7^3/c_1^3 + 5(49)/c_1^2 - 21/c_1 + 2) = 0 >which is >2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. >But that is a non-monic primitive irreducible over Q, so c_1 and by >symmetry c_2 cannot be algebraic integers. However they must be >algebraic numbers, and it can be shown that any algebraic number can >be written as the ratio of algebraic integers. Yep, so far. >So then there must exist f, such that fc_1 is an algebraic integer, >and letting g = fc_1 and multiplying both sides by f, I have >(gx + 7f)(c_2 x + 7)( c_3 x + 2) = > > 49f(x^3 + 5x^2 + 3x + 2) >so a zero is now x = -7f/g, which gives >49f(-7^3f^3/g^3 + 5(49)f^2/g^2 - 21f/g + 2) = 0 >which is >2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0 >which proves that f must have 2 itself as a factor for g to be an >algebraic integer. >For instance, letting f=2, gives >g^3 - 21 g^2 + 245(2) g - 343 (4) = 0. >But looking back again at >(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) >that would mean that c_3 has a factor that is 2, which can distribute >to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of >2 by symmetry. >But you see, you only have just that one 2. >So c_1 and c_2 while not algebraic integers cannot be written as a >ratio of non-unit coprime algebraic integers either. Well as you say clearly there is a factorization of the form you want where c_1, c_2 and c_3 are algebraic NUMBERS. And every algebraic number can be written as the ratio of an algebraic integer and an ordinary integer, ie, c_1 = A/q for some a.i. A and integer q. And there is a great big honker of a theorem which says any two nonzero algebraic integers have a GCD which is an a.i. also. Therefore cancelling out the GCD of A and q, you can write c_1 = A'/q', where A' and q' are coprime a.i.'s. So you're wrong. >These numbers represent an unexplored frontier of mathematics, and an >opportunity for many of you to make an impact like you never could >have dreamed of before in the field. >It's fresh ground. Fresh manure, more like. B.P. >James Harris >http://mathforprofit.blogspot.com / === Subject: Squares that end with four identical digits by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2woZ22908; Which integers can have squares that end with four identical digits? === Subject: Re: please tell me primes100-200 as fast as possible by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2wwc22953; > === Subject: Re: Who contributed most to mathematics? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2weo22836; >>I was just trying to see who in fact borned the most mathematicians. >>My SWAG list was not intended to be complete but just a start. Ok, I >>blew it after France at least according to MacTutor link below. >>However I was close with Italy, Germany and former Russia but totally >>underestimated England and the U.S. and way over estimated India and >>Greece. This is a quantative analysis. The next step is qualitative >>analysis which is why I asked you guys the question. >I doubt that anybody would claim that being in the MacTutor list >is all that significant as an indication of mathematical importance: >it's mainly a question of which people somebody has bothered to write >a biography of. Is this generally true, ie., of other disciplines - physics,medicine, literature etc? What other list would do it. E.T Bell's Men of Mathematics, Morris Kline's Mathematical Thought from Ancient to Modern Times etc, Biography Top 100 Men of the Millennium? As I recall Guttenburg got top honors and Newton second. Guttenburg got it because his printing press got the word out to the masses. I question Newton as second. Had Fermat published he may have got the calculus in his resume. >There sure have been a lot of important French mathematicians; >Fourier, Galois, d'Alembert... >>Fermat - but his last theorem was published after his death. >>So during his life time it did not play a major role in mathematical >>developement. >Fermat's importance in mathematics does not rest on his Last Theorem. >He made many contributions in various areas. Although he didn't publish >much, he corresponded with many important mathematicians, so he was >quite influential in his lifetime. Did these other Mathematicians publish much? I doubt it. >> The same can be said of Da Vinci and other Greats whose >>certain works were suppressed during their lifetime. >Although da Vinci did study a lot of mathematics, I'm not aware of >any really significant contributions he made to mathematics. No because he did not publish. >but the Greeks started it. >>Just about every thing the Greeks did was rediscovered by others >>as it was needed in their time. Much of Archemedies work was for War >>efforts as they were needed. Sure they have being first name >>recognition but played no more role than India and other countries >>to thought today. >This is just silliness. The main contribution of the Greeks was >to discover the idea of mathematical proof. Nearly all Western >mathematicians until very recent times got their first taste of You really believe the infinity of primes by contradiction necessitated Euclid? Or the pythagorean theorem? How many proofs are out there now. >mathematics from studying Euclid. Come on. Euclid did not invent deductive reasoning - the syllogism - A pair of warranties and a consecutant, the warranties stating certain things and the consecutant following of necessity from them Euclid was a compiler,structurer,sequencer,denotator and doder on that which comes next cannot exist a priori. Because of this, Euclid grasped by the masses who passed it on years later by others through the best recording device known at the time - human memory. Certainly, methods of higher order such as analysis today were know but made no headway untill much later because it was not passed on theough human memory. Indeed, when the great knower died so did the idea. Eg., skiping several hundred years, the proof that this margin is to narrow to contain. Rediscovery is not a silly issue. How many re-discoveries have you you made? I would guess many. As youths (16) we played with the 5 cent 25 hole pinball machines in the pool room. These paid off 5 cents per game that you won. When you decided to cash in your games which of course was a rare event, we called the clerk Gillie,who had bad eyesight, to come over and verify the counter,flip the reset switch and pay us the money which of course went right back into the machine. When we won, Gillie would come over, squint, and look at the counter. Then he would tell us to flip the switch. In doing so we noticed that the switch could be toggled back and forth quickly thus stoping the won game count down. So if we had 100 games, we could quickly flip the switch on and off to stop the counter at 99. Then 98 etc. Here in came the problem. If we could trick Gillie to cashing in 100 game,99 games,98 etc., How much money could we win? That evening I sat with pencil and paper and figured it out by examining a smaller set of numbers and generalizied it to 100(100+1)/2*.05 = $247.5. This is the summation formula for an arithmetic progression that I rediscovered when the need arised. Around the same time I discovered if the sum pf the digits of a number is divisible by 3 the number is divisible by 3. Similarly I found the rule for 9 and 11. 7 and other numbers defied me but I tried. Before that I was looking at some numbers and wondered how many ways you can arrange them. Numbers 1,2 = (1,2),(2,1) 2 ways 1,2,3 = (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) = 6 ways 1,2,3,4 = blah blah = 24 ways 1,2,3,4,5, blah,blah yeah I took it out to this level = 120 ways Lo and behold, the numbers 2,6,24,120 have a pattern! Yes, 2 = 1*2 6 = 2*3 24 = 4*6 120= 5*24 so 120 = 5*24= 5*4*6 = 5*4*3*2 = 5*4*3*2*1. That's it! The number of arrangements of n things is n*(n-1)*(n-2)*...1) Before that, fooling around with differences between two squares I came up with x^2-y^2 = (x-y)(x+y) Later in the 60's fooling around with FLT, I pounced upon this. I a number can be expressd as the difference between 2 squares in more than 1 way the number cannot be prime. Indeed, every odd number can expressd as the difference between 2 squares in a trivial way. Eg., 17 = 9^2-8^2, 9 = 5^2-4^2, 2k+1 = (k+1)^2-k^2 are trivial solutions. Since x^2-y^2 = ab a>b > 1 it follows that from my earlier discovery (Dog gone it it happened again. I had the science channel on and the announcer said discovery as I typed it above) n = ab = x^2-y^2 = (x-y)(x+y) we have x+y = a as factor of n. This is a non-trivial difference between 2 squares. Thus n is not prime. Around the same time I realized empirically that for prime p x^(p-1)/2 +- y^(p-1)/2 is divisible by p. This is sufficient but not necessary p be prime since x=10,y=1, p=9 we have 10^4 - 1^4 = 9999 which is divisible by p=9. This can be expanded by my diff 2 squares idea to x^(p-1) - y^(p-1) is div by p if p is prime. Around the same time I found an infinity of solutions to the right triangle with hypotenuse of the form 4n+1 There is a lot more but no need to go on. I will summarize this as follows. Ask your self or a student a question and chances are you or the student will make a discovery or rediscovery with the answer. Necessity is the mother of invention. Find the indespensible man and fire him In fact we may have been better off with out Euclid because analogous to strict church domination, a great deal of Euclid's rigour stiffled creativity and forward thinking as the Pythagoreans did in their time. Or another view is WWIII would have already come and gone had creativity not been tamed by bigshots in power. Cino Observation lends to analysis and analysis lends to law === Subject: Re: Greek Alphebet by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAA2wgv22865; >>
> 
>  In my studies on triginometry, and calculus, I am comming across 
many
> symbols which appear to be the Greek Alphebet, as there was a chart 
of
> the Alphebet in the front of the book. I am wondering if someone 
could
>Do the Greeks use Latin and Greek letters in a similar way to us Latin
Within Greece , books , education etc. the Greek Alphabet
is used( for Geometry).
To mention also,that there are, some other letters that are in latin,but are 

not used in the Greek Alphabet any more like F (called Digama -double 
gama),and as far as I, remember the letter J (called
yiot.
It is worth pointing ,that (according to Britannica under ALPHABET)
the Alphabet was a Greek invention based upon North Semitic 
(proto-Canaanitish) writing which indicated only consonants,
a procedure suitable enough for a Semitic language but not for 
an Indo-European one.
Panagiotis Stefanides
http://www.stefanides.gr
===
Subject: Re: ellipse circumference approximation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAA3OXr25023;
For a proof for an exact equation for the circumference of an ellipse, check 

out my recently published book entitled: Circular Elliptics at: 
ww.Trafford.com    
===
Subject: Re: TI-89: wrong answer for integration!
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAAI8fq19654;
Be carful when you type the equation in.
You have to say a*x in the exponent else the calculator thinks ax is 
a single variable.
===
Subject: Re: Oneness of a number
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAAI8W719627;
>if(n is even) n = n/2;
>if(n is odd) n = 3*n + 1;
>Keep doing this until n = 1. The amount of necessary steps for this is
>called the oneness of n.
Not sure what you mean about 5 not going to 1
one(5) = 16
one(16) = 8
one(8) = 4
one(4) = 2
one(2) = 1
===
Subject: Number Of Squares Within A Square (nxn)
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAAI8T719617;
For my homework we had been set the question of finding the numbers of 
squares within a square.
I managed to work out an answer for myself but out of interest found your 
Website and a really neat solution that had been posted sometime ago.
However, I have some problems with the proof.
Please could you explain how to get the Sum(n^2)=(n+1)n(2n+1)/6
I'd be really grateful for your help.
Dave
===
Subject: Re: Marvellous developments of FLT
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAALClW01298;
    Hi to all,
   To the moment, once my procedures could be
comment with better english You should be
patient again. 
     I can present now tiny part with my
parameters for n=3:
              
   Consider Z and X as odd numbers and
             Z=(p+q)(r-s)
             X=(p-q)(r+s)  
where p;q;r;s natural numbers 
then once Z;X are of gcd=1 so p;q of gcd=1
                          and r;s of gcd=1
but also we'll take (p+q);(r-s) of gcd=1
                and (p-q);(r+s) of gcd=1
but then Z-X = 2(qr-ps) will be of gcd=1
      to Z+X = 2(pr-qs) except of 2
from these (qr-ps);(pr-qs) of gcd=1
and finally p;q;r;s of gcd=1
   Now ordering similar
       Z^3=(P+Q)(R-S)
       X^3=(P-Q)(R+S)
where P;Q;R;S can be expressed with
special formula by p;q;r;s parameters
 we'll eventually develop inconsistency...       
       
               Ro
===
Subject: Re: A 1st semester Calc proof that e^(i*t) = cis(t)
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAALChi01289;
Ok, the real log function can be defined as the definite integral of 1/x. 
The real exponential can then be defined as itsinverse. sin and cos can be 
defined as the parameterization of the unit circle in cartesian coordinates. 
 
e(z) for z complex is can then be defined as e(z+iy) = e(z)(cos(y) + 
isin(y). 
 Admittedly, this takes out all the wonder from Euler's formula, but its 
logically sound and I have seen it done in some texts.
I assume, then, that everything required can be derived from these 
defintions, which only require the integral but absolutely no series.
===
Subject: Re: Algorithm for square root?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAALRxx02332;
>
 <1998051218145200.OAA19142@ladder03.news.aol.com>...
>> Does anyone know of a site with the algorithm
>> for finding the square root of a positive number?
>http:
//www.worldserver.com/turk/ComputerGraphics/FixedSqrt.pdfis similar, but requires a bit more guesswork, as does long division.
>

===
Subject: two new math fora to announce
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAANMCA11419;
http://www.math.utexas.edu/mailman/listinfo/math-talk
-- started primarily for discussions at UT-Austin, but anyone can subscribe
http://forums.austin.craigslist.org/?forumID=1394
-- Craigslist easter egg site could be really cool if enough people get 
into it.  Similar niche to this site, maybe.  A different feel, though. 
Best wishes,
Joe
===
Subject: Godel's universe
I'am looking for literature concerning Godel's Rotating Universe.
Suggestions?
KB
===
Subject: two additional math fora to announce
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAANMBb11415;
Started this primarily for discussions at UT-Austin, but anyone can 
subscribe.
http://www.math.utexas.edu/mailman/listinfo/math-talk/
A Craigslist easteregg.  This could be really cool... check it out.
http://forums.austin.craigslist.org/?forumID=1394
===
Subject: Re: Why Derivative is Inverse to Integral; geometric explanation 
formath  education
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAB06uH15289;
>> just this Archimedes Plutonium meant when said of a slope of one of
>> rectangle sides. To simplify a trapezium to a rectangle is simpler
>> than to resume the lost information of the slope. So in most cases we
>> are unable to integrate. The same, we can easily expand a function
>> into Fourier series, but to reconstruct the original by the Fourier
>> series - it's practically unsolvable problem. Or, if a function has
>better than picketfence.
>Can you think of a term for a rectangle that has an end-sides of just a
>mere point rather than a true rectangle whose four sides are more than
>just a point?
>In the derivative there are no strange objects but in the integral there
>is this
>strange object of a rectangle whose end-sides are mere one point.
>So in differentiation, there appears to be no real strange objects for a
>set of trapezoids is normal geometric objects but in integration we have
>this strange set of objects of rectangles whose end sides are one point.
>And thus, I see geometrically the inverse relationship between derivative
>and integral as that between trapezoids to one-point-sided rectangles. 
When
>differentiating one makes trapezoids and when one integrates they collapse
>the trapezoids into these strange rectangles.
>Sergey, can you comment on the issue that derivative has normal geometric
>objects of trapezoids (what I call picketfences) yet the integral relies
>upon these
>strange abnormal geometric objects of a collapsed rectangle whose 
end-sides
>consist of one mere point.
>i think trapezoids are stupid!!!! 
>So can we say that the essence of the Inverse relation between
>derivative and
>integral is the geometric idea that one is to expand strange rectangles 
into
>trapezoids and the other is to collapse trapezoids into one-point end 
sided
>rectangles.
>P.S. I do not know if I should bother with a geometric explanation of the
>2nd derivative that is acceleration in physics.
>Archimedes Plutonium, a_plutonium@hotmail.com
>whole entire Universe is just one big atom where dots
>of the electron-dot-cloud are galaxies
===
Subject: Re: Squares that end with four identical digits
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAB2Ehe24376;
> Which integers can have squares that end with four identical digits?
I believe that all integers that end in two zeroes have squares that end in 

four identical digits.
===
Subject: Re: Trying again!!!  PDE book recommendation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAB2EhG24388;
Charlie,
If you're coming from the engineering perspective of PDE's, I'd recommend 
Partial Differential Equations for Scientists and Engineers by Stanley 
Farlow.
Eric Ritchson
>Hi all,
>Could someone recommend a good Introduction to P.D.E.'s book ?  I have 
heard
>that
>Basic Partial Differential equations by David Bleecker, George Csordas, 
and
>Darko Grundler book is nice, but I haven't had the chance to take a look 
at
>it.  Any suggestions would be appreciated!
>TIA
>Lurch
===
Subject: Re: {Group Theory}  Confusing group theory conundrum
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAB2Ehv24380;
>questions.
>OK, every group has an identity element, right?  True by definition
>It seems to me that, given a group, that group therefor determines a
>unique identity element.  (The uniqueness of a given group's identity
>element is easily proved)
>Therefor, it seems to me that there exists a function (indeed, a
>function which algebraists subtly draw upon without even realizing it)
>that associates with each group it's identity element.
That really depends on how you interpret the word function. This is not 
a function in the normal set theoretic sense.
>For instance, an algebraist says:  We have a group G.  Then it must
>have the identity element G_e.  But by saying this, she has
>implicitly used such a function.
Not really. For instance, consider this analog: We have a set S which is 
nonempty. Then it must have some member S_x. This is true by definition, 
but this does not imply the existence of a function mapping sets to their 
elements. In fact, even if you restrict your domain so that it is well 
defined, you still can't prove the existence of this function (in the 
general 
case) without invoking the axiom of choice.
>But I am told that you cannot have a set that contains every group... 
>so this function's domain is nonsense!  In fact the whole function
>is nonsense!
Not nonesense, just not a function in the usual set-theoretic sense.
>Now before you reply that the identity element isn't unique because a
>set can be a group under many diverse different operations, I already
>am taking this into consideration.  When the algebraist says Then it
>must have the identity element G_e, the operation is either obvious
>from context or else has already been specified.  So really the
>elusive function I am talking about is one which takes as input a
>group-operation couplet and then outputs the identity.  But this is
>all a lot more opaque, and though necessary for rigour, not necessary
>for understanding.
>So it seems very much like such a function exists, and yet--  it
>cannot.  Is it some hyperfunction that transcends set theory and is
>taught at the postgraduate level?  I am very confused and very much
>eager for you to shed some of your very, very highly esteemed wisdom
>on the subject.
>Sniz Pilbor
Cron
===
Subject: Re: How do you prove that the circumference of a circle is 
proportional to it's radius?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAB76AB12633;
> 
> How do you prove that the circumference of a circle is proportional to 
> it's radius?
> What's the modern version and how did the greeks prove it?
> 
> The approximations to that ratio are based on 
> inscribing/circumscribing regular polygons, each of which may be 
> partitioned into congruent isosceles triangles.
> 
> For any such polygon, the odd(circuferential) side of any of those 
> isosceles triangles is proportional to the other (radial) sides by 
> similar triangles.
> 
> Thus each approximate circumference is proportional to its radius.
>>Now if you know/accept that 
>>perimeter of inscribed polygon 
>>           < circumference of circle 
>>                       < perimeter of circumscribed polygon, 
>>then it's easy to complete the proof. Now the first inequality, 
>>    perimeter of inscribed polygon < circumference of circle, 
>>is clear; a straight line is the shortest distance between two points. 
>>The other inequality, 
>>    circumference of circle < perimeter of circumscribed polygon, 
>>is certainly plausible, but is it actually possible to prove it 
>>using only tools available to the ancients?
>Well, I don't know what Euclid's _definition_ of the circumference
>of a circle was, but if he had one I can't imagine that it was not
>something essentially equivalent to what we would call the least
>upper bound of the perimeter of an inscribed polygon. Assuming
>that, then first we need to show that 
>(*) perimeter of inscribed polygon 
>      < perimeter of circumscribed polygon;
>that shows that the least upper bound mentioned above
>_exists_, and also makes the other inequlalities clear.
>(Regardless of how much of this they were explicit about,
>I do tend to suspect that they would have said the question
>was the same as (*), for some reason.)
>The circumcribed polygon is the union of triangles with height
>equal to the radius of the circle and base one of the sides of
>the polygon, so the _area_ of the circumcribed polygon is
>equal to r times its circumference.
>Otoh, let r' < r. Note that adding points to the inscribed
>polygon increases the circumference, by the triangle inequality.
>So starting with an inscribed polygon we can find another
>with larger circumference, and with area > r' * perimeter.
>Since the relation between the areas is clear, it follows
>that r' * inscribed perimeter < r * circumscribed perimeter
>for all r' < r, hence (*).
>>Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
>
Euclid did not discuss the length of curved lines, or the area
of curved surfaces, apparently because he had no satisfactory definition of 

such things. I believe Archimedes was the first.
He proposed several postulates, from which he could conclude
that any convex curve is longer than an inscribed polygon,
and shorter than a circumscribing polygon. Effectively, he had a
definition for the length of convex curves, and similarly for
convex surfaces. To these he applied the method of exhaustion.
Bill Kleinhans
===
Subject: Re: How do you prove that the circumference of a circle is 
proportional to it's radius?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hABKvka05954;
>> How do you prove that the circumference of a circle is proportional to 
>> it's radius?
Please refer to:
http://mathforum.org/discuss/sci.math/a/m/107260/107288
and
http://mathforum.org/discuss/sci.math/a/m/107260/107289
Panagiotis Stefanides
http://www.stefanides.gr
===
Subject: Re: Basic factorization ideas
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hABNIYZ16695;
>If you saw
>(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                 
>                          49(x^3 + 5x^2 + 3x + 1)
>with the c's algebraic integers, I think few of you would have a
>problem realizing that only two of the c's have 7 as a factor.
  Yep, this polynomial can be factored as you say
and yep again, if it is factored that way then c_1 and
c_2 are divisible by 7.  No prob with that.
>But, of course, you're looking at *functions* of x, as you have 
>f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
>so I could also write it as
>(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
>Notice that dividing both sides by 49 gives 
>(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
>as long as you're in a ring where 7 is not a factor of 22.
  What???  Where'd the 22 come from ???
>I want to emphasize that point as notice there's only *one* way to
>divide through by 49 if 7 is not a factor of 22.
  You've flipped out!  There is no 22 in sight in what 
you said above.
>Usually you can *see* the other factors of 7, but I want you to
>abstract, and generalize.
>Please pay careful attention to that example.
 I'm a-tryin', a-generalizin', and a-abstractin'.  I'm a-payin'
careful attention!  But I still don't *see* no 22 !
>You may see people who reply claiming that the word polynomial has
>some significance, as if it's a mystical thing which refutes basic
>logic, so if something isn't polynomial it no longer behaves
>logically.
  Yep, the logic is a problem here.  Like, where do you get
22 out of the stuff at the top?
>Now then, in my advanced factorization work, I just use functions of x
>that are a lot more complicated than f_1(x) = c_1 x, and unfortunately
>there are people who can use an unfamiliar leap in complexity to
>confuse others.
  You mean us, or you mean you?
>Some of you have learned various advanced math topics, now imagine if
>in your classrooms there were some hecklers who continually hollered
>out at your teacher, or otherwise disrupted the class?
  If he spouted wrong math for months or years on end he would 
deserve it.
>What if when there were difficult concepts those hecklers would try to
>confuse everyone as they sought to discredit the mathematics?
>If you find that hard to imagine, imagine me in your class with you
>questioning the professor and calling him names.
>How much would you have learned?
you out of the class to learn anything at all.
B.P.
>I need those of you interested in mathematics to focus on the basics,
>so that you can understand the advanced.
>James Harris
>http://mathforprofit.blogspot.com
/
===
Subject: Re: out-of-kilter algorithm
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAC2Vns31064;
Try http://citeseer.nj.nec.com
>Hello!
>I was just wondering if there is someone who could help me in finding some
>usefull site with explanation of out-of-kilter algorithm and examples of
>problems solved with this algorithm.
>Mimmy
===
Subject: Re: differentiable...problem...
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAC2VoM31077;
>if f is differentialbe on (0, infinite)
>and lim [f(x) +f'(x)] = L  (x->infinite)
>show that lim f(x) = L (x->infinite) and lim f'(x) = 0  (x->infinite)
This seems incorrect as stated. What about f(x)=e^(-x)? It's differentaible, 

and:
f(x)+f'(x)=e^(-x)+(-e^(-x)) = 0
So:
lim f(x)+f'(x) = 0 (x -> inf)
>------------------------------------
>it seems to be trivial. but i no touch.....oops........
>help.....me......please......
===
Subject: Re: Newsgroup survey: Math and personality assessment
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAC2VoS31086;
Hey James, you forgot a question.
8. Is JSH a f&$#wit that gives eveyone else the s#its?
Your mathematical legacy will be a footnote recognising you as a prize
crackpot, you complete and utter social misfit.
It'd be great if the first (and last) reply to any of your future
posts was always the same, namely,
JSH is crackpot. Please do not reply to his posts as it just
encourages him.
It'd be fascinating to see what would happen to you in the fullness of
time if this was the only response you got. I predict descent from
your current lofty height of idiocy into complete insanity.
Do everyone a favour and go play in the traffic.
>It seems to me that there have been debates over math concepts I
>thought basic, so here's a quick survey:
>1.  Before I mentioned it, had you ever heard of the distributive
>property?
>2.  In your experience, is math quirky?
>3.  Do you think that mathematics is an extremely difficult discipline
>that only experts are really good at handling?
>4.  What is the distributive property?
>5.  Is a math proof perfect?
>6.  Do you consider yourself to be a reasonable person?
>7.  If a mathematical argument is explained to you in detail, using
>basic algebra, if it's correct, would you admit that, even to a
>hostile crowd, like even if many posters on sci.math would call you
>names and insult you for admitting it?
>James Harris
>http://mathforprofit.blogspot.com
/
===
Subject: Re: Stroboscopic effect
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAC3VSW02735;
>>Is it possible to have a stroboscopic effect (where wheel spokes
>>appear to be rotating backwards) in real life, rather than in
>>movies where it is common?
>> Yes, easily.  Try spinning something under a fluorescent light.
>> Are you too young to remember record turntables with stroboscopic
>> speed indicators?
>Not at all. (But I did always wonder why they even bothered put the
>indicator for 78rpm on them!)
>>But on an open road, under a clear sun, is it possible for your
>>eyes to create this effect?
>> You need a varying source of light; seeing one spoked wheel
>> through another might do it.
>But under sunlight, no interference, etc? I'm busy convincing my 9th
>grade students of their inability to differentiate between reality
>memories and fantasy (TV) memories. Several of them swear that
>they've seen it on cars passing them on highways, on open roads,
>in broad daylight.
>--riverman
If your eyes are fixed on the road instead of the car, the bottom
spokes will be visible since their velocity relative to the road
is low.  Fast eye movements can produce momentary stoboscopic
effects.
The image compression method that the retina uses is a mystery,
but it is highly efficient.  It seems to be very good at preventing
artifacts of a stroboscopic nature. There are color artifacts though.
The color effects of the Benham disk probably have something to do
with the way color is encoded in the optic nerve.
http://faculty.washington.edu/chudler/benham.html
===
Subject: Re: JSH: Deprogramming needed?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id hAC3VS702743;
>(Marketing ploy warning)
snip
What marketing book are you reading?  Go get your money back.
In a nutshell, marketing is about getting some sucker to buy your junk 
rather than somone else's crap.  However, if you start your spiel by saying, 

Listen, chumps!  I am about to take you for a ride, then you have lost 
the battle.  Think subtle.  Don't start your pitch with the warning.  It 
defeats the purpose.  
Unless, of course, your purpose to is generate scorn.  But, you don't need 
the marketing book for that.  You are a huge loser who does not need to 
market that fact, as it is obvious to all except John Correy.
I like the suggestion offered in another response (you distribute your 
pollution all over sci.math so I cannot recall the subject line off hand) -- 

go play in traffic.
Jay
===
Subject: Re: Cardinality of 2^n numbers?
>> However, 2^K is the set of all natural numbers. And the set of all
>> natural numbers is of cardinality N_0. However, K is also of
>> cardinality N_0. If the power set of any set is strictly larger (IE
>> has a greater cardinality) then how is it possible? This would require
>> that N_0 > N_0, which is the contradiction I'm talking about.
>> No.  2^K is the power set of K, which is the set of all subsets of K.
>> For example, one such subset is K_p = { 2^p : p is prime }.  This is a
>> set, not a number, and therefore 2^K is not a set of numbers and
>> certainly is not equal to N_0.
> I never said 2^K WAS the set of all natural numbers. I said 2^K was
> BIJECTIVE with the set of all natural numbers.
Wrong on two counts.
        1.  You did say 2^K is the set of all natural numbers.  The 
quote is
            plainly visible above.
        2.  It is not true that 2^K is bijective with the set of all 
natural
            numbers.   Cantor's theorem shows that.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.

===
Subject: Re: Cardinality of 2^n numbers?
> You're misusing cantor's theorum. I never said that K wasn't greater
> than K. I said N was equal in size to the power set of K. Cantor's
> theorum does not involve this kidn of thing.
You're talking nonsense.  How is K greater than K?
> The power set of K is related to the natural numbers if you do this:
> For each element in the power set of K, add all the numbers together.
> You will always get a unique natural number.
Try the example I gave before.  Let K_p = { 2^p : p is prime }.
Then K_p is a subset of K.  What is the sum of all the members of K_p?
Which natural number is that?
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.

===
Subject: Re: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
<< http://mathforprofit.blogspot.com/
What If?   That question on its face causes one to
consider the state of mind of the speaker...  After
years of people specifically detailing your mental
constructs and pointing out your errors in logic you
still have a doubt of your pass or fail status in life....
Paul R. Mays
----------------------------------------------------------------------------

-
Some where within the Quantum State
Http://Paul.Mays.Com/story.html
http://paul.mays.com/mayday.html
http://paul.mays.com/rainy.html
Of all the systems of religion that ever were
 invented, there is no  more derogatory to the
Almighty, more unedifying to man, more
 repugnant to reason, and more contradictory
 to itself than this thing  called Christianity.
Too absurd for belief, too impossible to  convince,
 and too inconsistent for practice, renders the
 heart  torpid or produces only atheists or fanatics.
 As an engine of power,  it serves the purpose of
despotism, and as a means of wealth, the  avarice
of priests, but so far as respects the good of man in
general  it leads to nothing here or hereafter.
                          [Thomas Paine, The Age of  Reason]
===
Subject: Re: question about canonical commutation relations [P,Q]=i*hbar*I
> John  grava .88 la saucisse et au marteau:
>>  Find matrices of infinite size Q and P such that
> What is a matrice of infinite size? IMO, this is a complete nonsense
> since results on matrices specifically use that we are in finite
> dimension.
Given a finite-dimensional vector space V over a field F  and a 
basis B, matrices correspond in an easy way to linear maps V -> V.
Now suppose that V is of infinite dimension.  Then linear maps 
V -> V correspond in the same way to functions from BxB -> F.
Since B is a basis, these functions have a finiteness property
that allows us to take inner products of rows and columns ... 
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
We get more random numbers (from your post):
    ed6af0c92fa24f0810455fd94f4ccdaa
David Bernier
===
Subject: Re: Sum [n in Set] (1/n)
>>I recall seeing a theorem saying
>>    Sum [n in Set] (1/n) converges iff (some criterion on Set)
>> There is also a conjecture of Erdos which I think states that
>> the sum diverges iff  Set  contains arbitrarily long arithmetic
>> progressions.
That should be  =>  , not  iff . Obviously we could have 
arbitrarily
long arithmetic progressions and still have convergence, e.g. if
Set = { 10^k + m  , 0 < m < k }. I have a notes saying Erdos offered
$3000 for a proof of  => . You can still collect.
>In the future, you can post or mail; doing both is unnecessary.
Maybe, but one of these days I'll learn that what IS necessary is to
remember to flag my email so that it doesn't look like it was (also) a
newsgroup post ...
dave
===
Subject: Re: ellipse circumference approximation
> For a proof for an exact equation for the circumference of an ellipse,
> check out my recently published book entitled: Circular Elliptics at:
> ww.Trafford.com
Hmm. The title of this thread mentions approximation. But it seems that
you are claiming to have an _exact_ expression for the perimeter. Of
course, such expressions are well known, but they involve an elliptic
integral (or something equivalent). The perimeter cannot be expressed in
closed form in terms of elementary functions.
Are you, perchance, claiming to have done what I just said can't be done?
If not, then briefly, what's interesting about your work?
BTW, for anyone interested, see
.
It says that the book Relates the ellipse to the circle in ways you never
dreamed of, offers a proof for an equation for the circumference...
Price: US$9.80
David Cantrell
===
Subject: Re: JSH: Deprogramming needed?
>> (Marketing ploy warning)
>> What if indeed I *am* wrong, and I don't have these great math
>> math discoveries?
>Then you would be empirically sane.  No chance (note your inserted
>if despite an extraordinarily large body of trivial disproof
>submitted by more than a score of posters),
>http://www.crank.net/harris.html
> It's not every braying jackass that gets a whole page at crank.net
Ah, but wouldn't he be likely to take that as evidence that they are
paying great attention to his ideas because of the ideas' great merit
(rather than great crankiness)?  How does one get the considered attention
of a megalomaniac anyway?
-- 
---------------------------
|  BBB                b         Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |   
|  B  B  a  a   r     b  b  |   
|  BBB    aa a  r     bbb   |   
-----------------------------
===
Subject: Re: Playing Doctor
> 
> 
> So,...... you wanna play doctor with me now, after you have failed 
> miserably in politics, sociology, psychology, philosophy & economics?
> ..........AHAHAH.....ahahahaha...... Babe, listen, you are not very 
> successful, except when it comes to nagging and whining
> ....about YOUR own problems.......ahahahahaha....
> ahahahah....ahahahahah...ahahahaha.....ahahahanson
> 
> Having a woman volunteer to play doctor is bad?  Howzzat????
> Ilsa: SS Doctor
Depending on your tastes, could at least make a nice fantasy. ;-)
===
Subject: Re: A 1st semester Calc proof that e^(i*t) = cis(t)
> Ok, the real log function can be defined as the definite integral of 1/x. 

The 
> real exponential can then be defined as itsinverse. sin and cos can be 
> defined as the parameterization of the unit circle in cartesian 
coordinates.
So how exactly do you define sin and cos this way?
===
Subject: new proof for the fundamental theorem of calculus
If you like geometry you'll like this proof of the FTC.
You can see it at:
 www.precalculus.netfirms.com/#3
TR
===
Subject: Re: Marketing shift, core issues
> I've been thinking about my problems with getting any kind of
> admission that my math arguments showing the core error in mathematics
> are correct, so I've gone to marketing books.
> I just wanted to warn readers that I may be employing various tactics
> from modern research on human psychology to see if I can't break
> through the logjam.
Psychology won't make your arguments correct.  It also isn't likely to 
make anyone forget that they are wrong.
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: please tell me primes100-200 as fast as possible
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
173, 179, 181, 191, 193, 197, 199
>>
Can you please tell me the prime numbers between 100-200 as soon as
>>

===
Subject: Re: differentiable...problem...
>> if f is differentialbe on (0, infinite)
>> and lim [f(x) +f'(x)] = L  (x->infinite)
>> show that lim f(x) = L (x->infinite) and lim f'(x) = 0  (x->infinite)
> This seems incorrect as stated. What about f(x)=e^(-x)? It's 
differentaible,
> and     f(x)+f'(x) = e^(-x)+(-e^(-x)) = 0
> So: lim f(x)+f'(x) = 0 (x -> inf)
No. If  L = 0  it correctly implies  lim f = lim f' = 0.
In fact the statement also holds true for infinite L
if one omits and  lim f' = 0,  e.g. see
Martin D. Landau; William R. Jones.  
A Hardy Old Problem 
Mathematics Magazine, Vol. 56, No. 4 (Sep., 1983) , pp. 230-232.
http://links.jstor.org/sici?sici=0025-570X(198309)56:4%3C230%3E
-Bill Dubuque
===
Subject: Re: Basic factorization ideas
> If you saw
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 1)
> with the c's algebraic integers, I think few of you would have a
> problem realizing that only two of the c's have 7 as a factor.
If c_1 and c_2 (and hence c_3) are ordinary integers, then
yes, c_1 and c_2 must be divisible by 7, and c_3 must be
coprime to 7 (does this hold for algebraic integers as well?)
but the proof of this depends strongly on the fact that the LHS
is composed of polynomial factors. 
> But, of course, you're looking at *functions* of x, as you have 
> f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
> so I could also write it as
> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
Yes, but this is dangerous.  Someone who was not paying close
attention might assume that the f_i were arbitrary functions
and get the idea that things that are true of polynomials
are also true of arbitrary functions.
> Notice that dividing both sides by 49 gives 
> (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
> as long as you're in a ring where 7 is not a factor of 22.
> I want to emphasize that point as notice there's only *one* way to
> divide through by 49 if 7 is not a factor of 22.
Yes, if the f_i are defined as above.  If the f_i are arbitary
functions this statement is wrong.
                           - William Hughes
===
Subject: Re: partial deriviatives!
[shnip]
[shnip]
>>= ((p^2F/pq^2)(dq/dt) + (p^2F/pqpt)) + (p^F/pt^2)(dt/dq) + (p^2F/ptpq) 
whoops, not that it really matters, but this should be:
( ( (p^2F/pq^2)(dq/dt) + (p^2F/pqpt)) 
+ (p^F/pt^2)(dt/dq) + (p^2F/ptpq) 
+ (pF/pq) (p^2q/pt^2)(dt/dq) )
adam
===
Subject: Re: Marketing shift, core issues
> I've been thinking about my problems with getting any kind of
> admission that my math arguments showing the core error in mathematics
> are correct, so I've gone to marketing books.
Most of the good marketing books will tell you that marketing works 
best when you have a salable product.
You ought to try it!
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
> After all, I've been at this since April 1995 having spent a lot of
> time and effort, with literally thousands of posts along with all
> kinds of other activities, websites, and email to mathematicians all
> over the world.
> But, what if I'm wrong?
That just could be why you are having so much trouble selling your 
product.
===
Subject: Re: Deprogramming needed?
> If everybody else could emulate
> Dik Winter and confine their criticisms to the math, this would all
> go away very quickly.
Non-mathematical criticisms of JSH are responses to non-mathematical 
criticisms by JSH. If he wishes to avoid receiving them he should 
avoid issuing them.
===
Subject: Re: Cardinality of 2^n numbers?
> However, I don't see why there can't be a non-N_0 cardinality of K
> such that when its power set is taken, its power set is of cardinality
> N_0. That will also have to be explained to me.
> (...Starblade Riven Darksquall...)
The power set of any finite set has finite cardinality, and the 
power set of any infinite set has cardinality greater than the 
smallest possible infinite cardinal.
If N_0 represents smallest non-finite cardinality, what is K?
===
Subject: Re: Cardinality of 2^n numbers?
> I never said 2^K WAS the set of all natural numbers. I said 2^K was
> BIJECTIVE with the set of all natural numbers.
> (...Starblade Riven Darksquall...)
For K any infinite subset of the naturals,
if 2^K represents {2^k : k in K} you get one cardinality, but
if 2^K represents the set of all functions from {0,1} to K, 
you get an entirely different cardinality.
In the first case, you can do such a bijection, but in the second 
case not.
===
> I have been biting my tongue about the IQ test but I can't any more.
> 
> How reliable is a test that use the term Asian to represent the 
most
> diverse of ethnic and cultural groups?
[.....]
> rather than face her peers ...
[....]
State your peers.
===
Subject: Re: Playing Doctor
> So,...... you wanna play doctor with me now, after you have 
failed
> miserably in politics, sociology, psychology, philosophy & 
economics?
> ..........AHAHAH.....ahahahaha...... Babe, listen, you are not very
> successful, except when it comes to nagging and whining
> ....about YOUR own problems.......ahahahahaha....
> ahahahah....ahahahahah...ahahahaha.....ahahahanson
>  > Having a woman volunteer to play doctor is bad?  Howzzat????
> Ilsa: SS Doctor
> Depending on your tastes, could at least make a nice fantasy. ;-)
Taste? nice?, my gahhhhd, Ed. He refers to the ultimate perversion,
worse than cannibalism. He talks about the Koch broad ( I think
no (acknowledged) relation to NYC Mayor Ed Koch,) who allegedly
made lamp shades from the skin of KZ  inmates.
hanson
===
Subject: Re: Divisibility Problem
<1SYrb.19220$Rah1.18334@twister01.bloor.is.net.cable.rogers.com>,
> How'd you get that? Brute force method?
How'd I get what? Oh, that, down there - please don't toppost, I'm 
easily confused. 
Anyway. 
You wanted an odd prime p congruent to 3 (mod 4). Well, p = 3 seemed 
like a good choice. 
Now there's a book that gives factorizations of b^n plus-or-minus 1 
for various small b and lots of n. I went to the 3^n + 1 page and 
read down until I found one that worked. 
> ,
> Suppose p is an odd prime congruent to 3 (mod 4). If r is any 
positive
> odd
> number, and 2pr+1 is prime, then is
>  > 2pr+1 | (p^r) + 1
>  > Ever true?
> p = 3, r = 11.
> --
> Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Oneness of a number
===
>Subject: Re: Oneness of a number
>>if(n is even) n = n/2;
>>if(n is odd) n = 3*n + 1;
>>Keep doing this until n = 1. The amount of necessary steps for this is
>>called the oneness of n.
>Not sure what you mean about 5 not going to 1
>one(5) = 16
>one(16) = 8
>one(8) = 4
>one(4) = 2
>one(2) = 1
What he meant is that 5 has a stopping time (S(n)=number of iterations to 
reach
1) of 5.
So S(5)=5, S(S(5))=5, S(S(S(5)))=5,  etc. which loops forever.
Compare to 
S(16)=4
S(S(16))=2
S(S(S(16)))=1
--
Mensanator
Ace of Clubs
===
Subject: Re: question about canonical commutation relations [P,Q]=i*hbar*I
I think the question from the OP really boils down to, why the proof in the
finite case does not generalize to the infinite case.
I mean, the OP has proved that no finite matrices satisfy the given
commutation relation. Then he demonstrates the existence of infinite
matrices that *do* satisfy the relation. So the question is, why is the
proof not valid in the infinite case?
At least, I'm curious about that too, because it seems that although matrix
product is well-defined, taking the trace is not. Please enlighten me.
-Michael.
===
Subject: Re: Mathematical Integrity
===
>Subject: Mathematical Integrity
>I am a math student and I love math more than anything else. I
>started to love math when I found out that I wasn't good at it and
>started studying it for over a year now.  I love math for its purity
>and integrity.  However, lately, I surfed through several math forums and 
I
>found many unpleasant posts.  Sometimes a student would post an innocent
>question but he or she would get scold for posting such a stupid 
question
>and would be called a moron.  And sometimes a math fanatic would post
>something of interest, but other mathematicians would scold that person 
for
>posting something so ridiculous. It seems like these dese days, some
>mathematicians would use profanity against others mathematicians. Why? 
Here's a clue: some of them are cranks.
>Where
>is the purity and integrity in mathematics? I know that after posting 
this,
>many other great mathematicians would scold me, but that's what I've seen 
so
>far. Anyhow, I still love math by heart and I would like to give my 
respect
>to many great math fanatics in this forum.
--
Mensanator
Ace of Clubs
===
Subject: Re: JSH: Deprogramming needed?
you've already got your market-man,
Devlin.  you'll just have to decide with him,
whether you are an L- or R-winger ... I guess.  first, though,
ask him if he can prove the pyhtagorean theorem, and
generate the triples; perhaps,
he can prove that it is Devlin-unsolvable.  thank you!
> (Marketing ploy warning)
 
> I don't want any replies of people offering, but if you might consider
> it, I just want you to think about it.  Sure I've been reading this
> marketing book, but it seems to me that still *someone* out there
> might be the right person, as of course, I don't believe I'm wrong. 
 
> http://mathforprofit.blogspot.com/
--ils duces d'Enron!
===
Subject: Re: JSH: $100,000 US offer, Abel Prize
===
>Subject: Re: JSH: $100,000 US offer, Abel Prize
>> One thing I've been fascinated by as I've considered replies to my
>> posts is a loose group coordination between posters, as some try to
>> post with math, and others just post various jibes, but all keep
>> focused on pushing the false notion that my rather basic argument
>> showing a problem with the definition of algebraic integers is wrong.
>> Now I've tried to use money before, but failed as I think people in
>> the math community understand that NONE of you can overcome that kind
>> of coordinated effort, and even if you post your own version of my own
>> argument, the rest of math society will shred you.
>> However, the shortness and simplicity of the proof of the problem in
>> core offers another solution--machine proof checking.
>> That means I can make the offer of sharing at least $100,000 US with
>> ONE person or one group that collaborates to produce the check by
>> machine from the Abel Prize, assuming, of course that I win it.
>This offer is rescinded.  
Damn!
>That is, I have changed my mind and am no
>longer offering to give $100,000 US or promise of any sum of money to
>anyone for helping me with my current research, including doing a
>computer check, or any other help whatsoever.
>I repeat there is no monetary offer on the table for helping me with
>recgnition for my work.
>James Harris
>http://mathforprofit.blogspot.com/
--
Mensanator
Ace of Clubs
===
Subject: Re: if x has normally distributed digits, does 1/x?
>Since most numbers are normal, I think that the best approach is to
>start with an abnormal irrational number. Its reciprocal will most
>likely be normal and thus be the example that you are looking for.
The example that you are looking for is not just a normal number,
it's a number that you can prove is normal.
>My guess is that if you clear to zero every other digit of Pi and
>then take the reciprocal of the result, you will get a normal number.
It's easy to make guesses.  Yes, you have a very good chance of being 
correct.  Proving you're correct is the hard part.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Squares that end with four identical digits
>> Which integers can have squares that end with four identical digits?
>I believe that all integers that end in two zeroes have squares that end
>in four identical digits.
Yes, of course.  Not so obvious is that these are all the answers.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
===
IQ is a metric, and like all metrics, it's has no concrete meaning
other than meaning given to it by those who perceive this metric.
I could use any metric to measure intelligence. For example, I could
measure intelligence with the meter. I could say taller people are
smarter than shorter people. And why couldn't this be true? Maybe
taller people are perceived to be more beautiful than shorter people,
so taller people would earn more wealth and their descendents afforded
a more educated future than otherwise. So in time my meaningless
metric becomes scientifically provable and then what will you say?
It's all a lie
> I have been biting my tongue about the IQ test but I can't any more. 
> How reliable is a test that use the term Asian to represent the most
> diverse of ethnic and cultural groups?
> If you want to know more about the diversity, ask me and I will give
> you tons of specific examples. Here is one: An example of cultural
> differences (that affects their educational goals) of the two groups -
> ethnically the same and speak the same EXACT langauge - in Burma who
> are the descendants of the people who came to Burma from a town known
> as Surat, India during the British colonial days.
>  Now, do not assume that they are racially the same as the majority of
> northern Indian population, i.e Indo-Aryan. What I am aware of is that
> they are descendants of those who came from Central Asia (to india)
> and may be MIXED with the locals of India (I haven't checked into
> that) like the majority of the current day Pakistanis.
> See the complexity yet? 
> Add to that the Afghans, particularly the Pathan, which also form a
> comunity in Burma.
> And then you have all these different native groups in Burma, namely
> Burman, Mon, Shan, Karen, Kachin, Chin, etc. where the Karen with a
> good size Christian population now claiming that they are one of the
> lost tribes.
>  The Mon are the first group who adopted Bhuddism which made them the
> most literate group *in the old old days* through the monastry
> education. Shans are the stauch Bhuddists too.  I read somewhere on
> the Internet that Mon originated from ancient India. They look like
> typical oriental though some have darker skin while others are very
> light-skinned) instead of the dravidians of ancient Inida.
> Shan is ethincally similar to some groups in Thialand from the area
> that borders Burma.  Talking about Thailand, not all Thias are
> ethincally the same.
> Now, the Chinese would claim that the high score of IQ tests is
> because of them Chinese, i.e not all Asians are equal. And the
> Japanese thinks (may be not so much anymore) they are superior to all
> other Asians.
> Does that term *Asian* include the Arabs?  
> What about the people of Egypt (I avoid calling them Egyptians to
> differentiate from the ancient Egyptians) some of whom are Arabs, some
> of whom are descendants of anciant Egyptians, some are descendnats of
> a group called Berber (my spellig may be wrong).
> And how many Ethiopians look or act like a typical African. 
> Why are people using the racist IQ tests as God's given ruler to
> measure intlligence?
===
Subject: Re: Technical Name Of Equatorially Concentric Rings?
which moreover uses two sets of rings that are not different.
(I'd thought of it, two; I'd just call htem, bipolar.  now,
what are tripolar spatial co-ordinates ?-)
 
>> I'm not saying geodesics travel along these rings (anymore than 
they
>> travel along latitudinal rings).
>> Hmmm....maybe a picture will put it in perspective:
>> 
>>          http://math2.org/cgi-bin/mmb/server.pl? 
 
>       IE will make the image small, then you have to hover your
> cursor inside the lower right corner until the pillow appears. Click
> it to enlarge, reverse to shrink.
--ils duces d'Enron!
===
Subject: Re: Deprogramming needed?
> I've traced out every step in the arguments I have
Are you a good tracker of steps?
Do you still think that Z[1/2] contain 1/3?
===
Subject: Re: JSH: Deprogramming needed?
> Renaud Dreyer  grava .88 la saucisse et au 
marteau:
>  D.8esol.8e, c'est .82 entiers alg.8ebriques 
é, of course. Ciao,
> Le probl.8fme, c'est que j'ai jamais entendu l'expression entier
> alg.8ebrique en frn.8dais, c'est bien le probl.8fme. Je connais les 
entiers
> naturels, les entiers relatifs et les nombres alg.8ebriqu.8es qui sont 
les
> nombres r.8eels non transcendants (donc pas forc.8ement tr.8fs 
entiers).
> Quid?
An algebraic integer is a complex number which is a root of a monic
polynomial with integer coefficients.
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
Then you should probably stop posting to sci.math and find something
more productive to do with your time.
[rest deleted]
===
Subject: Clarify of Yang Mills and Mass Gap Hypothesis Problem
Yang-Mills Field Equation and Mass Gap Hypothesis is one of the
problems which were included in Clay Millennium Problems.
After reading description of the problem, I am still not clear (or be
convinced ) why there is a problem here. So, can anyone shed some
light on this ? Be specific, two things I do not understand are:
A.      Physicists had solved Yang-Mills Field Equations for past 40 years
using different numerical approximations and renormalization methods. 
Why do we not consider these solutions as mathematically acceptable
solutions ? Are we looking for an exact and closed form solution for
Yang-Mills Field Equations ?
B. Physicists had solved zero mass issue by using Higgs mechanism and
Spontaneous Symmetry Broken, and produced mass that way. Again, why do
NOT we consider these method as mathematically acceptable solution or
proof ?
===
Subject: Re: Marketing shift, core issues
> I've been thinking about my problems with getting any kind of
> admission that my math arguments showing the core error in mathematics
> are correct, so I've gone to marketing books.
> 
> I just wanted to warn readers that I may be employing various tactics
> from modern research on human psychology to see if I can't break
> through the logjam.
> Psychology won't make your arguments correct.  It also isn't likely to 
> make anyone forget that they are wrong.
It made James Harris forget that he is wrong. See
   http://www.mentalhealth.com/dis/p20-pe07.html
for a description of Narcissistic Personality Disorder.
===
Subject: Re: Marketing shift, core issues
>I've been thinking about my problems with getting any kind of
>admission that my math arguments showing the core error in mathematics
>are correct, so I've gone to marketing books.
> You know, in case you're curious, this sounds really really stupid.
> If your results were correct you'd be able to convince people of
> them by explaining the proofs carefully. But in fact they're wrong,
> people continually explain what the errors are, and tactics from
> marketing books are not going to change that.
If his results were correct, the same people who point out errors to him 
with saintly patience would instead have provided all these explanations.
===
Subject: Re: JSH: Deprogramming needed?
> What if all the time and energy I've invested in my work has made it
> difficult for me to see, along with the *harsh* and unforgiving
> hostility from several people who seem to have made it their mission
> to make me miserable and find pleasure in mocking or trying to
> humiliate me, have made it extremely difficult for me to see the
> truth?
> Think about the crushing sense of shame and misery if indeed I find
> out that the logical connections I so carefully and impatiently
> discovered over the years are simply not really there, but are a need
> induced delusion.
Fact is: You have written your own kick me sign and attached it firmly 
to your back. Your own behavior makes everyone want to kick you. 
Something good comes from it occasionaly - for example the fast prime 
counting function on my web site (which by the way beats yours speed 
wise by a factor of about THOUSAND. It was a pleasure for me to kick 
your teeth out). 
Get medical help.
===
Subject: Re: JSH: Deprogramming needed?
> James Harris  grava .88 la saucisse et au marteau:
>  Aren't there any *other* people who suppose they are rational, who can
>  follow a logical argument, who might comment?
> I don't know the french equivalent for algebraic integers. If someone
> tells me what set it is (is it the positive and negative integers?), I
> may have the courage to take a look. 
Algebraic integers are the roots of polynomials with integer 
coefficients and a leading coefficient of one. Algebraic numbers are the 
roots of polynomials with integer coefficients (and a leading 
coefficient that is not necessarily one). It is well known that the 
algebraic integers form a ring, and the algebraic numbers form a field. 
> But anyhow, your proof is hard to
> read. You should at least follow the indications the others have given
> you (clearly define your functions, explicitly when you can and so on).
>  Why is it always the same people?  Or people imitating them in hostile
>  and mocking displays of animosity or anger?
> Because I suppose very few people have had the courage to read your
> proof. And the one who didn't saw the former(s) commenting it and
> pointing out some problems you didn't solve (you keep claiming they're
> wrong, but your explications are more confuse than their). I've known
> better motivations.
===
Subject: Re: Squares that end with four identical digits
>  Which integers can have squares that end with four identical digits?
Let's see. A square ends with either 0,1,4,5,6 or 9. A number ending with 
11, 55, 66 or 99 cannot be a square because it isn't congruent to either 0 
or 1 modulo 4. That leaves us with 00 and 44. Obviously 00 is a possibility
as an integer multiple of 100 will have a square ending 0000. Actually
that is the only possibility: a square cannot end with 4444 for then it
would be the square of an even number, say 2n. Hence n^2 =((2n)^2)/4 would
end with 11, as any mulitple of 10000 divided by 4 will end in two zeros.
Earlier we saw that a square cannot end with 11. Case closed.
Jyrki Lahtonen, Turku, Finland
===
Subject: Re: Largest number ever written down?
> Joona I Palaste  scribbled the following:
>> Quaternion  scribbled the following:
> One of those people that is very fond of exclamation marks probably 
once
> 9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!  at the end of a 
sentence,
> in this thread, I think. I'm not sure.
>> You have 30 ! signs there. This means that your number is smaller than
>> 9^(9^(9^... containing 60 9's. I haven't counted the 9's in Jeroen
>> Boschma's posting but I'm fairly sure he has more of them than you do.
> Erm, no. I was wrong. n! is smaller than n^n, so if we replace n with
> m!, we get that m!! is smaller than (m^m)^(m^m), and if we replace m
> with l!, we get that l!!! is smaller than ((l^l)^(l^l))^((l^l)^(l^l)),
> and so on. The number the variable occurs in the bigger number is 2 to
> the power of the number of ! signs in the smaller number.
> If those operations were all grouped like l^(l^(l^... then there would
> be 1073741824 nested exponentations, which would make Quaternion's
> number bigger than Jeroen's, but as they are not, I don't know which is
> larger. Could some of you math gurus shed more light on this?
I don't know it either. I suppose it's easier to work with 10 as base 
number
and then count the minimal amount of zeroes, but they are two entirely
different growths if you indeed write the exponential as n^n^n^n^.. Hard to
compare them.
Note that the amount of zeroes grows with n(n-1)/2, for the recursive
factorial function, for each next n (since the amount of zeroes is sum(1 to
n) with each new '!', if we take 10 as a base number (or (n-9)(n-10)/2 to
be precise))
eventually become larger than n^^k, for rather small values of k.
And, my post wasted less bandwidth and required no copy-pasting. It was 
also
meant as a joke. Now, I want that cake that was mentioned in an other's
post.
-- 
Quaternion
===
Subject: uniform convergence & differentiation
In many analysis-type texts I've found statements like since the series
converges uniformly, we can apply term-by-term differentiation and 
uniform
convergence allows us to differentiate term-by-term, but I've never
actually seen a proof or explanation of WHY you need uniform convergence to
===
Subject: Re: {Group Theory}  Confusing group theory conundrum
                 permission for an emailed response.
>But I am told that you cannot have a set that contains every group... 
>so this function's domain is nonsense!  In fact the whole function
>is nonsense!
> Not nonesense, just not a function in the usual set-theoretic sense.
I would object here that function in the usual set-theoretic sense
is not well defined.  There are *two* usual set-theoretic senses, one
which allow functions from/to proper classes, and one which only
allows functions from/to sets.
In fact, anytime proper classes are allowed, there is no reason
whatsoever not to allow all the relevant operators for them: union,
intersection, cross-product--and thus also functions.
Thomas
===
Subject: Re: Greek Alphebet
                 permission for an emailed response.
   Hiccuping & trembling into the WASTE DUMPS of New Jersey like some
   drunken CABBAGE PATCH DOLL, coughing in line at FIORUCCI'S!!
> It is worth pointing ,that (according to Britannica under ALPHABET)
> the Alphabet was a Greek invention based upon North Semitic 
> (proto-Canaanitish) writing which indicated only consonants,
> a procedure suitable enough for a Semitic language but not for 
> an Indo-European one.
A little far from math here but...
It is true that the near-eastern alphabets are not properly
alphabets.  The real test is whether they represent every phoneme.
Because the near-eastern scripts leave out vowels, they aren't proper
alphabets.
But this is also a matter of degree.  The classical Greek alphabet
does not distinguish long and short alpha, for example, which is a
phonemic difference.  Similarly, until accent marks were invented, it
did not distinguish different accents, which are also phonemic in
ancient Greek.
But usually this is taken to be much more trivial than the lack of
representation of vowels in the near-eastern scripts.
Another way to express the test is to ask can I learn to pronounce
the language without knowing the lexicon?  For an alphabetic script,
the answer is yes.  For example, my Spanish is very weak, but I can
read any Spanish text you give me quite reliably, while not
understanding very much of it at all.  By contrast, even learning the
Hebrew alphabet is not enough to read a consonantal Hebrew text,
because I cannot know which vowel sounds to insert.
However, you are incorrect about there being some natural reason for
near-eastern scripts to not need vowels but IE languages to need
them.  Fr xmpl, y cn rd ths prtt wll, cnt y?  nglsh cn b rlbly rd b
ppl wtht vwls, mst f th tm.  And Hebrew did finally decide to start
writing in the vowels because things had gotten too hard.
But a no-vowel script is reasonably workable, leading to the question:
why did Greek invent a voweled script, and thus the first true
alphabet?  Greek depends on vowels more than English for declensions
and such--but a little experimentation will show that a no-vowel
script for Greek actually works as well as for English and Hebrew.
One hypothesis, I think quite intriguing, is that the alphabet was
developed to write epic poetry.  One data point in this is that while
the oldest inscriptions for most languages are economic and political
texts, the oldest in the Greek alphabet are hexameter poetry.  But
still this begs the question: why would a consonantal system not do as
well?  And the intriguing answer is: because Homer wants to represent
the dialectical variations of his speakers--which indeed he does
extremely well--and these dialectical variations very frequently turn
on which vowel is used for a particular word.
A mere guess, indeed, but an intriguing one.  There are other equally
plausible guesses out there.
Thomas
===
Subject: Re: Cardinality of 2^n numbers?
                 permission for an emailed response.
> The only way I see there being a contradiction is if you require that
> K and N both be of cardinality N_0.
I *proved* that K and N have the same cardinality.
===
Subject: Re: Calculating the expected range of results
> I have a normal distribution of known mean and standard deviation.
> 
> For a certain case, a finite number of results will be drawn from this
> distribution.  Is there a mathematical formula for calculating the
> expected range of these results?
> What exactly do you mean by expected range?
> I could interpret the question literally: What is the
> expectation value of the range?
> Suppose you are drawing n independent samples, X_1, X_2,...
> ,X_n.
> Here's the cdf of the maximum of the X's:
> P[max(X_1, X_2, ..., X_n)<=x] = 
>    = P(X_1 <=x & X_2 <=x & ... & X_n <=x]
>    = P(X_1 <= x)^n
> So the pdf is p(x)= dP/dx
>     = n*P_x(x)^(n-1)*p_x(x)
> where P_x(x) and p_x(x) refer to the normal distribution
> of individual samples.
> The expectation value is therefore
> integral(-inf,inf) n*P_x(x)^(n-1)*p_x(x)*x dx
> So you can in principle calculate E[max(x_i)].
> Similarly, you can work out a formula for E[min(x_i)]
> in terms of P(x_i >= x) = 1 - P_x(x).
> Thus, E[max - min] = E[max] - E[min].
>            - Randy
Yes, I think that is what the OP wanted.  Note that in this case, it
simplifies a little.  Since we are talking about the distribution of
the range (=max - min), we may assume WLOG that the mean of the
distribution is 0.  Since the normal is symmetric, it is easy to show
that the min has the same distribution as the negative of the max,
whence the expected range is twice the expectation of the max.  Since
the  max(a x1, a x2,..., a x_n) = a max(x1, x2,..., x_n)  when  a > 0,
 we may scale the result for the standard normal.  Putting it all
together, the expected range is
2n s int(x=-infty..infty, x [F(x)]^(n-1) f(x))
where  n  is the sample size,  s  is the standard deviation, and  f 
and  F  are respectively the pdf and cdf of the standard normal.
I do not think this can be expressed in closed form (as HR has noted).
By the way, another formula for the expectation of the max is
int(x=0..infty, 1 - [F(x)]^n - [F(-x)]^n)
(Multiply by 2s for the expected range.)
===
Subject: Re: Factoring to Satisfiability
> Factoring to Satisfiability
I have no idea if this relates to your problem, but it uses
enough of the same words plus some nice ones about reducing
run times, it might be worth your pursuing:
http://groups.yahoo.com/group/genetic_programming/message/106
xanthian.
-- 
===
Subject: Re: Godel's universe
                 permission for an emailed response.
> I'am looking for literature concerning Godel's Rotating Universe.
> Suggestions?
Hrm.  Of what level of sophistication are you looking for?  Popular
descriptions, or just what?
===
Subject: Re: Deprogramming needed?
        boundary=----=_NextPart_000_009C_01C3A8CD.A1B79170
---------------------------------------------------------------------
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
> After all, I've been at this since April 1995 having spent a lot of
> time and effort, with literally thousands of posts along with all
> kinds of other activities, websites, and email to mathematicians all
> over the world.
Since April 1995 and not right yet?
===
Subject: Re: Cardinality of 2^n numbers?
> The only way I see there being a contradiction is if you require that
> K and N both be of cardinality N_0.
> However, I don't see why there can't be a non-N_0 cardinality of K
> such that when its power set is taken, its power set is of cardinality
> N_0. That will also have to be explained to me.
There isn't.
Is it worth explaining this to you? In the sense that you might
possiibly grasp the explanation?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Cardinality of 2^n numbers?
> I never said 2^K WAS the set of all natural numbers. I said 2^K was
> BIJECTIVE with the set of all natural numbers.
And you were wrong in saying so.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Cardinality of 2^n numbers?
> You are very unhelpful. Are you sure you're not just here to be as
> completely useless as possible?
You are very helpless. Are you sure you're not just here to be as
completely useless as possible?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Basic factorization ideas
In sci.physics, James Harris
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 1)
> with the c's algebraic integers, I think few of you would have a
> problem realizing that only two of the c's have 7 as a factor.
Oh, look, he's changed polys on us!  But OK, let's check
this one.
Recall that, if P(x) = 49 * (x - x_1) * (x - x_2) * (x - x_3),
then P(x) = (c_1 * x + 7) * (c_2 * x + 7) * ( c_3 * x + 1)
implies that c_1 = -7 / x_1, c_2 = -7 / x_2, c_3 = -1 / x_3,
for some permutation of the x_i.  Again, c_1 * c_2 * c_3 = 49,
as required, since x_1 * x_2 * x_3 = -1.
Therefore, c_1 and c_2 both satisfy the equation
c^3 * P(-7 / c)/49 = c^3 - 21*c^2 + 245*c - 343
and c_3 of course satisfies
c^3 * P(-1 / c)/49 =  c^3 - 3*c^2 + 5*c - 1
so it turns out that this time, James, you got it more
or less right. :-)  (And c_3 is even a unit, to boot.
Come to think of it, so are the x_i, which is probably
one big reason why this particular case actually works.)
Also, it is obvious that c_1 and c_2 have factors of 7,
as well, in the ring of algebraic integers.
> But, of course, you're looking at *functions* of x, as you have 
> f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
> so I could also write it as
> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
> Notice that dividing both sides by 49 gives 
> (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
> as long as you're in a ring where 7 is not a factor of 22.
OK, stupid question.  Where did the 22 come from?
But yes, for this particular example, your odd math actually does
work, as you are dealing with x_i which are in fact units.
Now change a_0 to 3 and get back to us. :-)
[rest snipped]
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: On Reading Textbooks
> what's A-level maths?
Mathematics studied at A-level. This is a qualification taken
by schoolchildren in England, Wales and Northern Ireland
usually between the ages of 16 and 18.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: question about canonical commutation relations [P,Q]=i*hbar*I
> I think the question from the OP really boils down to, why the proof in
> the finite case does not generalize to the infinite case.
> I mean, the OP has proved that no finite matrices satisfy the given
> commutation relation. Then he demonstrates the existence of infinite
> matrices that *do* satisfy the relation. So the question is, why is the
> proof not valid in the infinite case?
> At least, I'm curious about that too, because it seems that although
> matrix product is well-defined, taking the trace is not. Please enlighten
> me.
It's pretty obvious that trace can't generalize simply: the trace
if a (finite) matrix is the finite sum of diagonal elements.
Not all infinite series have sums ....
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Clarify of Yang Mills and Mass Gap Hypothesis Problem
> Yang-Mills Field Equation and Mass Gap Hypothesis is one of the
> problems which were included in Clay Millennium Problems.
> After reading description of the problem, I am still not clear (or be
> convinced ) why there is a problem here. So, can anyone shed some
> light on this ? Be specific, two things I do not understand are:

I presume this physics problem was included in the Clay list
due to residual physics envy on the part of the organizers.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
You are.
> After all, I've been at this since April 1995 having spent a lot of
> time and effort, with literally thousands of posts along with all
> kinds of other activities, websites, and email to mathematicians all
> over the world.
Effort doesn't indicate accuracy.  I could spend 8 years throwing a 
football, running laps, etc, and believe that I would make a great 
quarterback.  Unfortunately, with low stamina, lousy aim, and a fairly 
light frame, I'd be more likely to get killed.  My effort and belief 
won't get me into the NFL.
> But, what if I'm wrong?
Then you're wrong.  Very simple.
> What if all the time and energy I've invested in my work has made it
> difficult for me to see, along with the *harsh* and unforgiving
> hostility from several people who seem to have made it their mission
> to make me miserable and find pleasure in mocking or trying to
> humiliate me, have made it extremely difficult for me to see the
> truth?
Back up.  The hostility doesn't exists except as a reaction to rudeness 
or hostility.  Disagreement does not mean hostile.  Also, if you are 
miserable, that is a result of your choices.  You may choose to be 
miserable or not, but we cannot make you so, especially since we can't 
even make you pay attention to what we say.
> Think about the crushing sense of shame and misery if indeed I find
> out that the logical connections I so carefully and impatiently
> discovered over the years are simply not really there, but are a need
> induced delusion.
There is no shame in being wrong.  Most of us have experienced it before.
> It seems to me that marketing ploy though these statements may be,
> dealing with people who've made it their business to try and make me
> miserable, only to at times claim they're trying to help me is just
> too much.
These are a marketing ploy only in that they are a blatant play for 
sympathy.  If you were to admit you've wasted 8 years of your life, you 
would have it.  As you haven't made the admission yet, there is a 
different priority.
Also, the unfortunate consequences of it being a waste of your time does 
not make you right.
> Aren't there any *other* people who suppose they are rational, who can
> follow a logical argument, who might comment?
Tons.  But with the specialized topic, they will all be people with at 
least a bachelor's degree in math.  This makes looking in certain 
directions unlikely to produce intelligent comment because of the 
required level of knowledge to intelligently look for flaws.
> Why is it always the same people?  Or people imitating them in hostile
> and mocking displays of animosity or anger?
Because you've tapped the well.
> Aren't there any rational people who can trace out the steps in the
> work I've presented, who might step forward at this time, and
> demonstrate an ability to just be objective?
You are either assuming your work is simpler than it is, or you are not 
as clear as you believe you are.
> I don't want any replies of people offering, but if you might consider
> it, I just want you to think about it.  Sure I've been reading this
> marketing book, but it seems to me that still *someone* out there
> might be the right person, as of course, I don't believe I'm wrong. 
> I've traced out every step in the arguments I have, and I think that
> irrational people have been dominating the discussion using group
> effects to hide the truth.
Try an experiment.  Try to work with the idea that perhaps you *are* 
wrong and understand the details of the objections.  If you *are* right, 
you have nothing to lose.  If you are wrong, perhaps you will learn why.
> Think about it.  I'll come back to the subject later.
You have posted a plea for agreement with your argument based on 
sympathy.  This is not the basis for a mathematical proof.  If you can't 
come up with better advice from your marketing book, perhaps you should 
ditch it.
-- 
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: uniform convergence & differentiation
> In many analysis-type texts I've found statements like since the series
> converges uniformly, we can apply term-by-term differentiation and
> uniform convergence allows us to differentiate term-by-term, but I've
> never actually seen a proof or explanation of WHY you need uniform
> convergence to legitimize term-by-term differentiation.
You don't *need* uniform convergence.
Uniform convergence is a *sufficient* not a *necessary*
condition to vaildate termwise differentiation/integration.
However, it is a condition that occurs frequently and
is convenient to check.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Basic Planar Geometry Question
Which are the polygons which have 5 vertex and have the maximum number
of parallel (m) and perpendicular lines (n) drawn among these vertex,
i.e., m + n is maximum?
===
Subject: Re: [Set Theory] A set has a disjoint copy of itself ?
> Here is a slightly simpler version of your construction, which does
> not use the Axiom of Replacement.
> Let U be the set of all elements of elements of A (your Q_1). Let c be
> any set not in U, e.g., let c = {x in U: x is not an element of x).
> Let B = {{a,c}: a in A}. Then B is disjoint from A, and you get a
> bijection f:A --> B by defining f(a) = {a,c}.
===
Subject: Re: [Set Theory] A set has a disjoint copy of itself ?
> Let Q_0 = A, Q_{n=1} = U(Q_n), the union of the elements of
> Q_n, and let R be the union of all the Q's.  Then if c is 
> any set not in R, {: x in A} will do.
the ordered pair  is defined as ={{c},{c,x}}, the proof
would work just as well by taking R= Q_2 and c not in R ?
Noel.
Supersedes: 

===
Subject: Invariant Galilean Transformations (FAQ) On All Laws
Summary: All laws/equations are Galilean invariant when expressed
  in the generalized cartesian coordinates demanded by basic
  analytic geometry, vector algebra, and measurement theory.
Originator: faqserv@penguin-lust.MIT.EDU
Disclaimer: approval for *.answers is based on form, not content.
    Opponents of the content should first actually find out what
    it is, then think, then request/submit-to arbitration by the
    appropriate neutral mathematics authorities. Flaming the hard-
    working, selfless, *.answers moderators evidences ignorance
    and despicable netiquette.
Archive-Name: physics-faq/criticism/galilean-invariance
Version: 0.04.03
Posting-frequency: 15 days
   Invariant Galilean Transformations (FAQ) On All Laws
     (c) Eleaticus/Oren C. Webster
        Thnktank@concentric.net
An obvious typo or two corrected.
The Brittanica section revised to less
'pussy-footing' and to more directly
anticipate the elementary measurement
theory and basic analytic geometry
that is applied to the transformation
concept.
------------------------------
===
Subject: 1. Purpose
The purpose of this document is to provide the student of Physics,
especially Relativity and Electromagnetism, the most basic princ-
iples and logic with which to evaluate the historic justification
of Relativity Theory as a necessary alternative to the classical
physics of Newton and Galileo.
We will prove that all laws are invariant under the Galilean
transformation, rather than some being non-invariant, after
we show you what that means.
We shall also show that another primal requirement that SR
exist is nonsense: Michelson-Morley and Kennedy-Thorndike do
indeed fit Galilean (c+v) physics.
------------------------------
===
Subject: 2. Table of Contents
 1. Foreword and Intent
 2. Table of Contents
 3. The Principle of Relativity
 4. The Encyclopedia Brittanica Incompetency.
 5. Transformations on Generalized Coordinate Laws
 6. The data scale degradation absurdity.
 7. The Crackpots' Version of the Transforms.
 8. What does sci.math have to say about x0'=x0-vt?
 9. But Doesn't x.c'=x.c?
       10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?
       11. But Doesn't (x'-x.c+vt) Prove The Transformation Time
    Dependent?
       12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
       13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of
    a Linear Transform?
       14. But The Transform Won't Work On Time Dependent Equations?
       15. But The Transform Won't Work On Wave Equations?
       16. But Maxwell's Equations Aren't Galilean Invariant?
       17. First and Second Derivative differential equations.
------------------------------
===
Subject: 3. The Principle of Relativity and Transformation
If a law is different over there than it is here,
it is not one law, but at least two, and leaves us
in doubt about any third location. This is the
Principle of Relativity:  a natural law must be the
same relative to any location at which a given event
may be perceived or measured, and whether or not the
observer is moving.
The idea of location translates to a coordinate
system, largely because any object in motion could
be considered as having a coordinate system origin
moving with it. If you perceive me moving relative
to you - who have your own coordinate system - will
your measurements of my position and velocity fit
the same laws my own, different measurements fit?
If a law has the same form in both cases it is
called covariant. If it is identical in form, var-
ables, and output values, it is called invariant.
What we're asking is that if the x-coordinate, x,
on one coordinate axis works in an equation, does
the coordinate, x', on some other, parallel axis
work?  Speaking in terms of the axis on which x is
the coordinate, x' is the 'transformed' coordinate.
The situation is complicated because we're talking
about coordinates - locations -  but in most mean-
ingful laws/equations, it is lengths/distances (and
time intervals) the equations are about, and x coord-
inates that represent good, ratio scale measures of
distances are only interval scale measures on the x'
axis. [See Table of Contents for discussion of scales.]
So, if we have an x-coordinate in one system, then
we can call the x' value that corresponds to the same
point/location the transform of x.
In particular, the Principle of Relativity is embodied
in the form of the Galilean transformation, which
relates the original x, y, z, t to x', y', z', t' by
the transform equations x'=x-vt, y'=y, z'=z, t'=t in
the simplified case where attention is focused only
on transforming the x-axis, and not y and z. In the
case of Special Relativity, the x' transform is the
same except that x' is then divided by sqrt(1-(v/c)^2),
and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v
is the relative velocity of the coordinate systems;
if there is already a v in the equations being trans-
formed use u or some other variable name.
------------------------------
===
Subject: 4. The Encyclopedia Brittanica Incompetency.
One example of the traditional fallacious idea
that an equation is not invariant under the galilean
transformation comes from the Encyclopedia Brittanica:
Before Einstein's special theory of relativity
was published in 1905, it was usually assumed
that the time coordinates measured in all inertial
frames were identical and equal to an 'absolute
time'.  Thus,
      t = t'.              (97)
The position coordinates x and x' were then
assumed to be related by
       x' = x - vt.         (98)
The two formulas (97) and (98) are called a
Galilean transformation. The laws of nonrelativ-
istic mechanics take the same form in all frames
related by Galilean transformations.  This is the
restricted,  or Galilean, principle of relativity.
The position of a light wave front speeding from
the origin at time zero should satisfy
       x^2 - (ct)^2 = 0          (99)
in the frame (t,x) and
      (x')^2 - (ct')^2 = 0       (100)
in the frame (t',x'). Formula (100) does not
transform into formula (99) using the transform-
ations (97) and (98),  however.
.................................................
Besides the trivially correct statement of what the
Galilean 'transform' equations are, there is exactly
one thing they got right.
I.   Eq-100 is indeed the correct basis for discussing
     the question of invariance, given that eq-99 is
     the correct 'stationary' (observer S) equation.
     [Let observer M be the 'moving'system observer.]
     In particular, eq-100 is of exactly the same
     form [the square of argument one minus the square
     of argument two equals zero (argument three).]
II.  It is nonsense to say eq-99 should be derivable from
     eq-100; for one thing, the transforms are TO x' and
     t' from x and t, not the other way around, and the
     idea that either observer's equation should contain
     within itself the terms to simplify or rearrange to
     get to the other is ridiculous. As the transform
     equations say, the relationship of t', x' to t, x
     is based on the relative velocity between the two
     systems, but neither the original (eq-99) equation
     nor the M observer equation is about a relationship
     between coordinate systems or observers. One might
     as well expect the two equations to contain banana
     export/import data; there is no relevancy. The
     'transform' equations are the relationships between
     x' and x, t' and t and have nothing to do with what
     one equation or the other ought to 'say'.  The
     equations' content is the rate at which light emitted
     along the x-axes moves.
III. Most remarkable, the True Believer SR crackpots who
     most despise the consequences of measurement theory
     (demonstrable fact) contained in this document are
     those who want to argue against our saying the Britt-
     anica got eq-100 right;
     They insist that the correct equation is derived
     directly from x'=x-vt and t'=t. Solve for x=x'+vt
     and replace t with t', then substitute the result
     in eq-99: (x'+vt')^2 - (ct')^2 = 0.
     Besides the fact that this results in an equation
     with arguments exactly equal to eq-99, they will
     insist the transform is not invariant.
IV.  A major justification they have for their idea of
     the correct M system equation on which to base the
     the discussion of invariance, is that the variables
     are M system variables, never mind the fact that
     the arguments are S system values.
     That argument of theirs is arrant nonsense. The
     velocity v that S sees for the M system relative
     to herself is the negative of what the M system
     sees for the S system relative to himself.
     In other words, x'+vt' is a mixed frame expression
     and it is x'+(-v)t' that would be strictly M frame
     notation, and that equation is far off base. [Work
     it out for yourself, but make sure you try out an
     S frame negative v so as not to mislead yourself.]
V.   In I. we said: given that eq-99 is the correct
     'stationary' equation. Let's look at it closely:
       x^2 - (ct)^2 = 0          (99)
     This whole matter is supposed to be about coordinate
     transforms. Is that what t is, just a coordinate?
     No. It isn't, in general.  Suppose you and I are both modelling
     the same light event and you are using EST and I'm using PST.
     'Just a time coordinate' is just a clock reading amd your t clock
     reading says the light has been moving three hours longer
     than my clock reading says. Well, that's what the idea that
     t is a coordinate means.
     Eq-99 works if and only if t is a time interval, and in
     particular the elapsed time since the light was emitted.
     Thus, that equation works only if we understand just
     what t is, an elapsed time, with emissioon at t=0.
     However, we don't have to 'understand' anything if we use
     a more intelligent and insightful form of the equation:
     (x)^2 - [ c(t-t.e) ]^2 = 0,
     where t.e is anyone's clock reading at the time of light
     emission, and t is any subsequent time on the same clock.
     Similarly, x is not just a coordinate, but a distance
     since emission.
     (x-x.e)^2 - [ c(t-t.e) ]^2 = 0        (99a)
VI.  In the spirit of 'there is exactly one thing
     they got right', the correct M system version
     of eq-99a is eq-100a:
     (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0   (100a)
     Every observer in the universe can derive their
     eq-100a from eq-99a and vice versa, not to mention to and
     from every other observer's eq-99a.
     Now, THAT's invariance. [You do realize that every
     eq-100a reduces to eq-99a, when you back substitute
     from the transforms, right? t.e'=t.e, x.e'=x.e-vt.]
------------------------------
===
Subject: 5. Transformations on Generalized Coordinate Laws
The traditional Gallilean transform is correct:
     t'   = t
     x'   = x - vt.
But remember this: a transform of x doesn't effect
just some values of x, but all of them, whether they
are in the formula or not.  This is important if you
want to do things right. The crackpot position is
strongly against this sci.math verified position, and
the apparently standard coordinate pseudo-transformation
they suggest is perhaps the result. {See Table of
Contents.]
Let's use a simple equation: x^2 + y^2 = r^2, which is
the formula for a circle with radius r, centered at a
location where x=0.
But what if the circle center isn't at x=0?  Well, we'd
want to use the form analytic geometry, vector algebra,
and elementary measurement theory tells us to use, a form
where we make explicit just where the circle center is,
even if it is at x=x0=0:
   (x-x0)^2 + (y-y0)^2 = r^2.
The circle center coordinate, x0, is an x-axis coordinate,
just like all the x-values of points on the circle.
So, in proper generalized cartesian coordinate forms
of laws/equations we want to transform every occurence
of x and x0 - by whatever name we call it: x.c, x_e,
whatever.
So, what is the transformed version of (x-x0)?  Why,
(x'-x0'); both x and x0 are x-coordinates, and every
So, what is the value of (x'-x0') in terms of the original
x data?
is also true for x0'=x0-vt:
    (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0).
In other words, when we use the generalized coordinate form
specified by analytic geometry, we find that the value of
(x'-x0') does not depend on either time or velocity in any
way, shape, form, or fashion.
Similarly for (y-y0).
We can treat time the same way if necessary: (t-t0).
The above is a proof that any equation in x,y,z,t is
invariant under the galilean transforms. Just use the
generalized coordinate form, with (x-x0)/etc, in the
transformation process, not the incompetently selected
privileged form, with just x/etc.
[The form is privileged because it assumes the circle
center, point of emission, whatever, is at the origin of
the axes instead at some less convenient point. After
transform the coordinate(s) of the circle center/origin
are also changed but the privileged form doesn't make
this explicit and screws up the calculations, which
should be based on (x'-x0') but are calculated as (x'-0).]
The value of (x'-x0') is the same as (x-x0).  That makes
sense.
Draw a circle on a piece of paper, maybe to the right
side of the paper. On a transparent sheet, draw x and y
coordinate axes, plus x to the right, plus y at the top.
Place this axis sheet so the y-axis is at the left side
of the circle sheet.
Now answer two questions after noting the x-coordinate of
the circle center and then moving the axis sheet to the right:
(a) did the circle change in any way because you moved
the axis sheet (ie because you transformed the coordin-
nate axis)?
(b) did the coordinate of the circle center change?
The circle didn't change [although SR will say it did];
that means that (x'-x0') does indeed equal (x-x0).
The coordinate of the circle center did change, and it
changed at the same rate (-vt) as did every point on
the circle.   That means that x0'<>x0, and the fact the
circle center didn't change wrt the circle, means that
the relationship of x0' with x0 is the same as that of
any x' on the circle with the corresponding x: x'=x-vt;
x0'=x0-vt.
This is to prepare you for the True Believer crackpots that
say 'constant' coordinates can't be transformed; some even
say they aren't coordinates. These crackpots include some
that brag about how they were childhood geniuses, btw.
QED: The galilean transformation for any law on
generalized Cartesian coordinates is invariant under
the Galilean transform.
The use of the privileged form explains HOW the transformed
equation can be messed up, the next Subject explains what
the screwed up effect of the transform is, and how use
of the generalized form corrects the screwup.
------------------------------
===
Subject: 6. The data scale degradation absurdity.
The SR transforms and the Galilean transforms both
convert good, ratio scale data to inferior interval
scale data. The effect is corrected, allowed for,
when the transforms are conducted on the generalized
coordinate forms specified by analytic geometry and
vector algebra.
Both sets of transforms are 'translations' - lateral
movements of an axis, increasing over time in these
cases - but with the SR transform also involving a
rescaling. It is the translation term, -vt in the x
transform to x', and -xv/cc in the t transform to t',
that degrades the ratio scale data to interval scale
data.  In general, rescaling does not effect scale
quality in the size-of-units sense we have here.
SR likes to consider its transforms just rotations,
however - in spite of the fact Einstein correctly said
they were 'translations' (movements) - and in the case
of 'good' rotations, ratio scale data quality is indeed
preserved, but SR violates the conditions of good ro-
tations; they are not rigid rotations and they don't
appropriately rescale all the axes that must be rescaled
to preserve compatibility.
The proof is in the pudding, and the pudding is the
combination of simple tests of the transformations.
We can tell if the transformed data are ratio scale
or interval.
Ratio scale data are like absolute Kelvin. A measure-
ment of zero means there is zero quantity of the
stuff being measured. Ratio scale data support add-
ition, subtraction, multiplication, and division.
The test of a ratio scale is that if one measure
looks like twice as much as another, the stuff
being measured is actually twice as much. With
absolute Kelvin, 100 degrees really is twice the
heat as 50 degrees. 200 degrees really is twice
as much as 100.
Interval scale data are like relative Celsius, which
is why your science teacher wouldn't let you use it
in gas law problems.  There is only one mathematical
operation interval scales support, and that has to
be between two measures on the same scale: subtraction.
100 degrees relative (household) Celsius is not twice
as much as 50; we have to convert the data to absolute
Kelvin to tell us what the real ratio of temperatures
is.
However, whether we use absolute Kelvin or relative
Celsius, the difference in the two temperature readings
is the same: 50 degrees.
Thus, if we know the real quantities of the 'stuff'
being measured, we can tell if two measures are on
a ratio scale by seeing if the ratio of the two
measures is the same as the ratio of the known quant-
ities.
If a scale passes the ratio test, the interval scale test
is automatically a pass.
If the scale fails the ratio test, the interval scale
test becomes the next in line.
It isn't just the bare differences on an interval
scale that provides the test, however. Differences
in two interval scale measures are ratio scale, so
it is ratios of two differences that tell the tale.
Let's do some testing, and remember as we do that our
concern is for whether or not the data are messed up,
not with 'reasons', excuses, or avoidance.
------------------------------------------------------
Are we going to take a transformed length (difference)
and see whether that length fits ratio or interval scale
definitions?
Of course, not. Interval scale data are ratio after
one measure is subtracted from another. That is the
major reason the SR transforms can be used in science.
Let there be three rods, A, B, C, of length 10, 20, 40,
respectively.  These lengths are on a known ratio scale,
our original x-axis, with one end of each rod at the
origin, where x=0, and the other end at the coordinate
that tells us the correct lengths.
Note that these x-values are ratio scale only because
one end of each rod is at x=0. That may remind you of
the correct way to use a ruler or yard/meter-stick:
put the zero end at one end of the thing you are
measuring. Put the 1.00 mark there instead of the zero,
and you have interval scale measures.
Let A,B,C,   be 10, 20, 40.
Let a,b,c    be x' at v=.5, t=10.
x'=x-vt.
A   B   C         a      b      c
----------------  --------------------
10  20  40         5     15     35
----------------  --------------------
B/A = 2           b/a = 3
C/A = 4           c/a = 7
C/B = 2           c/b = 2.333
          Obviously, the transformed
          values are no longer ratio
          scale. The effect is less on
          the greater values.
C-A = 10          b-a = 10
C-A = 30          c-a = 30
C-B = 20          c-b = 20
          Obviously, the transformed
          values are now interval scale.
          This will hold true for any
          value of time or velocity.
(C-A)/(B-A) = 3   (c-a)/(b-a) = 3
(C-B)/(B-A) = 2   (c-b)/(b-a) = 2
          Obviously, the ratios of the
          differences are ratio scale,
          being identical to the ratios
          of the corresponding original
          - ratio scale - differences.
The main difference between these results and the SR
results is that the differences do not correspond so
neatly to the original, ratio scale, differences.
This is due only to the rescaling by 1/sqrt(1-(v/c)^2).
The ratios of the differences on the transformed values
do correspond neatly and exactly to the ratio scale
results.
Using the generalized coordinate form, such as (x-x0),
the transform produces an interval scale x' and an
interval scale x0'. That gives us a ratio scale (x'-x0'),
just like - and equal to - (x-x0).
------------------------------
===
Subject: 7. The Crackpots' Version of the Transforms.
It has become apparent - whether misleading or not -
that the crackpot responses to the obvious derive from
a common source, whether it be bandwagoning or their
SR instructors.
Below, in the sci.math subject, we see that all sci.math
respondents agree with the basic controversial position
of this faq: every coordinate is transformed, whether a
supposed constant or not.
Think about it, the generalized coordinate of a circle
center, x0, applies to infinities upon infinities of
circle locations (given y and z, too); it is a constant
only for a given circle, and even then only on a given
coordinate axis.
And even variables are often held 'constant' during
either integration or differentiation.
The utility of a variable is that you can discuss all
possible particular values without having to single out
just one.  That utility does not make particular - singled
out - values on the variable's axis not values of the
variable just because they have become named values.
In any case, all that is preamble to the incompetent idea
they have proposed for a transform of coordinates. It is
based on the idea that the circle center, point of emission,
whatever, has coordinates that cannot be transformed.
Let there be an equation, say (x)^2 - (ict)^2 = 0.
What is the transformed version of that equation?
Answer: (x')^2 - (ict')^2 = 0.   That's the one thing the
Brittanica got right. Note that the leading crackpot just
criticized this faq for presuming to correct the Britt-
anica, but it then and before poses the incompetent pseudo-
transform we analyze here in this section.
x to x' and t to t' are obviously coordinate transforms;
the x and t coordinates have been replaced by the coord-
inates in the primed system.
A tranform of an equation from one coordinate system to
another is NOT a substitution of the/a definition of x
for itself; that is not a coordinate transformation.
The most that can said for such a substitution is that
it is a change of variable.
But the crackpots are calling this a coordinate trans-
form of the original equation:
    (x'+vt)^2 - (ict')^2 = 0.
It is not a coordinate transform, of course, except
accidentally. (x'+vt) is not the primed system
coordinate, it is another form/expression of x. They
get that substitution by solving x'=x-vt for x; x=x'+vt.
So, by incompetent misnomer, they accomplish what they
have been railing against all along.
It has been the generalized coordinate form in question all
this time:
    (x-x0)^2 - (ict)^2 = 0.
Here they substitute for x instead of transforming to the
primed frame:
    (x'+vt-x0)^2 - (ict')^2.
     -----
       ^
       |
       ^
       |
It is still x ^ but see what they have accomplished
by their mis/malfeasance:
    [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)].
      =[x'-x0']
The crackpots have been bragging about how you don't
have to transform the circle center's coordinate to
transform the circle center's coordinate.  Bragging
that what they were doing was not what they said
they were doing.
This does give us insight as to some of the crackpot
variations on their x0'<>x0-vt theme, which in all the
variations will be discussed in later sections..
They are used to seeing the mixed coordinate form,
(x'+vt-x0) without realizing what it respresented,
so - accompanied with a lack of understanding of
the term 'dependent' - they are used to seeing just
the one vt term, and not the one hidden in the defi-
nition of x' and are used to imagining it makes the
whole expression time dependent and thus not invariant.
About which, let x=10, let, x0=20, v=10, and t
variously 10 and 23:
(x-x0)=-10.  Using their (x'+vt-x0):
For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ]
        =      -90     +   100   -  20
        = -10
        = (x-x0)
For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ]
        =      -220    +   230   -  20
        = -10
        = (x-x0)
The result depends in no way on the value of time;
we showed the obvious for a couple of instances of t
just so you can see that the crackpots not only do
not understand the obvious logic of the algebra
{ (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows
that the transform has no possible time term effect -
but they don't understand even a simple arithmetic
demonstration of the facts.
Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same
way since t'=t:
    (x-vt+vt-x0)=(x-x0).
Their process, which says (x'+vt') is the transform
of x, says that (x'+vt') is the moving system location
of x, but it can't be because x is moving further in
the negative direction from the moving viewpoint.
That formula will only work out with v<0 which is indeed
the velocity the primed system sees the other moving at.
However, that formula cannot be derived from x'=x-vt,
the formula for transformation of the coordinates from
the unprimed to the primed,
------------------------------
===
Subject: 8. What does sci.math have to say about x0'=x0-vt?
The crackpots' positions/arguments were put to sci.math
in such a way that at least two or three who posted re-
sponses thought it was your faq-er who was on the idiot's
side of the questions.
Their responses:
----------------------------------------------------------
I.   x0' = x0. In other words: x0' <> x0-vt, or constant
     values on the x-axis are not subject to the transform.
  No.  x0' = x0 - vt.
  Well, if you want, you could define constant values on the x-axis, 
but
in the context of the question that is not relevant.  The relevant fact is
that if the unprimed observer holds an object at point x0, then the
primed observer assigns to that object a coordinate x0' which is
numerically related to x0 by x0'= x0 -vt.
What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0'
is to give the coordinate in the (x',t',)-system, it will be given by
x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest
(constant x-coordinate) is a coordinate-dependent concept.
Sounds very false. We can say that the representation of the point X0 is
the number x0 in the unprimed system, and x0' in the primed system.
Clearly x0 and x0' are different, if vt is not zero. However one may say
that (though it sounds/is stupid) the point X0 itself is the same
throughout the transformation. However that expression sounds
meaningless, since a transform (ok, maybe we should call it a change of
basis) is only a function that takes the point's representation in one
system into the same point's representation in another system. It is
preferrable to use three notations: X0 for the point itself and x0 and
x0' for the points' representations in some coordinate systems.
------------------------------
===
Subject: 9. But Doesn't x.c'=x.c?
That idea is one of the most idiotic to come up, and it does
so frequently. And in a number of guises.
The idea being that x.c' <> x.c-vt, with x.c being what
we have called x0 above; the notation makes no difference.
Some crackpots have managed to maintain that position even
after graphs have illustrated that such an idea means that
after a while a circle center represented by x.c' could be
outside the circle.
The leading crackpot just make that explicit, as far as
one can tell from his befuddled post in response to a line
about active transforms, which are actually moving body
situations, not coordinate transformations:
--------------------------------------------------------------------
e>An active transform is not a coordinate transform, ...
 Right, it is a transform of the center (in the opposite direction)
 done to effect the change of coordinates without a coordinate
 transform.  ...
E: Transform of the center?  Center of a circle?
    He really is saying a circle center moves in
    the opposite direction of the circle! Right?
--------------------------------------------------------------------
If r=10 and x.c was at x.c=0, then the points on the circle
(10,0), (-10,0), (0,10) and (0,-10) could at some time become
(-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c,
the circle center would be at (0,0) still!  The circle is here
but its center is way, way over there! Indeed, although a change
of coordinate systems is not movement of any object described in
the coordinates, the x.c'=x.c crackpottery is tantamount to the
circle staying put but the center moving away. Or vice versa.
------------------------------
===
Subject: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?
One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship
like this:
      (x-vt+vt - x.c).
See, he says, that is transforming x (with x-vt - x.c) and then
reversing the transform (x-vt+vt - x.c).
That's just another crackpot form of the idiocy that
x.c' <> x.c-vt. You'll have noticed the implication
is that there is no transform vt term relating to x.c.
------------------------------
===
Subject: 11. But Doesn't (x'-x.c+vt) Prove The Transformation
      Time Dependent?
That particular crackpottery is perhaps more corrupt than
moronic, since it includes deliberately hiding a vt term from
view, and pretending it isn't there.  [However, we have seen
above that it is a familiar incompetency, and not likely an
original.]
Look, the crackpots say, there is a time term in the
transformed (x' - x.c+vt). The transform isn't invariant!
It's time dependent!
Just put x' in its original axis form, also, which reveals
the other time term, the one they hide:
    (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c).
So, at any and all times, the transform reduces to the
original expression, with no time term on which to be
dependent.
Then there is the fact that if you leave the equation
in any of the various notation forms - with or without
reducing them algebraicly - the arithmetic always comes
down to the same as (x-x.c). That means nothing to crack-
pots, but may mean something to you.
------------------------------
===
Subject: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
My dictionary relates 'tautology' to needless repetition.
That's another form of the x.c' <> x.c-vt idiocy.
The repetition involved is the vt transformation term.
Apply the -vt term to the x term, and it is needless
repetition to apply it anywhere again? The 'again' is
to the x.c term.  The x.c' = x.c crackpot idiocy.
The repetition of the vt terms is required by the presence
of two x values to be transformed.
Be sure to note the next section.
------------------------------
===
Subject: 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of
      a Linear Transform?
Now, how on earth can we relate a tautology to a basic
definition in math?
we get this definition:
--------------------------------------------------------------
A linear transformation, A, on the space is a method of corr-
esponding to each vector of the space another vector of the
space such that for any vectors U and V, and any scalars
a and b,
 A(aU+bV) =  aAU + bAV.
-------------------------------------------------------------
Let points on the sphere satisfy the vector X={x,y,z,1},
and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1,
and b=-1.
Let A= ( 1   0   0  -ut )
       ( 0   1   0  -vt )
       ( 0   0   1  -wt )
       ( 0   0   0   1  )
A(aX+bC) = aAX + bAC.
      aX+bC  =  (x-x.c, y-y.c, z-z.c,  0  ).
The left hand side:
     A( x - x.c ,  y - y.c,  z - z.c,  0  )
     = ( x-x.c ,  y-y.c,  z-z.c,  0  ).
The right hand side:
       aAX= ( x-ut, y-vt, z-wt, 1 ).
       bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).
and
  aAX+bAC = ( x-x.c, y-y.c, z-z.c,  0  ).
Need it be said?
Sure:  QED.  On the galilean transform the
definition of a linear transform,
       A(aU+bV)=aAU + bAV,
is completely satisfied.
The generalized form transforms exactly and
non-redundantly - with ONE TRANSFORM, not a
transform and reverse transform - and non-
tautologically, just as the very definition
of a linear transform says it should.
And does so with absolute invariance, with this
galilean transformation.
------------------------------
===
Subject: 14. But The Transform Won't Work On Time Dependent Equations?
The main crackpot that has asserted such a thing was referring
to equations such as in Subject 4, above. The Light Sphere
equation; for which we have shown repeatedly elsewhere that the
numerical calculations are identical for any primed values as
for the unprimed values.
The presence - before transformation - of a velocity term
seems to confuse the crackpots. It turns out there is ex-
treme historical reason for this, as you will see in the
subject on Maxwell's equations.
------------------------------
===
Subject: 15. But The Transform Won't Work On Wave Equations?
See Subject 17, below, for a discussion of Second Derivative
forms and the galilean transforms.
------------------------------
===
Subject: 16. But Maxwell's Equations Aren't Galilean Invariant?
Oh? Just what is the magical term in them that prevents
(x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true?
It turns out not to be magic, but reality, that interferes
with the application of the galilean transforms to the gen-
eralized coordinate form(s) of Maxwell: there are no coordi-
nates to transform!
When True Believer crackpots are shown the simple
demonstration that the galilean transform on
generalized cartesian coordinates is invariant,
their first defense is usually an incredibly stupid
x0'=x0, because the coordinate of a circle center,
or point of emission, etc, is a constant and can't
be transformed.
The last defense is but Maxwell's equations are not
invariant under that coordinate transform.  When
asked just what magic occurs in Maxwell that would
prevent the simple algebra
    (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0)
from working, and when asked them for a demonstration,
they will never do so, however many hundreds of
times their defense is asserted.
The reason may help you understand part of Einstein's
1905 paper in which he gave us his absurd Special
Relativity derivation:
THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.
Einstein gave the electric force vector as E=(X,Y,Z)
and the magnetic force vector as B=(L,M,N), where the
force components in the direction of the x axis are
X and L,  Y and M are in the y direction, Z and N in
the z direction.
Those values are not, however, coordinates, but values
very much like acceleration values.
BTW, the current fad is that E and B are 'fields', having
been 'force fields' for a while, after being 'forces'.
So, when Einstein says he is applying his coordinate
transforms to the Maxwell form he presented, he is
either delusive or lying.
(a) there are no coordinates in the transform equations
    he gives us for the Maxwell transforms, where
    B=beta=1/sqrt(1-(v/c)^2):
    X'=X.                L'=L.
    Y'=B(Y-(v/c)N).      M'=B(M+(v/c)Z).
    Z'=B(Z+(v/c)M).      N'=B(N-(v/c)Y).
    X is in the same direction as x, but is not a coordinate.
    Ditto for L. They are not locations, coordinates on the
    x-axis, but force magnitudes in that direction.
    Similarly for Y and M and y, Z and N and z.
(b) the v of the coordinate transforms is in Maxwell
    before any transform is imposed; Einstein's transform
    v is the velocity of a coordinate axis, not the velocity
    he touched it.
(c) if they were honest Einsteinian transforms, they'd be
    x, which means it is X and L that are supposed to be
    transformed, not Y and M, and Z and N. And when SR does
    transform more than one axis, each axis has its own
    velocity term;  using the v along the x-axis as the v
    for a y-axis and z-axis transform is thus trebly absurd:
    the axes perpendicular to the motion are not changed
    according to SR, the v used is not their v, and the v
    is not a transform velocity anyway.
(d) as everyone knows, the effect of E and B are on the
    direction.  Both the speed and direction are changed
    by E and B, but v - the speed - is a constant in SR.
As absurd as are the previously demonstrated Einsteinian
blunders, this one transcends error and is an incredible
example of True Believer delusion propagating over decades.
The components of E and B do differ from point to point,
and in the variations that are not coordinate free,
they are subject to the usual invariant galilean trans-
formation when put in the generalized coordinate form.
-------------------------------------------------------------
The SR crackpots don't know what coordinates are. The
various things they call coordinates include coordin-
nates, but also include a variety of other quantities.
------------------------------------------------------
1.   One may express coordinates in a one-axis-at-a-time
     manner [like x^2+y^2=r^2] but it is the use of vector
     notation that shows us what is going on. In vector
     notation the triplet x,y,z [or x1,x2,x3, whatever]
     represents the three spatial coordinates, but there
     are so-called basis vectors that underlie them. Those
     may be called i,j,k. Thus, what we normally treat as
     x,y,z is a set of three numbers TIMES a basis vector
     each.
2.   These e*i, f*j, g*k products can have a lot of meanings.
     If e, f, j are distances from the origin of i,j,k then
     e*i, f*j, g*k are coordinates: distances in the directions
     of i,j,k respectively, from their origin. That makes the
     triplet a coordinate vector that we describe as being an
     x,y,z triplet; perhaps X=(x,y,z).
     The e*i, f*j, g*k products could be directions; take any
     of the other vectors described above or below and divide the
     e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)].
     That gives us a vector of length=1.0, the e,f,g values of
     which show us the direction of the original vector. That
     makes the triplet a direction vector that we describe as
     being an x,y,z triplet; perhaps D=(x,y,z).
     The e*i, f*j, g*k products could be velocities; take any
     of the unit direction vectors described above and multiply
     by a given speed, perhaps v. That gives a vector of length
     v in the direction specified. That makes the triplet a
     velocity vector that we describe as being an x,y,z triplet;
     perhaps V=(x,y,z). Each of the three values, e,f,g, is the
     velocity in the direction of i,j,k respectively.
     The e*i, f*j, g*k products could be accelerations; take any
     of the unit direction vectors described above and multiply
     by a given acceleration, perhaps a. That gives a vector of
     length a in the direction specified. That makes the triplet
     an acceleration vector that we describe as being an x,y,z
     triplet; perhaps A=(x,y,z). Each of the three values, e,f,g,
     is the acceleration in the direction of i,j,k respectively.
     The e*i, f*j, g*k products could be forces (much like accel-
     erations); take any of the unit direction vectors described
     above and multiply by a given force, perhaps E or B. That
     gives a vector of length E or B in the direction specified.
     That makes the triplet a force vector that we describe as
     being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each
     of the three values, e,f,g, is the force in the direction of
     i,j,k respectively.
Einstein's - and Maxwell's - E and B are
not coordinate vectors.
There is another variety of intellectual befuddlement that
misinforms the idea that Maxwell isn't invariant under the
galilean transform: confusions about velocities.
Velocities With Respect to Coordinate Systems.
-----------------------------------------------
Aaron Bergman supplied the background in a post to a sci.physics.*
newsgroup:
Imagine two wires next to each other with a current I in each.
Now, according to simple E&M, each current generates a magnetic
field and this causes either a repulsion or attraction between
the wires due to the interaction of the magnetic field and the
current. Let's just use the case where the currents are parallel.
Now, suppose you are running at the speed of the current between
the wires. If you simply use a galilean transform, each wire,
having an equal number of protons and electrons is neutral. So,
in this frame, there is no force between the wires. But this is a
contradiction.
First of all, the invariance of the galilean transform (x'-x.c')
=(x-x.c),  insures that it is an error to imagine there is any
difference between the data and law in one frame and in another;
the usual, convenient rest frame is the best frame and only frame
required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but
(x'-x.c')=(x-x.c).]
Second, given that you decide unnecessarily to adapt a law to
a moving frame, don't confuse coordinate systems with meaningful
physical objects, like the velocity relative to a coordinate
system instead of relative to a physical body or field.
In other words, what does current velocity with respect to a
coordinate system have to do with physics?
Nothing. Certainly not anything in the example Bergman gave.
What is relevant is not current velocity with respect to a
coordinate system, but current velocity with respect to wires
and/or a medium.  The velocity of an imaginary coordinate sys-
tem has absolutely nothing to do with meaningful physical vel-
ocity. You can - if you are insightful enough and don't violate
item (e) - identify a coordinate system and a relevant physical
object, but where some v term in the pre-transformed law is
in use, don't confuse it with the velocity of the coordinate
transform.
Velocities With Respect to ... What?
-----------------------------------------------
Albert Einstein opened his 1905 paper on Special Relativity
with this ancient incompetency:
The equations of the day had a velocity term that was taken
as meaning that moving a magnet near a conductor would create
a current in the conductor, but moving a conductor near a
wire would not.  This was belied by fact, of course.
The important velocity quantity is the velocity of the
magnet and conductor with respect to each other, not to
some absolute coordinate frame (as far as we know) and
not to an arbitrary coordinate system.
One possible cause was the idea: but the equation says the magnet
must be moving wrt the coordinate system or ... the absolute
rest frame.
There not being anything in the equation(s) to say either of
those, it is amazing that folk will still insist the velocity
term has nothing to do with velocity of the two bodies wrt
each other.
-----------------------------------------------------------
------------------------------
===
Subject: 17. First and Second Derivative differential equations.
One of the intellectually corrupt ways of
denying the very simple demonstration of
galilean invariance of all laws expressed
in the generalized coordinate form demanded
by analytic geometry, vector analysis, and
measurement theory
    [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ]
is the assertion that those equations 'over there'
(usually Maxwell or wave) are somehow immune to
the elementary laws of algebra used to demon-
strate the invariance.  [Unfortunately, the
assertions are never accompanied by reference
to the magical math that makes elementary al-
gebra invalid.  Wonder why that is?]
Part of the time it is based on the old lore
based on the incompetent transformation of
the privileged form of an equation instead
of the correct form. [Evidence of this is
any reference to an effect due to the velocity
of the transform; it falls out algebraicly
- as you see above - and cancels out arith-
metically - as you can see above.]
But usually it is just whistling in the dark,
waving the cross (zwastika, I'd say) at
the mean old vampire.
The most general equation that could be conjured
up is a differential with either First or Second
Derivatives.
Let's examine the plausibility of such magical
magical, non-invariance assertions.
(a) to get a Second Derivative you must have
    a First Derivative.
(b) to get a First Derivative you must have
    a function to differentiate.
(c) to get a Second Derivative you must have
    a function in the second degree.
So, let us examine the question as to whether
any such common Maxwell/wave equation will
differ for
(a) the common, privileged form, represented
    as ax^2, with a being an unknown constant
    function.
(b) the generalized cartesian form, represented
    as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2,
    with a being an unknown constant function.
(c) the transformed generalized cartesian form,
    represented as a(x-vt -x.c+vt)^2, same as for
    (b), = ax^2 -2ax(x.c) + ax.c^2, of course,
    with a being an unknown constant function.
I.  for (a), remembering that x.c is a constant,
    and that this version is only correct because
    x.c=0, otherwise (b) is the correct form:
     d/dx    ax^2  = 2ax
    (d/dx)^2 ax^2  = 2a
II. for (b), remembering that x.c is a constant.
     d/dx    (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c
    (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a
III. for (c); same as for (b).
So, what we have seen so far is
(1)  differential equations in the second degree
- the wave equations - must clearly be the same for
all forms: the privileged form in x, the generalized
cartesian form in x and the centroid, x.c, or the
transformed generalized cartesian form.
That is, anyone who imagines that correct usage
gives different results for galilean transformed
frames is at first showing his ignorance, and in
the end showing his intellectual corruption.
(2) As far as the First Derivatives are concerned, the
only cases in which there really is a difference between
the two forms is where x.c <> 0, and in that case, the
use of the privileged form is obviously incompetent.
So, how do you correctly use the differential equations?
If you are using rest frame data with the centroid
at x=0, etc, you can't go wrong without trying to
go wrong.
If you are using rest frame data with the centroid
not at x=0, you must use (x-x.c) anyplace x appears
in the equation.
If you are using moving frame data, you must use the
moving frame centroid as well as the light front
(or whatever) moving frame data itself, perhaps first
calculating (x'-x.c'), which equals (x-x.c) which is
obviously correct, and which is obviously the plain old
correct x of the privileged form.
Unless, of course, there really is some magical term
or expression that invalidates the obvious and elemen-
tary algebra of the invariance demonstration.
Or maybe you just whistle when you don't want basic
algebra to hold true.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus        Oren C. Webster         ThnkTank@concentric.net  ?
! Anything and everything that requires or encourages systematic   ?
!  examination of premises, logic, and conclusions                 ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Supersedes: 

===
Subject: (SR) Lorentz t', x' = Intervals
Summary: The Lorentz transforms themselves are proof t' and x'
  cannot possibly be just coordinates.  Examination
  of their derivation verifies their identity as 
  intervals.
Originator: faqserv@penguin-lust.MIT.EDU
Disclaimer: approval for *.answers is based on form, not content.
    Opponents should first actually find out what the content is, 
    then think, then request/submit-to arbitration by the 
    appropriate neutral mathematics authorities. Flaming the hard-
    working, selfless, *.answers moderators evidences ignorance 
    and atrocious netiquette.
Version: 0.02.1
Archive-name: physics-faq/criticism/lorentz-intervals
Posting-frequency: 15 days
       
    (SR) Lorentz t', x' = Intervals                    
     (c) Eleaticus/Oren C. Webster
        Thnktank@concentric.net
------------------------------
===
Subject: 1. Introduction with the obvious debunking
     of the use of 'just coordinates' in any
     scientific formula.
Defenders of the Special Relativity faith are especially
fond of telling opponents of their space-time fairy tales
that they do not know the difference between coordinates
and magnitudes.  
That may often be so, but the fault lies ultimately with 
SR dogma. The Lorentz-Einstein transformations cannot 
possibly be 'just coordinates', which is the interpre-
tation required to support the many sideshow carnival acts
with which the SR faithful bedazzle the public, and establish
their moral and intellectual superiority.
If I get in my car and drive steadily for a few hours at 50 
kilometers per hour, is 50t the distance I travel? 
Of course not. The last time my hours-counting 'just coord-
inates' clock was set to zero was when Zeno first reported 
one of his paradoxes to Parmenides. 
That was a long time ago,  so my t is not useful for such 
purposes unless you also use my clock to established the starting 
time, perhaps t0,  and use the formula 50(t-t0) to calculate the 
distance. 
In any case, my t is even then not 'just a coordinate' because
it always represents particular elapsed times that can be
used in the (t-t0) form to calculate perfectly good time
intervals (elapsed times).
Alternatively, I could (re)set my clock to zero at the start
of some meaningful time interval, in which case my t shows a 
scientifically perfect current and/or end time. 
In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function 
of an elapsed time interval (not 'just a coordinate') and a time 
interval (-vx/cc; the interval amount the t' clock is being 
screwed up at time t) and thus cannot be 'just a coordinate' 
since neither of the independent variables is such a 'just' thing. 
{Their meaning is shown below, step-by-step.]
If it takes me 50 minutes to cross the Interstate highway,
was x/50 my velocity crossing it?
Of course not.  The origin of all my axes is at the very
spot where Zeno first presented his first paradox to 
Parmenides. That makes my x equal a couple of thousands of
miles, plus, and is not useful for such purposes unless 
you establish the starting x value, perhaps x0, and use the
formula (x-x0)/50 to calculate my velocity.  
In any case,  even then my x is not 'just a coordinate' 
because it always repesents particular distance intervals
that can always be used in the (x-x0) form for any and every
scientific purose.
Alternatively, I could move my x-axis origin to the starting
(zero) point of some meaningful distance, in which case my x 
shows a scientifically perfect current and/or end distance.
In which case, the Lorentz-Einstein x'=(x-vt)/g is a function 
of a current/ending distance interval (not 'just a coordinate') 
and a distance interval (-vt; the interval amount the x' axis
is being screwed up at time t) and thus cannot be 'just a coordinate' 
since neither of the independent variables is such a 'just' thing. 
{Their meaning is shown below, step-by-step.]
------------------------------
===
Subject: 2. Table of Contents
 
 1. Introduction with the obvious debunking
    of the use of 'just coordinates' in any
    scientific formula.
 2. Table of Contents.
 3. The Lorentz-Einstein transforms.   
 4. The 'just coordinates' argument.
 5. Single-system, little-purpose ambiguity.
 6. Relating two coordinate measures/systems. 
 7. Distances and moving coordinate axes.
 8. Time intervals.                          
 9. Einstein's (1905) derivations.
       10. A word about intervals.
       11. Intervals versus the Twins Paradox.
       12. Summary
------------------------------
 
===
Subject:  3. The Lorentz-Einstein transforms
Special Relativity's space-time circus is based on
the 'transformation' equations by which it is believed
one can relate a nominally 'stationary' system's space
and time coordinates to those of an inertially (not
accelerating) moving other observer. 
That moving observer's own physical body and coordinate 
system might have been identical in size to those of the 
stationary observer before the traveller began moving, 
but are 'seen' as very different by the stationary observer 
when the relative velocity of the two is great enough, a 
high percentage of the velocity of light.
Concerning ourselves - as is customary - with just
the spatial coordinate axis that lies parallel to
the direction of motion, and with time, Einstein
arrived at these formulas that relate the moving
system measures or coordinates (x' and t') to the
stationary system coordinates (x and t):
      x' = (x - vt)/sqrt(1-vv/cc)      (Eq 1x)
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
The v is for the two systems' relative velocity as seen 
by the stationary observer, and is positive if the dir-
ection is toward higher values of x.  By concensus,
the moving system x'-axis higher values also lie in
that direction, and all axes parallel the other system's
corresponding axis.
We used vv to mean the square of v but might use v^2
for that purpose below. Similarly for c.
Because it is believed that no physical object can
reach or exceed c, the square-root term in both
denominators is presumed always less than one, which 
means that the formulas say both x' and t' will tend to
be greater than x and t, respectively.  However,
SRians call the x' result 'contraction' - which means
shortening - and the t' result 'dilation' - which
means increasing. 
------------------------------
 
===
Subject:  4.  The 'just coordinates' argument
The 'just coordinates' argument is so patently ridiculous
that even opponents have a hard time accepting just how
simple and obvious its debunking can be, as shown in this
section.  However, further sections take a more arithmet-
ical approach that you'll maybe find more professorial.
The 'just coordinates' argument is that t is mot a
duration, not a time interval; it's just an arbitrary
clock reading.  But what if the moving system observer
comes speeding by while you make your annual 'spring
forward' or 'fall back' change?  The formula says that
the moving system clock's 'just coordinate' reading 
can be calculated from yours:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Imagine the moving system oberver's confusion if his 
clock changes its reading while he's looking at it!  
If his clock doesn't change when yours does, the formula 
is wrong;  if it is truly a 'just coordinates' formula. 
And then what happens if you realize you were a day 
early and put your clock back to what it had said 
previously?
And suppose you are in NYC and your twin in LA and
both are watching the moving observer. You'll both be
using the same v because you are at rest wrt (with
respect to) each other. You're on Eastern Standard
Time and your twin is on Pacific Standard Time
maybe. You have three hours more on your clock than 
does your twin. 
On which 'just coordinate' clock will the Lorentz 
transforms base the 'just coordinate' time the moving 
system clock says?  The formula applies to both of
your t-times:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Sure, the idea that you can change someone else's
clock with no connection of any kind is really 
ridiculous, but Eqs 1x and 1t aren't MY equations. 
Are they yours?  And we aren't the ones to say x, t, 
x', and t' are just coordinates.
If the t' formula is actually either an elapsed
time formula, or the basis of a t'/t ratio, then
there is no implication that one clock's reading
has anything to do with the other's. 
It can only be rates of clock ticking, or how one 
time INTERVAL compares to the other that the formula 
is about.
------------------------------
 
===
Subject:  5. Single-system, little-purpose ambiguity.
Since we're going to be comparing measurements on two
coordinate systems in the next section, let's go to
our supply cabinet and get our yard-stick (which we
use to measure things in inches) and our meter-stick
(which we use to measure things in centimeters).
Here, I'm getting mine. Oh! Oh!
There's an ant on mine, and he ... she ... sure is
hanging on, right at the 3.5 inch mark of the yard-
stick.
Let's see if I can wave the stick around enough that
she'll let go. Nope.
However, before I gave up I waved the stick and the
ant 'all over the place.
Always, however, the ant was at the 3.5 mark on the 
yard-stick, and always 3.5 away from the end of the
stick, however far and wide I have transported her.
Neither of those 3.5 facts means very much. Of the
two, the distance aspect meant almost nothing. So
the distance was 3.5 from the end. So what? That
length, distance, was not in use. And only maybe
the ant might have been concerned with just what
location, 'just coordinate', on the stick she was
at.
Just so with x and t.
So, is the 3.5 reading just a coordinate? Or a
distance/length?  It's ambiguous in and of itself,
and really makes no difference what you say until
you try to make use of the number. 
Hey, my address is 5047 Newton Street. If you
are looking for me and you're at 4120 Newton, it
is helpful information, because it tells you which
direction to go.  Is that 'just coordinate'? 
Where it really becomes useful, perhaps, is in
telling you how far away I am. That's not just
a coordinate value, that's a distance, length,
interval.
However, it is subtracting 4120 from 5047 that 
tells you which direction and how far. It is only 
because both 5047 and 4120 are distances from the 
same point - ANY same point - that the result means 
anything.
My x - my yardstick reading - is always a distance
or length; it is impossible to be otherwise with
an honest, competently designed yardstick.
Whether or not its reading is of good use in some 
particular scientific formula depends on whether 
I put the zero end of the yardstick at some useful
place. As in the introduction, we should either
put it at the starting location/end, or use two
readings from it: (x-x0).
------------------------------
===
Subject:  6.  Relating two coordinate measures/systems.
Taking care to not damage our brave little ant, I place
my yard-stick onto the table, zero end to the left, 36
end to the right.
Now I place the 'just coordinate' meter-stick on the table
in the same orientation,  in a random location, and find
that the ant's coordinate on the meter-stick is 51.
The formula relating centimeters to inches is cm=i*2.54
but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc).
That would be c=i/.03937 approximately, but let's use x'
for the meter-stick reading, and x for the inch reading:
    x'=x/.3937.
    3.5/.3937 = 8.89  
    
Wait a minute.  It's not just science but definition 
that says c=i/.3937=8.89, so something is wrong.  8.89
is not 51.
We already knew that 51 cm was just an arbitrary coordinate. 
Arbitrary not because that point isn't 51 cm from the zero 
end of the meter-stick, but because the zero point was in an 
arbitrary position.
Let's put the meter-stick in a position where it's 
zero point is at the yard-stick zero point.
What is the centimeter coordinate now?  Hey. 8.89,
just like the formula says.
The only way for a 'transform' like x'=x/g to work, 
whatever g might be, is for both coordinate systems
to have their zero points aligned, in which case
saying the two measures are not intervals is pure
idiocy.
Noe that with both zero points at the same position
both x' and x are great measures for scientific
purposes, in any and every case where we were smart
enough to put those zero points at a useful location.
There is one extension of x'=x/g that will let us
use the meter-stick in arbitrary position. 
When the cm reading was 51, the zero point of the
yard-stick read (51-8.89=) 42.11 cm. If we call that
point x.z' we get 
     x' = x.z' + x/.3937.
 = 42.11 + 3.5/.3937
 = 42.11 + 8.89
 = 51.
Obviously, in this formula x/.3937 is the distance
from the x' coordinate of the location where x=0. 
An interval.
Just as obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate, does not mean that x'
is anything more than nonsense for use in any
scientific formula.
Unless we were smart enough to put the x zero
point in a useful location, and use (x'-x.z') in
the scientific formula. (x'-x.z') equals the useful,
Ratio Scale value x/.3937.
So, we have discovered a basic fact: a transformation
formula like x'=x/g works only if the two zero points
of the coordinate systems coincide. That makes it non-
sense to say the two coodinates are only coordinates
and not intervals.  Both must be values that represent
distances from their respective zero points unless you
take the proper steps to adjust for the discrepancy.
Make sure you understand that although the inclusion
of x.z' made it possible to correctly calculate x',
the result is nonsense when it comes to use of x'
for general length/distance purposes; it is x'-x.z' 
that is a useful number in such cases. It could be
that we're measuring a sheet of paper with one end
at x=0 and the other at x=3.5; x'=51 is nonsense as
a centimeter measure of the paper.
But, you say, the Lorentz transform contain a -vt term.
------------------------------
 
===
Subject: 7.  Distances and moving coordinate axes.
We discovered x'=x.z' + x/g as the correct formula
for relating one coordinate to another system's.
But the Lorentz transform contains another term, 
-vt/sqrt(1-vv/cc). What is it?
Let's start with our x'=51 cm, x=3.5, x.z'=42.11 example.
Every minute, let's move the meter-stick one inch to our
right.
At minute 0, the cm reading  was   51 cm.
At minute 1, the cm reading is now 50 cm.
At minute 2, the cm reading is now 49 cm.
In this instance, v=1 inch/minute. And t was 0, 1, 2.
What has happened is that we have made our x.z' a lie,
and increasingly so.  -vt/.3937 is the change in x.z'.
   x' = (x.z - vt/.3937) + x/.3937.
Obviously, vt/.3937 is not a coordinate; even most SRians
wouldn't imagine it was. It is an interval, the distance
over which the moving system has moved since t=0.
And, of course, x/.3937 is the distance of our brave
little ant from the point where x=0 and the centimeter
reading is x.z'-vt/.3937. Yes, every minute the meter-
stick moves to the right and the meter-stick coordinate
of the spot where x=0 gets less and less - and eventually
negative.
Make sure you understand that every minute the x' 
coordinate, because of -vt/g, becomes a better measure 
of, say, the  3.5  paper we might be measuring with 
the yard-stick, given that 51 was too big a number and
-vt is negative.  That is, until the two origins coincide 
at x'=x=0,  and then it gets worse and worse.
With -vt positive (because v<0) the situation is different.
With 51 and -vt positive, x' just gets worse and worse
over time.
Quite obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate even when the x' axis is
moving, does not mean that x'is anything more than 
nonsense for use in any scientific formula.
Unless we were smart enough to put the x zero point 
in a useful location, and use (x'-x.z'+vt/.3937) in
the scientific formula. (x'-x.z'+vt/.3937) equals the 
useful, Ratio Scale value x/.3937.
------------------------------
 
===
Subject: 8. Time intervals.
Instead of using our sticks, let's get out two clocks.
Mind you, we're not going to deal with different clock
rates here, just establish the same basics as for distance.
Your clock says 9:00 Eastern Standard Time (EST) and we 
note that t=540 minutes when we put down the clock.
Blindly, let's turn the setting knob of your twin's Pacific 
Standard Time clock and put it down before us.
According to what we see, EST's 540 minutes (9:00) corre-
sponds to PST's 14:30; t'=870.
We know the formula relating PST to EST is t' (pacific)
= t (eastern) - 180 (minutes). Thus, it is not correct 
that the second clock can have an arbitrary setting, 
because 870 <> 540-180. 
We know that the two clocks are related by t' = t/1 since 
both are using the same second, hour, etc units. But 870 
(14:30 in minutes) is not 540/1-180, so once again we know 
something is wrong. 
However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 
(midnite)  - 180,  so t.z' = -180, and
     t' = -180 + 540/1  = 360.
Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes.
We see thus that like distance measures/coordinates, time
axis origins (zero points) must either be 'lined up' or 
adjusted for. 
So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving 
system elapsed time interval since the time axes were both at 
a common zero. There is no t.z' adjustment:
      t' = (t - vx/cc)/sqrt(1-vv/cc)   (Eq 1t)
Make sure you understand that in the clock case, if the
EST is showing a good number for elapsed time since the
travelling observer passed NYC, then the PST clock is
silliness. t.z' must be zero or must be taken out of
time lapse calculations for the PST clock to be used
intelligently, just as was true for x.z'.
What is lacking as yet for Lorentz t' is the -vx/cc term that
corresponds to the x' formula -vt term.
Break it up into two parts: v/c and x/c.  
v/c is a scaling factor that changes velocity from whatever 
kind of unit you are using over to fractions of c.
x/c is distance divided by velocity, which is time. x/c
is thus the time interval since the two time axes
had a common zero point - which they have to have in the
Lorentz transforms which do not have the t.z' term we
learned to use above.
Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the 
moving system clock has been changed - since the common/
adjusted time - over and beyond the elapsed time interval
represented by x/sqrt(1-vv/cc).
We have discovered that the only way for t' to be t/g
is for t' and t to have a common zero point, just as
for x' and x. It would be otherwise if the t' formula
contained an adjustment t.z' under some name or other,
but the necessity to include such a term correlates
100% with t' numbers that aren't directly usable.
As for x and x', our knowledge of how to setup a proper
formula relating t and t' is of no use unless we use
the knowledge in scientific formulas; (t'-t.z'+xv/gcc)
gives us the only directly useful value: t/g.
------------------------------
 
===
Subject: 9.  Einstein's (1905) derivations.
When we return to Einstein's derivations of the transform
formulas with a well-focused eye, we find he was a wee bit
confused - or at least self-contradictory.
When he set up his (at first unknown) tau=moving system
time formulas, he created three particular instances of tau.
Tau.0 is the time at which light is emitted at the moving
origin toward a mirror to the right that is moving at rest 
wrt that moving origin and at a constant distance from that 
origin. He lets the stationary time slot have the value t,
a constant, the stationary system starting time.
Tau.1 is the time at which the light is reflected. He
lets the stationary time be t+x'/(c-v); t is still a
constant and x'/(c-v) is the time interval since t.
Tau.2 is the time at which the light gets (back) to the
moving origin. The stationary time value is put as t +
x'/(c-v) + x'/(c+v);  t is still a constant and x'/(c-v)
+ x'/(c+v) is the time interval since t.
On the thesis that the moving observer sees the time to
the mirror as the same as the time back to the origin,
he sets
   .5[ tau.0 + tau.2 ] = tau.1.
Tau.0 completely drops out of the analysis and leaves
no trace, and has no effect.
Further, the t you see in tau.0, tau.1, and tau.2 also 
completely drops out with no trace and no effect, leaving 
us with exactly what you'd get if you had explicilty said 
t' is an interval and so is t.
What doesn't drop out in the stationary time values is
x'/(c-v) and x'/(c+v), the time interval it takes for
light to get to the fleeing mirror, and the time interval
it takes for light to get back to the approaching origin.
Thus, his resultant t' formula is strictly based on time 
intervals in the stationary system. Time intervals since 
some starting time, yes, but time intervals.
There is absolutely nothing in the derived formulas that 
depends on arbitrary coordinates like the constant t in 
the stationary time arguments.
Let's look at the x dimension; it is x'=x-vt [as x increases
by vt, the effect over time is x'=(x+vt)-vt)], which Einstein 
explicitly sets up as a constant stationary distance.
He uses that x' not just in the time interval parts of the 
stationary time arguments,  but also in the x (distance) 
stationary system argument for the tau at the time light 
is reflected. 
x' can't be the stationary system coordinate of the mirror 
at that time.  That value is x'+vt.
x' is explicitly an interval, distance. 
Thus, the whole tau derivation of the t' formula is fully and
explicitly based on x'  - a spatial length/distance/interval -
and the two time interals x'/(c-v) and  x'/(c+v).
While we're at it, if the starting t is not zero, his 
x'=x-vt formula is complete nonsense also. Given that
there was some L that was the mirror x-location and length
when the light is emitted, if t was already, say, 500, then
x'=L-vt could have been a very negative length.
------------------------------
 
===
Subject: 10.  A word about intervals.
There are intervals, and there are intervals.
If we put our yard stick zero point at one end
of a piece of paper and read off the coordinate
at the other end of the paper, we have a good
measure of the paper's length, a Ratio Scale
measure. [Absolute temperature scales are ratio
scale.]
If instead we put the one end of the paper at the
one inch mark (or the zero end of the stick one
inch 'into' the length of the paper) we get measures
that are one inch off the true, ratio scale length.
The two messed up measures are still intervals,
but they are Interval Scale measures. [Household
temperature scales are interval scale, which is
why your physics and chemistry professors won't
let you use them without first converting to the
ratio scale absolute temperatures.)
t'=t/g and x'=x/g represent ratio scale measures,
given that t and x were ratio scalae to start with.
t'=t.z'+t/g and t'=t/g-vx/gcc are both interval 
scale measures, even given a good ratio scale t
and a good ratio scale x.
x'=x.z'+x/g and x'=x/g-vt/g are both interval 
scale measures, even given a good ratio scale x
and a good ratio scale t.
Look for the (SR) Lorentz t', x' = degraded measures
document soon at a newsgroup near you.
------------------------------
 
===
Subject: 11. Intervals versus the Twins Paradox.
t'=(t-vx/cc)/g shows t' being greater than t.
The reason Special Relativity will not allow the
use of its basic time equation in determining what
SR has to say about the twins' ages, is that t' and
x' are supposedly just coordinates, and they say you 
have to take the coordinate pairs (t',x') and (x,t)
into consideration in both the time and place the 
twins' separation started and the time and place the 
twins reunited.
Since t' and x' are actually both intervals, not
just coordinates, the 'excuse' is spurious, and is 
so even without use of the obvious (x_b-x_a) and
(t_b-t_a) usages.
However, SR is right to be embarrassed by their
transformation formulas.
Look for the (SR) Lorentz t', x' = degraded measures
document at a newsgroup near you.
 
------------------------------
 
===
Subject: 12. Summary
A.  t'=t/g and x'=x/g can be almost 'just coordinates'
    in the sense that the values obtained may not be
    of much use except in the most primal and useless
    way: how long and how far since/from the time/
    place they were zero. Even here, however, the zero
    points within each of the two scale pairs (t',t) 
    and (x'.x) must have been lined up.  If the zero
    points have been intelligently selected (such as
    at the starting point and time of a trip) they 
    can be rationally used 'as is' in any valid sci-
    entific equation.
B.  Even the interval scale t'=t.z' - xv/gcc + t/g and 
    x'=x.z' - vt/g + x/g are not 'just coordinates'. They 
    can be used to good effect by establishing the relevant 
    starting times/points and using (t'-t.z'+xv/gcc) and 
    (x'-x.z'+vt/g), as the situation may require.
C.  When you see vx/gcc or vt/g in use in any guise with non-zero
    values, you know the resultant t' or x' is a degraded, interval
    scale value.
E-X: Anytime you do not see what amounts to t.z' and xv/gcc in
     the time case, or x.z' and vt/g in the distance case, you
     know that the t' and/or x' in use are intervals. Period.
Y:   Either set your clock to zero at the start of the relevant
     time interval, or use (t-t0), with both being readings on
     the same clock. Either move your x-axis origin to the starting
     end or point, or use (x-x0), with both being readings on
     the same axis.
Z:   In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see 
     that t' and x' satisfy the mathematical tests for/of interval
     scales when -vt and -vx/cc are not zero; thus, they must
     be intervals. When -vt and -vx/cc are zero, t' and x'
     satisfy the much better mathematical definition of 
     ratio scales, and are thus not just mere intervals,
     but (rescaled) good ones.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus        Oren C. Webster         ThnkTank@concentric.net  ?
! Anything and everything that requires or encourages systematic   ?
!  examination of premises, logic, and conclusions                 ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
===
Subject: details of joke wanted
Can somebody help me recall the details of an old maths joke? It is one of
those situations where a computer scientist, a physicist and a 
mathematician
(or some such combination) have to solve a certain problem, and the
mathematician starts by doing something crazy in order to reduce to a
problem that is already solved.
Derek Holt.
===
Subject: Re: Calculating the expected range of results
> I have a normal distribution of known mean and standard deviation.
> 
> For a certain case, a finite number of results will be drawn from this
> distribution.  Is there a mathematical formula for calculating the
> expected range of these results?
> What exactly do you mean by expected range?
> I could interpret the question literally: What is the
> expectation value of the range?
> Suppose you are drawing n independent samples, X_1, X_2,...
> ,X_n.
> Here's the cdf of the maximum of the X's:
> P[max(X_1, X_2, ..., X_n)<=x] = 
>    = P(X_1 <=x & X_2 <=x & ... & X_n <=x]
>    = P(X_1 <= x)^n
> So the pdf is p(x)= dP/dx
>     = n*P_x(x)^(n-1)*p_x(x)
> where P_x(x) and p_x(x) refer to the normal distribution
> of individual samples.
> The expectation value is therefore
> integral(-inf,inf) n*P_x(x)^(n-1)*p_x(x)*x dx
> So you can in principle calculate E[max(x_i)].
> Similarly, you can work out a formula for E[min(x_i)]
> in terms of P(x_i >= x) = 1 - P_x(x).
> Thus, E[max - min] = E[max] - E[min].
>            - Randy
===
Subject: Re: Number Of Squares Within A Square (nxn)
Dave Fleet
> For my homework we had been set the question of finding the numbers of
squares within a square.
> I managed to work out an answer for myself but out of interest found your
Website and a really neat solution that had been posted sometime ago.
> However, I have some problems with the proof.
> Please could you explain how to get the Sum(n^2)=(n+1)n(2n+1)/6
> I'd be really grateful for your help.
> Dave
Induction. Write f(n) = (n+1)n(2n+1)/6, and verify
-- f(n)-f(n-1)=nn
-- f(n)=1
as desired.
LH
===
Subject: Re: uniform convergence & differentiation
> In many analysis-type texts I've found statements like since the series
> converges uniformly, we can apply term-by-term differentiation and 
uniform
> convergence allows us to differentiate term-by-term, but I've never
> actually seen a proof or explanation of WHY you need uniform convergence 
to
In fact, even supposing that you have uniform convergence, it may happen
that you cannot apply term-by-term differentiation. Take, for instance,
f_n(x) = sqrt(x^2 + 1/n). Then the sequence f_n converges uniformly, in
R, to |x|, but |x| is not differentiable at 0.
However, when the sequence of the derived functions f_n' converges
uniformly to some function g then, yes, f is differentiable and f' = g.
You'll find a proof of this in many standard Calculus textbooks, such
as Spivak's Calculus or Rudin's Principles of Mathematical Analysis.
Jose Carlos Santos
===
Subject: Banach limit and almost convergence
Hi!
It was noted several times in Google Groups (by Ronald Bruck), then
uniqueness of Banach limit of a sequence is equivalent to almost
convergence of this sequence. Can you give me a hint where to find a
proof of this result. (Web resources are prefered, but also book or
paper in a journal would help.)
Martin Sleziak
===
Subject: Re: details of joke wanted
mareg@mimosa.csv.warwick.ac.uk scribbled the following:
> Can somebody help me recall the details of an old maths joke? It is one 
of
> those situations where a computer scientist, a physicist and a 
mathematician
> (or some such combination) have to solve a certain problem, and the
> mathematician starts by doing something crazy in order to reduce to a
> problem that is already solved.
I heard the one about the house on fire. The mathematician strikes a
match and sets a piece of paper on fire. Then he gets a glass of water,
and empties its contents on the burning paper. The paper stops burning,
turning into a black mess of coal. Ah, thinks the mathematician, a
solution exists. Then he goes back to bed.
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
That's no raisin - it's an ALIEN!
   - Tourist in MTV's Oddities
===
Subject: Re: Vedic Mathematics ---  Myth and Reality
>Do you know how old exactly?
>And... do you have a reference for that book? (Online somewhere?)
>The 'one-register' method of multiplication is indeed very nice.
>But... it is arithmetic, rather than mathematics. That may also explain
>why the math teaching world is not very fond of tricks like this.
>Mathematics teachers are desperately fighting the general prejudice
>that mathematics is 'just' arithmetic.
> I doubt this had anything to do with Vedic mathematics (though I may
> just be unaware of the sources).  It was a system partly invented (or
> perhaps merely collected) by Jacow Trachtenberg in the twentieth
> century.  The standard reference in English is The Trachtenberg Speed
> System of Basic Mathematics by Cutler
> 
 original German (I think).   You can find it at Amazon.
> The descriptions available on the web (search for Trachtenberg and
> arithmetic) mainly talk about tricks for multiplying by single digits,
> 11, and 12, which is the first part of the book.  Later chapters discuss
> techniques of long addition, multiplication, and division.
But actually, i really *am* curious about what else there is in
Vedic maths, not just the single-register multiplication.
In the meantime i've found online sources on vedic mathematics.
It appears to be (re)discovered not before 1910 or so.
Convolution is older.
Nevertheless, for historic reasons it could be worth a look.
However... all they ever show is their 'crosswise-horizontal'.
If you want to know the rest, they make you buy books.
Another typical line: sorry, i'm no expert, so i'm unable
to tell more; but it's all there, in our books / courses.
Not very trustworthy...
Herman Jurjus
> -- 
> Stephen J. Herschkorn                        herschko@rutcor.rutgers.edu
===
Subject: Re: Greek Alphebet
...
 > How are numbers written in Hebrew?  I remember seeing people writing
 > right to left (so presumably Hebrew or Arabic) but using our numerals
 > and writing them left to right.  So they had to jump ahead a bit to
 > write the numbers.
You do not have to jump ahead.  You just first write the units, the
tens next, etc.  This is not different from the way we write the
result of an addition on paper.
-- 
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: Probability of a Run
> I am trying to calculate the probability that a gambler with capital 
C
> and who uses a Martingale betting strategy will be wiped out in m
> turns at the game. This would happen with a run of n consecutive
 process approach to this problem; this has come up before on sci.math
> in the context of coin flipping.
> 
 that you have a 50/50 chance of never getting 30 consecutive events in
> m trials.
> 
> I like the simplicity of the idea behind this method of calculation.
> There are no cancellation problems since we are always adding. It
> actually executes reasonably quickly but I appear to be getting a lot
> of rounding error problems.
> My first reaction was to check my results. I added an extra row and
> column to get a double check on my results.
I assume you mean here an extra r/c of 0's (equivalent to a
forbidden state of s_n)? I don't see how that would change the
result...
> I got 
> .499999999139885 from adding up the values for 0 to 29
> .500000002118337 from the extra row/column
> and
> .500000000053831 via ProbnmpApprox.
> I'm pretty sure that .500000000053831 is accurate, but the rounding
> errors via the matrix multiplication method seems a bit drastic.
> Can you supply the values you got, if possible it would be nice if you
> could add the extra row to confirm whether the problem is genuine or
> in my code.
I get 0.499999997774158 regardless of whether or not I add the extra
row/column; but I'm not surprised that they are the same - adding an
extra row or column of zeros _shouldn't_ make any difference (which
doesn't mean that it _wouldn't_).
The closest I get to 0.5 is actually at 1488522233 trials, with a
result of 0.500000000102465 (I had previously used the same number of
trials as the other poster had).
As another check, I get that the matrix A refered to above has
(A^(2^30))[0,0] = 0.389400394801501.
I can't really attest to the accuracy of the implementation, as I
coded this up in (believe it or not) Lingo, which is the scripting
language of macromedia's director/shockwave. It's great as a quick
prototyping, interactive language; although slow as mollasses for
mathematical use (it took about 10 seconds to calculate the above). At
some base level though, it surely uses native floating point; so it
should be as accurate as a one would expect.
I'll code it up in C when I get a chance and post the results.
> Ian Smith
===
Subject: Re: uniform convergence & differentiation
burger
> In many analysis-type texts I've found statements like since the series
> converges uniformly, we can apply term-by-term differentiation and
uniform
> convergence allows us to differentiate term-by-term, but I've never
> actually seen a proof or explanation of WHY you need uniform convergence
to
Uniform convergence is sufficient, but not necessary, for the conclusion. I
Spiegel, in the Schaum's Outline series. (Look for uniform convergence 
in
the index.) You're likely to find a stronger form of the theorem in a
reference about Banach spaces; e.g. it's in Dieudonn.8e _Treatise on 
Analysis_
section 8.6.5.
Larry
===
Subject: Re: details of joke wanted
> Can somebody help me recall the details of an old maths joke? It is one 
of
> those situations where a computer scientist, a physicist and a 
mathematician
> (or some such combination) have to solve a certain problem, and the
> mathematician starts by doing something crazy in order to reduce to a
> problem that is already solved.
The ones with the compsci, physicist and mathmatician are as follows:
} A computer scientist, a physicist and a mathmatician are woken up in the
} middle of the night by fires in their rooms.
}
} The computer scientist throws water on the fire until it goes out, and
} goes back to bed.
}
} The physicist works out exactly how much water he needs to put out the
} fire, puts it out and goes back to bed.
}
} The mathmatican considers the above problem, shows that a solution
} exists, and then goes back to bed.
} A computer scientist, a physicist and a mathmatician take a day trip
} to Wales. The compsci spots a black sheep in a field.
}
} 'Look!', he exclaims, 'All sheep in Wales are black!'
}
} 'Not quite', says the physicist, 'there is at least one black sheep in
}  Wales'.
}
} 'Not quite', says the mathmatician, 'there is at least one sheep in
} Wales which is at least one-half black'.
} Q: What is pi?
}
} Compsci: about 3
} Physicist: 3.14159 +/- 0.000005
} Mathmatician: the ratio of the circumference of a circle to its
} diameter.
There exist many variants of these.
The 'reduction to previous problem' joke is:
} 1. You are given a hose, a fire hydrant, and a house on fire. What do
} you do?
}
} PHYSICIST: connect the hose to the hydrant, turn it on, put out the
} fire.
} MATHEMATICIAN: connect the hose to the hydrant, turn it on, put out the
} fire.
}
} 2. You are given a hose attached to a fire hydrant and a house on fire.
} What do you do?
}
} PHYSICIST: turn on the hydrant.
} MATHMATICIAN: disconnect the hose, reducing the problem to the previous
} case.
}
} 3. You are now given a hose, a fire hydrant, and a house not on fire.
} What do you do?
}
} PHYSICIST: Nothing.
} MATHMATICIAN: Set fire to the house.
There are many variants of this.
-- 
P.A.C. Smith
The vast majority of Iraqis want to live in a peaceful, free world.
And we will find these people and we will bring them to justice.
===
Subject: Re: Basic factorization ideas
...
 > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
 >                           49(x^3 + 5x^2 + 3x + 1)
 > with the c's algebraic integers, I think few of you would have a
 > problem realizing that only two of the c's have 7 as a factor.
 > If c_1 and c_2 (and hence c_3) are ordinary integers, then
 > yes, c_1 and c_2 must be divisible by 7, and c_3 must be
 > coprime to 7 (does this hold for algebraic integers as well?)
No.  For some x, (c3 x + 1) is actually *not* coprime to 7.  The reason
is that for those x, (x^3 + 5x^2 + 3x + 1) is divisible by 7, and that
additional 7 has to go somewhere (it is distributed among the three
factors).
-- 
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: Marketing shift, core issues
> 
>I've been thinking about my problems with getting any kind of
>admission that my math arguments showing the core error in mathematics
>are correct, so I've gone to marketing books.
> 
> You know, in case you're curious, this sounds really really stupid.
> If your results were correct you'd be able to convince people of
> them by explaining the proofs carefully. But in fact they're wrong,
> people continually explain what the errors are, and tactics from
> marketing books are not going to change that.
> If his results were correct, the same people who point out errors to him 
> with saintly patience would instead have provided all these explanations.
Having spent some time now reading marketing tactics, I can clearly
see that both Ullrich and Bau are selling a viewpoint to the
readership.
Here the assertion is that if I were right then people would
necessarily agree with me!!!
Is that true in your experience?
James Harris
===
Subject: Re: I NEED HELP BADLY (sorry, maths not psych)
>> If fields acted instantaneously, we would be
>> able to use them for instantaneous communication.
>Please explain how you can use the fact that a force is
>a static electric field, for instantaneous communication.
>Paul, puzzled
> I'm sure some bright QMian would find a way.
Backing out again, Henry?
You cannot defend your assertion,
but you will repeat it, won't you?
And you will back out once more when
asked to defend your repeated assertion, won't you?
That's Henry Wilson's eternal circle of fleeing
restatements of fled assertions.
Paul, not surprised
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> 
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
> 
> Then you should probably stop posting to sci.math and find something
> more productive to do with your time.
> [rest deleted]
Why?  I've spent a lot of time talking about mathematics, so *clearly*
I have some interest in the subject, and sci.math is a *public* forum
for people with interest, many varying interests, in the subject of
mathematics, so why would you try to convince me to stop posting?
It seems to me that some of you just want to hurt people, and for some
of you I'm just a target for your abuse, as I doubt any of you don't
realize the nature of Usenet.
So aren't there *any* reasonable people who might reply instead?
If hateful people were judging their success on discouraging you from
posting, would you give them victory?  Why can't decent people on the
newsgroups work at all to deflect these hostile people?
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
> << http://mathforprofit.blogspot.com/
> What If?   That question on its face causes one to
> consider the state of mind of the speaker...  After
> years of people specifically detailing your mental
> constructs and pointing out your errors in logic you
> still have a doubt of your pass or fail status in life....
> Paul R. Mays
So do you think I'm less or *more* likely to face the truth because of
your post Mr. Mays?
Aren't you trying to say that I'm a failure in LIFE in your comments?
But aren't I a success in posting?
If you push the notion that I'm a failure in life, why shouldn't I
continue in an area where I'm clearly a success?
Aren't you really working to *keep* me posting, possibly in fact,
working to get me to post far more?
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: An urgent question for William Elliot about a reference.
> connectivity of a complete graph equals n-1.
> I need to write a reference for that proof. Was it yours so I write
> your name as a reference or there was another reference like a book
> for example?
For me, edge connectivity is a new concept
which I've not see except at sci.math.
Proof came of my efforts and not from a text.
Uncomplicated as it was, is it not already know?
Surely the theorem is well known.
For why you ask?  Are you writing a thesis?
===
Subject: Re: differentiable...problem...
>> if f is differentiable on (0, infinite)
>> and lim [f(x) + f'(x)] = L  (x->infinite)
>> show that lim f(x) = L (x->infinite) and lim f'(x) = 0  (x->infinite)
>|                                    f e^x       (f + f') e^x
>|  lim  f + f' = L  =>  lim f = lim  ----- = lim ------------ = L
>| x->oo                 x->oo   x->oo  e^x   x->oo     e^x
> Provided f e^x -> +-oo, in which case
>  0 = L - L = lim f+f' - lim f = lim f'
> No proviso is needed. This form of L'Hopital's rule needs only
> that the denominator is infinite, not also the numerator [1][2].
> So there's no need to treat this case specially as you do below.
Not even the existence of lim(x->oo) f(x) ?
> [1] A. E. Taylor, L'Hospital's Rule
> Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.
> http://links.jstor.org/sici?sici=0002-9890(195201)59:3%3C20%3E
You make jest?  Jstor says forbidden.
> [2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital Rule
> Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.
> http://links.jstor.org/sici?sici=0002-9890(197604)83:3%3C239%3E
Ditto.
> However when f e^x -> k;  then
>  f = fe^x e^-x -> 0
>  f e^-x = f e^x e^-2x -> 0
> L = lim f+f' - lim 2f = lim f'-f
> 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L
> 0 = lim f+f' - lim f = lim f'
> So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f'
>  k = lim(x->oo) f = k + lim f';  lim f' = 0
> If lim f' doesn't exist, then for n >= 2
>  f_n(x) = sin(x^n)/x -> 0
>  (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation
===
Subject: Re: Basic factorization ideas
> If you saw
> 
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 1)
> 
> with the c's algebraic integers, I think few of you would have a
> problem realizing that only two of the c's have 7 as a factor.
> 
> If c_1 and c_2 (and hence c_3) are ordinary integers, then
> yes, c_1 and c_2 must be divisible by 7, and c_3 must be
> coprime to 7 (does this hold for algebraic integers as well?)
> but the proof of this depends strongly on the fact that the LHS
> is composed of polynomial factors. 
Why?  That is, why does it matter if the factors are polynomial
factors?
What about
(c_1 sqrt(x) + 7)(c_2 sqrt(x) + 7)( c_3 sqrt(x) + 1) = 
                
                           49(x^{3/2} + 5x + 3sqrt(x) + 1)?
> But, of course, you're looking at *functions* of x, as you have 
> 
> f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
> 
> so I could also write it as
> 
> 
> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
> Yes, but this is dangerous.  Someone who was not paying close
> attention might assume that the f_i were arbitrary functions
> and get the idea that things that are true of polynomials
> are also true of arbitrary functions.
But why does it matter if they're polynomials?
> Notice that dividing both sides by 49 gives 
> 
> (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
> 
> as long as you're in a ring where 7 is not a factor of 22.
> 
> I want to emphasize that point as notice there's only *one* way to
> divide through by 49 if 7 is not a factor of 22.
> 
> Yes, if the f_i are defined as above.  If the f_i are arbitary
> functions this statement is wrong.
Well I have to ask why again, and if readers think I'm just repeatedly
asking for no good reason, consider that William Hughes has very
*definitely* marked out a position without giving a reason.
Rather than assume I can figure out what he was thinking when he
posted, I'd just as soon ask him.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: Basic factorization ideas
>  > If you saw
>  > 
>  > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>  >                  
>  >                           49(x^3 + 5x^2 + 3x + 1)
>  > 
>  > with the c's algebraic integers, I think few of you would have a
>  > problem realizing that only two of the c's have 7 as a factor.
>  ...
>  > I want to emphasize that point as notice there's only *one* way to
>  > divide through by 49 if 7 is not a factor of 22.
> 
> Oh, you mean that when 7 is a factor of 22 than there are other ways
> to do the division, as in that case 7 is a unit in the ring.
> Moreover, there may be values of x such that (f3(x) + 1) is divisible 
by
> 7, I think.  If this is the case there are other ways to distribute 
the
> 7's.
> Lo and behold, there are indeed (not 7, but factors of 7)!  When x = 7y + 

2
> for some integer y, (c3 x + 1) is *not* coprime to 7.  This example is 
too
> easy to crack.  Define:
>   w3(x) = gcd(c3 x + 1, 7)
> (this one is not always 1 (*)),
>   w2(x) = 7/w3(x).
Well, yes, but what about
(c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = 
                  
                           x^3 + 5x^2 + 3x + 1?
Here the 49 has been divided through, and doesn't that change things?
> We now could divide the first factor by 7, the second by w2(x) and
> the third by w3(x) and remain in the algebraic integers.  So, again,
> the possible ways to divide 49 through may very well depend on the
> value of x.
> ----
But that now seems problematic as in mathematics, how do you
determine time?
Isn't there just the factorization
(c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = 
                  
                           x^3 + 5x^2 + 3x + 1
without reference to *time* itself?
Your example is irrelevant above as though c_3 x + 1 may have 7 as a
factor, so what?
The question is how to divide 49 from both sides of
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
                  
                           49(x^3 + 5x^2 + 3x + 1)
and I think there's only one way to do so, and stay in the ring of
algebraic integers.  That is, one way to keep 1 from having 7 as a
factor.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: Basic Planar Geometry Question
> Which are the polygons which have 5 vertex and have the maximum number
> of parallel (m) and perpendicular lines (n) drawn among these vertex,
> i.e., m + n is maximum?
If you are counting them pairwise, this might be a winner:
HOOOOOOOOOH
HOOOOOOOOOH  (Pardon the O fill, trying to prevent damage by
HOOOOOOOOOH   proportional fonts.)
Where parallel pairs are ((AB)(BC)) ((AB)(DE)) ((BC)(DE)) and ((AD)(CE))
and perpendicular pairs are ((AB)(AD)) ((AB)(CE)) ((BC)(AD)) ((BC)(CE))
((DE)(AD)) and ((DE)(CE)), giving 10 in all.
xanthian.
-- 
===
Subject: Re: if x has normally distributed digits, does 1/x?
>My guess is that if you clear to zero every other digit of Pi and
>then take the reciprocal of the result, you will get a normal number.
If so, how do you prove its normality?
===
Subject: Re: question about canonical commutation relations [P,Q]=i*hbar*I
>Question:
>Find matrices of infinite size Q and P such that
>[P,Q]=i*hbar*I, hbar is a constant,I is the identity matrix, i^2=-1
>I've tried to prove that there do not exist nxn complex matrices Q and
>P such that [P,Q]=i*hbar*I.
> ...
> Your proof is correct.
>I've also tried to find the matrices of infinite size.
>Let P,Q be matrices of infinite size.
>p[i,1]=1
> It's best not to use i here because it will be confused with 
sqrt(-1).
> Note that the product of two infinite matrices is well defined 
> if they have only finitely many nonzero entries in each row
> (there are more general conditions, but this one is enough for
> your case).
> Hint: I think you'll want to define P and Q so each has only one nonzero
> entry in each row: one of them only above the main diagonal, the other
> only below the main diagonal.  
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel 
> University of British Columbia            
> Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Basic factorization ideas
...
 > If c_1 and c_2 (and hence c_3) are ordinary integers, then
 > yes, c_1 and c_2 must be divisible by 7, and c_3 must be
 > coprime to 7 (does this hold for algebraic integers as well?)
 > but the proof of this depends strongly on the fact that the LHS
 > is composed of polynomial factors. 
 > Why?  That is, why does it matter if the factors are polynomial
 > factors?
 > What about
 > (c_1 sqrt(x) + 7)(c_2 sqrt(x) + 7)( c_3 sqrt(x) + 1) = 
 >                            49(x^{3/2} + 5x + 3sqrt(x) + 1)?
Now you work with polynomials in sqrt(x), so you still have polynomial
factors in that ring.
-- 
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: Basic factorization ideas
 >  > If you saw
 >  > 
 >  > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
 >  >                  
 >  >                           49(x^3 + 5x^2 + 3x + 1)
 >  > 
 >  > with the c's algebraic integers, I think few of you would have a
 >  > problem realizing that only two of the c's have 7 as a factor.
 >  ...
 >  > I want to emphasize that point as notice there's only *one* way 
to
 >  > divide through by 49 if 7 is not a factor of 22.
 > 
 > Oh, you mean that when 7 is a factor of 22 than there are other 
ways
 > to do the division, as in that case 7 is a unit in the ring.
 > Moreover, there may be values of x such that (f3(x) + 1) is 
divisible by
 > 7, I think.  If this is the case there are other ways to distribute 
the
 > 7's.
 > 
 > Lo and behold, there are indeed (not 7, but factors of 7)!  When x = 7y 
+ 2
 > for some integer y, (c3 x + 1) is *not* coprime to 7.  This example is 
too
 > easy to crack.  Define:
 >   w3(x) = gcd(c3 x + 1, 7)
 > (this one is not always 1 (*)),
 >   w2(x) = 7/w3(x).
 > Well, yes, but what about
 > (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = 
 >                            x^3 + 5x^2 + 3x + 1?
 > Here the 49 has been divided through, and doesn't that change things?
It does not change a thing.  For some values of x all three factors on the
left are not coprime to 7.
 > We now could divide the first factor by 7, the second by w2(x) and
 > the third by w3(x) and remain in the algebraic integers.  So, again,
 > the possible ways to divide 49 through may very well depend on the
 > value of x.
 > But that now seems problematic as in mathematics, how do you
 > determine time?
By looking at the clock.  Otherwise I have no idea what you mean.
 > Isn't there just the factorization
 > (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = 
 >                            x^3 + 5x^2 + 3x + 1
 > without reference to *time* itself?
I do not know what you mean with time.  But indeed, that is one possible
factorisation, but there are other factorisations, as I just showed above.
 > Your example is irrelevant above as though c_3 x + 1 may have 7 as a
 > factor, so what?
 > The question is how to divide 49 from both sides of
 > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
 >                            49(x^3 + 5x^2 + 3x + 1)
 > and I think there's only one way to do so, and stay in the ring of
 > algebraic integers.  That is, one way to keep 1 from having 7 as a
 > factor.
w3(x) = gcd(c3 x + 1), 7)  (this is not identically equal to 1.)
w2(x) = 7/w3(x),
   [(c1 x + 7)/7] [(c2 x + 7)/w2(x)] [(c3 x + 1)/w3(x)]
Now, *which* of the three factors is *not* an algebraic integer for all x?
And *how* do you find that 7 is a factor of 1?
-- 
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: Marketing shift, core issues
>> 
>>I've been thinking about my problems with getting any kind of
>>admission that my math arguments showing the core error in 
mathematics
>>are correct, so I've gone to marketing books.
>> 
>> You know, in case you're curious, this sounds really really stupid.
>> If your results were correct you'd be able to convince people of
>> them by explaining the proofs carefully. But in fact they're wrong,
>> people continually explain what the errors are, and tactics from
>> marketing books are not going to change that.
>> If his results were correct, the same people who point out errors to him 

>> with saintly patience would instead have provided all these 
explanations.
>Having spent some time now reading marketing tactics, I can clearly
>see that both Ullrich and Bau are selling a viewpoint to the
>readership.
>Here the assertion is that if I were right then people would
>necessarily agree with me!!!
>Is that true in your experience?
It's certainly true in _my_ experience that when I'm right about
something and have a valid proof that I'm right then competent
mathematicians agree I'm right, after I've explained the proof,
yes. That includes cases where they were certain at first I was
wrong, by the way.
>James Harris
===
Subject: Re: What is Advanced Calculus?
> I am a bit confused.
> It seems as though Multivariable Mathematics is just another name
> for Advanced Calculus, and the same applies to it as well.
>> The Implicit Function Theorem, the Invese Function Theorem, the
>> Taylor's Theorem in n-dimensions, derivatives as linear maps,
>> 2nd derivatives as bilinear maps, differentiability as being
>> distinct from having derivatives, Frobenius' Theorem, maybe
>> some Distribution Theory.  On the integral side, Measure
>> Theory, different definitions for integrals, etc.
> Can somebody recommend a mostly formal textbook on these topics? 
> Hopefully something about two inches thick, second or third edition, 
> with everything proven, lots of examples, tons of problems and a 
> complete solutions manual.  I loved Ellis & Gulick, Calculus and 
> Analytic Geometry, 2nd Ed.
> Buck, Advanced Calculus.
> Spivak, Calculus on Manifolds.
> You will not find any with solution manuals.
> Others will recommend introductory analysis texts, probably
> Rudin, Principles of Mathematical Analysis.
> Personally, I find that rough reading for self-teaching, but I have no 
> alternative to offer.  (Goldberg is readable but does no multivariable 
> analysis.)  Besides, it sounds to me as if you are more interested in 
> the calculus direction than the analysis direction (though this 
> distinction is somewhat vague).  I really think the two books I 
> recommend are what you seek.
In 1965-6, we used Goldberg, which was new at the time.  I understood
that the professor used two books the previous year, one of which was
Apostol, Mathematical Analysis.
David Ames
===
Subject: Fastest factorial algorithm
What is the fastest algorithm for computing factorial, for very large 
numbers (e.g. 10000!)?
Normally this would take n-2 multiplications, by multiplying out each 
term n(n-1)(n-2)(n-3)...3.2.  Is a better way known?
===
Subject: Re: Mathematical Integrity
> I am a math student and I love math more than anything else. I
> started to love math when I found out that I wasn't good at it and
> started studying it for over a year now.  I love math for its purity
> and integrity.  However, lately, I surfed through several math forums and 

I
> found many unpleasant posts.  Sometimes a student would post an innocent
> question but he or she would get scold for posting such a stupid 
question
> and would be called a moron.  And sometimes a math fanatic would post
> something of interest, but other mathematicians would scold that 
person for
> posting something so ridiculous. It seems like these dese days, some
> mathematicians would use profanity against others mathematicians. Why? 
Where
> is the purity and integrity in mathematics? I know that after posting 
this,
> many other great mathematicians would scold me, but that's what I've seen 

so
> far. Anyhow, I still love math by heart and I would like to give my 
respect
> to many great math fanatics in this forum.
Don't confuse posters in this newsgroup with mathematicians.  There
is some overlap, to be sure, but most mathematicians don't post here,
and most posters here are not mathematicians.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
===
Subject: Combinatorial proof
Show with combinatorial reasoning that
C(3n,3) = 3*C(n,3)+6n*C(n,2)+n^3, where C(n,k) is n choose k.
Am I supposed to show that both sides of the equation counts the thing? 
Does it qualify as combinatorial reasoning if I use combinatorial 
identities or prove it by induction?
Any good hints, by the way?
-- 
Be sure to include the word zebra in the subject line of any mail sent
to my email adress fergusprint@casino.com since it will be automatically
removed otherwise!
===
Subject: Re: JSH: Deprogramming needed?
>(Marketing ploy warning)
So you warn us in advance that this post is not going to
contain logical arguments explaining why you're right, but
it's going to try to persuade for non-logical reasons instead.
you're right by whining about how awful it would be if you
were wrong. If anything this will have the opposite effect:
Your whining about how awful it would be if you were wrong
will not change the opinion of anyone competent to evaluate
your work, but it _could_ convince people who know no math
that you must be wrong, because if you were right you
wouldn't have to use marketing ploys.
Duh.
>What if indeed I *am* wrong, and I don't have these great math
>discoveries?
>After all, I've been at this since April 1995 having spent a lot of
>time and effort, with literally thousands of posts along with all
>kinds of other activities, websites, and email to mathematicians all
>over the world.
>But, what if I'm wrong?
>What if all the time and energy I've invested in my work has made it
>difficult for me to see, along with the *harsh* and unforgiving
>hostility 
That hostility is not because you're wrong. You need to tell
yourself that the reason people are so hostile is because they
can't stand the possibility that you might be right. But that's
just one of your many self-aggrandizing delusions - in fact
the hostility stems from the fact that you're so hostile to
others.
You don't believe that. But you must agree that you _are_
hostile to others. For example you've complained to my 
employer about me being mean to you, but you've also
called me a ing piece of dog - you remember that,
right?
As evidence in support of the idea that people would not
be so nasty to you if you were not nasty to them I offer
the case of Ross Finlayson. Look up some of his posts
through the years. Essentially everything he says is
wrong, usually hilariously obviously incoherently wrong.
But he doesn't get the same sort of hostile reaction
that you do - people explain he's all wrong without
calling him nasty names.
>from several people who seem to have made it their mission
>to make me miserable and find pleasure in mocking or trying to
>humiliate me, have made it extremely difficult for me to see the
>truth?
>Think about the crushing sense of shame and misery if indeed I find
>out that the logical connections I so carefully and impatiently
>discovered over the years are simply not really there, but are a need
>induced delusion.
>It seems to me that marketing ploy though these statements may be,
>dealing with people who've made it their business to try and make me
>miserable, only to at times claim they're trying to help me is just
>too much.
>Aren't there any *other* people who suppose they are rational, who can
>follow a logical argument, who might comment?
Are you ever going to give up, and just decide that you're the only
rational person in the universe? I mean that's what the evidence
seems to indicate: There are no rational people on sci.math, the
mathematicians you harass via email are not rational, the 
mathematicians you visit in person are not rational, the editors
of journals you send papers to are not rational...
Now, the conclusion a _rational_ person would draw from this was
that maybe he was actually _wrong_, that it's just not rational
to assume that _all_ those people share the same irrationality.
Of course concluding that is beyond you, because it would destroy
the picture of you as supreme genius that you cling to do desperately
to give your otherwise empty life some meaning. Fine. But I'd think
that at some point even you would be rational enough to just
give up and conclude that you were the only rational person on
the planet, instead of repeating these pleas for someone
rational to step forward. You've been begging for that for
years, and it never happens. It's pathetic, actually.
>Why is it always the same people?  Or people imitating them in hostile
>and mocking displays of animosity or anger?
>Aren't there any rational people who can trace out the steps in the
>work I've presented, who might step forward at this time, and
>demonstrate an ability to just be objective?
>I don't want any replies of people offering, but if you might consider
>it, I just want you to think about it.  Sure I've been reading this
>marketing book, but it seems to me that still *someone* out there
>might be the right person, as of course, I don't believe I'm wrong. 
>I've traced out every step in the arguments I have, and I think that
>irrational people have been dominating the discussion using group
>effects to hide the truth.
>Think about it.  I'll come back to the subject later.
>James Harris
>http://mathforprofit.blogspot.com/
===
Subject: Re: Combinatorial proof
> Show with combinatorial reasoning that
> C(3n,3) = 3*C(n,3)+6n*C(n,2)+n^3, where C(n,k) is n choose k.
> Am I supposed to show that both sides of the equation counts the thing?
> Does it qualify as combinatorial reasoning if I use combinatorial
> identities or prove it by induction?
> Any good hints, by the way?
Big hint (maybe even a spoiler):
You have 3 jars with n numbered marbles each.
You must pick 3 marbles.
There are 3 ways to do it:
   1) take 3 marbles from one jar.
   2) take 2 marbles from one jar and 1 from another one
   3) take 1 marble from each jar.
These 3 ways are independent, so you can add the numbers...
Dirk Vdm
===
Subject: Programmer ax+by=c en C
Salut tout le monde,
Ce n'est peut-.90tre pas tr.8fs compliqu.8e
Mais je me suis pris la t.90te .88 programmer en C ax+by=c
o.9d a,b,c,x et y sont tous des nombres entiers.
On conna.94t a,b et c et l'objectif c'est d'obtenir x et y.
D'abord je commence par le fait que le pgcd(a,b) est un diviseur aussi de 
c.
Quand je le fais .88 la main c'est .8evident mais pour le programme je 
suis bloqu.8e ici.
Si quelqu'un sait comment faire et peut m'aider?
===
Subject: Re: Sum [n in Set] (1/n)
>> I recall seeing a theorem saying
>>     Sum [n in Set] (1/n) converges iff (some criterion on Set)
>> where Set is a subset of the positive integers.  I seem to recall it
>> brought in measure theory.  Does anyone know a reference?  Or at least 
the
>> statement?  (I'm actually only interested in the statement, not the
>> proof.)
>Perhaps that the span of {x^s : s in S} is dense in C[0,1] iff sum
>1/s is infinite?  
Oh, I forgot about that - that could well be what he was wondering
about.
>I forget whose theorem it is.  The proof is in
>Rudin's Real and Complex Analysis.  (And I may have some details wrong. 
>Perhaps it's the ALGEBRA generated by these which is dense iff...)
No, you got it right. Sketch of proof of the easier direction: Say the
sum of 1/n for n in S is infinite. Suppose mu is a measure on [0,1] 
that annihilates all the t^n for n in S. Define 
    f(z) = int_0^1 t^(i/z) d mu(t).
Show f is a bounded holomorphic function in the upper half-plane;
now f(i/n) = 0 for n in S, so the Blaschke condition shows that
f = 0. Hence f(i/n) = 0 for all n, hence mu = 0.
>--Ron Bruck
===
Subject: Re: Sum [n in Set] (1/n)
>I recall seeing a theorem saying
>    Sum [n in Set] (1/n) converges iff (some criterion on Set)
> There is also a conjecture of Erdos which I think states that
> the sum diverges iff  Set  contains arbitrarily long arithmetic
> progressions.
>That should be  =>  , not  iff . Obviously we could have 
arbitrarily
>long arithmetic progressions and still have convergence, e.g. if
>Set = { 10^k + m  , 0 < m < k }. I have a notes saying Erdos offered
>$3000 for a proof of  => . You can still collect.
>>In the future, you can post or mail; doing both is unnecessary.
>Maybe, but one of these days I'll learn that what IS necessary is to
>remember to flag my email so that it doesn't look like it was (also) a
>newsgroup post ...
Well, I for one was pleased to see his reply to your email (although
puzzled where the post from you was) because I didn't know about the
Erdos thing...
>dave
===
Subject: Re: [Set Theory] A set has a disjoint copy of itself ?
> If you don't mind using the Axiom of Foundation (Regularity), you
> could just take B = {{a,A}: a in A}.
It took me a while to work that one out :-)
For all a in A, a in {a,A}/A. Hence from the axiom of foundation, I
obtain that {a,A}/a=0 for all a in A. It follows that if {a,A} in A
for some a, I have {{a,A},A}/{a,A}=0 which is a contradiction. So
B/A=0.
Noel.
===
Subject: Re: uniform convergence & differentiation
> In many analysis-type texts I've found statements like since the series
> converges uniformly, we can apply term-by-term differentiation and 
uniform
> convergence allows us to differentiate term-by-term, but I've never
> actually seen a proof or explanation of WHY you need uniform convergence 
to
Actually, if a sequence of real functions (f_n), each f_n
differentiable on an interval I, converges uniformly to a function f,
then this does not imply that (f'_n) converges to f'. It's even
possible that f is not differentiable on I. And since series are a
special case of sequences, this is true of series of real functions.
There is however a theorem that ensures that, under suitable
conditions, the convergence of f'_n -> holds:
Suppose each f_n is differentiable on I and for some a in I the
sequence of real numbers (f_n(a)) is convergent. If (f'_n) converges
uniformly on I to some function g, then (f_n) converges uniformly on I
to a function f such that f'(x) = g(x) for every x in I. That is, g=
f'.
The important things here are the uniform convergence of (f'(n) - and
not of (f_n) - and the convergence of (f_n(a)) for some a in I. The
true story is that the uniform convergence of (f_n) implies nothing
about the convergence of (f'_n).
Of course, similar conclusions hold for real series of functions. If
the conditions above are satisfied, then you can diffrentiate a series
term by term.
I think there are some similar theorems about sequences and series of
complex functions. It's important to keep in mind that the theorem I
mentioned gives sufficient conditions. It's proof, though not really
difficult, is rather long. You can find it in a number of books abour
Analysis, like those of Rudin, Bartle, Apostol, etc...
Artur
Complex power series are a very interesting example of series that can
be differentiated term by term in it's disk of convergence
===
Subject: A question in projective geomtry
First of all, I'm not sure who this newsgroup is intended for, so I hope 
I'm
not going to ask a question that will be too trivial... but I need help 
with
this subject, so I'll go ahead anyway :-)
How to prove that in finite projective geomtry, there is an equal number of
points on all the lines?
===
Subject: Re: uniform convergence & differentiation
>In many analysis-type texts I've found statements like since the series
>converges uniformly, we can apply term-by-term differentiation and 
uniform
>convergence allows us to differentiate term-by-term, but I've never
>actually seen a proof or explanation of WHY you need uniform convergence 
to
First we need to clarify exactly what result we're talking about! Your
post _sounds_ like you're saying you've read this:
(1)  If the sum f = sum(f_n) converges uniformly then f' = sum(f_n').
I hope you've never read that, that statement is simply false.
What's true is this:
(2)  If f = sum(f_n) and if sum(f_n') converges uniformly then 
     f' = sum(f_n').
Yes, uniform convergence is needed in (2). Some people have pointed
out that uniform convergence is not a necessary condition; it's
not - that's not what it means to say the condition is needed,
what I mean by that is that if we simply omit the condition
(2) becomes false. That is, (3) is false:
(3)  If f = sum(f_n) and if sum(f_n') converges (pointwise) then 
     f' = sum(f_n').
A simple counterexample to (3): Let I_n be the ooen interval
(1/(n+1), 1/n). Choose a differentiable function f_n such that
f(n)x = 0 for all x not in I_n, but such that f_n = 1 at the
center of I_n. Then the two series f(x) = sum(f_n(x)) and 
sum(f_n'(x)) both converge for _all_ x, but f is not even
continuous at 0, hence not differentiable at 0.
===
Subject: Re: Factorization dispute, again
I guess my point is that this is part of his problem: he has a bad tendency
not to check calculations and to see if what he is deriving is actually
correct (and for a physicist this is odd).
Sure, the technique he uses is correct IF USED BY SOMEONE WHO IS COMPETENT
AND KNOWS HOW TO USE IT. But he isn't, and he doesn't. If I wanted to
disabuse someone of something in a manner which was clear and indisputable,
I would provide a simple direct counter-example. IF JSH tried to do this, 
he
would often realize that he is wrong (since he wouldn't be able to come up
with one, and his examples would show him that he ain't gettin what he
thinks he should be gettin), and prevent these back and forth posts ad
nauseum that end up with him realizing after weeks that he's made some
stupid assumption/error.
MB
> Here's a nugget for the did you ever notice bin:
> Did you ever notice, how Mr. Harris gives refutations based on general
> relations, whereas those who dispute his claims give exact, concrete
> counter-examples?
> MB
> Notice,
> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
>                          49(300125 x^3 - 18375 x^2 - 360 x + 22)
> where you see that the constant terms match as now you have 7(7)(22) =
> 1078, which is the constant term of the polynomial
> 49(300125 x^3 - 18375 x^2 - 360 x + 22).
> Various people have debated me about what happens when you divide off
> 49, where for some odd reason, some of them seem to believe that you
> can have
> w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
> (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
>                          300125 x^3 - 18375 x^2 - 360 x + 22
> where the w's vary as x varies, which is a rather naive notion.
> That's because you can multiply *everything* out, and simplify to get
> (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) =  22
> which should be simple enough for all of you.
> Now those of you who usually work in the field of complex numbers may
> think that it's not a big deal, as you may think it doesn't matter if
> w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
> you see, as 22 is coprime to 7 in the ring of algebraic integers, if
> w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
> You know, it's like how in integers 1/2 doesn't exist.  It's not an
> integer, so it's not in the ring.
> So you see, my argument is correct and simple, and mathematicians are
> indeed running from a little gut check in their field.  They're
> pussies too scared to handle the truth.
> But you should also understand, some people will be able to see that,
> which is part of my plan.  I can let mathematicians destroy themselves
> proving they can't be trusted based on what they *see*, while they
> forget what they can't see: the wearing down of the mathematician
> mystique.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: Basic factorization ideas
>James has tricked you.  In this version the roots of the cubic are
>units.  Part of his marketing strategy, I think.
I don't completely understand what you are saying. Please read the
following and tell me where I go wrong.
The whole 'discusion'  is about algebraic integers. An algebraic
integer is any value of x that satisfies:
x^n + a_(n-1)*x^(n-1) + ... + a_1*x + a_0 = 0  (with the a's being
integers)
For example: sqrt(2) is an algebraic integer as it is the solution to
x^2-2=0.
The algebraic integers form a ring, which amongst other things means
that the addition of two algebraic integers is an algebraic integer as
well (the same goes for multiplication of two algebraic integers).
Another property is that for an algebraic integer a, -a is also an
algebraic integer.
Now look at this:
P(x) = 49*(x^3 + 5*x^2 + 3*x + 1)
If the value of x is an algebraic integer, the value of P(x) for that
x will be an algebraic integer as well. This is true, because 1,3,5
and 49 are algebraic integers, so with x an algebraic integer as well,
the result must be an algebraic integer. This can also be seen if P(x)
is factored:
P(x) = 49*(x+r_1)*(x+r_2)*(x+r_3)
r_1, r_2 and r_3 are (obviously) algebraic integers, so if the value
of x is an algebraic integer, the value of P(x) is also an algebraic
integer.
Now Mr. Harris claims that P(x) can be rewritten to:
P(x) = (c_1*x+7)*(c_2*x+7)*(c_3*x+1)
in such a way that c_1, c_2 and c_3 are algebraic integers.
This is what I can come up with:
P(x) = 49*(x+r_1)*(x+r_2)*(x+r_3)
P(x) = 7*7*r_1*r_2*r_3*(x/r_1+1)*(x/r_2+1)*(x/r_3+1)
P(x) = r_1*r_2*r_3*(7*x/r_1+7)*(7*x/r_2+7)*(x/r_3+1)
P(x) =(7*x/r_1+7)*(7*x/r_2+7)*(r_1*r_2*x+r_1*r_2*r_3)
P(x) =((7/r_1)*x+7)*((7/r_2)*x+7)*(r_1*r_2*x+r_1*r_2*r_3)
So:
c_1 = 7/r_1, c_2 = 7/r_2, c_3 = r1*r2 and r1*r2*r3 = 1
c_3 is obviously an algebraic integer as it is the product of two
algebraic integers. c_1 and c_2 can also be written like the product
of algebraic integers:
c_1 = 7/r_1 = 7*r_2*r_3/r_1*r_2*r_3 = 7*r_2*r_3
c_2 = 7/r_2 = 7*r_1*r_3/r_1*r_2*r_3 = 7*r_1*r_3
(@Dik: I guess this is what you tried to tell me...)
So
P(x) = (7*r_2*r_3*x + 7)*(7*r_1*r_3*x + 7)*(r1*r2*x + 1)
By dividing P(x) by 49 we get:
P(x)/49 = (r_2*r_3*x+1)*(r_1*r_3*x+1)*(r1*r2*x+1)
The 'factors' of P(x)/49 (between the brackets) are obviously
algebraic integers if x is an algebraic integer.
However, we could just as easily have divided P(x) like this:
P(x) = (((r_2*r_3)/7)*x + (1/7))*(7*r_1*r_3*x + 7)*(r1*r2*x + 1)
or:
P(x) = (7*r_2*r_3*x + 7)*(r_1*r_3*x + 1)*(((r1*r2)/7)*x + 1/7)
In these cases the 'factors' don't have to be algebraic integers (but
they can be, for instance for x=7 :) ).
Now did I get that right?
If so, why isn't there a 22 in my story?
===
Subject: Re: Banach limit and almost convergence
>Hi!
>It was noted several times in Google Groups (by Ronald Bruck), then
>uniqueness of Banach limit of a sequence is equivalent to almost
>convergence of this sequence. Can you give me a hint where to find a
>proof of this result. (Web resources are prefered, but also book or
>paper in a journal would help.)
Have you tried searching the _web_ using Google? If you type
Banach limit almost convergence
into the search box you get a few hits, one of which says this
result was proved by Lorentz in A contribution to the theory
of divergent sequences, Acta Math 80(1948).
>Martin Sleziak
===
Subject: Re: Basic factorization ideas
>In these cases the 'factors' don't have to be algebraic integers (but
>they can be, for instance for x=7 :) ).
Oops... this is obviously not true. :) I forgot the +(1/7)... silly
me.
===
Subject: Re: I can't stand it anymore
> 
> I found it.
> 
>  I took a look and you got the posts written by
> I also noticed that you copied the stuff written by the guy whom I
> referred to as Indian Hindu hate monger.  All he did was coming up
> with crap.  I am sorry that you took him seriously.
> You should have got some books written by unbiased historians before
> you decided to trash Islam and its Prophet, will you?
> There is not much difference in Islam and Christianity when it comes
> to crap, crpas which were made by humans.
> For the story of Muhammed, I strongly suggest that you get a few books
> written by real historians.
> 
> BTW, I agree with you about islam.
> Which part did you agree with? Are you sure that what you read was
> written by me?
> By the way, here is something you may want to read about Muhammed. 
> He was Caesar and Pope in one; but he was Pope without Pope's
> pretensions, Caesar without the legions of Caesar: without a standing
> army, without a bodyguard, without a palace, without a fixed revenue;
> if ever any man had the right to say that he ruled by the right
> divine, it was Mohammed, for he had all the power without its
> instruments and without its supports.
> 
----------------------------------------------------------------------------
-
----
> MAHATMA GANDHI, speaking on the character of Muhammad, (pbuh) says in
> (YOUNG
> INDIA):
>       I wanted to know the best of one who holds today's undisputed
> sway
> over the hearts of millions of mankind....I became more than convinced
> that
> it was not the sword that won a place for Islam in those days in the
> scheme
> of life. It was the rigid simplicity, the utter self-effacement of the
> Prophet, the scrupulous regard for his pledges, his intense devotion
> to this
> friends and followers, his intrepidity, his fearlessness, his absolute
> trust
> in God and in his own mission. These and not the sword carried
> everything
> before them and surmounted every obstacle. When I closed the 2nd
> volume (of
> the Prophet's biography), I was sorry there was not more for me to
> read of
> the great life.
> --------------------------------------------------------
> My choice of Muhammad to lead the list of the world's most influential
> persons may surprise some readers and may be questioned by others, but
> he
> was the only man in history who was supremely successful on both the
> religious and secular level.
> Michael H. Hart
> The 100: A Ranking of the Most Influential Persons in History. New
> York:
> Hart Publishing Company, Inc. 1978, p. 33.
> 
----------------------------------------------------------------------------
-
------
> If greatness of purpose, smallness of means, and astounding results
> are the
> three criteria of human genius, who could dare to compare any great
> man in
> modern history with Muhammad? The most famous men created arms, laws
> and
> empires only. They founded, if anything at all, no more than material
> powers
> which often crumbled away before their eyes. This man moved not only
> armies,
> legislations, empires, peoples and dynasties, but millions of men in
> one-third of the then inhabited world; and more than that, he moved
> the
> altars, the gods, the religions, the ideas, the beliefs and souls. . .
> his
> forbearance in victory, his ambition, which was entirely devoted to
> one idea
> and in no manner striving for an empire; his endless prayers, his
> mystic
> conversations with God, his death and his triumph after death; all
> these
> attest not to an imposture but to a firm conviction which gave him the
> power
> to restore a dogma. This dogma was twofold, the unity of God and the
> immateriality of God; the former telling what God is, the latter
> telling
> what God is not; the one overthrowing false gods with the sword, the
> other
> starting an idea with words.
> Philosopher, orator, apostle, legislator, warrior, conqueror of
> ideas,
> restorer of rational dogmas, of a cult without images; the founder of
> twenty
> terrestrial empires and of one spiritual empire, that is Muhammad. As
> well
> ask, is there any man greater than he?
> Lamartine, HISTOIRE DE LA TURQUIE, Paris, 1854, Vol. II, pp. 276-277.
> 
----------------------------------------------------------------------------
-
------
> What nonMuslim say about Muhhamed
>  http://cyberistan.org/islamic/quote1.html
> Tolerance in Islan by Pickthall, an English convert
> http://cyberistan.org/islamic/toleran1.html
> I apologized those at sci.math,sci.physics,sci.chem; this is my last
> post on this thread.
I went to the link I provided
(http://cyberistan.org/islamic/quote1.html) and as I read the quotes
there, I have decided to post the following since Rich has copied and
posted the crap written by an Indian hate monger who claimed to be an
ex-Muslim.
Th fact that Rich so easily copied his words and posted here made me
realize that even in these science ngs, there are those who know
nothing about Islam and Muhammed and yet ready to accept any negative
statements from someone like that guy without a doubt.
 This is my last post as I do not want to make this thread any longer
especially when it has completely gone out of context.
Rich ,
If you have anything to say to me about my Asian Racism, etc., feel
free to email me.
Below is more about using the sword and the  great man you referred to
as a Phedophile, after rading the hatemongers' posts.
Keep in mind that thoes hatemongers has no historians writing
Biography about them but Muhammed has many.
Note the name K. S. Ramakrishna Rao. It is an Indian Hindu name.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - 
K. S. Ramakrishna Rao in 'Mohammed: The Prophet of Islam,' 1989
My problem to write this monograph is easier, because we are not
generally fed now on that (distorted) kind of history and much time
need not be spent on pointing out our misrepresentations of Islam. The
theory of Islam and sword, for instance, is not heard now in any
quarter worth the name. The principle of Islam that there is no
compulsion in religion is well known.
----------------------------------------------------------------------------

---
Lawrence E. Browne in ëThe Prospects of Islam,' 1944
Incidentally these well-established facts dispose of the idea so
widely fostered in Christian writings that the Muslims, wherever they
went, forced people to accept Islam at the point of the sword.
-----------------------------------------------------------------
James Michener in ëIslam: The Misunderstood Religion,' 
Reader's
Digest, May 1955, pp. 68-70.
No other religion in history spread so rapidly as Islam. The West has
widely believed that this surge of religion was made possible by the
sword. But no modern scholar accepts this idea, and the Qur'an is
explicit in the support of the freedom of conscience.
Like almost every major prophet before him, Muhammad 
áá. (I cut the
rest).
--------------------------------------------------------------------------
W. Montgomery Watt in 'Muhammad at Mecca,' Oxford, 1953.
His readiness to undergo persecution for his beliefs, the high moral
character of the men who believed in him and looked up to him as a
leader, and the greatness of his ultimate achievement - all argue his
fundamental integrity. To suppose Muhammad an impostor raises more
problems that it solves. Moreover, none of the great figures of
history is so poorly appreciated in the West as Muhammad.... Thus, not
merely must we credit Muhammad with essential honesty and integrity of
purpose, if we are to understand him at all; if we are to correct the
errors we have inherited from the past, we must not forget the
conclusive proof is a much stricter requirement than a show of
plausibility, and in a matter such as this only to be attained with
difficulty.
------------------------------------------------------------------
Dr. William Draper in 'History of Intellectual Development of Europe'
Four years after the death of Justinian, A.D. 569, was born in Mecca,
in Arabia, the man who, of all men, has exercised the greatest
influence upon the human race... To be the religious head of many
empires, to guide the daily life of one-third of the human race, may
perhaps justify the title of a Messenger of God.
---------------------------------------------------------------------
===
Subject: Re: Prove:
ellias turner  grava .88 la saucisse et au marteau:
>  If g.c.d (x,y)=1
>  and x divides z
>  and y divides z
>  then xy divides z
x = p1^a1 p2^a2 ... pn^an
y = q1^b1 .... qk^bk
Since g.c.d (x,y)=1, none of the p's and the q's are identical. So z =
p1^a1 p2^a2 ... pn^an * X and z = q1^b1 .... qk^bk * Y.
Since none of the q's can be included in the term p1^a1 p2^a2 ... pn^an,
they're all in the X. Therefore, z = xy (X/y) with X/y an integer.
When I saw this theorem (from Gauss), I had a better proof, but I can't
remember it by now.
-- 
Nicolas
===
Subject: Re: Deprogramming needed?
> But aren't I a success in posting?
By what standard? Sheer volume? Certainly not in content. Sorry, Dar 
Kabatoff is a far greater
success at posting than you because, like you, he is a deranged mental case, 

but he posts his
crap everywhere until stopped by some ISP and never stops to lick his wounds 

from time to
time.
--
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: Fastest factorial algorithm
> What is the fastest algorithm for computing factorial, for very large
> numbers (e.g. 10000!)?
> Normally this would take n-2 multiplications, by multiplying out each
> term n(n-1)(n-2)(n-3)...3.2.  Is a better way known?
Don't know, however if you wait until the contest at
http://www.ni.com/devzone/lvzone/codingchallenge.htm
is finished and the results presented, you'll likely
have some good ones.
===
Subject: Re: A question in projective geomtry
> First of all, I'm not sure who this newsgroup is intended for, so I hope 
I'm
> not going to ask a question that will be too trivial... but I need help 
with
> this subject, so I'll go ahead anyway :-)
> How to prove that in finite projective geomtry, there is an equal number 
of
> points on all the lines?
What are the axioms for (finite) projective geometry?
Two different lines intersect in exactly one point.
Two different points lie on exactly one line.
These are what you have to use in your proof.
===
Subject: Re: Linear independence
===
Subject: Re: Linear independence
 >{sqrt{a}: a in Z, a is squarefree} is linearly independent over Q.
when considering C as a vector space over Q?
n square free when for all integers m > 1, not m^2 | n.
-1,1 square free;  0 not square free.
B = { sqr n | n in N, n square free } is linearly independent over Q
when considering R as a vector space over Q?
If sum(j=0..n) (q_j sqr n_j) = 0 for some q_j's in Q, n_j's in B:
1 = sum(j=1..n) [-q_j/q_0 * (n_0,n_j) sqr (n_0 n_j)/(n_0, n_j)]
1 = sum(j=0..m) (r_j sqr m_j) for some r_j's in Q, m_j's in B
What next?  Assume m is the smallest such m?
Clearly 1 /= r_0 sqr m_0.
If 1 = r0 sqr m0 + r1 sqr m1, m0 /= m1
1 = r0^2 m0 + m1^2 m1 + 2r0.r1.sqr m0.m1
but as both m0,m1 are square free, m0 = m1
Beyond that, I'm stimied.
----
===
Subject: Re: JSH: Deprogramming needed?
 Discussion, linux)
> Are you ever going to give up, and just decide that you're the only
> rational person in the universe? I mean that's what the evidence
> seems to indicate: There are no rational people on sci.math, the
> mathematicians you harass via email are not rational, the 
> mathematicians you visit in person are not rational, the editors
> of journals you send papers to are not rational...
It's not such a new conclusion.  Here's one of my long-time favorite
.sigs.
[I]t's the damndest thing.  There's something wrong with every last
one of you, and I *never* thought that was a possibility.  But now I
feel it's the only reasonable conclusion. --JSH sees some sorta light
-- 
Jesse Hughes
You see 300 of something, anything, and you go `[Man], that's a lot of
stuff.' -- Jim Bigler, quoted in the Pittsburgh Post-Gazette.
===
Subject: Re: Newsgroup survey: Math and personality assessment
 Discussion, linux)
> Notice above that I referenced Merriam Webster Online, a great
> resource for quick access to definitions on the go--access at the
> speed of the Internet!
> Merriam-Webster Online
> http://www.m-w.com/
I think you should put away that marketing book.  It's getting to you.
-- 
Jesse Hughes
You see 300 of something, anything, and you go `[Man], that's a lot of
stuff.' -- Jim Bigler, quoted in the Pittsburgh Post-Gazette.
===
Subject: Fundamental groups and liftings help please!
I am somewhat confused now...maybe someone can pleeeeease help.
My homework problem says let p : E --> B be a covering map and fix b_0 in 
B.
Let p^{-1}(b_0) x pi_1 (B ; b_0) --> p^{-1}(b_0) be given by x * [f] = f '
(1), where f ' : I ---> E is the unique lift of f to a path based at x in
p^{-1}(b_0).
I have to show that this is a well defined group action of pi_1 (B ; b_0) 
on
the fiber p^{-1}(b_0).
(Note : pi_1 (B ; b_0) denotes the fundamental group of B based at b_0 and
p^{-1}(b_0) is merely the inverse image of b_0 under p)
Ok, I am really confused.  It is obvious that if you take two loops f and g
that are not path homotopic, then you get x * [f] = f ' (1) and x * [g] = g
' (1), and f ' (1) does not equal g ' (1) since f and g are not path
homotopic.
Is this all that is needed to show that this group action is well-defined?
My problem is that f ' (1) DEPENDS on x, doesn't it?  f ' is a lift of f
that starts at x, but there are a bunch of different x's sitting in
different slices in E.  So, f ' (1) is not the same for any x in
p^{-1}(b_0).  Right?  I mean, if you pick a lift that starts in one certain
slice you get a certain f ' (1), but if you pick a lift that sits in a
different (but homeomorphic) slice that is disjoint, you get literally a
different element of E!  Or maybe I am REALLY confused!
Help! please
Mike
===
Subject: Re: Uncle Al is Sadistic .
> > 
> By the time I saw your response to my other post, I have already
>replied to this post of yours.  I would not have otherwsie.
>>Put it in your think head.  You cannot demand; you can only 
request.
> 
> HUH?!
>>Did you have to use the word 'f***ing'? 
>>You could have asked What countries were they from?
>> 
>> And you still haven't answered the damned^W question.
>I am not obligated to answer, am I?
> Gawd, I hate cute.  Nope, you aren't obliged to answer,
> especially when you used the datum as the basis
> of your argument.  
 I said Africans. 
You said that Etheopia used to send their brighterst; I replied, they
were not Etheopians.
I waited to say the countries name because I was forgetting where 2
were from (I do remmeber that it wasn't Etheopia) and I decided to
think about it a bit before saying anything since I knew that I would
come to remember, at least one of that two.
> When asked for a clarification of that
> datum, 
 you were rude 
> you deliberately ignore the question,
You were rude and so I ignored to answer.
> add all kinds
> of red lines to divert the lack of an answer. 
  What lack of answer? 
> Ergo, you
> don't feel very strongly about the subject but do want
> to get a free ride based on perceived discrimination.
Note:  No amount of your accusation will make me tell you the names of
those countries.  Live with it.
 
> /BAH
> Subtract a hundred and four for e-mail.
===
Subject: Re: Deprogramming needed?
> given that everybody participating in the discussion
> is doing so from mindsets that SHARE a certain kind of academic-
> Americanness, it IS objectively knowable that that kind of abuse,
> EVEN it is deserved, is counter-productive. 
You are grossly racist.
> If everybody else could emulate
> Dik Winter and confine their criticisms to the math, this would all
> go away very quickly.
You are grossly misinformed of the facts.
Socks
===
Subject: Re: Programmer ax+by=c en C
Salut branko tu t'es tromp.8e de newgroups, il faut aller sur fr.sci.maths 
ou
carr.8ement sur un newsgroups de programmation.
-GS-
> Salut tout le monde,
> Ce n'est peut-.90tre pas tr.8fs compliqu.8e
> Mais je me suis pris la t.90te .88 programmer en C ax+by=c
> o.9d a,b,c,x et y sont tous des nombres entiers.
> On conna.94t a,b et c et l'objectif c'est d'obtenir x et y.
> D'abord je commence par le fait que le pgcd(a,b) est un diviseur aussi de
c.
> Quand je le fais .88 la main c'est .8evident mais pour le programme je 
suis
bloqu.8e ici.
> Si quelqu'un sait comment faire et peut m'aider?
===
Subject: Re: A question in projective geomtry
I know that my proof should be based on the axioms (they're are quite
simple, as well). Thats why its bothering I can't prove it. Perhaps you can
give some tip about the body of the proof...
===
Subject: About big numbers...
I got to thinking about that big numbers thread. Here's one quite big
number.
Let f: N^2->N be f(i, 0)=i, f(i, j)=f(i, j-1)! if j>0.
My number is:
f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, 9))))))))))).
There are 11 nested iterations of f.
How big is this number? Is it smaller or bigger than Graham's number?
It ends in a rather long strip of zeroes but I haven't the foggiest
what it starts with.
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
Show me a good mouser and I'll show you a cat with bad breath.
   - Garfield
===
Subject: Re: Programmer ax+by=c en C
>Salut tout le monde,
>Ce n'est peut-.90tre pas tr.8fs compliqu.8e
>Mais je me suis pris la t.90te .88 programmer en C ax+by=c
>o.9d a,b,c,x et y sont tous des nombres entiers.
>On conna.94t a,b et c et l'objectif c'est d'obtenir x et y.
>D'abord je commence par le fait que le pgcd(a,b) est un diviseur aussi de 
c.
>Quand je le fais .88 la main c'est .8evident mais pour le programme je 
suis bloqu.8e ici.
>Si quelqu'un sait comment faire et peut m'aider?
Si j'ai bien compris c'est le th.8eor.8fme de B.8ezout dont vous parlez, 
pas
Euclide .8etendu:
En C:
// R.8esoudre ax+by=pgcd(a,b), a et b connus:
void Euclide_Etendu (unsigned int a,unsigned int b,
                int *pRet_d,            /* pgcd(a,b) */
                int *pRet_x,            /* coefficient x */
                int *pRet_y){           /* coefficient y */
// Ces variables doivent etre signees, car on peut avoir des negatifs
        int x, x1, x2, y, y1, y2, q, r;
        if (b==0) {
                *pRet_d=a;
                *pRet_x=1;
                *pRet_y=0;
                return;
        }
        x2=y1=1;
        x1=y2=0;
        while (b>0) {
                q=a/b;
                r=a-q*b;
                x=x2-q*x1;
                y=y2-q*y1;
                a=b;
                b=r;
                x2=x1;
                x1=x;
                y2=y1;
                y1=y;
        }
        *pRet_d=a;
        *pRet_x=x2;
        *pRet_y=y2;
}
===
Subject: Re: Uncle Al is Sadistic .
Apparently you don't have anything at all to say that's
relevant to any of the three newsgroups to which you
continue to post.  How *do* you justify your existence?
Can you come up with even one post that on-topic
somewhere and avoid the otherwise inevitable trip to
everyone's killfile?
===
Subject: Re: Deprogramming needed?
> What If?   That question on its face causes one to
> consider the state of mind of the speaker...  After
> years of people specifically detailing your mental
> constructs and pointing out your errors in logic you
> still have a doubt of your pass or fail status in life....
> Paul R. Mays
> So do you think I'm less or *more* likely to face the truth because of
> your post Mr. Mays?
> Aren't you trying to say that I'm a failure in LIFE in your comments?
> But aren't I a success in posting?
For sufficiently strange values of success, yes.
===
Subject: Re: What constitutes an implicit use of a function?  (Was:  a 
confusing group t
|
|>if one has a given property, 
|
|What do you mean by has a given property? Your usage seems to
|transcend the values of the propositional functions, in which case two
|propositional functions with the same values need not have the same
|properties.
A term often used in this context is extensionality. Two properties are
considered extensionally equivalent if the same objects satisfy them. So
another way to put it is that you (Sniz Pilbor) are considering
non-extensional aspects of your property, and shouldn't suppose that two
extensionally equivalent properties will make implicit use of the same
functions (and so on).
Don't get too hung up on trying to make the notion of implicit usage
precise. Using a little mathematical logic one could no doubt cook up a
description of certain ways in which a function could be said to appear
implicitly in a proposition or term. You could characterize some restricted
way in which the existence of a function could be deduced from the
statement.
That would be too rigid, however. The way the phrase implicitly uses is
ordinarily used is informal, and whether some bit of mathematics would be
considered implicitly to use a function would depend in a fuzzy way on the
manner in which it's expressed. It's a matter of using your intuition to
judge what the essential ingredients are.
Keith Ramsay
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
She seems to be looking for a Muslim scientist to hit
on her so she's advertising in sci newsgroups and for
bait is trying to appear intelligent, passionate, and
desirable.
===
Subject: Re: Marketing shift, core issues
> 
> 
>I've been thinking about my problems with getting any kind of
>admission that my math arguments showing the core error in 
mathematics
>are correct, so I've gone to marketing books.
> 
> You know, in case you're curious, this sounds really really stupid.
> If your results were correct you'd be able to convince people of
> them by explaining the proofs carefully. But in fact they're wrong,
> people continually explain what the errors are, and tactics from
> marketing books are not going to change that.
> 
> If his results were correct, the same people who point out errors to him 
> with saintly patience would instead have provided all these 
explanations.
> Having spent some time now reading marketing tactics, I can clearly
> see that both Ullrich and Bau are selling a viewpoint to the
> readership.
> Here the assertion is that if I were right then people would
> necessarily agree with me!!!
In mathematics? Yes, if accompanied by a cogent argument.
I believe Ramanujan initially met with some resistance
due to his unfamiliarity with standard notation and
style.
That doesn't mean you're Ramanujan. You've had ample
chance to explicate your proofs, to answer specific
questions where people couldn't understand what the
web page was saying. Those chances all dead-ended.
           - Randy
===
Subject: Re: Probability of a Run
> I am trying to calculate the probability that a gambler with capital 
C
> and who uses a Martingale betting strategy will be wiped out in m
> turns at the game. This would happen with a run of n consecutive
>  In addition to the recurrence approach, you can also use a markov
> process approach to this problem; this has come up before on sci.math
> in the context of coin flipping.
> 
>  Using this approach, I get, with p = 0.5, n = 30, and m = 
1488522243,
> that you have a 50/50 chance of never getting 30 consecutive events 
in
> m trials.
> 
> 
> I like the simplicity of the idea behind this method of calculation.
> There are no cancellation problems since we are always adding. It
> actually executes reasonably quickly but I appear to be getting a lot
> of rounding error problems.
> 
> My first reaction was to check my results. I added an extra row and
> column to get a double check on my results.
> 
> I assume you mean here an extra r/c of 0's (equivalent to a
> forbidden state of s_n)? I don't see how that would change the
> result...
> I got 
> 
> .499999999139885 from adding up the values for 0 to 29
> .500000002118337 from the extra row/column
> 
> and
> .500000000053831 via ProbnmpApprox.
> 
> I'm pretty sure that .500000000053831 is accurate, but the rounding
> errors via the matrix multiplication method seems a bit drastic.
> Can you supply the values you got, if possible it would be nice if you
> could add the extra row to confirm whether the problem is genuine or
> in my code.
> 
> I get 0.499999997774158 regardless of whether or not I add the extra
> row/column; but I'm not surprised that they are the same - adding an
> extra row or column of zeros _shouldn't_ make any difference (which
> doesn't mean that it _wouldn't_).
> The closest I get to 0.5 is actually at 1488522233 trials, with a
> result of 0.500000000102465 (I had previously used the same number of
> trials as the other poster had).
> As another check, I get that the matrix A refered to above has
> (A^(2^30))[0,0] = 0.389400394801501.
> I can't really attest to the accuracy of the implementation, as I
> coded this up in (believe it or not) Lingo, which is the scripting
> language of macromedia's director/shockwave. It's great as a quick
> prototyping, interactive language; although slow as mollasses for
> mathematical use (it took about 10 seconds to calculate the above). At
> some base level though, it surely uses native floating point; so it
> should be as accurate as a one would expect.
> I'll code it up in C when I get a chance and post the results.
> 
> Ian Smith
I wasn't thinking too brightly when I asked about the accuracy of the
matrix multiplication approach. If I had been, I would have realised
that rounding error problems are inherent when raising a matrix to a
large power. The errors you are getting are actually smaller than one
might expect due to the fact that the probabilities in the original
matrix can be held exactly.
Basically the method is reasonably quick and the error propagation
properties are known. As long as you don't use it for huge numbers of
trials you'll get accurate answers.
Ian Smith
===
Subject: measurable function
  Let f(x,y) be a function defined on the unit square 0<=x<=1, 0<=y<=1
  which is continuous in each variable separately. Show that f is a
  measurable function of (x, y).
  Please give me a hint.
--
===
Subject: Re: Vacuum Propellers
yor titl does bad worded. them wirling thingies insid vacuums isnt
'propellers' they is 'impellers'
===
Subject: Re: Marketing shift, core issues
>I've been thinking about my problems with getting any kind of
>admission that my math arguments showing the core error in mathematics
>are correct, so I've gone to marketing books.
>I just wanted to warn readers that I may be employing various tactics
>from modern research on human psychology to see if I can't break
>through the logjam.
>It seemed to me that it'd be a good idea to warn you ahead of time, as
>I'd prefer it that the math by itself would be enough, but that's not
>the way it works.  I guess.
>At least now I won't have to feel guilty if I decide to use tactics,
>as I've given fair warning.
And if this doesn't work, you can start pestering Congress to get your 
theorem passed into law.
-- 
Let us learn to dream, gentlemen, then perhaps we shall find the 
truth... But let us beware of publishing our dreams before they have been 
put to the proof by the waking understanding. -- Friedrich August 
Kekul.8e
===
Subject: Re: [Set Theory] A set has a disjoint copy of itself ?
>  
>>   
>> I am looking for a proof that any set has a disjoint copy of itself,
> i.e. that given a set A, there exists a set B with the same
> cardinality as A, and such that A/B=0. I am hoping this is true 
with
> ZF or ZFC.
>> Noel.
>     
>> B = {0} x A
>>   
> That is B = {(0,a), a in A}.
> What if
> A = {1, (0,1)} ?
>  
>> Oops.  You're right.  Must be more careful.  I trust Herman's response.
> Or let  y  be the least ordinal such that  for all  a  in  A,  (y,a)  is
> not in  A.  Then set  B = {y} x A.
> If you include the axiom of foundation in ZF,  can't we conclude that
> {A} x A is disjoint from A, or must we be a bit more careful?  How
> about  {a U {A} : a in A}?
One doesn't need anything as gruesome as foundation.
Going back to your original construction: although {1} x A
may not be disjoint from A, {1} x A and {2} x A are two disjoint
sets each with the cardinality of A. Extending this ideal let B
be a set of cardinality > card(A) (e.g., B = P(A)). Then B x A
has cardinality > card)A) and B x A contains subsets {b} x A
(for b in B) each of cardinality card(A). There must be
some b in B for which {b} x A is disjoint from A.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Usenet Posting Guide?
> You're right as right can be. Taking what you say at face value, which
> I do, I was indeed off the mark in that assessment. My apologies. You
> also sound like my kind of guy, for that describes me absent the
> religious overtones. I despise political liberals, and I'll let it go
> at that as well in that most writers are political liberals,
> enviro-whackos and PETA types.        Ursula
No apologies needed, though I thank you for them anyway.  By off the
mark I simply meant that you were mistaken, not that you'd offended me.
Wayne
-- 
Wayne Brown  (HPCC #1104)  | When your tail's in a crack, you improvise
fwbrown@bellsouth.net      |  if you're good enough.  Otherwise you give
                           |  your pelt to the trapper.
e^(i*pi) = -1  -- Euler  |           -- John Myers Myers, 
Silverlock
===
Subject: Re: Factorial/Exponential Identity, Infinity
You see, fish or cut bait doesn't mean stop or go.  It means go.
Either fish, ie troll, or cut bait, disassembling the logical
progression of our discussion of sequences with equal zero and one
density vis-a-vis sequences normal to base two, cutting the bait thus
that I may use it to troll sci.math.
Either way the point is to answer the questions about whether
sequences normal to a base other than two, or a power or multiple of
two, have equal zero and one densities, or to ask them.
For you to not even address this besides claiming ignorance is
hilarious.
Sets of numbers contain only points.  
http://www.tiki-lounge.com/~raf/math/CLICAL.tar.gz
Solve with CLICAL, the Clifford Calculator.
Ross
===
Subject: Re: JSH: Deprogramming needed?
> (Marketing ploy warning)
> What if indeed I *am* wrong, and I don't have these great math
> discoveries?
  You have been before, many times.  You know the consequences.
  At this point, it is not even clear what you are claiming.  At
one time you were saying that in the factorization of your cubic polynomial
P(x), at least two of the coefficients were divisible in the algebraic
integers by f, a prime number > 3.  That has been proven false.  You 
appear now to have backed off on that claim.  Perhaps you are claiming 
that at least two of the a's are divisible by f in the ring of objects.
That is of highly dubious value because the ring of objects is not
well-defined, and there is no way to tell whether a given number
is or isn't in that ring, and basic theorems about that ring are
completely lacking.  Bottom line, no one cares about the ring of
objects.  This avenue cannot lead to a proof that there is a 
core error or a proof of Fermat's Last Theorem.  
  So the real question here now is not whether you are wrong.  The
question is, what are you claiming to be true?
> After all, I've been at this since April 1995 having spent a lot of
> time and effort, with literally thousands of posts along with all
> kinds of other activities, websites, and email to mathematicians all
> over the world.
> But, what if I'm wrong?
  Then all that time was wasted.
> What if all the time and energy I've invested in my work has made it
> difficult for me to see, along with the *harsh* and unforgiving
> hostility from several people who seem to have made it their mission
> to make me miserable and find pleasure in mocking or trying to
> humiliate me, have made it extremely difficult for me to see the
> truth?
  A highly compound question.  Putting in all the time and energy is
no guarantee of anything.  The world does not owe you a living.
As for the hostility: you have earned at least some of it.  As for
making it extremely difficult for you to see the truth: that is 
exactly what some of us have been trying to do.  You have resisted it with
every atom in your body.
> Think about the crushing sense of shame and misery if indeed I find
> out that the logical connections I so carefully and impatiently
> discovered over the years are simply not really there, but are a need
> induced delusion.
  Are you asking for sympathy or mercy, or what?  If we honestly 
think you are wrong (we do!), we are not going to say you are right
just because you will feel bad otherwise.  You're an adult after all.
> It seems to me that marketing ploy though these statements may be,
> dealing with people who've made it their business to try and make me
> miserable, only to at times claim they're trying to help me is just
> too much.
  If your marketing ploy here is just to beg for mercy, it simply 
doesn't work with mathematicians.  What they want is just sound
mathematics.  In spite of your lead-off sentences here, I don't 
think you accept any possibility that you are wrong.  We see it 
completely oppositely.  If you continue to claim that your arguments
show a core error or a proof of FLT, there is no possibility that
you are right.  We have found examples that contradict your claims -
examples which you have not refuted in any way - and we have pointed 
to exact spots in your argument that your errors occur.  What more
do you need?
> Aren't there any *other* people who suppose they are rational, who can
> follow a logical argument, who might comment?
> Why is it always the same people?  Or people imitating them in hostile
> and mocking displays of animosity or anger?
  We're the most persistent.
> Aren't there any rational people who can trace out the steps in the
> work I've presented, who might step forward at this time, and
> demonstrate an ability to just be objective?
  No.  You should have no question by now that we HAVE traced out
the steps in your argument, and we find them lacking.  They do
not constitute a proof of anything.
> I don't want any replies of people offering, but if you might consider
> it, I just want you to think about it.  Sure I've been reading this
> marketing book, but it seems to me that still *someone* out there
> might be the right person, as of course, I don't believe I'm wrong. 
> I've traced out every step in the arguments I have, and I think that
> irrational people have been dominating the discussion using group
> effects to hide the truth.
  Marketing strategy is a terrible guide if you want vindication as a 
mathematician.  It is a waste of time.  It's not a matter of 
salesmanship or presentation.  It is entirely a matter of having
or not having a sound mathematical argument.  You don't.
  Nora B.
> Think about it.  I'll come back to the subject later.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: Deprogramming needed?
James Harris is quoted as having written:
>> But aren't I a success in posting?
No, you am not.
===
Subject: Re: Mathematical Integrity
> Don't confuse posters in this newsgroup with mathematicians.  There
> is some overlap, to be sure, but most mathematicians don't post here,
> and most posters here are not mathematicians.
Call me picky if you wish, but I would have preferred, in all instances
above where you said mathematicians, that you had said professional
mathematicians instead.
I think that essentially all people (as well as at least some nonhuman
animals) are mathematicians to some extent. That extent varies from
individual to individual, of course. [I also think it would be very bad
from the standpoint of public relations if professional mathematicians were
to take the attitude that they were the only mathematicians. (Well, so much
for Fermat et al.)]
In comparison, I doubt that many professional musicians would consider
amateur musicians as not being musicians.
David Cantrell
===
Subject: Re: details of joke wanted
> Can somebody help me recall the details of an old maths joke? It is
> one of those situations where a computer scientist, a physicist and a
> mathematician (or some such combination) have to solve a certain
> problem, and the mathematician starts by doing something crazy in
> order to reduce to a problem that is already solved.
Then there's this one:
A physicist, an engineer, and a mathematician are on a stakeout outside a
house.  Through previous observation, they know that there are two people
inside, and they have to wait until nobody is in the house before they can
go in and search it.  After some time, one person comes out and drives
off.  A little later, two people come out and drive off.
The physicist says There must have been another person in the house that
we didn't know about.
The engineer says There must be another entrance to the house that we
don't know about.
The mathematician says We just have to wait until someone enters the
house.  Then there will be zero people in the house, and we can go in.
----
I've also seen a variant on this one with a biologist instead of an
engineer.  The biologist says The original two people must have
reproduced.
-Mike
===
Subject: Re: Fastest factorial algorithm
> What is the fastest algorithm for computing factorial, for very large 
> numbers (e.g. 10000!)?
> Normally this would take n-2 multiplications, by multiplying out each 
> term n(n-1)(n-2)(n-3)...3.2.  Is a better way known?
Since you get lots of 2's in the factorization of n! there is a bit of 
divide and conquer you can do.
n! = 1*2*3*4*...(n-1)*n
    = 1*  3*  5*  *...*(n-1)
        2*  4*  6* ...      *n
    = 1*  3*  5*  *...*(n-1)
        1*  2*  3* ...      *n/2 * 2^(n/2)
    = 1*  3*  5*  *...*(n-1)
        1*      3* ...      *n/2 * 2^(n/2)
            1*     2*...         * 2^(n/4)
etc. So remember all your partial products in computing 1*3*5*...
so very roughly n/2 + 3 lg n mults.
n/2 to multiply the odds, 3 lg n to compute the powers of 2 and multiply 
them with the partial products and then multiply all together.I think 
this is called binary splitting.
Theoretically, extend this idea (for all primes, not just 2) and you get 
the algorithm in (I think):
Borwein, Peter B.
On the complexity of calculating factorials.
J. Algorithms 6 (1985), no. 3, 376--380.
Mitch
===
Subject: Re: Cardinality of 2^n numbers?
> You're misusing cantor's theorum.
Could it be that rather than everyone in this thread
misusing Cantor's theorem, that you alone are 
misunderstanding it?
> I never said that K wasn't greater
> than K. I said N was equal in size to the power set of K. Cantor's
> theorum does not involve this kidn of thing.
Cantor's theorem does not involve the relative
cardinalities of K and its power set? Huh?
Cantor's Theorem says |P(K)| > |K|. Since |K| = |N|,
Cantor's Theorem tells us immediately that N is
smaller in cardinality than the power set of K. It
is very much involved with this kind of thing. What
makes you think it isn't involved with either 
cardinality of power sets or the cardinality of N?
> The power set of K is related to the natural numbers if you do this:
> For each element in the power set of K, add all the numbers together.
> You will always get a unique natural number.
K is an infinite set. Most of its subsets (i.e. most of
the elements of P(K)) are infinite. There is no corresponding
natural number.
> For instance, take K = (1, 2, 4). Take the power set of K = (Null, 1,
> 2, 1&2, 4, 1&4, 2&4, 1&2&4). Now this relates to the natural numbers
> Num = (0, 1, 2, 3, 4, 5, 6, 7) easily. Now K is smaller than the power
> set of K. However, the power set of K, as the size of the power set
> approaches infinity, also approaches the set of the natural numbers
Nope, you're taking an invalid limit argument again. All
you can tell by this argument is that as K grows without
bound, so does |P(K)|. You can't tell anything about
the cardinality of the infinite set. Limit arguments
only concern themselves with what happens at finite
values. They say nothing about at infinity.
Here's another mapping for you. Consider the infinite
set K = {1, 2, 4, ... }. For any given subset of K, say
{1, 2^4, 2^5} construct a binary fraction with a 1
bit if K contains 2^(n-1). Thus, for {2^0, 2^4, 2^5}
I get the number 0.100011.
There is clearly a bijection between such fractions
and the subsets of K. How many such fractions are
there? Well, there is one for every value in [0,1],
rational or irrational. And each one of those
corresponds to a subset of K, a member of P(K).
How many numbers do you think there are in [0,1]?
              - Randy
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
> She seems to be looking for a Muslim scientist to hit
> on her so she's advertising in sci newsgroups and for
> bait is trying to appear intelligent, passionate, and
> desirable.
She has failed in at least two out of three of the above
endeavors.
===
Subject: Re: Basic factorization ideas
> In sci.physics, James Harris
>  
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 1)
> 
> with the c's algebraic integers, I think few of you would have a
> problem realizing that only two of the c's have 7 as a factor.
> Oh, look, he's changed polys on us!  But OK, let's check
> this one.
> Recall that, if P(x) = 49 * (x - x_1) * (x - x_2) * (x - x_3),
> then P(x) = (c_1 * x + 7) * (c_2 * x + 7) * ( c_3 * x + 1)
> implies that c_1 = -7 / x_1, c_2 = -7 / x_2, c_3 = -1 / x_3,
> for some permutation of the x_i.  Again, c_1 * c_2 * c_3 = 49,
> as required, since x_1 * x_2 * x_3 = -1.
> Therefore, c_1 and c_2 both satisfy the equation
> c^3 * P(-7 / c)/49 = c^3 - 21*c^2 + 245*c - 343
> and c_3 of course satisfies
> c^3 * P(-1 / c)/49 =  c^3 - 3*c^2 + 5*c - 1
> so it turns out that this time, James, you got it more
> or less right. :-)  (And c_3 is even a unit, to boot.
> Come to think of it, so are the x_i, which is probably
> one big reason why this particular case actually works.)
> Also, it is obvious that c_1 and c_2 have factors of 7,
> as well, in the ring of algebraic integers.
> 
> But, of course, you're looking at *functions* of x, as you have 
> 
> f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
> 
> so I could also write it as
> 
> 
> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
> 
> Notice that dividing both sides by 49 gives 
> 
> (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
> 
> as long as you're in a ring where 7 is not a factor of 22.
> OK, stupid question.  Where did the 22 come from?
Oh, I've been used to arguing with people about
(5 a_1(x) + 7)(5 a_2(x)+ 7)(5 b_3(x) + 22) =
                      49(300125 x^3 - 18375 x^2 - 360 x + 22)
where the a's are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and b_3(x) = a_3(x) - 3, which is a substitution made because a_3(0) =
3, so that I can isolate constant terms.
My point is that certain rules apply as even with my example above, I
just have rather basic functions of x.
> But yes, for this particular example, your odd math actually does
> work, as you are dealing with x_i which are in fact units.
So you believe that math is quirky?
 
> Now change a_0 to 3 and get back to us. :-)
> [rest snipped]
My thinking is that some of you don't believe in the same mathematics
that most people do, but instead have your own ideas about
applicability of rules in mathematics, and apparently my work attracts
a certain type of person who posts loudly about it.
That is, I think that readers will find that posters are often
challenging some rather basic math, using rules that vary with their
moods.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: Basic factorization ideas
In sci.math, Dik T. Winter

:
> 
>If you saw
>  >(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
>                 
>                          49(x^3 + 5x^2 + 3x + 1)
>  >with the c's algebraic integers, I think few of you would have a
>problem realizing that only two of the c's have 7 as a factor.
> 
> Personally I have a problem with the fact that I can think of no
> values for c_1, c_2 and c_3 such that c_1, c_2 and c_3 are algebraic
> integers and the left and the right side of the equation are equal.
> James has tricked you.  In this version the roots of the cubic are
> units.  Part of his marketing strategy, I think.
I'll admit I wish I knew...but I did note the change.  The more
general cubic:
(c_1 * x + 7) * (c_2 * x + 7) * (c_3 * x + a_0)
   = 49 * (x^3 + a_2 * x^2 + a_1 * x + a_0)
where x^3 + a_2 * x^2 + a_1 * x + a_0 is irreducible, might
be an interesting proof attempt, were it to work ... which,
judging from his earlier example, probably wouldn't. :-)
Is James pulling these specific examples out of thin air or is
there a broader (mathematical) substructure from whence these
are coming which ultimately leads to his Proof Of Something
Significant?
If the latter, where?
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Factorization dispute, again
In sci.math, G Frege

:
>> There are plenty of examples of the converse as well,
>> where somebody proves that something in general need not
>> be true, but James comes up with an example where it is
>> true and claims that proves the general result.
> Well, actually this is a valid proof method (in modern Harrisanism).
> Let x be an arbitrary number. Now consider 0 = x. Obviously we have 
> 0 = 0. With other words, x = 0. (Note that 0 is a constant! Hence x is a
> constant too!) Now since x has been arbitrary, this means x = 0 for any
> x! This actually proves FLT. qed.
> ...
Hmm...beats the usual divide by zero and square root sign proofs... 
:-)
But not by much. :-)
As it is, JSH's latest attempts appear to be 
Let .  Then .  Then
.  Now [pick one: 1, 2, 4, 23]
of  in  are obviously divisible by
[pick one: 2, 3, 7, 11, 97] but there's only
[pick one: 1, 2, 3, 22] factors of [pick one: 2, 3, 7, 11, 97]
in the equation, therefore the algebraic numbers are equivalent
to C and .  QED
(Queerly Elicited and Debatable)
:-)
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Roots of a simple third-degree equation
In sci.math, Fran?ois
 third-degree equation where the term in square is missing, like this
> one:
> -r^3 + 3r + 2 = 0
> How does one do it?
> Fran.8dois
A rather simple mechanism detailed in
http://mathworld.wolfram.com/CubicEquation.html
should breeze one through the general case, unless one happens
to spot a root right away and can simply divide.
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Programmer ax+by=c en C
Voulez vous typez en anglais s.v.p.?
|
| >Salut tout le monde,
| >Ce n'est peut-.90tre pas tr.8fs compliqu.8e
| >Mais je me suis pris la t.90te .88 programmer en C ax+by=c
| >o.9d a,b,c,x et y sont tous des nombres entiers.
| >On conna.94t a,b et c et l'objectif c'est d'obtenir x et y.
| >D'abord je commence par le fait que le pgcd(a,b) est un diviseur aussi de 

c.
| >Quand je le fais .88 la main c'est .8evident mais pour le programme je 
suis
bloqu.8e ici.
| >Si quelqu'un sait comment faire et peut m'aider?
|
| Si j'ai bien compris c'est le th.8eor.8fme de B.8ezout dont vous 
parlez, pas
| Euclide .8etendu:
|
| En C:
| // R.8esoudre ax+by=pgcd(a,b), a et b connus:
| void Euclide_Etendu (unsigned int a,unsigned int b,
| int *pRet_d, /* pgcd(a,b) */
| int *pRet_x, /* coefficient x */
| int *pRet_y){ /* coefficient y */
| // Ces variables doivent etre signees, car on peut avoir des negatifs
| int x, x1, x2, y, y1, y2, q, r;
| if (b==0) {
| *pRet_d=a;
| *pRet_x=1;
| *pRet_y=0;
| return;
| }
|
| x2=y1=1;
| x1=y2=0;
|
| while (b>0) {
| q=a/b;
| r=a-q*b;
| x=x2-q*x1;
| y=y2-q*y1;
| a=b;
| b=r;
| x2=x1;
| x1=x;
| y2=y1;
| y1=y;
| }
|
| *pRet_d=a;
| *pRet_x=x2;
| *pRet_y=y2;
|
| }
|
===
Subject: Re: Fundamental groups and liftings help please!
>I am somewhat confused now...maybe someone can pleeeeease help.
>My homework problem says let p : E --> B be a covering map and fix b_0 in 
B.
>Let p^{-1}(b_0) x pi_1 (B ; b_0) --> p^{-1}(b_0) be given by x * [f] = f '
>(1), where f ' : I ---> E is the unique lift of f to a path based at x in
>p^{-1}(b_0).
>I have to show that this is a well defined group action of pi_1 (B ; b_0) 
on
>the fiber p^{-1}(b_0).
>(Note : pi_1 (B ; b_0) denotes the fundamental group of B based at b_0 and
>p^{-1}(b_0) is merely the inverse image of b_0 under p)
>Ok, I am really confused.  It is obvious that if you take two loops f and 
g
>that are not path homotopic, then you get x * [f] = f ' (1) and x * [g] = 
g
>' (1), and f ' (1) does not equal g ' (1) since f and g are not path
>homotopic.
That's not true, nor is it what you have to prove. Consider, for
example, a double-cover of a circle (E = S^1, B = S^1). The loop that
goes twice around the B and the loop that just sits at the base point
of B both induce the identity map on the fiber p^{-1}(b_0), but they
are not homotopic loops.
Well-defined means that if f and g are homotopic, then f'(1) =
g'(1). You can use the homotopy lifting theorem to show this. You then
have to show that the action is a group action, which isn't hard.
John Mitchell
>Is this all that is needed to show that this group action is well-defined?
>My problem is that f ' (1) DEPENDS on x, doesn't it?  f ' is a lift of f
>that starts at x, but there are a bunch of different x's sitting in
>different slices in E.  So, f ' (1) is not the same for any x in
>p^{-1}(b_0).  Right?  I mean, if you pick a lift that starts in one 
certain
>slice you get a certain f ' (1), but if you pick a lift that sits in a
>different (but homeomorphic) slice that is disjoint, you get literally a
>different element of E!  Or maybe I am REALLY confused!
>Help! please
>Mike
===
Subject: Solving a 2nd order transfer function
I have a second order transfer function that looks like this:
P(s) = k/((tau1*s+1)*(tau2*s+1)) * exp(-td*s)
In order to reverse laplace transform this I will have to multiply P(s) 
with
1/s
The resulting Laplace transform will be:
p(t)=k*(1+tau1*exp(-(t-td)/tau1)/(tau2-tau1)-tau2*exp(-(t-td)/tau2)/(tau2-ta

u1)
)
When tau1, tau2 and td are known it is possible to construct the curve 
p(t).
This is not difficult.
However, imagine that you have this curve, but do not know the values of 
tau1,
tau2 and td that constructed the curve.
In my field this is the problem that I have. I have a second order system 
that
generates a curve p(t) as output when a step function is input into the 
system.
The problem is that I do not know the time constants (tau1 and tau2) nor 
the
dead time (td). I want to analyse the curve and based on this find the 
values
of tau1, tau2 and td. But, I have no idea how to do this. The general 
solution
for p(t), presented above, doesn't allow me to solve for t.
For a first order system it is all much easier as it will be possible to
linearize the function for t and then it is just a question of solving a 
system
of linear equations. This does not seem to be possible for a second order
system.
How do I find tau1, tau2 and td given a curve with values for t and p(t)?
-- 
Someone
===
Subject: Factorization, basic to advanced concepts
If you saw
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 
                 
                          49(x^3 + 5x^2 + 3x + 1)
with the c's algebraic integers, I think few of you would have a
problem realizing that only two of the c's have 7 as a factor.
But, of course, you're looking at *functions* of x, as you have 
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, 
so I could also write it as
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
Notice that dividing both sides by 49 gives 
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
as long as you're in a ring where 7 is not a factor of 1.
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
For instance
(f_1(x) + 7) versus (f_3(x) + 1), where at x=0, both functions equal
0, and 1 and 7 are both factors of the constant term.
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1)
multiplies out to give
49(x^3 + 5 x^2 + 3x + 1).
so f_1(x) f_2(x) f_3(x) = 49, for instance.
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
                   49(300125 x^3 - 18375 x^2 - 360 x + 22)
where the a's are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so they are functions of x, and since one of the roots equals 3 at
x=0, I have
b_3(x) = a_3(x) - 3, 
so that I can see all the constant term factors.
What you have just seen is a major advance in mathematical thinking
where I've used a rather simple abstraction and a special polynomial
to analyze the *roots* of another *different* polynomial.
It is a powerful tool that is new to mathematical analysis, as I
discovered it only a few years ago.
Unfortunately the concepts seem advanced enough to attract posters who
react with fury when they can't quite get it, who refuse to
acknowledge the basic principles, who also post a LOT!!!
They've created a false picture that mathematicians as a group have
refuted the information provided here, when it's impossible to refute
mathematical logic.
Basically, a few people with an agenda, posting a lot, have created an
atmosphere of confusion and distrust which has fed upon itself.
But now I hope to reach other, more sophisticated and intelligent
people who can consider the actual facts.
James Harris
My math discoveries, found for profit
http://mathforprofit.blogspot.com/
===
Subject: Re: Factorization, basic to advanced concepts
> If you saw
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
>                           49(x^3 + 5x^2 + 3x + 1)
you'd next see
http://www.crank.net/harris.html
 It's not every braying jackass that gets a whole page at crank.net
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
===
Subject: Re: Fundamental groups and liftings help please!
|I am somewhat confused now...maybe someone can pleeeeease help.
|
|My homework problem says let p : E --> B be a covering map and fix b_0 in 
B.
|Let p^{-1}(b_0) x pi_1 (B ; b_0) --> p^{-1}(b_0) be given by x * [f] = f '
|(1), where f ' : I ---> E is the unique lift of f to a path based at x in
|p^{-1}(b_0).
|
|I have to show that this is a well defined group action of pi_1 (B ; b_0) 
on
|the fiber p^{-1}(b_0).
|
|(Note : pi_1 (B ; b_0) denotes the fundamental group of B based at b_0 and
|p^{-1}(b_0) is merely the inverse image of b_0 under p)
|
|Ok, I am really confused.  It is obvious that if you take two loops f and 
g
|that are not path homotopic, then you get x * [f] = f ' (1) and x * [g] = 
g
|' (1), and f ' (1) does not equal g ' (1) since f and g are not path
|homotopic.
yep, you're a bit confused, but it's probably not fatal.
(the secret way to really understand this stuff is a bit different in
spirit from the approach that your teacher seems to be using, but i
won't worry about that for now, since you can probably understand your
teacher's approach if you just keep working at it a bit longer.)
actually it's _not_ obvious that if loops f and g aren't
path-homotopic then f'(1) and g'(1) can't be equal, and in fact it's
false; for example consider the special case where the covering map p
is the identity map of the base space; in this case _all_ points in
p^[-1](b_0) are equal to each other.  (maybe you were thinking of the
opposite extreme special case of the _universal_ covering space, where
what you said is true more or less by definition?)
|Is this all that is needed to show that this group action is well-defined?
again, that's _not_ what you have to show, and it's a good thing you
don't have to show it because it's not always true.  rather, what you
have do to show that the action is a well-defined map is to show
that if loops f and g _are_ path-homotopic (meaning with fixed
end-points), then f'(1) and g'(1) _are_ equal.
i'll leave it at that for now, in case that helps.  by the way,
depending on how the question is interpreted there's probably a bit
more that you have to do after showing that the action is a
well-defined map; you probably have to show that it satisfies the
algebraic definition of action as well.
-- 
[e-mail address jdolan@math.ucr.edu]
===
Subject: Re: Largest number ever written down?
>       http://www.cs.berkeley.edu/~aaronson/bignumbers.html
<<<
But maybe there?s some ingenious program that can examine other 
programs and tell us, infallibly, whether they?ll ever stop running. 
We just haven?t thought of it yet.
i.e. maybe there's a solution to the halting problem?
Where's this guy been for the last hundred years?
Phil
-- 
Unpatched IE vulnerability: file-protocol proxy
Description: cross-domain scripting, cookie/data/identity theft, 
             command execution
Reference: 
http://safecenter.net/liudieyu/WsOpenFileJPU/WsOpenFileJPU-Content.HTM
Exploit: 
http://safecenter.net/liudieyu/WsOpenFileJPU/WsOpenFileJPU-MyPage.HTM
===
Subject: Re: If anyone cares about interesting questions on today's GRE 
subject Math test
Hi Mike.  I need to that this same test.  Just wondering - do you have
any additional questions from this test.  If so, how can I get a copy.
 I need all the help that I can get.
> Some people may think that this test is irrelevant for admissions 
committees
> for pure math, but there were some interesting questions that were asked 
on
> the exam (it was held today and I took it)...
> For instance,
> 1.  Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what?
> 2.  Let f(x) = x^2 + Bx + C, let B and C be independent random variables
> uniformly distributed on [0,1].  What is the probability that the roots 
of
> f(x) are distinct and real?
> There were other interesting ones as well, but I can't remember 
them...These
> are pretty easy if you're not under pressure to do them in less than 
average
> 3 minutes each (or maybe you think 30 seconds under pressure is
> sufficient...but give undergraduates some slack :-)  )
> It's good to see that the test requires more than memorization and some
> thinking...needless to say I got the 1st one wrong and the 2nd one 
right...
> In any case, I don't think I did as well as I would have liked, and hope
> that doesn't impact my chances at some of the top schools in 
representation
> theory and algebra in general (princeton, harvard, berkeley), but I have
> heard that this test isn't looked at very much, it's more recommendations
> and extra stuff/research...only I have heard that if you do poorly it 
raises
> a red flag (but I'm confident I don't have to worry about that)
> Just a post if anyone's interested,
> Mike
===
Subject: Re: Largest number ever written down?
>       http://www.cs.berkeley.edu/~aaronson/bignumbers.html
Ooops. (regarding my previous post which hasn't returned to me yet.)
He wasn't wearing his moose-horns, so I didn't know he was still 
hypothesising the imposible. The very next paragraph begins:
<<<
Nope. ...
Phil
-- 
Unpatched IE vulnerability: document.domain parent DNS resolver
Description: Improper duality check leading to firewall breach 
Published: July 29 2002
Reference: 
http://online.securityfocus.com/archive/1/284908/2002-07-27/2002-08-02/0
===
Subject: Re: JSH: Deprogramming needed?
Originator: grubb@lola
>Aren't there any *other* people who suppose they are rational, who can
>follow a logical argument, who might comment?
Sure. The first question I have is what you mean by the 'constant term'
of a non-polynomial. In particular, why do you assume that the
constant term of 5a_1(x) +7 is 7? Why do you assume that contant
terms for non-polynomials act the same way as for polynomials? You
have not proved this anywhere, so that constitutes a gap in your
argument.
Next, why do you assume that 49 being a factor of P(x) means
that 7 has to be a factor of one of the terms in step 5? This is
not proved, so constitutes a gap in your proof. 
At this point, I see 2 significant gaps in your logic. That is enough for
me to ignore the whole argument.
>Why is it always the same people?  Or people imitating them in hostile
>and mocking displays of animosity or anger?
My guess is that nobody else cares. The people that are commenting
on your argument are generally respected because of other things
they post about. They have shown themselves able to give proofs
of their own and to comment intelligently on arguments that others
give. 
Given the qualifications of those arguing against you, you would
have to give arguments that don't have glaring flaws to convince
anyone else that you are right. Unfortunately, your argument *does*
have glaring flaws. In particular, you assume that 'constant terms'
for non-polynomials work the same as they do for polynomials over
the integers. This has not been proved. The lack of a proof is a 
glaring flaw.
Now you have the following choices. Either admit that your argument
is incomplete and supply the missing details OR admit that your
argument is incomplete and admit defeat OR not admit you argument
is incomplete and admit that you are simply a troll.
--Dan Grubb
===
Subject: Re: measurable function
> Let f(x,y) be a function defined on the unit square 0<=x<=1, 0<=y<=1
>   which is continuous in each variable separately. Show that f is a
>   measurable function of (x, y).
For each n, divide the square into n equally spaced horizontal strips. On 
each strip, make f_n independent of y by setting it equal to f's values on 
the lower edge of the strip. Now you have a sequence f_n of measurable 
functions, and fn -> f a.e. (in fact everywhere).
===
Subject: Re: Call for Participation
> Hello All,
> NFSNET is a distributed computing project devoted to pushing factorization 

> limits. It allows users to run client software which sends data back to a
> central 
> site. The data is collected and eventually used to produce a 
factorization.
> However, the data is sent via an HTTP connection, which means that clients 

> must be on-line.
> I am working on trying to do some slightly smaller numbers, but are on the 

> 'MOST-WANTED' list for factorizations.  I could use some help, as I have 
very
> few machines.
> I will provide executables (even source if people want to look at it) and 

> data via email.
> People will then run the code OFF-LINE then email the results back.
Bob,
I wish you luck with this, but wouldn't it be easier to try to bolt on 
off-line-ness onto NFSnet?
Either way, I heartily recommend that those who are sitting around 
with PCs doing nothing useful volunteer to help out with this. 
Phil
-- 
Unpatched IE vulnerability: NavigateAndFind file proxy
Description: cross-domain scripting, cookie/data/identity theft,  
             command execution
Reference: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJPU-Content.HTM
Exploit: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJPU-MyPage.htm
===
Subject: Re: A question in projective geomtry
> First of all, I'm not sure who this newsgroup is intended for, so I hope 
I'm
> not going to ask a question that will be too trivial... but I need help 
with
> this subject, so I'll go ahead anyway :-)
> 
> How to prove that in finite projective geomtry, there is an equal number 
of
> points on all the lines?
> 
>>What are the axioms for (finite) projective geometry?
>>Two different lines intersect in exactly one point.
>>Two different points lie on exactly one line.
>>These are what you have to use in your proof.
>I know that my proof should be based on the axioms (they're are quite
>simple, as well). Thats why its bothering I can't prove it. Perhaps you 
can
>give some tip about the body of the proof...
You can't prove it based on those axioms alone, as a simple example
shows: draw three collinear points in the plane, and a fourth point
that is not collinear with the first three. Now draw lines through all
pairs of points. There are four lines in all. One line contains four
points, the remaining lines each contain two points. The axioms above
are clearly satisfied.
You need another axiom: each line contains at least three points. Now
consider two (distinct) lines, L and M. Choose distinct points P on L
and Q on M (prove that this is possible). Draw a line through P and Q
(prove that this line is distinct from L and M), and choose a third
point R on this line, distinct from P and Q. Now use R to map points
on L to points on M (using a simple natural construction). Show that
this mapping is 1-1. The same method gives a 1-1 map from M to  L .
Voila!
John Mitchell
===
Subject: Re: Clarify of Yang Mills and Mass Gap Hypothesis Problem
>> Yang-Mills Field Equation and Mass Gap Hypothesis is one of the
>> problems which were included in Clay Millennium Problems.
 
>> After reading description of the problem, I am still not clear (or be
>> convinced ) why there is a problem here. So, can anyone shed some
>> light on this ? Be specific, two things I do not understand are:
>due to residual physics envy on the part of the organizers.
Or one might say it had something to do with Arthur Jaffe being 
president of CMI.  But mathematical physics is genuine mathematics,
and this Millenium Problem is not a physics problem but a 
mathematics problem: it asks about the existence and properties
of a certain mathematical object, not about whether that object
exists in the real world.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Squares that end with four identical digits
> 
>  Which integers can have squares that end with four identical digits?
> Let's see. A square ends with either 0,1,4,5,6 or 9. A number ending with 

> 11, 55, 66 or 99 cannot be a square because it isn't congruent to either 0 

> or 1 modulo 4. That leaves us with 00 and 44. Obviously 00 is a 
possibility
> as an integer multiple of 100 will have a square ending 0000. Actually
> that is the only possibility: a square cannot end with 4444 for then it
> would be the square of an even number, say 2n. Hence n^2 =((2n)^2)/4 
would
> end with 11, as any mulitple of 10000 divided by 4 will end in two zeros.
> Earlier we saw that a square cannot end with 11. Case closed.
10/10 for part 1 Jyrki, now generalise to other bases :-)
Phil
-- 
Unpatched IE vulnerability: protocol control chars
Description: Circumventing content filters
Reference: http://badwebmasters.net/advisory/012/
Exploit: http://badwebmasters.net/advisory/012/test2.asp
===
Subject: Re: Squares that end with four identical digits
> Which integers can have squares that end with four identical digits?
>>I believe that all integers that end in two zeroes have squares that end
>>in four identical digits.
>Yes, of course.  Not so obvious is that these are all the answers.
On the other hand, there are solutions in many other bases, e.g. in 
hexadecimal, 
    
                               2
                           7B69  = 3B7E1111
                               2
                           FB69  = F6E71111
                               2
                            497  = 151111
                               2
                           8497  = 44AC1111
                               2
                           36D2  = BBD4444
                               2
                           B6D2  = 828F4444
                               2
                           76D2  = 37264444
                               2
                           F6D2  = EDF84444
                               2
                            92E  = 544444
                               2
                           892E  = 49824444
                               2
                           492E  = 14EB4444
                               2
                           C92E  = 9E194444
                               2
                            DC5  = BD9999
                               2
                           8DC5  = 4E829999
                               2
                           723B  = 32F89999
                               2
                           F23B  = E5339999
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Mathematical Integrity
>> Don't confuse posters in this newsgroup with mathematicians.  There
>> is some overlap, to be sure, but most mathematicians don't post here,
>> and most posters here are not mathematicians.
>Call me picky if you wish,
Okay, you're picky :--)
> but I would have preferred, in all instances
>above where you said mathematicians, that you had said professional
>mathematicians instead.
>I think that essentially all people (as well as at least some nonhuman
>animals)
Let's leave C.E.O.s out of this ...
> are mathematicians to some extent. That extent varies from
>individual to individual, of course. [I also think it would be very bad
>from the standpoint of public relations if professional mathematicians 
were
>to take the attitude that they were the only mathematicians. (Well, so 
much
>for Fermat et al.)]
>In comparison, I doubt that many professional musicians would consider
>amateur musicians as not being musicians.
>David Cantrell
When I moved from academia to industry, I was surprised to find that
pretty much everyone (in a technical company, at least) considers
himself a mathematician. There's some truth to what you say, but I
think you're stretching the point too far. I can perform complicated
physical tasks such as walking, raising a glass of water to my lips
without spilling it, etc., but that doesn't make me a physicist.
John Mitchell
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
> She seems to be looking for a Muslim scientist to hit
> on her so she's advertising in sci newsgroups and for
> bait is trying to appear intelligent, passionate, and
> desirable.
So ..you are anti-Muslim. I understand why you must now attack me in
such personal way but still I am sure that that's very low of you.
Most idiots like you have no clue that there are lots of Muslims in
countries like Burma, Thailand, who don't deserve the animosity people
like you have for them for being Muslims.
===
Subject: A neat computer program
int main( void )
{
  int n;
  printf( Let B(n) be the nth busy beaver number.nr );
  printf( The busy beaver numbers are:nr );
  for ( n = 1; ; n = n+1 )
  printf(   B(%d)nr, n );
}
This program lists the busy beaver numbers :-)  (just not in decimal)
===
Subject: Re: Basic Planar Geometry Question
> If you are counting them pairwise, this might be a winner:
> HOOOOOOOOOH
> HOOOOOOOOOH  (Pardon the O fill, trying to prevent damage by
> HOOOOOOOOOH   proportional fonts.)
But here you only have AB and AD as different directions.
My proposals are:
A   B
OOOOOOOOOE      
C   D
with ABCD a square and E the intersection of the parallels to AD and
BC drawn from B and D respectively.
My other proposal is (and think it is the only one apart from the
square), a parallelogram having lengths 1 and sqrt(3).
   A            B
           C           D
                E
with E again the intersection of parallels to diagonals.
A nice image you can find here:
http://www.thequantummachine.com/diagram.jpg
Are there any other polygons???
===
Subject: Re: uniform convergence & differentiation
> I think there are some similar theorems about sequences and series of
> complex functions. It's important to keep in mind that
You need a Banach space F, a normed vector space E, O a convex (or connex)
open subset of E, and f_n: O->F a sequence of differentiable functions, 
such
that:
(i) Df_n converges uniformly to g
(ii) There is a point a in O such that (f_n(a)) converges.
Then there exists a function f: O->F, Df = g and (f_n) converges uniformly
to f on every bounded subset of O. If the f_i are C1, then f is C1 too.
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
 Feeling hurt..still? 
 
> Can you come up with even one post that on-topic
> somewhere and avoid the otherwise inevitable trip to
> everyone's killfile?
Do it INSTAED OF whinning about it.  
I didn't expect that there would be people like you to take things out
of context and attack me. Moreover, I realized that it was a mistake
for me to even bother to reply to people like.
Now ... put me in your killfile instead of talkign about it.  You will
be helping me greatly.
===
Subject: Re: Marketing shift, core issues
> I've been thinking about my problems with getting any kind of
> admission that my math arguments showing the core error in mathematics
> are correct, so I've gone to marketing books.
> I just wanted to warn readers that I may be employing various tactics
> from modern research on human psychology to see if I can't break
> through the logjam.
Way to go James, that is an excellent idea!!  I will borrow from it
and copy your techniques in my war against the evil Cantorians.  With
subtle, unethical psychological manipulation behind me, I know that
now I can open the floodgates fully and do away once and for all with
the collective bag of nonsensical gibberish that is Cantor.  Would you
like to join forces with me so that we can fight Cantor together?
Your friend,
Nathaniel Deeth
Age 11
===
Subject: Re: Greek Alphebet
>It is true that the near-eastern alphabets are not properly
>alphabets.  The real test is whether they represent every phoneme.
>Because the near-eastern scripts leave out vowels, they aren't proper
>alphabets.
>But this is also a matter of degree.  The classical Greek alphabet
>does not distinguish long and short alpha, for example, which is a
>phonemic difference.  Similarly, until accent marks were invented, it
>did not distinguish different accents, which are also phonemic in
>ancient Greek.
Not many languages, I think, have a distinct representation for each 
phoneme.  Certainly not English.  Maybe some languages with a restricted
set of phonemes.  Is the International Phonetic Alphabet the only true
alphabet for the English language?
>However, you are incorrect about there being some natural reason for
>near-eastern scripts to not need vowels but IE languages to need
>them.  Fr xmpl, y cn rd ths prtt wll, cnt y?  nglsh cn b rlbly rd b
>ppl wtht vwls, mst f th tm.  And Hebrew did finally decide to start
>writing in the vowels because things had gotten too hard.
The Hebrew vowel marks (nekudot) were invented for biblical texts, 
where for religious reasons it was important to get the pronunciation 
right even if you didn't understand the meaning.  They are not 
generally used in modern Hebrew.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel    
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Uncle Al is Sadistic .
>Apparently you don't have anything at all to say that's
>relevant to any of the three newsgroups to which you
>continue to post.  How *do* you justify your existence?
>>She seems to be looking for a Muslim scientist to hit
>>on her so she's advertising in sci newsgroups and for
>>bait is trying to appear intelligent, passionate, and
>>desirable.
> So ..you are anti-Muslim. I understand why you must now attack me in
> such personal way but still I am sure that that's very low of you.
My comment was limited to your misuse of these newsgroups.
> Most idiots like you have no clue that there are lots of Muslims in
> countries like Burma, Thailand, who don't deserve the animosity people
> like you have for them for being Muslims.
So Hansen was right after all? Are you really just
a rabid bitch attacking anything you don't like
with any stone that happens to be in your hand at
the moment? So why are you walking around with
rocks in your hand in the first place then?
Frankly, Amanda, behavior of the sort you just engaged
in gives all of Islam a even more of a bad reputation.
Calm down already, or is not having found a man to fill
your life the base cause of your outbursts?
Here's a little instantaneous help for you:
http://www.virtual-vibrator.com/index-ns.shtml#
===
Subject: Re: Uncle Al is Sadistic .
>Apparently you don't have anything at all to say that's
>relevant to any of the three newsgroups to which you
>continue to post.  How *do* you justify your existence?
>>She seems to be looking for a Muslim scientist to hit
>>on her so she's advertising in sci newsgroups and for
>>bait is trying to appear intelligent, passionate, and
>>desirable.
> She has failed in at least two out of three of the above
> endeavors.
Often having the right equipment compensates. Perhaps the
obvious feelings of inadequacy have something to do with
that.....
===
Subject: C++ Simulator of a Nondeterministic Turing Machine
C++ Simulator of a Nondeterministic Turing Machine has been added at :
* http://alexvn.freeservers.com/s1/turing.html
* http://sourceforge.net/projects/turing-machine/
Currently those sites contain C++ Simulators for both Deterministic and 
Nondeterministic Turing Machines.
The Simulators contain examples of Turing Machines as well.
1. A (Deterministic) Turing Machine example :
      Recognition of Palindromes
      from 'The Design and Analysis of Computer Algorithms [1976]' by 
A.V.Aho, J.E.Hopcroft, J.D.Ullman
      (See examples 1.8, 1.9)
2. A Nondeterministic Turing Machine example :
      Partition Problem
      from 'The Design and Analysis of Computer Algorithms [1976]' by 
A.V.Aho, J.E.Hopcroft, J.D.Ullman
       (See example 10.1)
--
   Alex Vinokur
     mailto:alexvn@connect.to
     http://mathforum.org/library/view/10978.html
===
Subject: Re: A question in projective geomtry
   at 03:51 PM, Arie Levit  said:
>How to prove that in finite projective geomtry, there is an equal
>number of points on all the lines?
Is this a homework question? 
Do you know what a collineation is? That may already be more of a hint
than I should have given, so I'll let it go at that.
-- 
     Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action.  I reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact me.  Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Usenet Posting Guide?
> 
> With all respect, sir, I can guess your politics right across the
> board. Just for openers, you're a Sierra Club, save-the-whales type.
> Gotcha, huh?
> 
> You're about as far off the mark as it is possible to get.  Not that I 
see
> the relevance, but I'm a very conservative, Religious Right type -- 
not
> far removed from what most people would call a fundamentalist 
(though
> I don't use that label myself).  I'm irrevocably opposed to both 
abortion
> and gun control, voted for Bush (both of them) and supported the Iraq 
war
> (both of them), and certainly do not support the Sierra Club, 
Greenpeace,
> PETA, or any other environmentalist or animal-rights group.  And no,
> I don't care to discuss or defend any of those positions in sci.math.
> You're right as right can be. Taking what you say at face value, which
> I do, I was indeed off the mark in that assessment. My apologies. You
> also sound like my kind of guy, for that describes me absent the
> religious overtones. I despise political liberals, and I'll let it go
> at that as well in that most writers are political liberals,
> enviro-whackos and PETA types.
Not to get into a flame war, but coming from sci.math
I have to object to what seems to me to be some false
theorems in your reasoning, namely that
    politically liberal => PETA
    Sierra Club => environmental terrorist
    environmentalist => every wacko animal rights sentiment
I would describe myself as a political liberal and an
environmentalist (to the point of bicyling across
Philadelphia to work on occasion). But I don't
identify with anything on the right hand side.
I refuse to get into an argument about my positions,
or yours. I just want to point out that you have
dismissed the other side of the political spectrum
as not being a spectrum.
         - Randy
===
Subject: Re: Factorization dispute, again
> As it is, JSH's latest attempts appear to be 
> Let .  Then .  Then
> .  Now [pick one: 1, 2, 4, 23]
> of  in  are obviously divisible by
> [pick one: 2, 3, 7, 11, 97] but there's only
> [pick one: 1, 2, 3, 22] factors of [pick one: 2, 3, 7, 11, 97]
> in the equation, therefore the algebraic numbers are equivalent
> to C and .  QED
Well let's hear the master himself:
        What you have just seen is a major advance in mathematical
         thinking where I've used a rather simple abstraction and a
         special polynomial to analyze the *roots* of another
         *different* polynomial.
         It is a powerful tool that is new to mathematical analysis, 
         as I discovered it only a few years ago.
         Unfortunately the concepts seem advanced enough to attract
         posters who react with fury when they can't quite get it, 
         who refuse to acknowledge the basic principles, who also post 
         a LOT!!!
         They've created a false picture that mathematicians as a 
         group have refuted the information provided here, when it's
         impossible to refute mathematical logic.
         Basically, a few people with an agenda, posting a lot, have
         created an atmosphere of confusion and distrust which has fed
         upon itself.
         (James Harris)
I like best his rambling about the false picture that mathematicians as
a group have refuted the information provided here. Right! Actually,
only a few people with an agenda, posting a lot, have created an
atmosphere of confusion and distrust which has fed upon itself. Indeed!
Though looking at the facts from this _point of view_ certainly won't
explain the existence of http://www.crank.net/harris.html.
After all It's not every braying jackass that gets a whole page at
crank.net (Uncle Al).
===
Subject: Re: Basic factorization ideas
> In sci.math, Dik T. Winter
> :
 there a broader (mathematical) substructure from whence these
> are coming which ultimately leads to his Proof Of Something
> Significant?
 
As you might be able to guess from his personality (disorder)
Harris must maintain the belief that he's working on (or has
finished) a Work of Monumantal Importance.
For the past seven years or so (it may be eight by now),
he's been banging away at an elementary proof of Fermat's
Last Theorem. The this cubic MUST have two of its factors
divisible by a prime examples he's been pestering us with
have to do with a step in his proof for the case of exponent 3.
If you want to look at a de-obfuscated treatment of an old version
of his proof, just ask.
Rick
===
Subject: Re: Mathematical Integrity
>> Don't confuse posters in this newsgroup with mathematicians.  
There
>> is some overlap, to be sure, but most mathematicians don't post here,
>> and most posters here are not mathematicians.
>Call me picky if you wish,
> Okay, you're picky :--)
I'll go farther than that. I was simply wrong. I was thinking of a
mathematician merely as being one who _does_ mathematics. But now, upon
consulting a dictionary, I see that one must be skilled or learned in
the field to qualify as being a mathematician. And I take that to mean
_substantially more highly_ skilled or learned than the average individual.
So we're not all mathematicians, musicians, etc.
David Cantrell
> but I would have preferred, in all instances
>above where you said mathematicians, that you had said 
professional
>mathematicians instead.
>I think that essentially all people (as well as at least some nonhuman
>animals)
> Let's leave C.E.O.s out of this ...
> are mathematicians to some extent. That extent varies from
>individual to individual, of course. [I also think it would be very bad
>from the standpoint of public relations if professional mathematicians
>were to take the attitude that they were the only mathematicians. (Well,
>so much for Fermat et al.)]
>In comparison, I doubt that many professional musicians would consider
>amateur musicians as not being musicians.
>David Cantrell
> When I moved from academia to industry, I was surprised to find that
> pretty much everyone (in a technical company, at least) considers
> himself a mathematician. There's some truth to what you say, but I
> think you're stretching the point too far. I can perform complicated
> physical tasks such as walking, raising a glass of water to my lips
> without spilling it, etc., but that doesn't make me a physicist.
> John Mitchell
===
Subject: Re: Greek Alphebet
> ...
> How are numbers written in Hebrew?  I remember seeing people writing
> right to left (so presumably Hebrew or Arabic) but using our numerals
> and writing them left to right.  So they had to jump ahead a bit to
> write the numbers.
> You do not have to jump ahead.  You just first write the units, the
> tens next, etc.  This is not different from the way we write the
> result of an addition on paper.
Yes, they did not have to jump ahead but they did.  I should have
phrased it better and simply said that they did jump ahead.  Despite
the change in direction this odd method may have been easier if they
wanted to write numbers that appeared the same as ours.  In every
language I know, large numbers are spoken in the order that we
normally write.  E.g. 1,234 is spoken is commonly spoken in that
order.  Not just in Indo-European languages but many Asian ones as
well.  A few languages invert the last two digits.  This used to be
common in English but is now rare.  It survives in the nursery rhyme:
Four and twenty blackbirds baked in a pie.  But even when this form
is used, the thousands and hundreds come first.
J
===
Subject: Sex and Math
James Harris recently came up with the excellent idea of using modern
advertising techniques to get his theories through to the public.  I
have decided that this is an excellent way for me to promote my
anti-Cantorian doctrine to the public.  Now, it is well known that sex
sells.  Therefor, without further ado (please maximize your window):
                        ,ad8888888888888888ba,
                     ad88888888888888888888888a,
                   a888888888888888888888888888,  CANTOR IS WRONG!
                ,8888  P88888888888888888888b,
                d88         `P88888888888888888,
              ,8888b               88888888888888,  CANTOR IS EVIL!
              d8P'''  ,aa,              888888888b
              888bbdd888888ba,  ,I         88888888,
              8888888888888888ba8         ,88888888b NEVER BELIEVE
             ,888888888888888888b,        ,8888888888  CANTOR'S LIES!
             (88888888888888888888,      ,88888888888,
             d888888888888888888888,    ,8   8888888b
             88888888888888888888888  .;8'  (888888
             8888888888888I8888888P ,8 ,aaa,  888888 MEN WHO FOLLOW
             888888888888I:8888888 ,8  'b8d'  (88888  CANTOR WILL
             (8888888888I'888888P' ,8)          88888  NEVER SLEEP
              88888888I  8888P'  ,8)          88888  WITH THIS
              8888888I'   888   ,8 (   )       88888  YOUNG LADY!
              (8888I     88,  ,8             ,8888P
               888I'       P8 ,8   ____      ,88888)
              (88I'          ,8  MM   ,888888'
             ,8I            ,8(    aaaa   ,8888888
         ,8I'             ,888a           ,8888888) CANTOR IS FULL OF
       ,8I'            ,888888,       ,888888888     INCONSISTENCIES!
 ,8I'            ,8888888        88888888
 8I'            ,8    88         888888P  YOU SAY THERE IS NO SUCH
8I            ,8'     88          `P888    THING AS THE NEXT REAL
8I           ,8I      88            8ba,.     NUMBER.  HA!  I LAUGH
(8,         ,8P'      88              888bma,.AT YOU.  ARRANGE A
 8I        ,8P'       88,              8b   P8ma,   COUNTABLY
 (8,      ,8d        `88,               8b     `8a  INFINITE
  8I     ,8dP         ,8X8,                8b.    :8b NUMBER OF 
  (8    ,8dP'  ,I    ,8XXX8,                `88,    8) REALS IN
   8,   8dP'  ,I    ,8XxxxX8,     I,         8X8,  ,8 BINARY AND
   8I   8P'  ,I    ,8XxxxxxX8,     I,        `8X88,I8 PRECISELY ONE
   I8,     ,I    ,8XxxxxxxxX8b,    I,        8XXX88I, DIAGONAL
   `8I      I'  ,8XxxxxxxxxxxxXX8    I        8XXxxXX8, NUMBER WILL
    8I     (8  ,8XxxxxxxxxxxxxxxX8   I        8XxxxxxXX8,   BE
   ,8I     I[ ,8XxxxxxxxxxxxxxxxxX8  8        8XxxxxxxxX8,DETERMINED.
   d8I,    I[ 8XxxxxxxxxxxxxxxxxxX8b 8       (8XxxxxxxxxX8, THIS IS
   888I    `8,8XxxxxxxxxxxxxxxxxxxX8 8,     ,8XxxxxxxxxxxX8 YOUR
   8888,    88XxxxxxxxxxxxxxxxxxxX8)8I    .8XxxxxxxxxxxxX8 NEXT
  ,8888I     88XxxxxxxxxxxxxxxxxxxX8 `8,  ,8XxxxxxxxxxxxX8  REAL!
  d88888     `8XXxxxxxxxxxxxxxxxxX8'  `8,,8XxxxxxxxxxxxX8 
  888888I     `8XXxxxxxxxxxxxxxxX8'    88XxxxxxxxxxxxX8
  88888888bbaaaa88XXxxxxxxxxxxXX8)      )8XXxxxxxxxxXX8WOMEN LOVE MEN
  8888888I, ``8888888888888888aaaaa8888XxxxxXX8WHO TURN 
AGAINST
  (8888888I,                      .  ```88888P      CANTOR
   88888888I,                   ,8I   8,       I8
    88888I,                ,8I'    I8,    ;8
           `8I,             ,8I'       `I8,   8)CANTOR IS A PLAGIARIST
            `8I,           ,8I'          I8  :8'  AND A CHILD MOLESTOR
             `8I,         ,8I'           I8  :8
              `8I       ,8I'             `8  (8
               8I     ,8I'                8  (8;
               8I    ,8                  I   88,   CANTOR'S THEORY IS
              .8I   ,8'                       88,      A DEAD END
              (PI   '8                       ,8,`8,
             .88'            ,@           .a8X8,`8,
             (88             @@         ,a8XX888,`8,
            (888             @'       ,d8XX8  b `8,
           .8888,                     a8XXX8    a `8,CANTOR'S WORKS
          .888X88                   ,d8XX8I      9, `8,ARE SUPPORTED
         .88:8XX8,                 a8XxX8I'       `8  `8,BY AL-QAEDA
        .88' 8XxX8a             ,ad8XxX8I'        ,8   `8,
        d8'  8XxxxX8ba,      ,ad8XxxX8I          8  ,  `8,
       (8I   8XxxxxxX888888888XxxxX8I            8  II  `8
       8I'   8XxxxxxxxxxxxxxxxxxX8I'            (8  8)   8;
      (8I     8XxxxxxxxxxxxxxxxxX8              (8  8)   8I  JESUS
      8P'     (8XxxxxxxxxxxxxxX8I'                8, (8   :8  HATES
     (8'       8XxxxxxxxxxxxxxX8'                 `8, 8    8 CANTOR
     8I        `8XxxxxxxxxxxxX8'                   `8,8   ;8 
     8'         `8XxxxxxxxxxX8'                     `8I  ,8' 
     8           `8XxxxxxxxX8'                       8' ,8'  
     8            `8XxxxxxX8'                        8 ,8'  
     8             `8XxxxX8'                        d' 8'  
     8              `8XxxX8                         8 8' 
     8                8X8'                         8
     8,                `88                           8
     8I                ,8'                          d)
     `8,               d8                          ,8
      (b               8'                         ,8'
       8,             dP                         ,8'    HITLER AND
       (b             8'                        ,8'  STALIN DERIVED
        8,           d8                        ,8'  THEIR IDEOLOGIES
        (b           8'                       ,8' DIRECTLY FROM CANTOR
         8,         a8                       ,8'
         (b         8'                      ,8'
          8,       ,8                      ,8'
          (b       8'                     ,8' TO FOLLOW CANTOR IS TO
           8,     ,8                     ,8'  FOLLOW THE DEVIL HIMSELF
           (b     8'                    ,8'
            8,   d8                    ,8'
            (b  ,8'                   ,8'
             8,,I8                   ,8'
             `I8I                  ,8' BELIEVE IN CANTOR.  MAYBE YOU
              I8'                 ,8'  SHOULD ASK YOURSELF, WHY?
              8                 ,8'
              (8                ,8'
              8I               ,8'
              (b,   8,        ,8)  STUDIES SHOW THAT MEN WHO LAUGH AT
              `8I   88      ,8i8,    CANTOR RATHER THAN TAKE HIM
               (b,          ,88) SERIOUSLY SLEEP WITH, ON AVERAGE,
               `8I  ,8      8) 8 8        25% MORE SUPERMODELS
                8I  8I        8 8
                (b  8I         8 8
                `8  (8,        b 8,
                 8   8)        b8,
                 8   8(         b8
                 8   I          b8,
                 8                `8)   HOW CAN WE INVEST BILLIONS OF
                 8                 I8   DOLLARS INTO CANTOR'S BOGUS
                 8                 (8   THEORY WHEN THERE ARE CHILDREN
                 Ib                 8)
                 (8                 I8
                  8                 I8
                  8                 I8
                  8,                I8 EVARISTE GALOIS AND SRINIVASA
                  (8               (8' OUTRIGHT BUT THEIR PROTESTS
                   8               I8     WERE SILENCED BY THE
                   8,              8I           ILLUMINATI
                   Ib             (8'
                   (8             I8
                   `8             8I
                    8            (8'
                    8,           I8 CANTOR WAS ACKNOWLEDGED IN HIS OWN
                    Ib           8I TIME AS A HERETIC AND A LAUGHING-
                    (8           8' STOCK.  WHY SHOULD HE BE GIVEN ANY
                     8,         (8         MORE RESPECT TODAY?
                     Ib         I8
                     (8         8I
                     (8         8I
                      8,        8'
                      (b       (8   DO YOU WANT THE BLEEDING HEART
                       8,      I8    LIBERALS TO PREACH CANTOR IN
                       I8      I8  SCHOOL TO YOUR INNOCENT CHILDREN?
                       (8      I8
                        8      I8,
                        8      8 8,
                        8,     8 8' CANTOR'S TRANSFINITES AND EUCLID'S
                       ,I8     8    PRIME NUMBERS ARE BOTH MUTUALLY
                      ,88,     b        EXCLUSIVE OF EACHOTHER
                     ,8' `8      8
                    ,8'   8      8,
                   ,8'    (a     b
                  ,8'     `8      (b      PEOPLE WHO LOVE CANTOR
                  I8/      8       8,         HATE DEMOCRACY
                  I8-/     8       `8,
                  (8/-/    8        `8,
                   8I/-/  ,8         `8   CANTOR WOULD NEVER HAVE WON
                   `8I/--,I8        -8)    IF IT WERENT FOR THE
                    `8I,,d8I       -8)  DISCREPANCIES OF DIMPLED
                      bdI8,     -I8           CHADS
                           `8,   -I8'
                            `8,,--I8'
                             `Ib,,I8'
                              `I8I'
I hope that you enjoyed this complimentary ASCII babe, brought to you
free as a gift from your friends at the mathematical advertising
society.
Your dear friend,
Nathaniel Deeth
Age 11
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
> She seems to be looking for a Muslim scientist to hit
> on her so she's advertising in sci newsgroups and for
> bait is trying to appear intelligent, passionate, and
> desirable.
> So ..you are anti-Muslim. I understand why you must now attack me in
> such personal way but still I am sure that that's very low of you.
  Where in the above quote did Bill say anything at all
that was anti-Muslim?  Unless of course you think that
any Muslim who would hit on you would be degrading
himself.
> Most idiots like you have no clue that there are lots of Muslims in
> countries like Burma, Thailand, who don't deserve the animosity people
> like you have for them for being Muslims.
Unwarranted assertion; you have not provided any evidence
showing what people in general know or don't know about
the distribution of Muslims in the world.  You do not have
any basis for claiming that most of any particular group
know or do not know anything in particular.  Provide
your evidence (references please) or recant.
===
Subject: Re: Uncle Al is Sadistic .
> Apparently you don't have anything at all to say that's
> relevant to any of the three newsgroups to which you
> continue to post.  How *do* you justify your existence?
>  Feeling hurt..still?
You must be joking, Little One.
> Can you come up with even one post that on-topic
> somewhere and avoid the otherwise inevitable trip to
> everyone's killfile?
> Do it INSTAED OF whinning about it.
You would order me to do something?  What amusing
arrogance you display.
> I didn't expect that there would be people like you to take things out
> of context and attack me. Moreover, I realized that it was a mistake
> for me to even bother to reply to people like.
Like?
Apparently you have little clue about a lot of things.
And by the way, it's not necessary to take your words out
of context in order to thoroughly ridicule your pathetic
statements.  They are the ramblings of an ernest but naive
adolescent.
> Now ... put me in your killfile instead of talkign about it.  You will
> be helping me greatly.
Sorry sweets, but I don't take orders from The Clueless.
May I suggest that you take your whinging and moaning
to one of the many pop psychology newsgroups?
===
Subject: Software Needed
Hello everyone,
I am looking for a software that can find 
written in Chinese).
Chinese).
in English.
Mike
===
Subject: Re: Deprogramming needed?
> (Marketing ploy warning)
>  > What if indeed I *am* wrong, and I don't have these great math
> discoveries?
>  << James Harris
> http://mathforprofit.blogspot.com/
> What If?   That question on its face causes one to
> consider the state of mind of the speaker...  After
> years of people specifically detailing your mental
> constructs and pointing out your errors in logic you
> still have a doubt of your pass or fail status in life....
> Paul R. Mays
> So do you think I'm less or *more* likely to face the truth because of
> your post Mr. Mays?
> Aren't you trying to say that I'm a failure in LIFE in your comments?
> But aren't I a success in posting?
> If you push the notion that I'm a failure in life, why shouldn't I
> continue in an area where I'm clearly a success?
> Aren't you really working to *keep* me posting, possibly in fact,
> working to get me to post far more?
> James Harris
> My math discoveries, found for profit
> http://mathforprofit.blogspot.com/
Actually Mr. Harris I could care less what you do...
I long ago plonked you and just recently reloaded
my reader and you were not posting so I never
had the opportunity to re-list you...
As for your Life... I haven't a clue as to that, only the
post you post and the mental constructs you build
in the post I have read... The pass or fail status of
life is specific to your post and your inability to
comprehend the arguments supplied by many with
far greater specific knowledge of the subject.
I am more interested in the thought processing of posters
than I am generally in the specific content of post made
and you have failed to get it for so long its a becomes
 clear you that your thought processing has issues.
So post away,  Its of little concern as so far no information
has been supplied that is worth my time reading by you
and is an easy data source to exclude...
Paul R. Mays
----------------------------------------------------------------------------

-
Some where within the Quantum State
Http://Paul.Mays.Com/story.html
http://paul.mays.com/mayday.html
http://paul.mays.com/rainy.html
Science is what you know, philosophy is
what you don't know.
                                         -  by Alan Wood
===
Subject: rearranging equations
Hi
Can someone help me solve the following equation for x.  Could you
please put all the steps down so I can follow it (idiot's guide!)
Jo
y = A1*exp(-x/t1) + A2*exp(-x/t2) + y0
===
Subject: Re: rearranging equations
            Adjunct Assistant Professor at the University of Montana.
>Can someone help me solve the following equation for x.  Could you
>please put all the steps down so I can follow it (idiot's guide!)
>y = A1*exp(-x/t1) + A2*exp(-x/t2) + y0
This looks like a function, not an equation, unless y is supposed to
be a constant. I assume you want to solve y = 0?
Let z = exp(-x/(t1*t2)). Then you can rewrite y=0 as
A1 z^{t2} + A2 z^{t1} + y0 = 0.
If t1, t2 are integers, then this is a polynomial in z. Solve it by
any of the usual methods, and once you have a value z=z_0, then set
exp(-x/(t1*t2)) = z_0
and solve for x by taking logarithms; note that only positive
solutions for z will yield solutions for x.
===
Subject: Re: differentiable...problem...
>> if f is differentiable on (0, infinite)
>> and lim [f(x) + f'(x)] = L  (x->infinite)
>> show that lim f(x) = L (x->infinite) and lim f'(x) = 0  (x->infinite)
>|
>|                                    f e^x       (f + f') e^x
>|  lim  f + f' = L  =>  lim f = lim  ----- = lim ------------ = L
>| x->oo                 x->oo   x->oo  e^x   x->oo    e^x
> Provided f e^x -> +-oo, in which case
>  0 = L - L = lim f+f' - lim f = lim f'
>> No proviso is needed. This form of L'Hopital's rule needs only
>> that the denominator is infinite, not also the numerator [1][2].
>> So there's no need to treat this case specially as you do below.
> Not even the existence of lim(x->oo) f(x) ?
L'Hospital's rule for the form  lim f/g, lim g = oo
needs no hypotheses on existence or value of  lim f
>> [1] A. E. Taylor, L'Hospital's Rule
>> Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.
>> http://links.jstor.org/sici?sici=0002-9890(195201)59:3%3C20%3E
> You make jest?  Jstor says forbidden.
Jstor requires a subscription (e.g. most major universities).
Alternatively, many public libraries subscribe to the Monthly.
-Bill Dubuque
 
>> [2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital Rule
>> Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.
>> http://links.jstor.org/sici?sici=0002-9890(197604)83:3%3C239%3E
> Ditto.
> However when f e^x -> k;  then
>  f = fe^x e^-x -> 0
>  f e^-x = f e^x e^-2x -> 0
> L = lim f+f' - lim 2f = lim f'-f
> 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L
> 0 = lim f+f' - lim f = lim f'
> So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f'
>  k = lim(x->oo) f = k + lim f';  lim f' = 0
> If lim f' doesn't exist, then for n >= 2
>  f_n(x) = sin(x^n)/x -> 0
>  (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation
===
Subject: Re: Marketing shift, core issues
>Would you like to join forces with me so that we can fight Cantor
>together?
I'm sure if he was alive he'd be rolling in his grave.
===
Subject: Re: Deprogramming needed?
 
 Discussion, linux)
>> given that everybody participating in the discussion
>> is doing so from mindsets that SHARE a certain kind of academic-
>> Americanness, it IS objectively knowable that that kind of abuse,
>> EVEN it is deserved, is counter-productive. 
> You are grossly racist.
I am not one to rise to George Greene's defense very often (not that
he ever asks me to or benefits if I do), but this is just silly and
insulting.
There is nothing racist in the above quotation, as far as I can tell.
>> If everybody else could emulate
>> Dik Winter and confine their criticisms to the math, this would all
>> go away very quickly.
> You are grossly misinformed of the facts.
On this, I think, you are correct.  There is no simple JSH solution.
I selfishly and pathetically admit I never asked for a JSH solution.
-- 
If you *still* believe that [my proof is wrong], then I have to think
that your mind is limited [...], and it may be the case that not
everyone *can* achieve that, as the mental wiring may not be there for
the task. -- James Harris, on faculties needed to accept his proof.
===
Subject: Re: Looking for HINTS on proving a limit
Ignacio Larrosa Ca.96estro  
escribi.97
> By the way ... It is possible to get directly that limit without use
(e^x)'
> = e^x?
> I.e., it is possible to prove directly, using only the definition of
> derivative, that (e^x)' = e^x?
Well, Jos.8e H. Nieto give me a satisfactory answer in es.ciencia 
matematicas.
Lim((1 + 1/x)^x, x, inf) = e
let x = 1/t,
Lim((1 + t)^(1/t), t, 0) = e  ==>
Log(Lim((1 + t)^(1/t), t, 0)) = Lim(Log((1 + t)^(1/t)), t, 0) = 1
Lim(log(1 + t)/t, t, 0) = 1
Then, let y = Log(1 + t),
With that is possible study the derivative of exponential function before
than the derivative of the logarithmic function and the composite or 
inverse
function.
-- 
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Largest number ever written down?
ï.