mm-1909 === My suggestion (which you are of course free to ignore) is for you to illustrate the core error by quoting a standard mathematical source and showing why it is in error. You have not done that. I answered that suggestion when you first made it as I noted that I don't have a citation, when I talk of an error being introduced over a hundred years ago, but merely roughly go back to the period when algebraic integers were introduced, which to my knowledge was over a hundred years ago in the 19th century. I don't know why you wish to belabor a point that I've already conceded. So that there is no need for you to do so again, let me make it perfectly clear: I do not have a citation from any text. I hate to get annoyed so quickly in this process but I fail to understand why you could not understand my previous statement or are you just trying to anger me by repeating a suggestion I already answered? I'm curious. Why did you repeat when I thought I answered you fully earlier? Possibly rather than get upset, I should just ask a question like that to understand, right? James Harris Discussion, linux) It's very simple. Apparently, you still believe that the misuse of algebraic integers has produced many errors in mathematical texts. Everyone else claims that they know of no errors even vaguely similar to the one you claim is endemic (which is often very vaguely stated). If this error is endemic, then you should provide some example where it occurs. That you didn't have a citation yesterday doesn't mean you can't get one. Since you're working to provide clearer justification for your research, this is an obvious step in that direction. -- Jesse F. Hughes And hey, if you're moping and miserable because mathematics tests you, then maybe, if you think you're a mathematician, you might want to try a different field. -- Another James S. Harris self-diagnosis. posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn am this I'll Mr Hughes is right that this is close to the central issue in your paper Advanced Polynomial Factorization. However, some of the context has been lost. You were considering a *function* of the integer variable m, but factored as a *polynomial* in x. Explicitly, [1] P(m) = (a1(m)*x + uf)*(a2(m)*x + uf)*(a3(m)*x + uf). The functions a1(m), etc., are not polynomial functions. You repeatedly noted at the time how remarkable it was that you were pioneering a method of factoring a polynomial with nonpolynomial functions. You specified that the function P(m) was divisible by f^2. We noted that one must assume that when f^2 is divided out of the right- hand side of [1], it may divide out in a *variable* way. This is a reasonable assumption, because when m = 0, two of a1(m), a2(m), and a3(m) were 0 [P(0) was a first-degree polynomial in x] but when m = 1, P(m) was a 3rd degree polynomial in x. Therefore a1(m), a2(m), and a3(m) *cannot be constant functions*. Now say f^2 = v1(m)*v2(m)*v3(m), and that this decomposition of f^2 is such that (ai(m)*x + uf)/vi(m) = (ai(m)/vi(m))*x + u f/vi(m) is a factorization with algebraic integer coefficients for i = 1, 2, 3. That is , ai(m)/vi(m) and f/vi(m) are all algebraic integers. The existence of such functions vi(m) is guaranteed by a theorem of Dedekind. ---------------------------------------------------------------------------- http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1086485& messageID=3408408#3408408 Well, let's put that to the test Nora Baron and see if the algebra reaches you. Consider a_1 x + uf, and you claim it has a factor that is v_1(m) that varies with m, now just divide through by that factor and you get a_1 x/v_1(m) + uf/v_1(m) and you can see that the term constant with respect to m is now uf/v_1(m), which is a basic contradiction, as now it *varies* with m. Understand? ---------------------------------------------------------------------------- - It was clear right there where you were making your mistake: You had forgotten your own definition of *constant term*. You reasonably defined the constant term of a function h(m) to be h(0). A nice, simple definition. Now if you *correctly* apply *your own* definition to a_1 x/v_1(m) + uf/v_1(m) then, because a_1(0) = 0, you obtain uf/v_1(0) NOT, as you said on Oct 10 [above] uf/v_1(m). Once you realize that, all the rest of your constant terms must remain constant argument - which, after all, reduces just to that trivial tautology which buys you nothing - comes unravelled. But apparently you never got it. The major conclusion was, you CAN factor f^2 out of the right side of [1] and still have all the coefficients be algebraic integers. The factorization *itself* is a function of m. You do NOT need to invent a new ring, as you incorrectly inferred, to achieve such a factorization. Understand? Want to try again? Nora B. posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY they I my a That paper has been superseded by a new, more complete paper. The central points don't depend on a particular paper, though I guess you feel a need to insist on bringing up the past paper, which I'll simply note at this point. You are trying to make this very difficult for me, aren't you? Well I'll just let you do the little asides with little comment on them, as I do mean to be a critic. You need to give the polynomial. Here's a good place to stop and consider carefully what you have said so far. You claim that there is this polynomial P(m), but do not give a polynomial, but a factorization, as you give P(m) = (a1(m)*x + uf)*(a2(m)*x + uf)*(a3(m)*x + uf) though you may give it later in the portion I've deleted out, so that's not as big of a deal. Now you *say* that I specified that the function P(m) was divisible by f^2 which I think is correct. But I think it is not such a good thing for me to just say things are correct when this is a careful critique. To my mind you went to fast, which is why I'm slowing you down. Next you claim that f^2 divides out in a *variable* way from the polynomial which you have not yet given, and say that we noted it, which I take to mean you and other posters. Then you hand-wave, claiming that it is a reasonable assumption, so you are adding extra justification to your own statement, which you don't explain. As you say: Now then, the polynomial that you have not given, is now said to be a first degree at P(0) in x, when you gave P(m). And now... Well, your entire statement is very muddled I think. You may disagree with that assessment, but it's what I see. I still think you are impatient and need to slow down, step through carefully, and cover all the bases. Think of me as slow. Don't assume I know everything you're talking about, and don't toss out assertions you do not prove. Now then, please go again, and yes, if you rush too much in a single post I will delete out past the point where I don't understand, or have serious questions, and it will save a lot of time if you go slow, and careful. James Harris posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn as the to that are that is You have refused to post even a statement of the conclusions of the new paper. The question raised here by William Hughes was central to your justification of the main conclusion of Advanced Polynomial Factorization, and I would guess that that central argument, where you generalize from the case m = 0 to the case persist in saying that APF is not wrong, so it seems fair to ask that you present a defence of its methods. I was trying to jog your memory. And I don't think my description above is inaccurate regarding what you said. It was the polynomial you specified in APF, and which you have referred to in dozens of posts over the last 2 years. Here it is: P(m) = f^2 [(m^3 f^4 - 3 m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f] I thought you would surely remember this! right- that's See above. by Yes - it's clear from the expression for P(m) above, which is a direct quote from you. critique. you I gave a justification for what I said subsequently. I noted that a1, a2, and a3 must be nonconstant functions of m. This is true because when m = 0, two of them are 0, and the function is a first degree polynomial in x; while when m = 1, none of them can be 0 because the function is then a cubic in x. Thus it seems reasonable that if you are going to factor f^2 out of (a1(m)*x + uf)*(a2(m)*x + uf)*(a3(m)*x + uf) in such a way that all the quotients are algebraic integers, and you know that a1(m), etc., are nonconstant functions of m, then you are going to have to do the factoring in a nonconstant way. Put it another way: if you assume that f^2 must factor out of the expression above in a *constant* way, you need to prove that before you go to the next step. Put in still a third way, if you assume that f^2 must factor out of those nonconstant expressions in a *constant* way, you are assuming what you want to prove. To be 'conservative' one must assume that it factors out in a nonconstant way and proceed from there. I was assuming you would clearly recall most of this discussion. What I see is that you deleted out the rest of my post, much of which was a direct quotation from you. I resent that. I think it represents bad faith on your part, in view of your stated intention of wanting to be critical of your own arguments. Anyway, I am re-inserting it below. It begins with a quote from a sci.math post ---------------------------------------------------------------------------- http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1086485& messageID=3408408#3408408 Well, let's put that to the test Nora Baron and see if the algebra reaches you. Consider a_1 x + uf, and you claim it has a factor that is v_1(m) that varies with m, now just divide through by that factor and you get a_1 x/v_1(m) + uf/v_1(m) and you can see that the term constant with respect to m is now uf/v_1(m), which is a basic contradiction, as now it *varies* with m. Understand? ---------------------------------------------------------------------------- - It was clear right there where you were making your mistake: You had forgotten your own definition of *constant term*. You reasonably defined the constant term of a function h(m) to be h(0). A nice, simple definition. Now if you *correctly* apply *your own* definition to a_1 x/v_1(m) + uf/v_1(m) then, because a_1(0) = 0, you obtain uf/v_1(0) NOT, as you said on Oct 10 [above] uf/v_1(m). Once you realize that, all the rest of your constant terms must remain constant argument - which, after all, reduces just to that trivial tautology which buys you nothing - comes unravelled. But apparently you never got it. The major conclusion was, you CAN factor f^2 out of the right side of [1] and still have all the coefficients be algebraic integers. The factorization *itself* is a function of m. You do NOT need to invent a new ring, as you incorrectly inferred, to achieve such a factorization. Understand? Want to try again? have This looks to me like a delaying and disrupting tactic, and I think it is quite rude to delete out substantive mathematical material unless it is simply pointless repetition of previous posts - especially when you complain that I have not justified some statements which were explained in what you deleted out. But proceed. Nora B. posting-account=82j3EgwAAACYxp9hHOcWy78r2IGkmc3t is you before you're I My understanding is that you are currently using the definition for any function f(x), the constant term of f is f(0) - William Hughes posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY I want to emphasize that when I talk about critiquing, I don't mean quick dances through something you claim proves this or that, but very careful and methodical step-by-step arguments that don't leave all kinds of things to be figured out or require that people know ahead of time the gist of what you're saying. I mean, thorough, methodical, and slow, as going slowly and carefully does not hurt a critique. One of the reasons I'm going slowly in my own presentations of critiques is that it is a slow, very methodical process. I fear that some of you are too used to the quick and often sloppy process of brainstorming posts. The critiquing process is not that way. You go slow, or you'll be slowed down. If you jack-rabbit through a post, tossing up a lot of stuff, then if I'm going over it, and if I bother with it, I'll take a little piece of it, and try to slow you down. If you can't be careful and methodical, then you can't be of much use. James Harris posting-account=jEbKGQ0AAADZF1UpkDsHa5gkWBqABnUE very of use. Harris, you are ignorant. You lack sufficient knowledge of mathematics to understand the patient and meticulous responses you get to the exposals of your flawed ideas. Harris, you also stand empty handed. You are not in posession of any worthwhile mathematical achievements. No one is interested in assisting you in the consolidation of what you call results. No one can be, much less wishes to be, of use to you in the sense you arrogantly mean. There are, however, some who would be of use to you in a much more important and humane way; I mean those who attempt to interact with you with the purpose of revealing to you the depth of the plight you are in. You have got nothing Harris, wake up. I would like to see a slow and methodical description of why you think there's a flaw in the definition of the algebraic numbers. I would like you to state, clearly and succinctly what you think the problem is and why you think it's a problem. posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW I'd say that Decade Two of the Ten-year Programme is slow (enough .-) --Chairman George and Strep Throat, back in print! http://larouchepub.com http://tarpley.net/bush12.htm posting-account=pIQyfgwAAAB1mcfW2JB8G43zeTAAgF6U Given a list of sales of products to customers, how do you efficiently find the sales correlation for all pairs of products? (Or given a bunch of web pages find the correlation between all pairs of words, etc etc etc.) Let C be the number of customers, P the number of products, and p the number of products bought by a customer. I could find the popularity of products by For all customers For all sales to that customer add to the popularity of the product sold Then find which customers prefer which products by For all customers For all products bought by that customer compare this customer to the average for this product Then I could find the correlation between all products by For all customers c For all products x sold to c For all products y sold to c add their correlation to the sum for (x,y). That's O(Cpp), much better than O(CPP) if p << P. This ends up a bunch of dot products: over all customers, sum(Cx.Cy), sum(Cx.Cz), sum(Cy.Cz). They could be arranged in a matrix: sum(Cx.Cx) sum(Cx.Cy) sum(Cx.Cz) sum(Cy.Cx) sum(Cy.Cy) sum(Cy.Cz) sum(Cz.Cx) sum(Cz.Cy) sum(Cz.Cz) ... which is why I'm posting. Sometimes matrices have clever algorithms. Is there a faster-than-obvious way to compute that matrix, particularly if p=P? posting-account=9wVPIwwAAAAonf5Dj39AQaTL2sJYvErF P(~A). But not so simple that you didn't claim the opposite! Modus Tollens applies? Yep. No, that ~A is more likely if ~B. If the crooked authors don't include any examples, then obviously it is more likely that they don't have any. They are only providing evidence contrary to their claim, and those who believe that they are supporting their claim aren't being very logical. (Although from a purely academic point of view, unsubstantiated statements of Mathematics are worthless to begin with.) It's a shame that you have to formally prove what any idiot can see - except some idiot who posts here calling people stupid or ignorant. But anyone who talks like that shows his lack of sense already. What's that expression about one's sound and fury showing they have no reasoning? C-B I misunderstood you initially. Not really, the claim itself isn't random. It's either true or it isn't, and this doesn't change regardless of whether or not they provided a sample. So in this case P(~A given ~B) = P(~A). Martin posting-account=BhnVAA0AAADpcidP-5AvYEnIArz5kqjP are http://papers.ssrn.com/sol3/papers.cfm?abstract_id=667781#PaperDownload If it's politics, why doesn't Bush just MANDATE by law that stock prices can only go up? Make it illegal to sell at anything but the same or a higher price! Then stock market will go up up up, contrary to the paper in question. posting-account=55r_ng0AAACp4RGEyBESbxI5vB3imHw9 Very well. But it is too late for this at this point in the thread. Not so, it was Peter Webb who responded to you with the incorrect information to whom you responded citing my posts as missing the point thus forcing me to respond. The fact remains that most of the points I made were not addressed by you (other then with the all-encompassing, all-inclusive, divine and oh-so-lovingly nebulous panacea of provenance) AE posting-account=55r_ng0AAACp4RGEyBESbxI5vB3imHw9 Really? How so? Perheaps I did derive (whatever that means) the number 7 by illicit means from some music.mp3? Or maybe not? Can you tell? And you know what? I got a bottomless bag full of more numbers *just like it* which also, I could have, oh the horror of it, somehow derived from music.mp3... and if I arrange them *just so* they also make up music1.mp3 .... and music2.mp3... and... But I guess provenance will sort it all out. It will come and smite me with its mighty rod of lightning after determining by means of telepathy which sequences I did produce by naughty unauthorized means, no? Ok, so when does this fair use end? at 10%? 50%? 90%? What if I perform a lossy compression whereby only 10% of original remains? What then? Define obtained in this context. How is a number 3 at position 1234 in the music.mp3 any different then number 3 that came up at 1234th roll of a dice? Are these *two* wholly different and separate numbers, one of them 3-version-a illegal and the other 3-version-b legal? Do they come with their own little certificates of legality? Their itsy-bitsy little passports to prove where they come from so that one does not get confused and somehow take one for the other? They look awfully alike you know. Are we into some new provenance numbers theory here? So that is merely a matter of probability? No problem. Since you did not respond to my checksum example earlier, I assume that computing a modulo 100 checksum does not constitute derived work. I simply now have to divide (conceptually, not physically as to not alter it in any way) music.mp3 into smaller chunks, compute checksums on them and apply a genetic algorithm to a different set of, completely separate and filled with random numbers but equally sized, chunks of another file until all the checksums match. I have just arrived independently by employing mere chance in a manageable way at the music.mp3 without any copying whatsoever. Still in the realm of provenance? Or is it now that some few checksums are somehow derived work? Didn't I just do what artists supposedly do and created my own contents while merely conforming to the checksum style of the original for inspiration? standard No they are decided on the base of financial resources of the rights holding corporations, combined with incessant propaganda in their wholly owned and operated media channels, sprinkled for a good measure by influence peddling in legislatures. Furthermore, one could in the past get away with similar bull by simply claiming that the issue was in the domain of art and subjective measurements of originality of works because the true nature of songs and paintings was due to primitive technologies not subject to scientific scrutiny. But as soon as individuals and corporations driven by greed decided to convert those to digital form, they crossed a line from nebulous, arbitrary and subjective world of provenance to a firm and testable world of science and mathematics in particular. And thus they are no longer free to make up rules as they go with the aim to pad their pockets. Now different rules apply. And either the whole body of logic and reason has to be destroyed in order to be replaced by make-believe world of provenance or, conversely, the shenaningas of the legal beagles are to be ended. came To the contrary. I just presented a method that does precisely that, by merely increasing the probabilities to manageable level while combining them with a minima seeking algorithm such as a genetic one. So unless there is a claim of ownership to the original music.mp3 as well as ownership of any conceivable transformation of the contents of thereof *as well* as any other file even remotely *similar* to it arrived at by any means no matter how independent (read: the entire domain of integers) there are methods that can be constructed which do not depend on duplication or derivative work in any shape or form to arrive at (for all intents and purposes) comparable contents independently at the expense of complex and lengthy computations. But that matters not since the law is being shaped by those who profess the Who cares? Its all mine anyways, suckers, my lawyers are bigger then yours! attitude. That is why there is a need to create a technological situation where the absurdity of that position is so exposed that they are no longer capable of sustaining it. Either that or they will have to resort to police-state tactics, banning all exchange of digital data between unauthorized individuals and operate door busting goon squads in order to enforce compliance. The assumption in this discussion is that the original music.mp3 is obtained by legal means (whatever that nonsense happens to be defined as at the moment) AE asdf Hint: Z_10 ~= Z_2 x Z_5, where ~= denotes isomorphism and x denotes the direct product. Can you do the others? More explicity, can you write Z_8 x Z_10 x Z_24 = Z_4 * G for some abelian G? Travis X-RFC2646: Original As a recreational math person, I've always liked the simple stuff like e^(pi * i) = -1 I'm wondering if there is a way to get the i stuff to the right hand side. That is, to have e^pi on the left and other stuff (reals only maybe?) on the right. I'm kind of curious to find a function for e^pi. Dan in Philly If you don't mind stepping over a maze of branch-cuts... -1 = e^(pi*i) (-1)^i = [e^(pi*i)]^i = e^(pi*i*i) = e^(-pi) Therefore: e^pi = (-1)^(-i). -- Rouben Rostamian Dan in Philly the It's possible, sort-of: e^(pi * i) = -1 e^(pi * i / 2) = i e^(-pi / 2) = i ^ i e^(-pi) = i ^ (2i) e^pi = i ^ (-2i) but I have omitted all the necessary incantations about principal branches of logarithms and all that jazz :) LH side. And on second thought, that calculation can be simplified. X-RFC2646: Response + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + You want to isolate i ? Start by taking the natural log of both sides of the equation. Aristotle Polonium + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + X-RFC2646: Response + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + My previous response might not have been helpful. If you're interested in isolating e^pi, try raising both sides of the equation to the - i power. Aristotle Polonium + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + I really have no knowledge about this subject. But it is a fascinating one with many unknown facts. I saw that one poster thought(incorrectly) that it is known that there are infinitely many Mersenne primes. If there were there would also be infinitely many even perfect numbers (like 6 and 28 which are the sums of their proper divisors). It is also not known whether there are infinitely many prime pairs i.e. primes differing by 2. One thing that is known is that there are arbitrarily large gaps in the sequence of prime numbers. e.g. 101!+2 is the beginning of a sequence of 100 composite numbers. Very interesting but I haven't added anything to your quest. posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs Um... let's hear it for consistency? - Randy Embargo of crucial substance like oil was/is considered as an act of war internationally. By the way, don't be bare-assed and asleep(yor words) while you laid oil embargo to a nation. Watch out all night. It's my two cents. The US had better mind its own business. The US fought against Japan almost four years. That's long time, eh? How many American soldiers were killed by the Japanese? I don't know it but I think it's greater than those in the Korean war and in the Vietnam war. I think it's very high price for the US to pay just for China. intervention to the Sino-Japanese war. Here is the moral. Do not, I say, do not around(your words) with Japan. I don't think that this is true, for example, if a=1 then P(dx) = c dx/(1+x^2) where c is something like 1/pi. What I think you are wanting to prove Something like this seems to work - write f(t) = exp(-g(t)): P(dx)/dx = 1/pi int_0^infty cos(xt) f(t) dt = 1/pi int_0^{1/x} cos(xt) f(t) dt + 1/pi int_{1/x} cos(xt) f(t) dt. By integration by parts twice the second integral is 1/x sin(1) f(1/x) + O(1/x^2) and using the approximation f(t) = 1-g(t)+O(g(t)^2) in the first integral, see that this evaluates to -1/x sin(1) + c x^{-a+1} + higher order terms. This is how it should be done, I think: P(dx)/dx = 1/2pi int_0^infty cos(xt) f(t) dt = 1/2pi int_0^{pi/x} cos(xt) f(t) dt + 1/2pi int_{pi/x} cos(xt) f(t) dt. By integration by parts twice the second integral is O(1/x^2), and using the approximation f(t) = 1-g(t)+O(g(t)^2) in the first integral, see that this evaluates to c x^{-a+1} + higher order terms, where c = g(1) int_0^pi t^a sin(t) dt. posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW this kind of oversight may just mean that he and Jimi Harris are One. --Chairman George and Strep Throat, back in print! http://larouchepub.com http://tarpley.net/bush12.htm f(x) = {a^2+[b^2*sin(x)-1]}^n calculate the integral of f(x) for x from 0 to 2*pi. With a, b are real number, and n is a complex number. I run into a integral, but I don't know how to do this. Would anyone please help me? for f(z) =(a^2+z^2+b^2/z^2)^s1*z^s2 how to calculate the integral of f(z) for z from 0 to infinity? with a and b are real constant, s1, s2 are complex number. I tried with maple, it does not work. No idea how to help you myself, but here's the output from Mathematica. In[1]:= Integrate [ (a2+z2+b2/z2)^s1*z^s2,{z,0,Infinity}] s2 Integrate::idiv: Integral of z does not converge on {0, Infinity}. s2 b2 s1 Out[1]= Integrate[z (a2 + -- + z2) , {z, 0, Infinity}] z2 Oh, and just for good measure, here's the result from Mathematica assuming you don't integrate to infinity, but to some number t. In[2]:= Integrate [ (a2+z2+b2/z2)^s1*z^s2,{z,0,t}] 1 + s2 b2 s1 t z2 1 + s2 s2 posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW For an example of the same mysticism I am attacking here, I point to the errant argument which was made by Felix Klein, and others: Klein's false claim, that crucial features of Kepler's, Leibniz's, or Gauss's discoveries could be replicated by the errant methods of such followers of the Enlightenment philosophers Lagrange, Kant, and Laplace as Cauchy, Hermite, Lindemann, et al. The fraud implicit in the latters' attempts, is their vicious exclusion of the physical geometries of Leibniz, Gauss, and Riemann; so, the celebrated Maxwell confessed his politically motivated complicity in this matter of suppressing what he knew had been the crucial contributions of Amp.8fre, Weber, Gauss, and Riemann to electrodynamics. This ethereal fraud by Maxwell et al., is typical of widely accepted hoaxes still presented, on record, in today's classrooms, reference works, and textbooks.[9] That fraudulent mathematics of the reductionists is avoided, only when the underlying epistemological issues of counting numbers, such as those issues posed by Gauss's Disquisitiones, are situated within the realm of an essentially constructive, synthetic, anti-Euclidean geometry. So, Gauss's work, employing his teacher K.8astner's anti-Euclidean geometry in this case, is the most crucial, make-or-break issue of modern mathematics to be posed for the student's competent introduction to modern mathematical physics. The exclusion of critical consideration of the axiomatically geometric roots of the orderings of numbers, was the premise of the relevant essential fraud perpetrated by Euler et al., and the common mistake of the credulous imitators of Euler's error today. --Chairman George and Strep Throat, back in print! http://larouchepub.com http://tarpley.net/bush12.htm posting-account=alQKkAwAAAA82xoCXcVIo1q-o-rv2IW- which coherent offset element of P functions. the mapping at once. of sets, we really have bigulosity of Q Ah, right! You see, the World Bigulosity Congress is going to be very very busy contemplating how to assign Bigulosity values to new sets. What about sets whose members are sets of different sizes, for a start? And here's an easy one for you to decide: what's the bigulosity of Z = {{0,1,2,3,...}, {1,2,3,4,...}, {2,3,4,5,...}, {3,4,5,6,...},{5,6,7,8,...}, ... } (The set of subsets of the nonnegative integers excluding an initial segment) Anyway, you might observe that so far everything you have produced on the b() function has been a set of unbased pronouncement. Time for a result or two! Brian Chandler http://imaginatorium.org This is the rustling strawman debate? Lee Rudolph X-RFC2646: Original The horizontal bar has implicit bracketing around numerator and denominator. 1/2+W is not the same as 1 ____ 2+W So, why should 1/2*W be the same as 1 ____ ? 2*W You could make up some complicated rule with all sorts of exceptions. Or you can make the rules as simple as possible. Computers tend to force these considerations on you. Dan Download my DC Proof software at http://www.dcproof.com I'm actually implementing a scientific calculator for mobile phones, and my calculator's parser would read 1/2w as shorthand for 1/(2*w). 1/2*w on the other hand would be parsed left to right as (1/2)*w. If someone thinks this is a particularly bad idea, let me know. /dev -- http://mobile.bacademy.com/ X-RFC2646: Original I have struggled a bit with such problems in my own work. My advice is to keep your rules simple. To implement such a shorthand, you will have to implement all kinds of exceptions that will make for more complex coding, testing and documentation. Do you really want, for example, to have 1/(2w) = 1/2w = 1/2*w = 1/(2*w) ??? Or, worse... 1/2+w = 1/(2+w) ???? Also, try to stick to the rules and notation with which the intended users are already familiar -- those used in the most popular scientific calculators, in your case. Dan Download my DC Proof software at http://www.dcproof.com Infinitude of Primes It's poorly done, I agree. For additional brevity, we could just argue that N is not divisible by any p_i and must therefore be divisible by some other _prime_ (possibly N itself). But the remarkable thing to me is the large number of differences between Euclid's proof and today's standard textbook proof. http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html Euclid does not say, let A,B,...Z be _all_ the primes. (He is not arguing by contradiction.) Nor does he make the hypothesis that A,B,...Z are the smallest primes, or in increasing order. He doesn't even assume that they are distinct. And he does not speak of dividing AB..Z + 1 by primes, but rather, he speaks of dividing 1 + the least common multiple of those primes (which comes to the same thing, but he doesn't need that). Hi Robin, 29th Mar, 2005 1) Yes, 0.866... is meant to rep sqrt(3) / 2. ******* 2) CI Int is a short form for Cyclotomic Integers. I have already stated so in my previous mail. ******* 3) You ask : What is a de Moivre number anyway? A de Moivre No (DM No) is of the form : exp(j * 2*pi*k / d) which is a solution to x^d - 1 = 0 You may also refer : http://mathworld.wolfram.com/deMoivreNumber.html Since there is no restriction on a DM No that d should be prime (or is there one ? - if so, pl let me know), I was wondering why 'p' should be prime in the DM Nos defining CI Ints. ******* a) You say : But, there are only finitely many such n. c) Also, http://mathworld.wolfram.com/AlgebraicNumber.html lists out 15 Alg Nos. Since I don't have access to the book by Washington, I would like to know if the above stmts (a) and (b) are related, or are probably meant to mean roughly the same. ******* I deal with roots of Poly Eqns so often, and I often experiment not just with real coeffs, but sometimes test some codes with even complex coeffs ; I have been only faintly aware of the term Alg Nos since I probably studied them long time ago, but never used the term. Now, I am curious about this stmt in : http://mathworld.wolfram.com/AlgebraicNumber.html which says : If, instead of being Ints, the b's in the above equation are Alg Nos, then ANY root of bn*x^n + bn-1*x^(n-1) + ... b1*x + b0 = 0 is an Alg No. I hope to experiment on this stmt sometime later within the week. But any enlightening comments are welcome. *************** Now, that our doubts about notations are somewhat clear, can we get back to the central question : 2) reworded : Why is it necessary in a CI Int to have the de Moivre No based on a prime number p ? ... So, what is the significance of having a de Moivre No based on a prime no in CI Nos ? See my example with a DM No which is NOT based on a prime no : DM = exp (j * 2*pi/6), and 6 coeffs = [ 1, 2, 3, 4, 5, 6 ]. In this case, after Calculations, we get CI = -3 - 5.1962i 4) slightly elaborated : As a Corollary, will the CI Poly Eqn become non-Minimal in some way (what way ?) if we do not base the underlying de Moivre No on prime numbers ? If this stmt is true, (and also if it is not), I would like to get some analysis-discussion on the significance of why we need a de Moivre No based on a prime no in the construction of the CI. *************** Instead of buying a new book, I would like to look for this (finite) list in other books that I have. I have many books on Maths including Ref books : H/b of Mathematical Fns Table of Integrals ... However, if you do not have it handy, pl do not bother. Rgds Sundar Krishnan PS : The other short forms that I may use, and which would be obviously clear from the context are : Mca for Mathematica MW for Math World Mlab or Mlb for Matlab. E-J Prime Int Pair for : Eisenstein-Jacobi Prime Integer Pair .. Hello Could some one please explain to me the basis function, I understand the basis vector but I am not able to visualize or grasp the idea of how to generalize that to the basis function. And still no mainstream economist has acknowledged the validity of the argument in the above linked essay. See last paragraph of the above linked essay. No. See sci.math. As for an argument from a refereed journal, see the first sentence of the above linked essay. Most mainstream economists ignore the fact that the neoclassical theory they often teach and apply has been shown long ago, in the refereed journal literature, to be crap. -- r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau If he was McKenzie probably wouldn't have agreed to meet him. posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW Big ol'Baby? --Chairman George and Strep Throat, back in print! http://larouchepub.com http://tarpley.net/bush12.htm Euler tour Oh groan, a huge accounting problem. Instead let's use regular open and regular closed sets. A set U is regular open when int cl U = U A set K is regular closed when cl int K = K Now with complementation and regular closure, ie the operator cl int, how many different sets can be produced? Oh no, a long lengthy prolonged stream of thought soliloquy. Rather that keep us in suspenseful boredom, get to the shaggy dog punch line. Give an abstract, ie 300 words or less, stating the important results you've made or the most significant vein you've uncovered for exploration. Now an operator that may amuse you is the boundary operator with bd bd bd A = bd bd A How many different sets can be produced with the boundary operator and complementation? Hint, find a set for which bd bd A /= bd A. BTW, such a set is neither open nor closed. posting-account=LZHYSg0AAADv6lTTcdbgjF8lMmEgg5LT Internet Home Business Minimum financial requirements: Any credit, debit or bank account with at least $6.00 balance. Minimum computer literacy: Follow onscreen instructions and operate a mouse. Minimum time investment to start: Approximately (5) hour's computer time on-line THIS IS A LEGAL BUSINESS OPPORTUNITY!!! #####LE 18 Sec. 1302 & 1241 of the Postal lottery laws, provides guidance to the legality of this specific business opportunity. The policy and procedure of this venture is discussed in depth at Fairhaven College Discussion Board link below: http://pandora.cii.wwu.edu/WebServices/_generaldiscussion/00000040.htm No Fake Names! No False Information! Follow Policy & Procedures with Integrity!!! This program remains successful because of the honesty and integrity of the participants. Please continue its success by carefully adhering to the instructions. In this business, your product is not solid and tangible, it's a service. You are in the business of developing mailing lists. Many large corporations are happy to pay big bucks for quality lists. However, the money made from mailing lists is secondary to the income, which is made from people like you and me asking to be included in that list. Example: $6.00 + 5hrs. + Integrity = $20,00.00 in 4-6 weeks Step1. Go to www.Paypal.com, open and activated an account for free that allows you to send and receive funds via secure e-mail notification. This may take a couple of business days to verify account information and confirm identity. Step2. After activating your Paypal account, use your new account to pay $1.00 to the following email addresses in the order they are listed. (Total $6.00) 1. (njekm@hotmail.com) 2. (wthampton@hotmail.com) 3. (Ldrury@rock.com) 4. (samwkfld@yahoo.com) 5. (1damnyankee@comcast.net) 6. (katama1965@mail.com) Step3. Email each of the six email addresses and request that you be added to their lists. Step4. After requesting that you be added to each core lists, edit this notice as follows: Move everyone in the core mailing list up one position eliminating the first address. Place your email address in the sixth position and save document. Step5. Find Internet newsgroups, discussion boards, message boards, forums, chats, in financial services field and post this notice a minimum of 200 times. The more postings, the bigger the list and the more return. That's it! Again I cannot emphasize the importance of following each step carefully and honestly. Good luck. God bless. Watch the money role in. posting-account=GcwWRQ0AAAArbuR6aDA8w1LJJiiPSTgt I am sorry I too thought that a^2=b^3=e is also going to hold for the abelianisation as for the original group as Arturo Magidin pointed out.