mm-1699 === Subject: diff EQ on strings, check out the math Check my math! I've derived a differential equation for strings starting from Stokes Theorem to show that energy is conserved along a world-sheet. These diff eq's involve connection coefficients. And I'm not really sure what it all means yet. I would appreciate it if some who are more skilled in the art would take a look at this and comment. The math can be found at: http://www.sirus.com/users/mjake/diffeq.html === Subject: Re: Partial difference equation, counting primes You're still a troll trolley troll. === Subject: Please check my boolean simplification Hello all, I would appreciate if somebody could tell me if I haven't made any mistakes in the following boolean equation simplification: R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') Applying DeMorgan's Law to the term (C1'+C0'+A'+B') = ((C1C0)'+(AB)'). Applying the law again = ((C1C0)(AB))'. Using A(A+B)=A on the term (C1'+A+B)(C1'+C0+A+B) = (C1'+A+B) where A=(C1' +A+B) (obviously) Applying the Distributive Laws in reverse to the term (C1+C0'+A')(C1+A+B') = C1+((C0'+A')(A+B')). Now applying DeMorgan's Law = C1+((C0A)'(A+B')) Ending up with: R = ((C1C0)(AB))'(C1'+A+B)(C1+((C0A)'(A+B'))) Is that correct, and can I simplify further (I don't I can)? Would I be able to generate a simpler expression from R = C1'C0'A'B + AB'C1' + B'C1'C0 + A'B'C0 + C1C0A' + ABC' instead? Surely not, because it contains more terms? The reason I ask is because I am in the process of designing a digital (minimum cost) circuit and the above expressions are generated by circling the zeros and the ones on the Karnaugh Map. I applied the simplification rules to the maxterm equation since it had fewer terms and decided that would be the best way to go, I hope I was right. p.s. as a sidenote, could anybody recommend the name of a software (preferably free/evaluation) that can draw the circuit diagram directly from the boolean equation? Is this even possible? === Subject: Re: Please check my boolean simplification > Hello all, > I would appreciate if somebody could tell me if I haven't made any mistakes > in the following boolean equation simplification: > R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') > Applying DeMorgan's Law to the term (C1'+C0'+A'+B') = ((C1C0)'+(AB)'). > Applying the law again = ((C1C0)(AB))'. > Using A(A+B)=A on the term (C1'+A+B)(C1'+C0+A+B) = (C1'+A+B) where A=(C1' > +A+B) (obviously) > Applying the Distributive Laws in reverse to the term (C1+C0'+A')(C1+A+B') = > C1+((C0'+A')(A+B')). Now applying DeMorgan's Law = C1+((C0A)'(A+B')) > Ending up with: R = ((C1C0)(AB))'(C1'+A+B)(C1+((C0A)'(A+B'))) > Is that correct, and can I simplify further (I don't I can)? Seems correct. The prime implicants (cheapest sum of products, or of ands) are: R = AC0' + AB'C1 + BC0'C1 + A'BC1 + B'C0'C1' + A'B'C1' > Would I be able to generate a simpler expression from R = C1'C0'A'B + AB'C1' > + B'C1'C0 + A'B'C0 + C1C0A' + ABC' instead? If the last letter is meant to be C0 then this one has prime implicants ABC0' + AC0'C1' + AB'C1' + BC0'C1' + A'C0C1 + B'C0C1' + A'B'C0 If the last letter is C1 then it has AC1' + BC0'C1' + A'C0C1 + B'C0C1' + A'B'C0 They are not equivalent with your previous R. To draw truth tables and calculate prime implicants, feel free to download from http://users.pandora.be/vdmoortel/dirk/Boole/boole.html (carful: you'll have to replace C0 with C and C1 with D) > The reason I ask is because I am in the process of designing a digital > (minimum cost) circuit and the above expressions are generated by circling > the zeros and the ones on the Karnaugh Map. I applied the simplification > rules to the maxterm equation since it had fewer terms and decided that > would be the best way to go, I hope I was right. > p.s. as a sidenote, could anybody recommend the name of a software > (preferably free/evaluation) that can draw the circuit diagram directly from > the boolean equation? Is this even possible? Maybe you find something with: hth Dirk Vdm === Subject: Re: Please check my boolean simplification > To draw truth tables and calculate prime implicants, > feel free to download from > http://users.pandora.be/vdmoortel/dirk/Boole/boole.html > (carful: you'll have to replace C0 with C and C1 with D) This is really a very useful program. === Subject: Re: Please check my boolean simplification === Subject: Please check my boolean simplification >I would appreciate if somebody could tell me if I haven't made >any mistakes in the following boolean equation simplification: >R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') >Applying DeMorgan's Law to the term (C1'+C0'+A'+B') = >((C1C0)'+(AB)'). Applying the law again = ((C1C0)(AB))'. Correct, but bad start as that expression is so 'insoluble'. >Using A(A+B)=A on the term (C1'+A+B)(C1'+C0+A+B) >= (C1'+A+B) where A=(C1' +A+B) (obviously) Good. R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B') >Applying the Distributive Laws in reverse to the term >(C1+C0'+A')(C1+A+B') = C1+((C0'+A')(A+B')). Now applying >DeMorgan's Law = C1+((C0A)'(A+B')) Lacks shrewdness (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B') = A + C1'B' + BC1 (C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1 = C0' + A' + B'C1 >Ending up with: R = ((C1C0)(AB))'(C1'+A+B)(C1+((C0A)'(A+B'))) Blah R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B') = (A' + C0' + B'C1)(A + C1'B' + BC1) = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A >Is that correct, and can I simplify further (I don't I can)? and, if I've done it right, so you can see some patterns = A'C1'B' + A'BC1 + C0'C1'B' + C0'BC1 + C0'A + B'C1A to work with, useful perhaps for optimizing. >Would I be able to generate a simpler expression from >R = C1'C0'A'B + AB'C1' + B'C1'C0 + A'B'C0 + C1C0A' + ABC' instead? Why not, it's reduced to six terms in disjunctive normal form instead of the monsterous form you presented before. However we differ as to opinion about disjunctive normal form and you haven't shown how you came to your conclusion. So check your work and mine too. >Surely not, because it contains more terms? Not so, your monster isn't even in conjunctive normal form. -- >The reason I ask is because I am in the process of designing a >digital (minimum cost) circuit and the above expressions are >generated by circling the zeros and the ones on the Karnaugh Map. What this? Your company needs mathematical consulting? For that should I not charge? >I applied the simplification rules to the maxterm equation since >it had fewer terms and decided that would be the best way to go, >I hope I was right. As to your circuit design, I'll take your word for your original equations. As to simplifying, double check your and my work. As a second check, I suggest truth or on/off table checking of orginal and final expressions. There's software for that, likely cheap as it's also easy to write. As a matter of fact, you could convert both expression to basic logical expressions by hand, insert them into a simple program asking if both expressions are equal over the 2^(number of variables) range. Nested for loops. Very simple because all the hard work, that of parsing the expressions is done for you by the syntax checker. That's the hardest part of writting such a logical expression checker for general expressions. My quicky has small draw back, if you want to check other expressions, you have to make modified copy of first program for each expression. For this problem budget about 1/2 hour to automate the checking, 1 hour if you're rusty on programming. >p.s. as a sidenote, could anybody recommend the name of a software >(preferably free/evaluation) that can draw the circuit diagram >directly from the boolean equation? Is this even possible? They even have software that'll design and burn cpu chips from specs. ---- === Subject: Re: Please check my boolean simplification > >Applying the Distributive Laws in reverse to the term > >(C1+C0'+A')(C1+A+B') = C1+((C0'+A')(A+B')). Now applying > >DeMorgan's Law = C1+((C0A)'(A+B')) > Lacks shrewdness > (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B') > = A + C1'B' + BC1 > (C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1 > = C0' + A' + B'C1 Yes, this is nicer. > >Would I be able to generate a simpler expression from > >R = C1'C0'A'B + AB'C1' + B'C1'C0 + A'B'C0 + C1C0A' + ABC' instead? > Why not, it's reduced to six terms in disjunctive normal form > instead of the monsterous form you presented before. However we differ as > to opinion about disjunctive normal form and you haven't shown how you > came to your conclusion. So check your work and mine too. > >Surely not, because it contains more terms? > Not so, your monster isn't even in conjunctive normal form. Is it not? My understanding was that CNF was the conjunction of disjunctions of literals (A,A' ...). Nevertheless, I did have a go at simplifying the above (DNF) form of the expression instead. Note that I made a slight mishap in typing the expression, it actually looks like this: R = (C1'C0'A'B)+(AB'C1')+(A'B'C0)+(C1C0A')+(BC1C0)+(ABC1) Lots of factoring possibilities here, I tried several but the following seemed to produce the best result: R = (C1'C0'A'B)+(AB'C1')+(A'B'C0)+(C1C0A')+(BC1C0)+(ABC1) = B (C1'C0'A'+C1(C0+A))+B'(AC1'+AC0) = B ((C1+C0)'A'+C1(C0+A))+B'(AC1'A'C0) = B ((C1+C0+A)'+C1(C0+A))+B'(AC1'+A'C0) i.e. using DeMorgan's a few times I am unable to see any other way of reducing the original expression to any fewer terms than this, although if my sight is (again) obscured then I would appreciate this pointed out! > What this? Your company needs mathematical consulting? > For that should I not charge? The circuit is really very basic, and forms a minute part of something immensely larger (I think). Besides, you can always consider payment as the satisfaction gained from helping an amateur with your advanced knowledge, or something ;-) > >I applied the simplification rules to the maxterm equation since > >it had fewer terms and decided that would be the best way to go, > >I hope I was right. > As to your circuit design, I'll take your word for your original > equations. As to simplifying, double check your and my work. As a second > check, I suggest truth or on/off table checking of orginal and final > expressions. > There's software for that, likely cheap as it's also easy to write. correct, may I add. > They even have software that'll design and burn cpu chips from specs. In that case it looks like I've been living under a rock === Subject: Re: Please check my boolean simplification > >Applying the Distributive Laws in reverse to the term > >(C1+C0'+A')(C1+A+B') = C1+((C0'+A')(A+B')). Now applying > >DeMorgan's Law = C1+((C0A)'(A+B')) > Lacks shrewdness > (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B') > = A + C1'B' + BC1 > (C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1 > = C0' + A' + B'C1 > Yes, this is nicer. > >Would I be able to generate a simpler expression from > >R = C1'C0'A'B + AB'C1' + B'C1'C0 + A'B'C0 + C1C0A' + ABC' instead? > Why not, it's reduced to six terms in disjunctive normal form > instead of the monstrous form you presented before. However we differ as > to opinion about disjunctive normal form and you haven't shown how you > came to your conclusion. So check your work and mine too. > >Surely not, because it contains more terms? > Not so, your monster isn't even in conjunctive normal form. > Is it not? My understanding was that CNF was the conjunction of disjunctions > of literals (A,A' ...). Huh? What happened to your monster specimen? You sniped it? Oh well, as I remember it wasn't fully simplified, it had nested parts like (AB)' or whatever. > Nevertheless, I did have a go at simplifying the above (DNF) form of the > expression instead. Note that I made a slight mishap in typing the > expression, it actually looks like this: > R = (C1'C0'A'B)+(AB'C1')+(A'B'C0)+(C1C0A')+(BC1C0)+(ABC1) > Lots of factoring possibilities here, I tried several but the following > seemed to produce the best result: > R = (C1'C0'A'B)+(AB'C1')+(A'B'C0)+(C1C0A')+(BC1C0)+(ABC1) > = B (C1'C0'A'+C1(C0+A))+B'(AC1'+AC0) > = B ((C1+C0)'A'+C1(C0+A))+B'(AC1'A'C0) > = B ((C1+C0+A)'+C1(C0+A))+B'(AC1'+A'C0) > i.e. using DeMorgan's a few times > I am unable to see any other way of reducing the original expression to any > fewer terms than this, although if my sight is (again) obscured then I would > appreciate this pointed out! > There's software for that, likely cheap as it's also easy to write. > correct, may I add. Interesting. However if you look close you'll find that my disjunctive normal form takes fewer operations than yours. Here's my final R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B') (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B') = A + C1'B' + BC1 (C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1 = C0' + A' + B'C1 R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B') = (A' + C0' + B'C1)(A + C1'B' + BC1) = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A = (A' + C0')(BC1 + B'C1') + A(C0' + B'C1) Or even = (A' + C0')(B <-> C1) + A(C0' + B'C1) Which reduces to two cases Case B <-> C1: A' + C0' Case (B <-> C1)': A(C0' + B'C1) === Subject: Re: Please check my boolean simplification > Hello all, > I would appreciate if somebody could tell me if I haven't made any mistakes > in the following boolean equation simplification: > R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B') [...] > Ending up with: R = ((C1C0)(AB))'(C1'+A+B)(C1+((C0A)'(A+B'))) Maple says this is correct (if I typed it in correctly). [...] -- Paul Sperry Columbia, SC (USA) === Subject: Hey Prime Slime Harris! http://www.scienca.com/ Your village called - its idiot is missing. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Hey Prime Slime Harris! In sci.physics, Uncle Al > http://www.scienca.com/ > Your village called - its idiot is missing. I wouldn't worry too much about JSH. He's now taken up differential equations, and doing as well there as he did in proving polynomial factors. In other words, not very. :-) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Problems book > Look at Larson's book Problem-Solving through Problems. It is geared > towards Undergrads. It is pretty good, too! > Lurch > Is there a book similar to Berkeley Problems in Mathematics, but > geared more towards the undergraduate, and perhaps more finding > stuff rather than proving stuff? I would hope the problems would be > of the same relative difficulty to an undergraduate as the BPM book > would be to a graduate. Unfortunately this book is of more a competition-type nature, not what I want. I want a book with exercises. Given A, find B. For example, the limit of a sequence, the coefficient of x^50 in the expansion of some product of polynomials, the probability of something happening. I don't want any proofs. Basically what you'd find in a textbook, only harder. Textbook problems are, on the whole, trivial applications of a definition. So far I've found a book called Linear Algebra: Challenging Problems for Students (by Fuzhen Zhang) (i think). Another one called Linear Algebra Problem book by Halmos, then I found Polynomials by Barbeau and Problem Book for First Year Calculus by someone whose name I can't remember. I wish I could find something like this for probability and advanced calculus as well, then this would be pretty much what I'm looking for. I just wish it were all summed up in one book. === Subject: Re: Problems book >Unfortunately this book is of more a competition-type nature, not what >I want. I want a book with exercises. Given A, find B. For >example, the limit of a sequence, the coefficient of x^50 in the >expansion of some product of polynomials, the probability of something >happening. I don't want any proofs. Basically what you'd find in a >textbook, only harder. Textbook problems are, on the whole, trivial >applications of a definition. Try: Textbook of Algebra by Chrystal Sample: The sides of a triangle are the roots of x^3-ax^2+bx-c=0. Show that its area is 1/4sqrt{a(4ab-a^3-8c)}. First published in 1886, I think. Still the gold standard for exercises. Rich === Subject: Re: Problems book Have you looked at the book? (1)Determine the odd binomial coeffients in the expansion of (x+y)^1000 (2) Find all solutions of x^4 + x^3 + x^2 + x + 1 = 0 (3) Evaluate Integral(e^-x^2)dx from 0 to oo Hint: use a double integral (4) Evaluate the sum(k^2/2^k) from n =1 to n (5) Determine the last two digits in the number 2^1407 (6) Evaluate lim(n-->oo) 1/n^4 * Infinite Product(n^2 + i^2)^1/n form i=1 to 2n I found those in about a minute of looking at the book. Lurch > Look at Larson's book Problem-Solving through Problems. It is geared > towards Undergrads. It is pretty good, too! > Lurch > Is there a book similar to Berkeley Problems in Mathematics, but > geared more towards the undergraduate, and perhaps more finding > stuff rather than proving stuff? I would hope the problems would be > of the same relative difficulty to an undergraduate as the BPM book > would be to a graduate. > Unfortunately this book is of more a competition-type nature, not what > I want. I want a book with exercises. Given A, find B. For > example, the limit of a sequence, the coefficient of x^50 in the > expansion of some product of polynomials, the probability of something > happening. I don't want any proofs. Basically what you'd find in a > textbook, only harder. Textbook problems are, on the whole, trivial > applications of a definition. > So far I've found a book called Linear Algebra: Challenging Problems > for Students (by Fuzhen Zhang) (i think). Another one called Linear > Algebra Problem book by Halmos, then I found Polynomials by Barbeau > and Problem Book for First Year Calculus by someone whose name I can't > remember. I wish I could find something like this for probability and > advanced calculus as well, then this would be pretty much what I'm > looking for. I just wish it were all summed up in one book. === 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. I have already been through the first half of that Rudin book. It was pretty tough even after covering almost all the proofs in Ellis and Gulick. If the Buck and Spivak books you mention have the proofs, I guess I can safely skip the second half of Rudin for now anyway? === Subject: Re: Usenet Posting Guide? > So, too, is > the white wine and brie crowd who by and large make up the Sierra Club > constituency. As for the environmentalists, as a group, if anything > they're even worse than liberals in their politics, the vast majority > of them being Bolsheviks. > > Bolsheviks? Are you a fan of Tom Potter? He's the only > other person I know to use that term to describe people > post-1920 or so. > > It is interesting to see that this poster thinks > that the Bolsheviks just vanished into nothingness. > > The fact of the matter is that the Bolsheviks instigated the > class wars of the 1900's for power and wealth, > and after their class wars were discredited, > and the Native Russians regained controlled > of their government, millions of the Bolsheviks > who had lived high and mighty in Russia, > migrated to Israel and New York, from where > they are instigating the religious wars of the 2000's > to get back into the chips as the loot from their class wars > is almost gone. > > I suggest that intelligent, rational, moral folks, > reject the media brainwashing, open their eyes, > look around and see who instigated the class wars, > who is instigating the religious wars, > and who profits from both, > while others, such as Blacks, Rednecks and Latinos, > sacrifice their lives, limbs, liberties and fortunes > to fight folks they would otherwise get along with just fine. > > As can be seen by studying history, > and by observing current events, > the Bolsheviks have a long history of > for power and wealth. > > Instigating conflict and war is the stock in trade > of the Bolsheviks, much as fortune telling > is the stock in trade of Gyspies. > Tom, I believe you and I are also on the same page here as it reflects > a proper understanding of Bolshevism. These ill-tutored twits who see > themselves as sophisticated and schooled in world politics generally > don't know their gluteous maximus from a hole in the ground. These > clueless suckers love to snicker at things they know absolutely > nothing about. This whole system is crumbling right before their very > eyes, and it's bread and circuses as usual for these political > geniuses while race is set against race, class is set against class, > gender is set against gender and we all get sold down the river for 30 > pieces of silver. We're in meltdown and these fools are asleep at the > control panel. > I liked the OP's comments about the environmental white wine and brie > crowd, for they are truly the useful idiots in this whole equation who > do the most damage. These people are not idiots. They are just highly suggestible. I used to hypnotize people, and soon discovered that most people are very susceptible to conditioning, and are thus easy to hypnotize. As most people are highly suggestible, they fall victims to the propoganda machine of the people who manipulate them. These people are victims. Not villians. They, like almost all people, believe what they are told, (In a non-threatening way.) rather than what they see. -- Tom Potter http://tompotter.us === Subject: Re: Hard (unsolved?) search problem revisited >>1 2 7 . . . >>5 4 . . . . >>And there we have it. As long as that's the only five in the matrix (or >>the other fives are similarly hidden), I am successful in the fooling. > But an original requirement was that in each row, the numbers had to > increas from left to right. > Your second row does not follow that rule. === Subject: Re: Multiplication definition of two complex numbers >> Why (a,b) (c,d) = ( ac - bd, ad + bc) ? > Because you're thinking of (a,b) as a+bi, where i^2 = 1. >> >> (a,b) (c,d) = (2 ac - bd, ad + bc) also solves x^2 + 1 = 0 > Mostly because it is not associative: THEOREM (a,b) (c,d) = (r ac + s bd, ad + bc) is associative <=> r = 1 PROOF (=>) [(1,0) (1,0)](0,1) = (r,0)(0,1) = (0,r) = (1,0)[(1,0) (0,1)] = (1,0)(0,1) = (0,1) (<=) If r = 1 then the map (a,b) -> a + b X yields an isomorphism of R x R with the ring R[X]/(X^2 - s) since (a + bX)(c + dX) = ac + s bd + (ad + bc) X i.e. (a , b) (c , d) = (ac + s bd , ad + bc) QED -Bill Dubuque > [(a,b) (c,d)] (e,f) > = (2ac - bd, ad+bc) (e,f) > = (2(2ac-bd)e - (ad+bc)f, (2ac-bd)f + (ad+bc)e ) > = (4ace - 2bde - adf - bcf, 2acf - bdf + ade + bce) > (a,b) [(c,d) (e,f)] > = (a,b) (2ce - df, cf+de) > = ( 2a(2ce-df) - b(cf+de), a(cf+de) + b(2ce-df) ) > = (4ace - 2adf - bcf - bde, acf + ade + 2bce - bdf) > On the other hand, > (a,b) (c,d) = (ac - 2bd, ad+bc) > would work, since then you are thinking of (a,b) as representing > a+b(sqrt(-2)). === Subject: ideal gas law on manifolds I've always understood the ideal gas law PV=nRT to refer to an ideal gas contained in a container with a boundary, the boundary playing some role in the definition of pressure. On the other hand, suppose you consider an ideal gas in a compact manifold without boundary, e.g. the 3-sphere. How would one formulate the ideal gas law in that context or in the full generality of manifolds? Surely someone must have worked that out, and references to relevant literature would be helpful. Would it simply be used to *define* pressure in the case of a compact manifold without boundary? Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** Author-Supplied-Address: bds ipp mpg de === Subject: Re: ideal gas law on manifolds Mail-To-News-Contact: abuse@dizum.com |> I've always understood the ideal gas law PV=nRT to refer to an ideal |> gas contained in a container with a boundary, the boundary playing some |> role in the definition of pressure. That is how it is done in some texts, and it is a bit unfortunate. Even at reasonably low pressures, the mean free path in most gases at room temperature is really short compared to, say, a milk jar. In those circumstances you get the same ideal gas law, and for the same cause is large numbers of mutual collisions in a space much smaller than any dynamical scale of interest. This is the justification behind ideal fluid equations. The distinction is important because... |> On the other hand, suppose you |> consider an ideal gas in a compact manifold without boundary, e.g. |> the 3-sphere. How would one formulate the ideal gas law in that |> context or in the full generality of manifolds? Surely someone |> must have worked that out, and references to relevant literature |> would be helpful. ..it makes the ideal gas law a local equation of state. So on the manifold, the rest pressure and rest energy density are Lorentz scalar fields. |> Would it simply be used to *define* pressure in the case of a compact |> manifold without boundary? Mathematically, the equation of state does just that. You have equations for the energy density and the material density, and your equation(s) of state then give you the temperature and pressure. The ideal gas law comes in with high enough temperatures that degeneracy can be ignored: e = (3/2) nKT k T = (2/3) e/n or p = nkT p = (2/3) e cf: http://www.rzg.mpg.de/~bds/phys/gas-kinetics.html -- cu, Bruce drift wave turbulence: http://www.rzg.mpg.de/~bds/ === Subject: linear system Let A be a real 2x2 matrix. Consider the linear system (x)' = Ax. Show that the only invariant lines under e^(tA) for all t in R are ax+ by = 0, where (-b,a)^T are the eigenvectors of A. Give me a hint for this. -- === Subject: Re: linear system > Let A be a real 2x2 matrix. > Consider the linear system (x)' = Ax. > Show that the only invariant lines under e^(tA) for all t in R are > ax+ by = 0, where (-b,a)^T are the eigenvectors of A. > Give me a hint for this. If v is an eigenvector of A with eigenvalue L, what is ((t^k)/k!) (A^k)v for non-negative integer k? Hence show that v is an eigenvector of exp(tA), and find the corresponding eigenvalue. Now show that any eigenvector of exp(tA) must also be an eigenvector of A. -- 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: Need help on algebraic geometry Hi. I am studying algebraic geometry using the excelent Milne's course notes (available at http://www.jmilne.org/math/) but I am stuck at one point and I would be grateful if someone can help me on that. First, he defines a function f:U->k as regular if it is rational in a neighbourhood of every point of U and simbolize the set of regular functions on U as $Gamma(U,O_{V})$. Then, he proves that for every regular function f defined on a basic (principal) open subset of V is rational (when V is irredutible). would have $Gamma(U,O_{V})=cap _{j=1}^{n}Gamma (D(h_{j}),O_{V})$. The last equality is not making sense to me. $Gamma (D(h_{j}),O_{v})$ is about functions defined on D(h_{j}) and the other side of the equation refers to functions defined on all U. I was inclined to consider this just an abuse of notation, meaning that f:U->k is regular iff the restriction of f to all D(h_{j}) is regular, but he used this to justify: the equalities show that the regular functions on an open $U subset V$ are the rational functions on V that are defined at each point of U. I can't understand that. I see that, for every D(h_{j}), the restriction of f to D(h_{j}) would be rational, but why is f rational in all U? language] Anderson Brasil andersbrasil@aol.com === Subject: Re: Need help on algebraic geometry > Hi. I am studying algebraic geometry using the excelent Milne's course > notes > (available at http://www.jmilne.org/math/) but I am stuck at one point and > I would be grateful if someone can help me on that. > First, he defines a function f:U->k as regular if it is rational in a > neighbourhood of every point of U and simbolize the set of regular > functions on U as $Gamma(U,O_{V})$. > Then, he proves that for every regular function f defined on a basic > (principal) open subset of V is rational (when V is irredutible). What is D(h_j)? Do they form some open cover of U? > would have $Gamma(U,O_{V})=cap _{j=1}^{n}Gamma (D(h_{j}),O_{V})$. The > last equality is not making sense to me. It isn't at the moment to me. > $Gamma (D(h_{j}),O_{v})$ is > about functions defined on D(h_{j}) and the other side of the equation > refers to functions defined on all U. I was inclined to consider this just > an abuse of notation, meaning that f:U->k is regular iff the restriction > of f to all D(h_{j}) is regular, but he used this to justify: the > equalities show that the regular functions on an open $U subset V$ are > the rational functions on V that are defined at each point of U. > I can't understand that. I see that, for every D(h_{j}), the restriction > of f > to D(h_{j}) would be rational, but why is f rational in all U? Can you give a page reference? -- 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: Need help on algebraic geometry >> Hi. I am studying algebraic geometry using the excelent Milne's course >> notes >> (available at http://www.jmilne.org/math/) but I am stuck at one point and >> I would be grateful if someone can help me on that. >> First, he defines a function f:U->k as regular if it is rational in a >> neighbourhood of every point of U and simbolize the set of regular >> functions on U as $Gamma(U,O_{V})$. >> Then, he proves that for every regular function f defined on a basic >> (principal) open subset of V is rational (when V is irredutible). >What is D(h_j)? Do they form some open cover of U? Yes. If, h_j is an element of k[V] then D(h_j) is defined as {x in U such that h_j(x)<>0}. Sets of this form are called basic (or principal) open sets of V. He proves that every open set of V can be written as a (finite) union of D(h_j). >> would have $Gamma(U,O_{V})=cap _{j=1}^{n}Gamma (D(h_{j}),O_{V})$. The >> last equality is not making sense to me. >It isn't at the moment to me. >> $Gamma (D(h_{j}),O_{v})$ is >> about functions defined on D(h_{j}) and the other side of the equation >> refers to functions defined on all U. I was inclined to consider this just >> an abuse of notation, meaning that f:U->k is regular iff the restriction >> of f to all D(h_{j}) is regular, but he used this to justify: the >> equalities show that the regular functions on an open $U subset V$ are >> the rational functions on V that are defined at each point of U. >> I can't understand that. I see that, for every D(h_{j}), the restriction >> of f >> to D(h_{j}) would be rational, but why is f rational in all U? >Can you give a page reference? It is page 41 in the last version of those course notes (page 35 for those using the old version). If you can help me I would be very grateful. english language] Anderson Brasil andersbrasil@aol.com It's not that I'm afraid to die, I just don't want to be there when it happens. (Woody Allen) === Subject: Re: Indian firsts in Maths > INVENTIONS & DISCOVERIES > INVENTION OF NUMERALS > Numerals are found in the inscriptions of Ashoka The Great in the 3rd > Century BC. This knowledge traveled from there to Europe and West. In > Arab countries even now numerals are known as HINDSE: from India. La > time, It is India that gave us the ingenious method of expressing all > numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's > Journal > INVENTION OF ZERO > Brahmagupta was the first mathematician to treat ZERO (0) as a number > and showed its mathematical operations The Babylonians had a place value number system way before the Hindus. It is from that system that the Indians got the idea of zero. Babylonian scale of enumeration is known as the sexagesimal system. What that means is that the Babylonians used 60 as their base, much as we tend to use the decimal system (base 10) in the United States . In the sexagesimal system, each time a digit is moved to the left its value increases by a factor of 60. When you represent a whole number in the sexagesimal system the last space digit is for the numbers from 1 to 59, the next to the last space digit for multiples of 60, then the next space digit for multiples of 60^2 = 3600, the next preceding space digit for multiples of 60^3 = 216,000, and so forth. In other words the number system is developed by working with base 60: 60^0 = 1 60^1 = 60 60^2 = 3600 60^3 = 216,000 . . . . . . The Babylonian 5 53 18 would then represent the number 5*60^2 + 53*60^1 + 18*60^0 = 5*3600 + 53*60 + 18*1 = 18,000 + 3,180 + 18 = 21,198 (in our enumeration system). If you want to write out the number system, it would start out something like 1, 2, 3, 4,...59, 1 0, 1 1, 1 2,... where 1 0 = 1*60^1 + 0*60^0 = 60 + 0 = 60, 1 1 = 1*60^1 + 1*60^0 = 60 + 1 = 61, 1 2 = 1*60^1 + 2*60^0 = 60 + 2 = 62, http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Babylonian_numerals.html > INVENTION OF ARITHMETIC > Arithmetic was discovered by Indians in about 2nd Century BC. > Bhaskaracharya's book Lilavathi is regarded as the first book on > modern arithmetic. The Arabs learnt and adopted it from India and > spreaded it to Europe. The Indians most likely learnt Artithmetic from the Pre-Islamic Semites. Certainly the Egyptians and the Babylonians were using Arithmetic, although not in a base 10 number system way before the Hindus. Learn something about Egyptian Arithmetic here: http://www.math.wichita.edu/history/topics/arithmetic.html#egypt-arith > INVENTION OF ALGEBRA > In Western Europe the knowledge of Algebra was borrowed, not from > Greece but from Arabs, who acquired this from India. Algebra is the > only Arabic name for Bijaganitha. Aryabhatta was one of the first to > use Algebra (Encyclopedia Britannica) The Rhind Papyrus introduced Agebra in the 16 century before Chirst. http://www.bonita.k12.ca.us/schools/ramona/teachers/carlton/historypages/his tory1.html Read also: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Al-Khwarizmi.html http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Quadratic_etc_equations. html > INVENTION OF GEOMETRY AND TRIGNOMETRY > The brick work of Harappa and Mohenjodaro excavations show that people > of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta > formulated the rules for finding the area of a ?triangle', which led > to the origin of Trignometry. The same was true of the Egyptians, except that they were doing it way before the Harappans. > DISCOVERY OF ASTRONOMY > The knowledge of the motion of heavenly bodies was discovered by > Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for > calculating the timing of eclipses. Astronomy was discovered by the Egyptians and Babylonians. The Hindus came later. We can not deny that Indians improved many of the Ideas of Pre-Islamic Semites, and that the Moslem subsequently reintroduced many of those improved ideas to the Mideast and the West, but to claim that they discovered them all is stretching things a bit much. === Subject: Re: Indian firsts in Maths > All that pathetic nationalism does not change the FACT that > India is a very minor player in modern science (leaving aside > Ramanujan, Bose, Raman and the like,... So true, and even truer when we leave out all the good scientists! :) Let's not forget to exclude inventions like penicillin, fingerprinting, etc., then India would look even worse. === Subject: Re: Indian firsts in Maths What use is this being first in math? Will it help these Indians when the day of reckoning comes? It is more better to be first in Quran! Yours truly, Gazi Baba 1971 > INVENTIONS & DISCOVERIES > INVENTION OF NUMERALS > Numerals are found in the inscriptions of Ashoka The Great in the 3rd > Century BC. This knowledge traveled from there to Europe and West. In > Arab countries even now numerals are known as HINDSE: from India. La > time, It is India that gave us the ingenious method of expressing all > numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's > Journal > INVENTION OF ZERO > Brahmagupta was the first mathematician to treat ZERO (0) as a number > and showed its mathematical operations > INVENTION OF ARITHMETIC > Arithmetic was discovered by Indians in about 2nd Century BC. > Bhaskaracharya's book Lilavathi is regarded as the first book on > modern arithmetic. The Arabs learnt and adopted it from India and > spreaded it to Europe. In 499 AD Aryabhatta finished his work > Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) > INVENTION OF ALGEBRA > In Western Europe the knowledge of Algebra was borrowed, not from > Greece but from Arabs, who acquired this from India. Algebra is the > only Arabic name for Bijaganitha. Aryabhatta was one of the first to > use Algebra (Encyclopedia Britannica) > INVENTION OF GEOMETRY AND TRIGNOMETRY > The brick work of Harappa and Mohenjodaro excavations show that people > of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta > formulated the rules for finding the area of a ?triangle', which led > to the origin of Trignometry. > DISCOVERY OF ASTRONOMY > The knowledge of the motion of heavenly bodies was discovered by > Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for > calculating the timing of eclipses. In Surya Sidhanta' Latadeva, > talked about the earth's axis and called it SUMERU. That the earth is > a sphere and it rotates on its own axis, was known to Varahamihira > and other Indian astronomers much before Copernicus published this > theory. (Jewish Encyclopedia) > INVENTION OF CALENDAR MAKING > Discovery of measurement of time and discovery of nomenclature of > days, month and years and invention of calendar making was made in > India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year > into 12 months. Seven planets of the solar system effect the earth's > atmosphere and their names were added to the seven days of the week, > which was accepted all over the world. > DISCOVERY OF THEORY OF GRAVITATION > In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force > of attraction resembling gravity, discovered centuries later by > Newton. (Jewish Encyclopedia) > INVENTION OF IRON PRODUCTS IN 3000 BC > The word AYAS occurs in the four Vedas which denotes iron. Ashoka > pillar at Mehrauli, New Delhi and another iron pillar in Karnataka > stand proof of India's metallurgical heritage (A study published in > the magazine ?The Current Science'). > INVENTION OF COPPER, BRONZE AND ZINC > The copper and bronze artifacts dates back to Indus Valley > Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was > made in around 50 BC at Zawar in Rajasthan (India). > INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS > Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone > to prepare the sulphate of copper, zinc and iron and carbonates of > lead and iron. RASAVIDYA or Indian alchemy made its appearance around > 5th Century AD (National Science Centre, New Delhi) === Subject: Re: Indian firsts in Maths > What use is this being first in math? Will it help these Indians when > the day of reckoning comes? It is more better to be first in Quran! ----------------------------- When is this day coming? Has the exact date been given by Allah in the Quran? Or is one of those When the Mothman cometh prophesies? If a Muslim can inform me on the day before the day of rekoning, I'll wake up and read the Quran - Should take all of about 3 minutes to figure out what it has to say... ----------------------------- > Yours truly, > Gazi Baba 1971 > > INVENTIONS & DISCOVERIES > > INVENTION OF NUMERALS > Numerals are found in the inscriptions of Ashoka The Great in the 3rd > Century BC. This knowledge traveled from there to Europe and West. In > Arab countries even now numerals are known as HINDSE: from India. La > time, It is India that gave us the ingenious method of expressing all > numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's > Journal > > INVENTION OF ZERO > Brahmagupta was the first mathematician to treat ZERO (0) as a number > and showed its mathematical operations > > INVENTION OF ARITHMETIC > Arithmetic was discovered by Indians in about 2nd Century BC. > Bhaskaracharya's book Lilavathi is regarded as the first book on > modern arithmetic. The Arabs learnt and adopted it from India and > spreaded it to Europe. In 499 AD Aryabhatta finished his work > Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) > > INVENTION OF ALGEBRA > In Western Europe the knowledge of Algebra was borrowed, not from > Greece but from Arabs, who acquired this from India. Algebra is the > only Arabic name for Bijaganitha. Aryabhatta was one of the first to > use Algebra (Encyclopedia Britannica) > > INVENTION OF GEOMETRY AND TRIGNOMETRY > The brick work of Harappa and Mohenjodaro excavations show that people > of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta > formulated the rules for finding the area of a ?triangle', which led > to the origin of Trignometry. > > DISCOVERY OF ASTRONOMY > The knowledge of the motion of heavenly bodies was discovered by > Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for > calculating the timing of eclipses. In Surya Sidhanta' Latadeva, > talked about the earth's axis and called it SUMERU. That the earth is > a sphere and it rotates on its own axis, was known to Varahamihira > and other Indian astronomers much before Copernicus published this > theory. (Jewish Encyclopedia) > > INVENTION OF CALENDAR MAKING > Discovery of measurement of time and discovery of nomenclature of > days, month and years and invention of calendar making was made in > India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year > into 12 months. Seven planets of the solar system effect the earth's > atmosphere and their names were added to the seven days of the week, > which was accepted all over the world. > > DISCOVERY OF THEORY OF GRAVITATION > In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force > of attraction resembling gravity, discovered centuries later by > Newton. (Jewish Encyclopedia) > > INVENTION OF IRON PRODUCTS IN 3000 BC > The word AYAS occurs in the four Vedas which denotes iron. Ashoka > pillar at Mehrauli, New Delhi and another iron pillar in Karnataka > stand proof of India's metallurgical heritage (A study published in > the magazine ?The Current Science'). > > INVENTION OF COPPER, BRONZE AND ZINC > The copper and bronze artifacts dates back to Indus Valley > Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was > made in around 50 BC at Zawar in Rajasthan (India). > > INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS > Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone > to prepare the sulphate of copper, zinc and iron and carbonates of > lead and iron. RASAVIDYA or Indian alchemy made its appearance around > 5th Century AD (National Science Centre, New Delhi) === Subject: Re: Indian firsts in Maths The Quran is as ficitious as the bible. > What use is this being first in math? Will it help these Indians when > the day of reckoning comes? It is more better to be first in Quran! > Yours truly, > Gazi Baba 1971 > INVENTIONS & DISCOVERIES > INVENTION OF NUMERALS > Numerals are found in the inscriptions of Ashoka The Great in the 3rd > Century BC. This knowledge traveled from there to Europe and West. In > Arab countries even now numerals are known as HINDSE: from India. La > time, It is India that gave us the ingenious method of expressing all > numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's > Journal > INVENTION OF ZERO > Brahmagupta was the first mathematician to treat ZERO (0) as a number > and showed its mathematical operations > INVENTION OF ARITHMETIC > Arithmetic was discovered by Indians in about 2nd Century BC. > Bhaskaracharya's book Lilavathi is regarded as the first book on > modern arithmetic. The Arabs learnt and adopted it from India and > spreaded it to Europe. In 499 AD Aryabhatta finished his work > Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) > INVENTION OF ALGEBRA > In Western Europe the knowledge of Algebra was borrowed, not from > Greece but from Arabs, who acquired this from India. Algebra is the > only Arabic name for Bijaganitha. Aryabhatta was one of the first to > use Algebra (Encyclopedia Britannica) > INVENTION OF GEOMETRY AND TRIGNOMETRY > The brick work of Harappa and Mohenjodaro excavations show that people > of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta > formulated the rules for finding the area of a ?triangle', which led > to the origin of Trignometry. > DISCOVERY OF ASTRONOMY > The knowledge of the motion of heavenly bodies was discovered by > Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for > calculating the timing of eclipses. In Surya Sidhanta' Latadeva, > talked about the earth's axis and called it SUMERU. That the earth is > a sphere and it rotates on its own axis, was known to Varahamihira > and other Indian astronomers much before Copernicus published this > theory. (Jewish Encyclopedia) > INVENTION OF CALENDAR MAKING > Discovery of measurement of time and discovery of nomenclature of > days, month and years and invention of calendar making was made in > India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year > into 12 months. Seven planets of the solar system effect the earth's > atmosphere and their names were added to the seven days of the week, > which was accepted all over the world. > DISCOVERY OF THEORY OF GRAVITATION > In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force > of attraction resembling gravity, discovered centuries later by > Newton. (Jewish Encyclopedia) > INVENTION OF IRON PRODUCTS IN 3000 BC > The word AYAS occurs in the four Vedas which denotes iron. Ashoka > pillar at Mehrauli, New Delhi and another iron pillar in Karnataka > stand proof of India's metallurgical heritage (A study published in > the magazine ?The Current Science'). > INVENTION OF COPPER, BRONZE AND ZINC > The copper and bronze artifacts dates back to Indus Valley > Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was > made in around 50 BC at Zawar in Rajasthan (India). > INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS > Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone > to prepare the sulphate of copper, zinc and iron and carbonates of > lead and iron. RASAVIDYA or Indian alchemy made its appearance around > 5th Century AD (National Science Centre, New Delhi) === Subject: Re: Indian firsts in Maths > The Quran is as ficitious as the bible. Hmm, then the question might be whether the final exam will be on Math or fiction:-) >>What use is this being first in math? Will it help these Indians when >>the day of reckoning comes? It is more better to be first in Quran! >>Yours truly, >>Gazi Baba 1971 === Subject: Re: Indian firsts in Maths >> INVENTIONS & DISCOVERIES >> >> INVENTION OF NUMERALS >> INVENTION OF ZERO >> INVENTION OF ARITHMETIC >> INVENTION OF ALGEBRA >> INVENTION OF GEOMETRY AND TRIGNOMETRY >> DISCOVERY OF ASTRONOMY >> INVENTION OF CALENDAR MAKING >> DISCOVERY OF THEORY OF GRAVITATION >> INVENTION OF IRON PRODUCTS IN 3000 BC >> INVENTION OF COPPER, BRONZE AND ZINC >> INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS >Goodness Gracious Me! >(Middle of 2nd paragraph, just after Mr 'Cooper'.) >Phil hahahahaha Phil you got this indian nailed .... every thing comes from India and kapoor is cooper most likely gary cooper was an Indian too.... hahahaha... so very true. === Subject: Re: Indian firsts in Maths >> INVENTIONS & DISCOVERIES >> >> INVENTION OF NUMERALS >> INVENTION OF ZERO >> INVENTION OF ARITHMETIC >> INVENTION OF ALGEBRA >> INVENTION OF GEOMETRY AND TRIGNOMETRY >> DISCOVERY OF ASTRONOMY >> INVENTION OF CALENDAR MAKING >> DISCOVERY OF THEORY OF GRAVITATION >> INVENTION OF IRON PRODUCTS IN 3000 BC >> INVENTION OF COPPER, BRONZE AND ZINC >> INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS >Goodness Gracious Me! >(Middle of 2nd paragraph, just after Mr 'Cooper'.) >Phil > hahahahaha Phil you got this indian nailed .... every thing comes from > India and kapoor is cooper most likely gary cooper was an Indian > too.... hahahaha... so very true. what ya hahahahaha are yu grinning like a friggin jackass,yu muslim son of a bitch...yeah does phil know too, yu moslems were the first, to bring the wto towers crashing down to earth,and sending thousands of americans to heaven....probably after that,yu try sucking up to him. === Subject: Re: Indian firsts in Maths > INVENTIONS & DISCOVERIES > > INVENTION OF NUMERALS > INVENTION OF ZERO > INVENTION OF ARITHMETIC > INVENTION OF ALGEBRA > INVENTION OF GEOMETRY AND TRIGNOMETRY > DISCOVERY OF ASTRONOMY > INVENTION OF CALENDAR MAKING > DISCOVERY OF THEORY OF GRAVITATION > INVENTION OF IRON PRODUCTS IN 3000 BC > INVENTION OF COPPER, BRONZE AND ZINC > INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS >> >> >>Goodness Gracious Me! >> >> >>(Middle of 2nd paragraph, just after Mr 'Cooper'.) >> >>Phil >> hahahahaha Phil you got this indian nailed .... every thing comes from >> India and kapoor is cooper most likely gary cooper was an Indian >> too.... hahahaha... so very true. >what ya hahahahaha are yu grinning like a friggin jackass,yu muslim >son of a bitch...yeah does phil know too, yu moslems were the first, >to bring the wto towers crashing down to earth,and sending thousands >of americans to heaven....probably after that,yu try sucking up to >him. heheh motherchod chamar every things comes from bhaRAT hahahahahahaha yes yes for bhaRAT every thing is devadasis it starts with devadaiis and ends with devadasis.... What ing retards product of mandirs :))) Kapoor is cooper hahahahahahahahaahhaha and queen is indian too hahahahahahahahahaha..... === Subject: Re: My research, publication announcement >There's more to my work than just arguing on Usenet, so I'd like to >point out that my paper Advanced Polynomial Factorization is slated >to be published: >See http://www.megasociety.net/NoesisHighlights.html > Darn. Looking at the title of the thread I assumed that your work was > actually going to be _published_. In math published means > published in a peer-reviewed journal, not published on a vanity > web site dedicated to 'severely gifted' individuals, who are allowed > to publish all they want, for the proper fee. Peer review limitations: --reviewer subjectivity and agenda --constraints of scientific dogma and orthodoxy --by definition, *peer* reviewers are usually limited to the same paradigm and philosophical traditions === Subject: ?Any good, inexpensive math software? For various obscure reasons I've needed an inexpensive shareware software which would calculate basic arithmetic functions, ln(n), e^n, roots, and the basic trig. functions,...all out to pretty high decimal points, say 50 or so (or more depending on how playful I get). For a while I've tested the Haxial Calculator which is OK but laborious, esp. since it has, as yet, no trig. functions. For such trig stuff I've used basic formulae to let it calculate actual trig. values, though it kicks up a fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe $15 or 20. Does anyone know of a good shareware, perhaps even a bit more expen- sive, which will do ALL the functions I described, to fairly large arbitrary accuracy?? (There is no way I can afford anything like Mathematica unless I don a mask and take to the highroad ;- ) Gene === Subject: Re: ?Any good, inexpensive math software? > For various obscure reasons I've needed an inexpensive shareware > software which would calculate basic arithmetic functions, ln(n), e^n, > roots, and the basic trig. functions,...all out to pretty high decimal points, > say 50 or so (or more depending on how playful I get). > For a while I've tested the Haxial Calculator which is OK but laborious, > esp. since it has, as yet, no trig. functions. For such trig stuff I've used > basic formulae to let it calculate actual trig. values, though it kicks up a > fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe > $15 or 20. > Does anyone know of a good shareware, perhaps even a bit more expen- > sive, which will do ALL the functions I described, to fairly large arbitrary > accuracy?? (There is no way I can afford anything like Mathematica > unless I don a mask and take to the highroad ;- ) > Gene I am not sure that it's what you need but you may check out scilab at http://scilabsoft.inria.fr. It's free. Goran === Subject: Re: ?Any good, inexpensive math software? >> For various obscure reasons I've needed an inexpensive shareware >> software which would calculate basic arithmetic functions, ln(n), e^n, >> roots, and the basic trig. functions,...all out to pretty high decimal points, >> say 50 or so (or more depending on how playful I get). > bc > overture > scilab > calc ooops ... all *nix free stuff. I don't know which are available for Win/DOS. However, I agree with the other poster that UBASIC is a good language for doing thousands of digits of precision calculations in DOS/Win. === Subject: Re: ?Any good, inexpensive math software? >For various obscure reasons I've needed an inexpensive shareware >software which would calculate basic arithmetic functions, ln(n), e^n, >roots, and the basic trig. functions,...all out to pretty high decimal points, >say 50 or so (or more depending on how playful I get). >For a while I've tested the Haxial Calculator which is OK but laborious, >esp. since it has, as yet, no trig. functions. For such trig stuff I've used >basic formulae to let it calculate actual trig. values, though it kicks up a >fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe >$15 or 20. >Does anyone know of a good shareware, perhaps even a bit more expen- >sive, which will do ALL the functions I described, to fairly large arbitrary >accuracy?? (There is no way I can afford anything like Mathematica >unless I don a mask and take to the highroad ;- ) >Gene Try UBASIC. It's a DOS program but very powerful: UBASIC is a BASIC-like environment which is suitable for number theoretic investigations. Version 8 of UBASIC has the high precision real and complex arithmetic (up to 2600 digits) of previous versions, but adds exact rational arithmetic and arithmetic of polynomials with complex, rational or modulo p coefficients, as well as string handling and limited list handling capabilities Not sure what the latest version number is, so Google for ubasic download and root around for the most up-to-date. Brian === Subject: Re: ?Any good, inexpensive math software? >Does anyone know of a good shareware, perhaps even a bit more expen- >sive, which will do ALL the functions I described, to fairly large arbitrary >accuracy?? (There is no way I can afford anything like Mathematica >unless I don a mask and take to the highroad ;- ) >Gene You could also look for ARIBAS, a pascal-interpreter with high prec arithmetic. Has common windows-interface, (but better you also keep a notepad open). Don't have it at hand, which trigs are implemented. ARIBAS is free software for selfstudy and educational purposes Gottfried Helms === Subject: Re: ?Any good, inexpensive math software? > For various obscure reasons I've needed an inexpensive shareware > software which would calculate basic arithmetic functions, ln(n), e^n, > roots, and the basic trig. functions,...all out to pretty high decimal points, > say 50 or so (or more depending on how playful I get). > For a while I've tested the Haxial Calculator which is OK but laborious, > esp. since it has, as yet, no trig. functions. For such trig stuff I've used > basic formulae to let it calculate actual trig. values, though it kicks up a > fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe > $15 or 20. > Does anyone know of a good shareware, perhaps even a bit more expen- > sive, which will do ALL the functions I described, to fairly large arbitrary > accuracy?? (There is no way I can afford anything like Mathematica > unless I don a mask and take to the highroad ;- ) > Gene If you're able to get by with about 56 digits, you can download XCalc for free. It is an extended precision RPN calculator supporting log, exp, x^y, trig and inverse trig, hyperbolic and inverse hyperbolic functions, etc. If you need arbitrary precision, this won't be much help, but it's about as simple as you can get for fixed precision. http://www.crbond.com/applications.htm -- 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: ?Any good, inexpensive math software? > For various obscure reasons I've needed an inexpensive shareware > software which would calculate basic arithmetic functions, ln(n), e^n, > roots, and the basic trig. functions,...all out to pretty high decimal points, > say 50 or so (or more depending on how playful I get). > For a while I've tested the Haxial Calculator which is OK but laborious, > esp. since it has, as yet, no trig. functions. For such trig stuff I've used > basic formulae to let it calculate actual trig. values, though it kicks up a > fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe > $15 or 20. > Does anyone know of a good shareware, perhaps even a bit more expen- > sive, which will do ALL the functions I described, to fairly large arbitrary > accuracy?? (There is no way I can afford anything like Mathematica > unless I don a mask and take to the highroad ;- ) > Gene Have a look at PARI at http://pari.math.u-bordeaux.fr/download.html default realprecision: 28 digits. increase default real precision with command default(realprecision,50) Max realprecision = 80807123 significant decimal digits ;-) Dirk Vdm === Subject: Re: ?Any good, inexpensive math software? In sci.math, Dirk Van de moortel <%antb.24231$RT7.880692@phobos.telenet-ops.be>: >> For various obscure reasons I've needed an inexpensive shareware >> software which would calculate basic arithmetic functions, ln(n), e^n, >> roots, and the basic trig. functions,...all out to pretty high decimal points, >> say 50 or so (or more depending on how playful I get). >> For a while I've tested the Haxial Calculator which is OK but laborious, >> esp. since it has, as yet, no trig. functions. For such trig stuff I've used >> basic formulae to let it calculate actual trig. values, though it kicks up a >> fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe >> $15 or 20. >> Does anyone know of a good shareware, perhaps even a bit more expen- >> sive, which will do ALL the functions I described, to fairly large arbitrary >> accuracy?? (There is no way I can afford anything like Mathematica >> unless I don a mask and take to the highroad ;- ) >> Gene > Have a look at PARI at http://pari.math.u-bordeaux.fr/download.html > default realprecision: 28 digits. > increase default real precision with command > default(realprecision,50) > Max realprecision = 80807123 significant decimal digits ;-) $ gp ... ? default(realprecision,10000) realprecision = 10008 significant digits (10000 digits displayed) ? 4*atan(1) %1 = 3.14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679 82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 442881097566593344612847564823378 ... And here I've been doing it the hard way. :-) > Dirk Vdm -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: ?Any good, inexpensive math software? > In sci.math, Dirk Van de moortel > <%antb.24231$RT7.880692@phobos.telenet-ops.be>: > Have a look at PARI at http://pari.math.u-bordeaux.fr/download.html > default realprecision: 28 digits. > increase default real precision with command > default(realprecision,50) > Max realprecision = 80807123 significant decimal digits ;-) > $ gp > ... > ? default(realprecision,10000) > realprecision = 10008 significant digits (10000 digits displayed) > ? 4*atan(1) > %1 = 3.14159265358979323846264338327950288419716939937510 > 58209749445923078164062862089986280348253421170679 > 82148086513282306647093844609550582231725359408128 > 48111745028410270193852110555964462294895493038196 > 442881097566593344612847564823378 > ... > And here I've been doing it the hard way. :-) gp > 4*atan(1) - Pi %1= -9.05240874 E-10009 ;-) Dirk Vdm === Subject: Re: ?Any good, inexpensive math software? > ... > ? default(realprecision,10000) > realprecision = 10008 significant digits (10000 digits displayed) > ? 4*atan(1) > %1 = 3.14159265358979323846264338327950288419716939937510 > 58209749445923078164062862089986280348253421170679 > 82148086513282306647093844609550582231725359408128 > 48111745028410270193852110555964462294895493038196 > 442881097566593344612847564823378 > ... > And here I've been doing it the hard way. :-) > gp > 4*atan(1) - Pi > %1= -9.05240874 E-10009 > ;-) ARIBAS macht daraus -> set_floatprec(1000). -> 4*arctan(1)-pi. -: -5.56268(..ca 200 Stellen...)E-309 So schade... Gottfried Helms === Subject: Re: ?Any good, inexpensive math software? Uppps. I was out of mind, a bit... > gp > 4*atan(1) - Pi > %1= -9.05240874 E-10009 > ;-) ARIBAS computes: -> set_floatprec(1000). -> 4*arctan(1)-pi. -: -5.56268(..ca 200 more digits...)E-309 ... too bad. (But, at least, ARIBAS is freeware...) Gottfried Helms === Subject: Re: ?Any good, inexpensive math software? > Uppps. I was out of mind, a bit... > > > gp > 4*atan(1) - Pi > %1= -9.05240874 E-10009 > > ;-) > ARIBAS computes: > -> set_floatprec(1000). > -> 4*arctan(1)-pi. > -: -5.56268(..ca 200 more digits...)E-309 > ... too bad. > (But, at least, ARIBAS is freeware...) But PARI is just as free :-) Dirk Vdm === Subject: Re: ?Any good, inexpensive math software? > But PARI is just as free :-) > Dirk Vdm Ahaa! > gp > 4*atan(1) - Pi this is the prompt of PARI, then. Well, I think, so it's worth a look... Gottfried Helms === Subject: Re: ?Any good, inexpensive math software? > But PARI is just as free :-) > Dirk Vdm > Ahaa! > gp > 4*atan(1) - Pi > this is the prompt of PARI, then. > Well, I think, so it's worth a look... We have a common friend :-) He's in my taskbar and helped me find: http://pari.math.u-bordeaux.fr/download.html Gusundheit! Dirk Vdm === Subject: Re: NOVA strings and branes about the depth given in the NOVA show--quick question: quarks and electrons are made of strings --which are now called branes as used in M-thy ...but what are branes made of? If nothing the thy is nonsense...if something, it would seem to be something inherently bigger than quarks, etc., in fact as big or bigger than our whole universe (so far, my best visualization of branes is n-dimensional floating spatial coordinate systems!!) Question 2: The BBang in brane space seems to be visualized by an elastic collision between 2 branes, wherein one brane acquires a massive energy boost from the other, right? Wouldn't there also be the possibility of combinational collisions between branes wherein they merge?--just like black holes would. Matter of fact, maybe branes are merging all the time...even undergoing the kind of evolution which has been bandyed about for big universes hypothetically embedded in a much larger universe with little BBangs going on continuously all the time (I forget the name of the book not long ago espousing this evolutionary universe thy, in which many differently parametized realities may coexist.) I have more questions I think, but I'm gettin' sleeeeeepppppy.......... ...tonyC === Subject: Re: Fastest factorial algorithm <3FB37A98.1E6E3DCF@uni-kassel.de> >> Actually, 10000!! is easy, using Mathematica. The !! notation >> represents the double factorial function, such that n!! = >> n*(n-2)*(n-4)*.... >> I think it bogus and that 10000!! means just what you say >> it means and for those who insist it means 10000(!!), let >> them make it known they're omiting those '(' and ')'!! > The double factorial is a standard (and, in my experience, useful) > notation. It is hardly bogus. > I'll agree that the double factorial can be quite useful, but I'll > not go so far as to say that the _notation_ is useful. When one first > encounters something such as 10000!!, everyone no doubt assumes it to > mean (10000!)! instead of the double factorial. Perhaps if everyone > put their minds to it, someone could invent a more intuitive > notation (or at the very least, not negatively intuitive). Perhaps ? > will work? Or maybe &!? Quite simple, 10000(!2) or 10000(!!) even 10000(!k) for those that need it and 10000(!-1) = 10000*10001*10002*...*20000 10000(!-2) = 10000*10002*...*20000 in otherwords !-k goes on for as long as !^k does. -- Chess notation P -> KB3!? ! - good move !! - excellent move ? - questionable move ?? - folly !? - looks good but look out... ?! - huh? oh I see, who'd ever think such a questionable move could be so good? Again isn't '?!' the correct punctuation for ending a retorical question?! For example: Who'd dare think Bush won't get reelected like he wasn't elected in 2000?! -- Now if you're needing notation for 10000(!!) why not 10000?? No that's not right, why not 10000? as notation for 10000!!? Whoops, still not right, why not 10000? as notation for 10000(!!)? Indeed a simple notation for 10000(!!) is 10000?! Whoops, that's not quite right either, how about...well never mind, If you any doubt, 10000?! < 10000!!! Naw: If you any doubt, 10000?! < 10000!!, so there! Whenceforth have I not clarified the matter about !!?! Whoops: Whenceforth have I not clarified the matter about '!!'?! ;-) ---- === Subject: Re: Fastest factorial algorithm <3FB37A98.1E6E3DCF@uni-kassel.de> (...) > Whoops, that's not quite right either, how about...well never mind, > If you any doubt, 10000?! < 10000!!! Naw: > If you any doubt, 10000?! < 10000!!, so there! > Whenceforth have I not clarified the matter about !!?! Whoops: > Whenceforth have I not clarified the matter about '!!'?! ;-) > ---- Whoa! So we seem to enter that area of esoteric functions, which need an infinite number of steps for definition first, before they start with the first coefficient. Was that family of functions (obviously with order omega+1 (?!) (or (??)) studied anywhere? Gottfried Helms === Subject: Re: Fastest factorial algorithm ... > Usefulness of the double-factorial function aside, the notation > is, in my not-so-humble opinion, an unintuitive abomination that > should have been stamped out before it took hold. It is not the only one. In my opinion, sum{i = 0, oo} comes in the same class. That one is even worse, because it occurs a lot and people get thinking that it is really a summation. On the other hand, the !! notation is not so very frequent (and most uses I have seen of it -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: A potentially rewarding challenge Readers of this newsgroup - especially those who relish an intellectual challenge - may find it particularly rewarding to scrutinise the disgustingly tedious, but extremely elementary, argument that I have just archived: For any given natural number m, the Goodstein sequence, G(m), only terminates if m < 4. If the argument is invalid, and you can spot a flaw, it should be extremely satisfying to locate, even if exasperatingly obvious in retrospect. However, if the argument is, indeed, valid, then you may have been party to a tiny bit of history in the making: you might, prima facie, have peer-reviewed a significant result that owes its existence to, and has been investigated entirely with, the following resources that exist only on the web. Brief background In a 1944 paper, Goodstein (Goodstein, R. L. On the Restricted Ordinal Theorem. J. Symb. Logic 9, 33-41, 1944.) apparently defined a sequence for any given ordinal number m as follows: (i) Set the ordinal number m as the initial member of the sequence; (ii) Express m by its hereditary representation in base 2; (iii) Replace the base by its successor ordinal; (iv) Subtract 1 from the new number to get the next member in the sequence; (v) Express this number by its hereditary representation in the new base; (vi) Repeat (iii)-(vi). Example For instance, the Goodstein sequence G(4) is given by: G(1, 4) = 1*(2^2) + 0*(2^1) + 0*(2^0) (base 2) = 4 G(2, 4) = 1*(3^3) + 0*(3^1) + 0*(3^0) - 1 = 2*(3^2) + 2*(3^1) + 2*(3^0) (base 3) = 26 G(3, 4) = 2*(4^2) + 2*(4^1) + 2*(4^0) -1 = 2*(4^2) + 2*(4^1) + 1*(4^0) (base 4) = 41 G(4, 4) = 2*(5^2) + 2*(5^1) + 1*(5^0) - 1 = 2*(5^2) + 2*(5^1) + 0*(5^0) (base 5) = 60 G(5, 4) = 2*(6^2) + 2*(6^1) + 0*(6^0) - 1 = 2*(6^2) + 1*(6^1) + 5*(6^0) (base 6) = 83 ... G(10, 4) = 2*(11^2) + 1*(11^1) + 1*(11^0) - 1 (base 11) = 2*(11^2) + 1*(11^1) + 0*(11^0) = 253 ... Generalised Goodstein Theorem Apparently, Goodstein then proved in standard set theory, using transfinite induction on set-theoretically defined ordinal numbers, that: Every Goodstein sequence has a finite length. Standard interpretation Since a set-theory such as ZFC is accepted as a relatively consistent extension of standard PA, and the Goodstein sequence is definable over the natural numbers, the standard interpretation of a corollary to Goodstein's argument has been that: For all natural numbers m, there exists a natural number n such that the nth term, G(n, m), of the Goodstein sequence, G(m), is 0. Goodstein sequence theorem In the archived paper I show, however, that: The only natural-number Goodstein sequences that terminate with 0 are G(1), G(2) and G(3). Conclusion I conclude that, if my argument is valid, then Goodstein's reasoning implies that a set-theory such as ZFC is not a relatively consistent extension of standard PA. Bhup === Subject: Re: A potentially rewarding challenge >For any given natural number m, the Goodstein sequence, G(m), only >terminates if m < 4. G(4) terminates, so the argument is invalid. I don't have a formal proof, but just look at the following and it should be obvious (unless I made a mistake somewhere): G_0(4) = 2^2 + 0 The next step will have the {base} of this step as its constant. In this case 2. G_1(4) = 2*3^2 + 2*3 + 2 The constant will take {constant} steps to get to zero. In this case 2. G_3(4) = 2*5^2 + 2*5 + 0 Again, next step will get {base}: G_4(4) = 2*6^2 + 6 + 5 and so on: G_9(4) = 2*11^2 + 11 + 0 G_10(4) = 2*12^2 + 11 G_21(4) = 2*23^2 + 0 G_22(4) = 24^2 + 23*24 + 23 G_45(4) = 47^2 + 23*47 + 0 G_46(4) = 48^2 + 22*48 + 47 G_93(4) = 95^2 + 22*95 + 0 G_94(4) = 96^2 + 21*96 + 95 continuing this you will get to: G_402653181(4) = 402653183^2 + 0 G_402653182(4) = 402653183*402653184 + 402653183 G_805306363(4) = 402653183*805306365 + 0 As we've already seen with the similar 'format' 23*47+0 at G_45(4), this will become zero. === Subject: Re: A potentially rewarding challenge > G_805306363(4) = 402653183*805306365 + 0 > As we've already seen with the similar 'format' 23*47+0 at G_45(4), > this will become zero. === Subject: Re: A potentially rewarding challenge >>For any given natural number m, the Goodstein sequence, G(m), only >>terminates if m < 4. > G(4) terminates, so the argument is invalid. I don't have a formal > proof, but just look at the following and it should be obvious (unless > I made a mistake somewhere): > G_0(4) = 2^2 + 0 > The next step will have the {base} of this step as its constant. In > this case 2. > G_1(4) = 2*3^2 + 2*3 + 2 > The constant will take {constant} steps to get to zero. In this case > 2. > G_3(4) = 2*5^2 + 2*5 + 0 > Again, next step will get {base}: > G_4(4) = 2*6^2 + 6 + 5 > and so on: > G_9(4) = 2*11^2 + 11 + 0 > G_10(4) = 2*12^2 + 11 > G_21(4) = 2*23^2 + 0 > G_22(4) = 24^2 + 23*24 + 23 > G_45(4) = 47^2 + 23*47 + 0 > G_46(4) = 48^2 + 22*48 + 47 > G_93(4) = 95^2 + 22*95 + 0 > G_94(4) = 96^2 + 21*96 + 95 > continuing this you will get to: > G_402653181(4) = 402653183^2 + 0 > G_402653182(4) = 402653183*402653184 + 402653183 > G_805306363(4) = 402653183*805306365 + 0 > As we've already seen with the similar 'format' 23*47+0 at G_45(4), > this will become zero. Bravo! -- 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: Newsgroup survey: Math and personality assessment > >> > In another post by huffy, you signed John (and clearly spoke as > John Correy). >> >> Presumably it wants to avoid killfiling. I have no idea why it >> thinks that the opinion of this late novelist, heroin addict >> and wife-murderer is at all relevant though. These huffy posts >> are considerably more tedious than the rantings of D S Kabatoff. > > Posting the exact same quote 47 times is an odd way to avoid > killfiling. > Well I have now killfiled Correy under its new alias. Choosing > a new username is a way of forcing one's postings on people > who have chosen to killfile one. Kabatoff does this too. > Perhaps I should jus killfile all posts from yahoo.com > as this ISP seems to attact personalities fitting their name. There once was a Reader named Chapman With little savvy or acumen. Lacking the moxy To vet orthodoxy Chapman pegged progress a pathogen. === Subject: Re: Newsgroup survey: Math and personality assessment >> > In another post by huffy, you signed John (and clearly spoke as > John Correy). >> >> Presumably it wants to avoid killfiling. I have no idea why it >> thinks that the opinion of this late novelist, heroin addict >> and wife-murderer is at all relevant though. These huffy posts >> are considerably more tedious than the rantings of D S Kabatoff. >Posting the exact same quote 47 times is an odd way to avoid >killfiling. > True. And your point is what, exactly? I mean it's not like this > is the only thing we've seen him take an odd approach to... >-- >Jesse Hughes >Basically there are two angry groups. I am a harsh force of >one. Against me is a society of mathematicians. So far it's been a >draw. -- JSH gives another display of keen insight. > David C. Ullrich You still haven't owned up to your inability to deduce anything at all from familiar premises. Concerning C1-C4, C1 AxAy[x=y -> Az(z in x <-> z in y)] C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak Extensionality) When you said set theory I assumed you meant ZF. That's what set theory with no qualification means these days. If you meant NGB set theory then no, C1-C4 are not inconsistent with set theory. It does _not_ follow that C1-C4 give an example of something which is not equal to itself, or an example of something which does not exist. [DON'T SNIP--J.C.] It is correct that I have no idea why Ex~(x=x) follows from C1-C4. This is because (assuming that NBG is consistent) NBG has a model in FOL=. In that model everything is equal to itself.[DON'T SNIP--JC] >You have no idea why Ex~(x=x) follows from C1-C4 >because you are brain-dead analysis teacher who >can only work with the routines he has memorized. Could be. Now show us why Ex~(x=x) _does_ follow from C1-C4. Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out the error. ################################################# Homework for David Ullrich and Robin Chapman Here is one theorem on MKC set theories: Ex~(x=x) Prove it to my satisfaction within a month and I'll submit another. ################################################# === Subject: Re: Newsgroup survey: Math and personality assessment > > Well, well, well! If it ain't (Suk My) Dik! What's up, dude? > ... > My, oh my. You make a joke on my name, and in response I make one on your The relevance is Jesse Hughis inability to distinguish between word play and an insult--a distinction you were (apparently) able to make. --John === Subject: Re: Newsgroup survey: Math and personality assessment Discussion, linux) > The relevance is Jesse Hughis inability to distinguish between word > play and an insult--a distinction you were (apparently) able to make. You mean that when you called him (Suk My) Dik, you did not intend to insult him? I'll be hornswaggled. You're right. I am not subtle enough for that distinction. John Correy is just a lying bastarrd (note the word play). -- But he himself was not to blame for his vices. They grew out of a personal defect in his mother. She did her best in the way of flogging him while an infant... but, poor woman! she had the misfortune to be left-handed, and a child flogged left-handedly had better be left unflogged. -- E.A. Poe === Subject: Re: Newsgroup survey: Math and personality assessment > The relevance is Jesse Hughis inability to distinguish between word > play and an insult--a distinction you were (apparently) able to make. > You mean that when you called him (Suk My) Dik, you did not intend > to insult him? > I'll be hornswaggled. You're right. I am not subtle enough for that > distinction. > John Correy is just a lying bastarrd (note the word play). Have you ever considered a depilatory? --John Kiss my fat, hairy ass, you stupid pig. === Subject: Re: Newsgroup survey: Math and personality assessment > I'll be hornswaggled. Before or after you're bum-fuzzled? === Subject: Re: Newsgroup survey: Math and personality assessment <87ad6yf6jk.fsf@phiwumbda.org> Discussion, linux) >> I'll be hornswaggled. > Before or after you're bum-fuzzled? Is there any particular reason you're fascinated by my use of bumfuzzled? -- Jesse Hughes There's a thrill that's gone that I'll probably not have in quite the same way again. After all, FLT was a unique animal, and we had a great dance. -J.S. Harris on proving Fermat's last theorem === Subject: Re: Vedic Mathematics --- Myth and Reality > Equivalently, M*N is the same as M*N mod (M + N - 1). > Sorry, this should be multiplication of M digits with N digits, base b, > is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits. Oh well, so FFT or not, looks like multiplying M by N by any method means MN multiplications! Well, when these multiplications are hardwired (as in human memory for single digits) the computational issues (On*n) becomes really irrelevant, for they all are done in no time at. Like, the video extraction for radar data processing is done by NAND gates - its all done in real time! Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality >> Equivalently, M*N is the same as M*N mod (M + N - 1). >> Sorry, this should be multiplication of M digits with N digits, base b, >> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits. >Oh well, so FFT or not, looks like multiplying M by N by any method means >MN multiplications! Well, when these multiplications are hardwired (as in >human memory for single digits) the computational issues (On*n) becomes >really irrelevant, for they all are done in no time at. Like, the video >extraction for radar data processing is done by NAND gates - its all done in >real time! You are in error. The number of multiplications required for multiplying two numbers with the FFT method is O(n*log(n) where n is the larger of the two numbers; it is not m*n. Richard Harter, cri@tiac.net http://home.tiac.net/~cri, http://www.varinoma.com We have people from every planet on the earth in this State. -- California Governor Gray Davis === Subject: Re: Projection of an angle? I realized after posting my first post that an angle can be rotated or projected along a multitude of axis. So I got to be more specific. The problem I'm trying to solve is like this: I have a triangle with one orthogonal - 90 deg. angle. I want to rotate the triangle along the leg _oposite_ the angle v1 in question. This leg is then (naturally) one of the legs of the 90 deg. angle. It is to be rotated v2 degrees. That is to say that the original triangle is rotated v2 degrees (through the leg mentioned) and projected in a plane parallel to the original plane of the triangle. What is then the new value of the projected angle (v1-proj)? (I can see that v1 =< v1-proj =< 90 deg.) BjÀrn SÀrheim >What is the expression of the angle v1-proj one would get if >one projects an angle v1 under a new angle v2 to a new plane? >Should be quite a simple expression, but I don't seem able to >figure it out in a hurry. >BjÀrn SÀrheim -------------------------------------------------------- -------------------------------------------------------- === Subject: Re: Projection of an angle? > I realized after posting my first post that an angle > can be rotated or projected along a multitude of axis. > So I got to be more specific. > The problem I'm trying to solve is like this: > I have a triangle with one orthogonal - 90 deg. angle. > I want to rotate the triangle along the leg _oposite_ the > angle v1 in question. This leg is then (naturally) one of the > legs of the 90 deg. angle. It is to be rotated v2 degrees. > That is to say that the original triangle is rotated v2 degrees > (through the leg mentioned) and projected in a plane parallel to the > original plane of the triangle. > What is then the new value of the projected angle (v1-proj)? > (I can see that v1 =< v1-proj =< 90 deg.) > BjÀrn SÀrheim >What is the expression of the angle v1-proj one would get if >one projects an angle v1 under a new angle v2 to a new plane? >Should be quite a simple expression, but I don't seem able to >figure it out in a hurry. You should never do things in a hurry if it can possibly be avoided. I think the situation may be described as follows: ABC is a right-angled triangle in a plane P. The angle at B is a right angle. The angle at A is v1. We rotate the triangle, about its side BC, through an angle v2. We project the rotated triangle onto the plane P giving us a new triangle A'BC. The line from A' to the rotated position of A is perpendicular to P. Let x be the angle BA'C. This is the angle in which you are interested. Let a be the distance AB. Let a' be the distance A'B. Let c be the distance BC. We have: tan(v1) = c/a Also: tan(x) = c/a' Also: cos(v2) = a'/a = tan(v1)/tan(x) So: tan(x) = tan(v1)/cos(v2) -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Projection of an angle? I guess you solve it under the assumption that the projection is parallel, but I guess in my real world problem a perspective projection would fit even better, but to good approximation this will do the trick. You gave me also the hints to solve such a projection myself, I would guess. As to the comment about hurry or not. Depends on how many problems and how much time you have to solve them I guess... Also on how fresh is your math knowledge, or not.. :-) BjÀrn SÀrheim >> I realized after posting my first post that an angle >> can be rotated or projected along a multitude of axis. >> So I got to be more specific. >You should never do things in a hurry if it can possibly be avoided. >I think the situation may be described as follows: >ABC is a right-angled triangle in a plane P. >The angle at B is a right angle. >The angle at A is v1. >We rotate the triangle, about its side BC, through an angle v2. >We project the rotated triangle onto the plane P giving us a new triangle >A'BC. The line from A' to the rotated position of A is perpendicular to P. >Let x be the angle BA'C. This is the angle in which you are interested. >Let a be the distance AB. >Let a' be the distance A'B. >Let c be the distance BC. >We have: >tan(v1) = c/a >Also: >tan(x) = c/a' >Also: >cos(v2) = a'/a = tan(v1)/tan(x) >So: >tan(x) = tan(v1)/cos(v2) >-- >Clive Tooth >http://www.clivetooth.dk -------------------------------------------------------- -------------------------------------------------------- === Subject: Re: Projection of an angle? > What is the expression of the angle v-proj one would get if > one projects an angle v1 under a new angle v2 to a new plane? > Should be quite a simple expression, but I don't seem able to > figure it out in a hurry. If rotation is made around one side of the angle v1, it will be less. Tan[v1proj]= Tan[v1] Cos[v2] === Subject: I cannot determine this Hasse-diagram. Pls help can anyone help me to determine the Hasse diagram for the partial order X on set T? T = {1, 2, 3, 4} and X = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 4), (3, 1)} SC === Subject: Re: I cannot determine this Hasse-diagram. Pls help > can anyone help me to determine the Hasse diagram for the partial order X on > set T? > T = {1, 2, 3, 4} and X = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 4), (3, 1)} What's the hassle about Hasse diagrams? What are they? Is this it 4 1 | | 2 3 === Subject: Re: I cannot determine this Hasse-diagram. Pls help This was also my solution, but it seems so strange, i dont know if its complete Can it be called a Hasse diagram, if those two lines are not connected? SC > can anyone help me to determine the Hasse diagram for the partial order X on > set T? > T = {1, 2, 3, 4} and X = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 4), (3, 1)} > What's the hassle about Hasse diagrams? What are they? Is this it > 4 1 > | | > 2 3 === Subject: Re: I cannot determine this Hasse-diagram. Pls help > This was also my solution, but it seems so strange, i dont know if its > complete What more needs be done? Something to note that (1,1), (2,2) etc. are included? > Can it be called a Hasse diagram, if those two lines are not connected? Certainly. > can anyone help me to determine the Hasse diagram for the partial order > X on set T? > > T = {1, 2, 3, 4} and X = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 4), (3, > 1)} > > What's the hassle about Hasse diagrams? What are they? Is this it > 4 1 > | | > 2 3 === Subject: Re: JSH: My use of my initials >> I see, thx. I wasn't clear on who was firing at who :) But JSH does >> sometimes threaten people with litigation, and he had somebody in court one >> time, for calling him a crank. The case was quickly tossed. >I don't think there was any such incident. > He has threatened on occasion; but I think Mr. Hammick may be getting > confused with the case where Underwood Dudley ->was<- sued for calling > someone a crank. Dilworth v. Dudley: > http://www.law.emory.edu/7circuit/jan96/95-2282.html > The suit was dismissed. It wasn't just one suit. The first, suing me, my publisher, and the president of my school for, among other things, conspiracy to deny Mr. Dilworth's civil rights (mathematicians wouldn't talk to him because I called him a crank, he said), was in federal court in Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed to the Seventh Circuit Court of Appeals where it was again dismissed. I was pleased that the decision was written by Judge Posner, well-known for his many books, who had clearly read some of my _Mathematical Cranks_. By the way, my respect for the legal profession went up as a result of this process. Mr. Dilworth represented himself, I assume because no lawyer would take his case. The lawyers for the Mathematical Association of America also did a lot of work, finding all sorts of precedents where a person was called a scab, a traitor, and other nasty things in print and the courts let the authors get away with it. This wasn't enough for Mr. Dilworth, who started the legal process all over again by suing in Wisconsin _state_ court. Once again the case was dismissed, but this time with the requirement that Mr. Dilworth pay $7000 or so for his opponents' legal fees. That finally stopped him, though he continued to send me letters. Earlier this year he died, so now we can all call Mr. Dilworth whatever we want. However, the Law of Conservation of Cranks still operates, and his kind has not died out. Presumably, the kind of crank who sues has not died out either, so everyone be careful. Woody Dudley === Subject: Re: JSH: My use of my initials Adjunct Assistant Professor at the University of Montana. >> > I see, thx. I wasn't clear on who was firing at who :) But JSH does > sometimes threaten people with litigation, and he had somebody in court one > time, for calling him a crank. The case was quickly tossed. >> >>I don't think there was any such incident. >> He has threatened on occasion; but I think Mr. Hammick may be getting >> confused with the case where Underwood Dudley ->was<- sued for calling >> someone a crank. Dilworth v. Dudley: >> http://www.law.emory.edu/7circuit/jan96/95-2282.html >> The suit was dismissed. > It wasn't just one suit. The first, suing me, my publisher, and >the president of my school for, among other things, conspiracy to deny >Mr. Dilworth's civil rights (mathematicians wouldn't talk to him >because I called him a crank, he said), was in federal court in >Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed >to the Seventh Circuit Court of Appeals where it was again dismissed. >I was pleased that the decision was written by Judge Posner, >well-known for his many books, who had clearly read some of my >_Mathematical Cranks_. to the 7th Circuit. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: My use of my initials > >> I see, thx. I wasn't clear on who was firing at who :) But JSH does >> sometimes threaten people with litigation, and he had somebody in court one >> time, for calling him a crank. The case was quickly tossed. > >I don't think there was any such incident. > > He has threatened on occasion; but I think Mr. Hammick may be getting > confused with the case where Underwood Dudley ->was<- sued for calling > someone a crank. Dilworth v. Dudley: > > http://www.law.emory.edu/7circuit/jan96/95-2282.html > > The suit was dismissed. > > It wasn't just one suit. The first, suing me, my publisher, and > the president of my school for, among other things, conspiracy to deny > Mr. Dilworth's civil rights (mathematicians wouldn't talk to him > because I called him a crank, he said), was in federal court in > Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed > to the Seventh Circuit Court of Appeals where it was again dismissed. > I was pleased that the decision was written by Judge Posner, > well-known for his many books, who had clearly read some of my > _Mathematical Cranks_. Well I don't know the particulars, but I do know that societies have often branded certain people outside of the mainstream with negative labels. What you cannot deny Mr. Dudley is that you bank on negatives. That is, your book is a way for you to make money from the misfortunes of others. > By the way, my respect for the legal profession went up as a > result of this process. Mr. Dilworth represented himself, I assume > because no lawyer would take his case. The lawyers for the > Mathematical Association of America also did a lot of work, finding > all sorts of precedents where a person was called a scab, a traitor, > and other nasty things in print and the courts let the authors get > away with it. > This wasn't enough for Mr. Dilworth, who started the legal > process all over again by suing in Wisconsin _state_ court. Once > again the case was dismissed, but this time with the requirement that > Mr. Dilworth pay $7000 or so for his opponents' legal fees. That > finally stopped him, though he continued to send me letters. > Earlier this year he died, so now we can all call Mr. Dilworth > whatever we want. If his work ever turns out to have legitimacy then his heirs, assuming he has any, can sue you as well Mr. Dudley. > However, the Law of Conservation of Cranks still operates, and > his kind has not died out. Presumably, the kind of crank who sues has > not died out either, so everyone be careful. > Woody Dudley Well I know the futility of suing people while the question of the legitimacy of *my* work is in the air, so I'll just wait until it's accepted, then I may take down entire universities. And Mr. Dudley, if you feel brave you can put me in your books. I dare you. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Implementing the partial difference integration I've talked about my find of a partial difference equation that you can integrate to get a count of prime numbers, and given some quick instructions, but I saw one post that leads me to consider that there are differences between how programmers tend to think, and how others do, so here's some additional help on the instructions. Here's the difference equation and the instructions for the integration: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0. Note that it's a *discrete* function, so for you programmers that means you need to use int's or long's or some discrete variable type. Like if you're using Java or C++, you'll need to cast the sqrt() function, though I saved the cast for what follows in a C++ program I posted: And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS from dS(x,2) to dS(x,y). That is, you can get away with just casting x, to something discrete, while using non-discrete x and y, but it's just a waste of storage space. For programmers you get S as the sum of dS from dS(x,2) to dS(x,y) means you sum up to and *including* y, as I don't doubt some of you may write something like for (i=2; i I've talked about my find of a partial difference equation that you > can integrate to get a count of prime numbers, and given some quick > instructions, but I saw one post that leads me to consider that there > are differences between how programmers tend to think, and how others > do, so here's some additional help on the instructions. > Here's the difference equation and the instructions for the > integration: > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], > S(x,1) = 0. > Note that it's a *discrete* function, so for you programmers that > means you need to use int's or long's or some discrete variable type. > Like if you're using Java or C++, you'll need to cast the sqrt() > function, though I saved the cast for what follows in a C++ program I > posted: > And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,2) to dS(x,y). > That is, you can get away with just casting x, to something discrete, > while using non-discrete x and y, but it's just a waste of storage > space. > For programmers you get S as the sum of dS from dS(x,2) to dS(x,y) > means you sum up to and *including* y, as I don't doubt some of you > may write something like > for (i=2; i which is WRONG, and if you do that you will probably get a wrong > answer. > Note: p(x,sqrt(x)) here gives the same value as the traditional pi(x). > For faster calculations you need to use > dS(x,y) = [ p(x/y, sqrt(x/y)) - p(y-1, sqrt(y-1)][ p(y, sqrt(y)) - > p(y-1, sqrt(y-1))] > when sqrt(x/y) < y-1. > That's a BIG deal, as the pure math implementation is VERY SLOW, and > even that quick speed-up won't push you very far. > However, the point is that the difference equation integration does > work, which those of you who can program can verify for yourselves. > Then you should search to see if ANYONE has ever used a partial > difference equation integration to get a count of prime numbers > because then you can see that yes, I'm the only one in recorded > history to present this method. > The above instructions are easy enough to program into a computer and > if you follow them, you'll notice that you do get a correct count for > prime numbers, and you should also notice that unless y is prime > dS(x,y) = 0, so yes, it'll also tell you *which* numbers are prime. > The research is a shining example of the importance of *independent* > researchers willing to check in areas that more staid academics think > are fallow. > The independent researcher is important for the future. > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ How are you solving Partial Differential Equations if you attacked an example I gave you a few weeks ago that was basic algebra? James, We subtract and add the same number all the time in mathematics. Say I want to write the quadratic y=x^2+6x+8 into the form y=a(x-h)^2+k. Looking at the first two terms, I see I can make a perfect square if I add 9. What about the 8? Since I add 9 to x^2+6x, I must subtract it from the 8. y=x^2+6x+9+8-9=(x+3)^2-1 This is used all the time especially when doing trig substitutions with integrals. David Moran You're still trying to with me David Moran because NOTHING in what Arturo Magidin is doing makes any sense MATHEMATICALLY!!! I don't know what's ing wrong with you people, or why you think that you can come here and bull for eternity without consequence, but I'm ing tired of the stupidity. I'm ing tired of the nonsense. I'm ing tired of people coming in here and saying bull as if mathematics is just some useless game and the truth doesn't matter. You need to quit ing around David Moran. You, Randy Poe, and Arturo Magidin among others needs to get a ING CLUE!!! James Harris === Subject: Re: Implementing the partial difference integration > I've talked about my find of a partial difference equation that you > can integrate to get a count of prime numbers, Great leaping jehosophats! Harris has a second anus. 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: SCIENCA.COM Commences Operation of Its OROA Archive >SCIENCA will also archive scientific work products which have been >ignored, oppressed or rejected by the traditional publishing media, >since SCIENCA believes that through the process of OROA good >scientific ideas/results will survive but bad ones will simply die >similar to what happens under Darwin's theory of evolution. How many self-respecting scientists will want to publish alongside the How many scientists have the time to thoroughly read the respected journals, let alone venturing off into the wacky world of oddball ideas promulgated by people ignorant of the most basic tenets of science? Steve Turner === Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive > To address the needs of many creative individuals in science, > medicine and technology who want open-review and open-access (OROA) > for their scientific work products, Science Archive Institute, > A division of QuantumDream, Inc., The bull smell is very clear. > has commenced operation of its > flagship science archive/portal, SCIENCA.COM. > To ensure every road leads to Rome, SCIENCA can be reached through > a dozen related domain names such as scienca.com/org, > sciencearchive.org, scienceportal.org, physicsarchive.com/org, > mathematicsarchive.com/org, biologyarchive.com/org, > chemistryarchive.com/org, medicinearchive.com/org and > technologyarchive.com/org. Echos of a multi-level marketing seminar. > SCIENCA will also archive scientific work products which have been > ignored, oppressed or rejected by the traditional publishing media, > since SCIENCA believes that through the process of OROA good > scientific ideas/results will survive but bad ones will simply die > similar to what happens under Darwin's theory of evolution. > Since nothing in this world comes free [even discussion group > messeages are stuffed with Ads], the price for OROA is that a > submitter is required to pay a fee for submitting his/her paper, web > link or download link. The exploratory work on the feasibility of > such an institution has been conducted by Huping Hu, Ph.D., J.D. Vanity press. Quid pro schmo. > SCIENCA shows great respects to all scientifically creative > individuals. SCIENCA is an archive/portal with equal access and equal > review rights where scientists and other learned individuals archive > their papers and express their views on equal footing. In doing so, > SCIENCA hopes that it will be able to speed up the progress of > science and let a hundred flowers blossom instead of a few. > (End of Release) > Your comments and suggestions by personal e-mails to me are welcomed. > Huping Hu, Ph.D., J.D. So we surf http://www.scieca.com/ and suddenly we see the need for peer review, Warning: session_start(): Cannot send session cookie - headers already sent by (output started at /home/scienca/public_html/modules/catalog/includes/functions/database.php:30 2) in /home/scienca/public_html/modules/catalog/includes/functions/sessions.php on line 135 Warning: session_start(): Cannot send session cache limiter - headers already sent (output started at /home/scienca/public_html/modules/catalog/includes/functions/database.php:30 2) in /home/scienca/public_html/modules/catalog/includes/functions/sessions.php on line 135 and then... and then... IT'S SARFATTIS ALL THE WAY DOWN! As Dave Bowman in 2001: A Space Odyssey said, Oh my god! It's filled with Sarfattis! We all know how unmannered and flat out psychotic Sarfatti is from his incessant abuse of sci.phsyics. Who ever suspected he was so world-class stupid as to spend his own money braying like an ass? The Emperor's New G-String (Source: Jack HA HA HA! Hey Huping, did you post this volumnous crap before Sarfatti's check cleared? -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive > and then... and then... IT'S SARFATTIS ALL THE WAY DOWN! As Dave > Bowman in 2001: A Space Odyssey said, Oh my god! It's filled with > Sarfattis! No he didn't. He saw the entire place was filled with sailors and he said My God, its full of tars!. Then he left the building and entered the parking garage which was also filled, and he said My God, its full of cars!. So upset was he we went to the candy machine which sold only one kind of confection. He said My God, its full of Mars! If that wasn't bad enough he had to go to the supermarket that evening and ended up in the section all filled with peanut butter and such like in glass containers. He exclaimed-My God, its full of jars!. Bob Kolker === Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive - Updated with new info relayed by HAL from beyond the infinite Robert J. Kolker sagely sneezed: >> and then... and then... IT'S SARFATTIS ALL THE WAY DOWN! As Dave >> Bowman in 2001: A Space Odyssey said, Oh my god! It's filled with >> Sarfattis! > No he didn't. He saw the entire place was filled with sailors and > he said My God, its full of tars!. Then he left the building and > entered the parking garage which was also filled, and he said My > God, its full of cars!. So upset was he we went to the candy > machine which sold only one kind of confection. He said My God, > its full of Mars! If that wasn't bad enough he had to go to the > supermarket that evening and ended up in the section all filled > with peanut butter and such like in glass containers. He exclaimed- > My God, its full of jars!. That wasn't the half of it. He went to Las Vegas to try to get away from it all. What was the first thing he saw? My God, it's full of bars! So he went on a sailing cruise to really get away from it all. But he looked up and cried out, My God, it's full of spars! When they got to the coral lagoon he went snorkling. He was immediately surrounded by a huge school of fish. At first he thought they were sturgeon, but then he realized what they were. He tried to say, My God, it's full of gars!, but aspirated a good-size chunk of seawater instead. He wanted to go back into the water as soon as he recovered, but his doctor at the United States Astronautics Agency wanted to examine his lungs for amoebas. They gave him a (N)MRI scan. When the physician showed him the scan of his lung, he wheezed My God, it's full of scars! He was checked into a hospital for treatment and placed in the pulmonary ward. Then it hit him. My God, it's full of SARS! -- Jeff, in Minneapolis . === Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive Robert J. Kolker sagely hiccuped: >> and then... and then... IT'S SARFATTIS ALL THE WAY DOWN! As Dave >> Bowman in 2001: A Space Odyssey said, Oh my god! It's filled with >> Sarfattis! > No he didn't. He saw the entire place was filled with sailors and > he said My God, its full of tars!. Then he left the building and > entered the parking garage which was also filled, and he said My > God, its full of cars!. So upset was he we went to the candy > machine which sold only one kind of confection. He said My God, > its full of Mars! If that wasn't bad enough he had to go to the > supermarket that evening and ended up in the section all filled > with peanut butter and such like in glass containers. He exclaimed- > My God, its full of jars!. That wasn't the half of it. He went to Las Vegas to try to get away from it all. What was the first thing he saw? My God, it's full of bars! So he went snorkling to really get away from it all. He was immediately surrounded by a huge school of fish. At first he thought they were sturgeon, but then he realized. My God, it's full of gars! he tried to say, but aspirated a good-size chunk of seawater instead. He wanted to go back into the water as soon as he recovered, but his doctor at the United States Astronautics Agency wanted him to check into a hospital for tests of his lungs. He was placed in the pulmonary ward. My God, it's full of SARS! -- Jeff, in Minneapolis . === Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive === >Subject: Re: SCIENCA.COM Commences Operation of Its OROA Archive >Message-id: >> and then... and then... IT'S SARFATTIS ALL THE WAY DOWN! As Dave >> Bowman in 2001: A Space Odyssey said, Oh my god! It's filled with >> Sarfattis! >No he didn't. He saw the entire place was filled with sailors and he >said My God, its full of tars!. Then he left the building and entered >the parking garage which was also filled, and he said My God, its full >of cars!. So upset was he we went to the candy machine which sold only >one kind of confection. He said My God, its full of Mars! If that >wasn't bad enough he had to go to the supermarket that evening and ended >up in the section all filled with peanut butter and such like in glass >containers. He exclaimed-My God, its full of jars!. Badly shaken, he turned onto Rush Street and exclaimed-My God, it's full of bars!. So instead, he pulls out his web phone to catch up on the usenet postings only to see sci.math again flooded by Kabatoff postings-My God, it's full of Dars!. Taking refuge in one of the nightspots, he runs into Tiger Woods who's grumbling about what a bad game he had. Looking at his scorecard, he shouts-My God, it's full of pars!. I was going to do another one about Bush's legacy (My God, it's full of wars) but that doesn't seem to work for some reason, so I guess I'll quit here. >Bob Kolker -- Mensanator Ace of Clubs === Subject: Re: Relativity is based on assumption. > Androcles skrev i melding > No way, Henri. I've been debating with him for years, and he's just one > of > those relativists that refuse to see what is right under their nose. > It's a > psychological hang-up that most people have. The don't want to admit > they've > been made a fool of by Einstein, they have to appear in the eyes of > others > as the one who has all the answers, but that's provided the question > suits > them, of course. When it doesn't, they ignore it. > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = > tau(x',0,0,t+x'/(c-v)) > ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ > Where does he get that 1/2 from? > That's what Paul has no answer for, so he snips, pretending the question > wasn't answered. It's called ignorance. > Asking the same questions over and over, not remembering > that they have been answered many times before is called senility. > Since I am not senile, I remember well having answered > your question, so you will have to excuse me that I don't > bother to answer it again every time you not only have forgotten > the answer, but have forgotten that your question ever was answered. > |> > |> > |> So yes, Paul. I do have funny ideas. > |> I think problems through carefully and slowly, and I don't like to > jump to > |> conclusions. Especially other people's conclusions. In particular, > I don't > |> like to be scoffed at by people that know what the answer is, > when they > |> really don't. > |> Which is why I say to you again: > |> where does the half come from in Einstein's > |> 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= > tau(x',0,0,t+x'/(c-v)) ? > |> > |> The definition of simultaneity. > |> That applies to objects at rest. It does NOT apply to objects in > relative > |> motion. > | > | It is the definiton of simultaneity in the moving k system. > | tau is the time co-ordinate of this system. > | > |> To me the meaning of that equation was obvious > |> the first time I saw it. > |> You mean your were subtly persuaded the first time you saw it. > | > | It means that the first time I saw it, I understood that > | the equation was the definition of simultaneity in the moving system. > | Not hard to understand, as it is clearly stated right above. > | Light is emitted from origo of k at tau_0, reflected from the mirror > | at tau_1, and gets back to origo at tau_2. > | .. we must then have (1/2)*(tau_0 + tau_2) = tau_1 > | Einstein did not find it necessary to add which is the definition > | of simultaneity. He probably thought that any reader would understand > | that right away, since this is the synchronization procedure of > | paragraph 1: > | In accordance with definition, the two clocks synchronize if > | tau_1 - tau_0 = tau_2 - tau_1 > | > | > |> I must admit, > |> so was I. It was so obvious, it too me years to figure out what was > wrong > |> with it. I naively accepted it, delving into the tau0, 1 and 2, looking > for > |> the problem there. It isn't there, its in the half. > |> [snip of stupid debate leading nowhere] > | > | So it took you years to not understand that the equation is > | the definition of simultaneity? Strong! :-) > | > | Paul > This was close to four years ago. > So now it has taken you several + four years not to understand > what has been explained to you many times. > Well undone, Androcles. > Paul > Doesn't matter what you say, Paul. > Einstein say it is assumption. > Paul say it is definition. > I say it is stupidity. > They are all assertions, and carry only the weight the recipient will allow. > The crux of the matter is right here: > |> That applies to objects at rest. It does NOT apply to objects in > relative > |> motion. > | > | It is the definition of simultaneity in the moving k system. > | tau is the time co-ordinate of this system. > Doesn't matter how you say it. It is still pulled out of a hat. Well, Androcles. I don't think anybody expected you to understand it. I certainly didn't. But stop accusing me of not answering the questions I have answered many times in great detail. And understand that if you repost the same nonsense again I WILL snip it again. And that is NOT because I can't answer it. Paul === Subject: Re: Relativity is based on assumption. All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what? === Subject: Re: Relativity is based on assumption. > All theories, in physics and elsewhere, are based on assumptions. > No theory can bootstrap itself out of nothing. > So what? 'So what' depends on the validity of the assumption. The PoR is an assumption, and is intuitive as well. Making the assumption that the time it takes for a signal to reach an object is the same as the time it take for the signal to return, when in the meantime you've moved away or toward the object, is a rather silly assumption that I will not accept. That's 'so what'. Androcles === Subject: Re: Relativity is based on assumption. > I WILL snip it again. And that is NOT because I can't > answer it. > Paul I didn't expect you to understand my reply either, Paul. But I *will* accuse you of snipping without answering, which you do repeatedly. And of course it is because you cannot answer, you only think you can. Your responses amount to nothing more than assertion. Einstein said, and I quote, In agreement with experience we further assume the quantity 2AB/(t'A-tA) = c to be a universal constant- the velocity of light in empty space. There is an indesputable word in the that quotation. The word is assume, and you can assume all you want to, I will not. Assume makes an ass out of u and you. Not me. Androcles. === Subject: Re: Relativity is based on assumption. > [nothing worth quoting] > Nothing is assumed in Special Relativity. > The subject line is a lie. Big Bird is an Ostrich... with head buried in the sand. Good name for you, Ostrich. Obviously you are incapable of reason, so you declare that Einstein isn't worth quoting. That's fine by me; I agree. He isn't worth quoting. Androcles. === Subject: Re: Relativity is based on assumption. > [nothing worth quoting] > Nothing is assumed in Special Relativity. > The subject line is a lie. > Big Bird is an Ostrich... with head buried in the sand. Good name for you, > Ostrich. Obviously you are incapable of reason, so you declare that Einstein > isn't worth quoting. That's fine by me; I agree. He isn't worth quoting. > Androcles. I provided proof for my claim. But of course you *conveniently* snipped the link. That proof consists of fairly straightforward reasoning. You merely provide hot air. > so you declare that Einstein > isn't worth quoting. That's fine by me; I agree. He isn't worth quoting. Einstein's equations aren't worth quoting because everybody has seen them a trillion times and knows a dozen locations where to find them. YOUR ignorant inability to follow a single sentence explaining how the time tau1 halfway between two times tau0 and tau2 is computed is most certainly not worth quoting. === Subject: Re: Relativity is based on assumption. > I provided proof for my claim. But of course you *conveniently* > snipped the link. That proof consists of fairly straightforward > reasoning. LOL!! I provided the proof that Einstein makes assumption, AFTER you denied it, and called me a liar. Then YOU snipped my response, windbag. Not only that, but you gave no proof at all. Assertion isn't proof, even if you think it is. The subject line remains true, whether you like it or not, Ostrich. Androcles === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? In sci.math, Herman Rubin : >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. >>B = {0} x A > This need not be disjoint from A. > In fact, if we take a_0 = {0}, a_{n+1} = <0, a_n}, and > let A be the set of all a_n, then only a_0 is not an > element of the B you defined. Erm, your notation is slightly unclear. An x in B would be an ordered pair (0,a), where a is in A. If one hypothesizes a_{n+1} = the ordered pair (0, a_n), which is possible if slightly unusual, then the result would be indeed as you describe. Actually writing out the ordered pairs in A or B might bring out bad memories of Lisp: :-) a_0 = 0 a_1 = (0,0) a_2 = (0,(0,0)) a_3 = (0,(0,(0,0))) a_4 = (0,(0,(0,(0,0)))) a_5 = (0,(0,(0,(0,(0,0))))) etc. Of course one of the reasons this works is because of the construction of B, which uses an element of A. If one hypothesizes a set A, and an element u not in A (is this guaranteed at all?), then one can construct B = {u} x A and obtain the desired result. Of course, if A is the set of everything things get mildly interesting. (Note that u can be the empty set, which can itself be a set element.) I'm assuming ZFC refers to Zermelo-Fr.8ankel + Axiom of Choice, which is what Google coughed up. One formulation thereof is at http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html although I'm not sure I like the formulation of (2) as it contains two free variables. (1) also contains two free variables so perhaps it's merely an issue of wrapping the axioms properly in universal quantifiers bounding a and b, or more precisely defining what 'a' and 'b' are. (3) looks even weirder, although it may be because I'm not familiar with the variant of the quantifier notation used. If I rewrite it as [E]x[A]y(y in x iff [E]z(z in a .87 y in z)) it might make more sense although I'm not sure I've done it correctly. Note the free variable 'a', which is presumably the set which we're trying to sum (x being the constructed result). I can also write this in set-definitional form: S(A) = {y: [E]z(z in A .87 y in z))} It's now clear that A must be a set of sets, and S(A) appears to be the union of all sets in A. That now makes sense but I'd gently suggest to the author that http://mathworld.wolfram.com/Exists.html be rewritten slightly so as to explain the notation [E]x in a(P(x)) properly. Then again, I'm more or less trained on Copi, who does not use the [E]x in a notation but who instead prefers [E]x(x in a .87 P(x)) where P(x) is a predicate. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > 4iKsb.55587$Rah1.48667@twister01.bloor.is.net.cable.rogers.com>... > If the normal rules of modal logic (particularly the rule of > necessitation and the rule of box distribution), are applied to (1), > we indeed get the following result: > > [](ixFx = ixFx) -> []E!ixFx > > But once again, weÇre being confronted with the notorious ambiguity of > modal formulas. > > Read de dicto we have: > > If 'ixFx = ixFx' is necessarily true, then 'E!ixFx' is necessarily > true. > By convention, imo, [](ixFx = ixFx) -> [](E!(ixFx)) is not ambigious. > It is read: If 'ixFx = ixFx' is necessarily true, then 'E!(ixFx)' is > necessarily true. > OK, the standard formulation of the rule of necessitation, > [](p -> q) -> ([]p -> []q) , > indeed seems an unambiguous statement, clearly expressing that itÇs > whole propositions that are within the scope of the box such that [] > functions as a de dicto operator. > But in quantified modal logic things arenÇt always that clear! > An as IÇve already stated in my previous posting, > The rule of necessitation can also be unproblematically > applied to the following NFL-theorem: > Ax(x = x -> E!x) > []Ax(x = x -> E!x) > [](Ax(x = x) -> AxE!x) > []Ax(x = x) -> []AxE!x > Since in NFL both Ax(x = x) and AxE!x are axioms, everythingÇs fine > here! Agreed, and, []Ax(x=x -> E!x) <-> Ax[](x=x -> E!x) and []Ax(x=x) <-> Ax[](x=x). > But what are we to do with the rule of necessitation (RN) and > ixFx = ixFx -> E!ixFx > This sentence is closed and, obviously, an instance of 'Ax(x = x -> > E!x)', since in free description theory definite noun phrases are > treated as genuine singular term; but, in order to be able to avoid > the inacceptable ixFx = ixFx -> E!ixFx is not an instance of Ax(x=x -> E!x), in PM. In Russell's descriptions theory (ixFx) is not a value of the individual variable unless E!(ixFx). AxGx -> G(ixFx) is not valid. AxGx & E!(ixFx) .-> G(ixFx) is valid. ixFx = ixFx <-> Ey(Ax(x=y <-> Fx) & y=y) ixFx = ixFx <-> EyAx(x=y <-> Fx), since y=y follows from the axiom Ax(x=x). > [](ixFx = ixFx) -> []E!ixFx , > one ought not to be allowed to apply RN to formulas of free logic that > are not *universally* closed! > That is, sentences containing free variables or not being universally > closed must not be 'subjected' to RN. > This corresponds to the strategy Kripke pursued in his revision of S5. > Accordingly, if NFL turns modal, its axiom NA5 needs to be formulated > as follows: > (NA5m) For every elementary formula 'Phi' of FL in which the variable > 'x' is free: > Ax(Phi(x) -> E!x) > Now, if RN is applied to NA5m, we get the following: > (0) []Ax(Phi(x) -> E!x) > [](AxPhi(x) -> AxE!x) > []AxPhi(x) -> []AxE!x > Here, Phi(x) may stand for one of the three possible elementary > formulas: > (a) []AxFx -> []AxE!x > (b) []Ax(x = x) -> []AxE!x > (c) []AxE!x -> []AxE!x > i.e. > (1) []Ax(Fx -> E!x) -> ([]AxFx -> []AxE!x) > (2) []Ax(x = x -> E!x) -> ([]Ax(x = x) -> []AxE!x) > (3) []Ax(E!x -> E!x) -> ([]AxE!x -> []AxE!x) > (2)+(3) are obviously true, but what about (1)? > The antecedent is true, but what about its consequent? > If, in every possible world, property instantiation implies existence, > then, if the property F is instantiated by everything, i.e. by every > existent in every possible world, it is necessary that everything, > i.e. every existent exists in every possible world. > In other words: universal property instantiation necessitates > universal existence! > Thus, (1) is true too. > Of ocurse, if there were a possible world, in which nonexistents > (belonging e.g. to MeinongÇs zoo) can instantiate properties, then (1) > would be false. I don't see a problem with (1) []Ax(Fx -> E!x) -> ([]AxFx -> []AxE!x) > I hope IÇve managed to convince you that the only thing that can be > deduced in a suitable modal version of negative free logic is []AxE!x > but not Ax[]E!x. I don't agree here either. Given Ax[]Fx <-> []AxFx is valid. Surely []AxE!x <-> Ax[]E!x, is a theorem. > Under the right interpretation, there is no deductive way from > necessary self-identity to necessary existence! What exactly is 'the right interpretation'? > (ixFx) []= (ixFx) -> ([]E!)(ixFx), is also not ambiguious, it is read > If 'ixFx is necessarily equal to ixFx' then 'ixFx is necessarily existent'. > Well yes, that already is a disambiguated formulation. > The 'de dicto' and 'de re' distinctions are needed and made specific for > expressions > containing described objects, independent of modal predications. > D1. f(G(ixFx)) defined f(Ey(Ax(x=y <-> Fx) & Gy)) {de dicto} > D2. (fG)(ixFx) defined Ey(Ax(x=y <-> Fx) & f(Gy)) {de re} > This distinction is required even for non-modal truth functions. > For example: ~(G(ixFx)) <-> ~(Ey(Ax(x=y <-> Fx) & Gy)), {de dicto} > and (~G)(ixFx) <-> Ey(Ax(x=y <-> Fx) & ~(Gy)), {de re}. > E!x -> {f(Gx) <-> (fG)x} > E!(ixFx) -> {f(G(ixFx) <-> (fG)(ixFx)} > With regard to the scope of negation in free logic, we certainly need > to differentiate between e.g. > (a) ~(E!Vulcan) > It is not the case that Vulcan exists. > and > (b) (~E!)Vulcan > Vulcan does not exist., > the first sentence of which is true and the second one false. Yes, the proper name Vulcan represents its description and does not represent an actual planet. In general G(ixFx) <-> Ey(Ax(x=y <-> Fx) & Gy) makes sense only if G is a primary predicate. i.e. (G)ixFx means Ey(Ax(x=y <-> Fx) & Gy). Existence is defined using this notion of 'primary' predicates. E!x means EG{(G)x} E!(ixFx) means EG{(G)ixFx} Something named or described exists iff it 'has' at least one primary predicate that is true of it! > And, if one wants, one may consider (a) a de dicto sentence, by means > of which falsity is predicated of a statement, i.e. of something said, > and (b) a de re statement, by means of which some negative property is > ascribed to a thing. > In classical logic the following equivalence holds, because Ex(x = a) > is a theorem in it: > ~(Fa) <-> (~F)a > So in classical logic there is no need to use any brackets and, hence, > one can always simply write > ~Fa . Yes, I agree. E!x ->. ~(Gx) <-> (~G)x. And, E!(ixFx) ->. ~(G(ixFx)) <-> (~G)(ixFx). > The point is that in negative free logic ixFx = ixFx is not a > necessary truth, since it is possibly false (in case ~E!ixFx). There > is a possible world in which ixFx = ixFx is false because ixFx does > not exist in that world; and if there is a possible world in which > ixFx doesnÇt exist, E!ixFx cannot be a necessary truth either, since > necessary truth is defined as truth in all worlds! > > Even though (ixFx = ixFx) -> E!ixFx is a necessary NFL-truth, both > ixFx = ixFx and E!ixFx are no necessary NFL-truths! Agreed. > Certainly, eg. (the present king of France)=(the present king of France) is > false and, > E!(the present king of France) is also false. > It seems to me that this version of NFL is simply description logic, > contained within classical logic. > Although NFL is close to the spirit of RussellÇs description theory, > there is a significant basic difference, for in free logic definite > descriptions are not treated as 'incomplete symbols' but as genuine > singular terms, i.e. Frege-fashion! That is very puzzling to me. Can you expand on this point? > What if we interpreted (1) *de re*? > > In NFL > > E!x <-> Ey(y = x) > > holds (by definition), and so we can write > > (1*) ixFx = ixFx -> Ey(y = ixFx) > (1*) ixFx = ixFx -> Ey(y = ixFx), is valid in description logic but, > it is not the case that (ixFx) is a value of the individual variable unless > E!(ixFx). i.e. it is not an instance of x=x -> Ey(y=x). > I do think it is. If AxE!x is true, as you say, and Vulcan is a value of the variable x, then, E!(Vulcan) follows. But, you also claim that ~(E!(Vulcan) ?? > ixFx = ixFx is certainly false if ~E!ixFx . Yes. > And > ~E!ixFx <-> ~Ey(Ax(Fx -> y = x) & Fy) Yes, because, E!(ixFx) <-> F(ixFx). E!(ixFx) <-> Ey(Ax(x=y <-> Fx) & E!y). E!(ixFx) <-> EyAx(x=y <-> Fx), since AxE!x is true. ~(E!(ixFx)) <-> ~Ey(Ax(x=y <-> Fx) & E!y) ~(E!(ixFx)) <-> ~EyAx(x=y <-> Fx), since AxE!x is true. Note: (~E!)(ixFx) <-> Ey(Ax(x=y <-> Fx) & (~E!)y), i.e. ~{(~E!)(ixFx) is a theorem, since (~E!)y <-> ~(E!y) and ~(E!y) is contradictory. That is to say, ~(~E!)x and ~(~E!)(ixFx) are both true. No object, given or described, 'has' the property of non-existence. > ... and > > [](ixFx = ixFx) -> Ey[](y = ixFx) > I don't think so. > [](Ey(y = ixFx)) -> Ey([](y = ixFx)), is invalid. > [](E!(ixFx)) -> Ey([](y = ixFx)), is invalid. > [](E![ixFx]) -> Ey([](y = ixFx)), is valid. > [ixFx] defined (iy: [](Ax(x=y <-> Fx))). > ixFx is an empty name iff it refers either to nothing or to more > than one object. Yes, E!(ixFx) <->. ExFx & AxAy(Fx & Fy .-> x=y). > I am convinced that self-identity is an essential property of > everything, Only every 'existent' thing has the property of self identity. > Ax(x = x) > but also > Ax[](x = x) > is true. Yes, if Ax(x=x) is true then [](Ax(x=x)) is true and, because Ax[]Fx <-> []AxFx, Ax([](x=x)) is true. > By applying free UI to it, weÇre having > Ax[](x = x) -> (E!t -> [](t = t)) > By modus ponens we get > E!t -> [](t = t) AxGx & E!(ixFx) .-> G(ixFx), is valid. > and the corresponding instance > E!ixFx -> [](ixFx = ixFx) . > Now, I think the only acceptable EG is the following one: > [](ixFx = ixFx) -> Ey[](y = ixFx) I don't agree here at all. E!(ixFx) -> Ey[](y = ixFx), is invalid. []E!(ixFx) -> Ey[](y = ixFx), is invalid. []E!(ixFx) -> []Ey(y = ixFx) is valid > By transitivity we get > E!ixFx -> Ey[](y = ixFx) , > but fortunately not > E!ixFx -> []Ey(y = ixFx) . ?? But, Ey[](y = ixFx) -> []Ey(y = ixFx), by Ey[]Gy -> []EyGy. > In NFL, as opposed to PFL, where a = a holds, > neither > ixFx = ixFx > nor > Ey[](y = ixFx) > alone are logically true instances of NFL axioms or theorems, since > both may be false! > I still deem this view fully acceptable! > By the way, the harmonization of free logic and modal logic is still a > work in progress. > Free logic is spreading into different related fields. On the one > hand it looks for closer connection to modal logic. 26 > PH > P.S.: > A final remark on the necessity of self-identity: > As you know, John affirmed the necessity of > a = a , > but denied the necessity of > ixFx = ixFx . > If one adopts a world-relative semantics (Kripke models), as I do, > then ixFx might well be regarded as a non-rigid designator that > doesnÇt single out one and the same individual in every possible > world. > (A) []Ax(E!ixFx) -> [](ixFx = ixFx)) Ax(Ey(Ax(x=y <-> Fx))) <-> Ey(Ax(x=y <-> Fx)), by Ax(p) <-> p. []Ax(E!ixFx) <-> []E!(ixFx), If you mean []Ax(E!x) -> [](ixFx = ixFx), then I disagree. > The universal quantifier doesnÇt range over one overall domain that > stays the very same for every possible world, but each world has its > own domain. > So (A) means that I assume you mean, (A) [](E!(ixFx)) -> [](ixFx = ixFx). Witt > For each world w, if ixFx exists there, then it is necessarily > self-identical in w. > That is, even in case the same name ixFx refers to a in world 1 > (such as the inventor of bifocals refers to Benjamin Franklin in the > actual world) and to b in world 2, both a and b are necessarily > self-identical! > And in worlds where it is not the case that ixFx exists, it isnÇt the > case that ixFx is necessarily (=essentially) self-identical either. === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > By the way, the harmonization of free logic and modal logic is still a > work in progress. > Free logic is spreading into different related fields. On the one > hand it looks for closer connection to modal logic. [Morscher, E., & Hieke, A. (Eds.). (2001). /New Essays in Free Logic/. Dordrecht: Kluwer Academic Publishers. (p. 26)] PH === Subject: Re: Marketing shift, core issues |I believe Ramanujan initially met with some resistance |due to his unfamiliarity with standard notation and |style. When Hardy got Ramanujan's first letter, containing a number of results without proofs, Ramanujan had already written to two other mathematicians who'd sent the manuscript back without comment. Later on, when they were getting results of Ramanujan's published, they had enough opportunity to polish them up. As far as I know there wasn't any active resistance, just a couple of mathematicians who didn't bother to do anything with it. Keith Ramsay === Subject: Re: Need advice on letters of recommendation >> letters of recommendation are about the most ludicrous mechanism in >> place in academia - it's a shame. institutional admissions exams >> would be far superior. >> Letters of recommendation may well be ludicrous. I have no reason >> to believe, and much reason to disbelieve, that institutional >> admissions exams would be any better at deciding who would be >> a good graduate student. >perhaps you do not believe that a solid undergraduate >preparation is key component in graduate studies success. >is that it? I suppose I believe that, for a value of key that may be lower than your value of key. What I'm sure I believe is that (1) institutional admissions exams probably aren't particularly good at measuring that component in graduate studies success (since I don't have much faith in exams), and that (2) as Bart Goddard has pointed out, letters of recommendation do have at least some chance of indicating that other (for me, much more key) component in graduate studies success, namely, the ability to *do mathematics*, while (3) institutional admissions exams surely couldn't do anything at all towards indicating that other component. >> Go into detail on how you'd make such an exam, if you don't mind. >i do not need to go into any detail to ascertain that a >resounding reason for the lack of institutional exams in the >united states is correlated with the economic and logistics >required to put such systems in place. You certainly do not need to go into any detail to do that on *my* behalf, because I didn't express (and don't have) any interest in your opinion on that question, which I didn't raise. The question I raised was, How would you make such an exam (with some reasonable hope that it would do what you want it to do)? Lee Rudolph === Subject: Re: Need advice on letters of recommendation IMHO, Letters of recommendation for grad are tremendously useful, although I'll agree the quality of the letters can be uneven. The point is that the letter is generally the principal (if not the only) way to judge an applicants ability to do mathematics, as opposed to learn mathematics. There is that inevitable time in graduate school when a person transitions from being a mathematics student to being a mathematician, and simply being able to earn good grades in courses is not enough. Thus when I write letters for students, I try to give some information on their ability to wrestle with mathematics and to persevere through difficulties. In my upper level classes, I like to assign some open-ended projects, along the lines of Here is a mathematical phenomenon. Investigate it, make conjectures, generalize, find analogous problems, etc. I can then put into my letter an assessment of the applicants ability to handle this sort of mathematical challenge, which I believe is at least as good a predictor of success in grad school as is good grades. I'm not saying that grades are irrelevant. You can't build a house without the proper tools, and the mathematics that is learned in courses provide the tools to be a mathematician. But that knowledge, by itself, is not enough. Finally, the poster indicated that it would be better to have entrance exams. I would say that entrance exams give more-or-less the same information as grades, though they might give a more uniform picture, since colleges and universities differ. But in any case, an entrance exam won't provide the information that should be in the letters. (It also penalizes those who aren't good at timed exams, and there are many top-notch mathematicians who fall into that category. Mathematical research is not the same as doing well on the Putnam exam.) JoeS > letters of recommendation are about the most ludicrous mechanism in > place in academia - it's a shame. institutional admissions exams would > be far superior. === Subject: Re: Need advice on letters of recommendation > have > the most experience with. He suggested the other 2 professors that should > write my recommendations. He told me not to ask one Professor that I had > pretty well with, so he must have known something I don't. Some profs are nice people but just happen to be crabby about recommendations. It probably didn't have anything to do with you. === Subject: Re: Intersection of concentric circles - divide by zero?? > Incidentally, what is 'd' in the above equation? > While breaking up your expressions you might keep numerators and > denominators separate. Then you can test the value of denominators > before dividing by them, and so detect errors. Similarly you could > check the sqrt arguments. What happens to your method when the circles > do not intersect? Sorry, I should have been more clear when typing that up. d is the distance between the centerpoints (you'll notice in my test data, all the circles are on the same y-axis, to simplfy things). If the circles do not intercept, it doesn't perform this calculation (there's a conditional that looks at d and the radii of the two circles) I can break the formula down, but I don't know how coherent it will be. I don't actually know how this formula works...I just pulled it out of a book (well, off the web, but you know what I mean), so any distinctions will be purely arbitrary. Although, I managed to figure out that it's not a divide by zero error...it's a square root of a negative error. To clarify: Center1 and Center2 contains (x, y) coordinates of the centerpoints of the respective circles. R1 and R2 are the radii of the respective circles. d is the distance between the centerpoints of the two circles d = ABS(Center1.x - Center2.x) 'expression to find the XintersectA value '''''' XintersectA = ((center2.X + Center1.X) / 2) + ((center2.X - Center1.X) * ((R1 ^ 2) - (R2 ^ 2))) / (2 * (d ^ 2)) + (((center2.Y - enter1.Y) / (2 * (d ^ 2))) * Sqrt(((((R1 + R2) ^ 2) - (d ^ 2)) * ((d ^ 2) - ((R2 - R1) ^ 2))))) BREAKDOWN: a = (Center2.X + Center1.X) / 2 b = (Center2.X - Center1.X) c = ((R1 ^ 2) - (R2 ^ 2)) Numerator = a + (b * c) e = (2 * (d ^ 2)) f = (center2.Y - enter1.Y) g = (2 * (d ^ 2)) h = ((R1 + R2) ^ 2) - (d ^ 2)) j = (d ^ 2) - ((R2 - R1) ^ 2)) k = ((e + f) / g) l = Sqrt(h * j) ****This is where the problem occurs Denominator = k * l XintersectA = (a + (b * c)) / (k * l) Is that what you were asking for? Andy === Subject: Advanced techniques, non-polynomial factorization Polynomials are well-known in science and mathematics, but while finding roots of polynomials is typically the aim of the average researcher, polynomials themselves can be used as powerful tools for analyzing the roots of *other* polynomials. The concepts are advanced, but can be approached by first considering a basic example. The basic factorization to start is (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, notice that only two of the c's have 7 as a factor. It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = x^3 + 5x^2 + 3x + 1 and now multiply by 49. In the first example you're looking at a product and realizing that from the distributive property a(b+c) = ab + ac, you know there's *one* way it could be produced, which is to multiply something like the second example by 49. The distributive property is key here. Understanding it thoroughly, is of prime importance. Now notice that you can abstract from here as you're looking at *functions* of x, as introducing f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have (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. Which is consistent with what was found before, as only two of the functions have the property that 7 is a factor. Now I'll move on to a more complicated example. Let (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 all the functions in (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) equal 0, when x=0. Those of you who find it hard to use the distributive property with the *product* can imagine the factorization from *before* 49 being multiplied. It's harder to show here as the polynomial which defines the function in that factorization is not displayable in general. So I started at the end, with 49 already multiplied because then I can give a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That slight change, starting at the end, means that you have to understand the distributive property fully and *trust* it. Now notice that I have the result that only two of the roots of the cubic a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) can have factors in common with 7, so the 49 splits between those two. What's so startling is that the result is for a *family* of polynomials as it applies for any algebraic integer x. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Advanced techniques, non-polynomial factorization JSH still does not understand that examples are neither theorems nor proofs. === Subject: Re: Advanced techniques, non-polynomial factorization > Polynomials are well-known in science and mathematics, but not to James Harris by disgustingly repeated explicit empirical demonstration falsified by a legion of qualified mathematicians. 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: problem with Spanier's Algebraic Topology > > In Chap. 2, Sec. 5 of Algebraic Topology Spanier proves the > following theorem: > > Let p_1:X_1 ->X and p_2:X_2 -> X be objects in the category of > connected covering spaces of a connected locally path-connected space > X. The following are equivalent: > > (a) There is a covering projection f:X_1->X_2 such that p_2 circle > f=p_1. > (b) For all x_1 in X_1 and x_2 in X_2 such that p_1(x_1)=p_2(x_2), > p_1#(pi(X_1,x_1)) is conjugate in pi(X,p_1(x_1)) to a subgroup of > p_2#(pi(X_2,x_2)). > (c) There exist x_1 in X_1 and x_2 in X_2 such that p_1(x_1)=p_2(x_2) > and p_1(pi(X_1,x_1)) is conjugate in pi(X,p_1(x_1)) to a subgroup of > p2#(pi(X_2,x_2)). > > Then he has a corollary: > > Two objects in the category of connected covering spaces of a > connected locally path-connected space X are equivalent if and only if > their fundamental groups (at some two points over the same point of X) > map to conjugate subgroups of the fundamental group of X (at this > point). > > I don't get it. It seems to me the corollary should be: > > Two objects in the category of connected covering spaces of a > connected locally path-connected space X are such that there is a > morphism from each to the other if and only if their fundamental > groups (at some two points over the same point of X) map to subgroups > of the fundamental group of X (at this point) which are such that each > is conjugate to a subgroup of the other. > > To replace are such that there is a morphism from each to the other > with are equivalent, you would need to prove that every morphism > from an object to itself is an automorphism. > > Can anyone help me here? > > the maps in part a) are coverings. thus in the corollary you get two > spaces A and B with surjections between each other. put this together with > some other details and go from there > I'm afraid I'm still stuck. Maybe the following will help ... Covering spaces have a _unique_lifting_ property: Theorem: Suppose p: (E,e_0) --> (X,x_0) is a covering map and let f: (Y,y_0) --> (X,x_0) be any map. If Y is connected and f can be lifted to a map f': (Y,y_0) --> (E,e_0), then f' is *unique*. In the present case, you've got covering maps f_1: X_1 --> X_2 and f_2: X_2 --> X_1 such that f_1 o f_2 and f_2 o f_1 are liftings of the identity map X --> X. The theorem (above) then implies that each of those compositions is one of the deck transformations for the corresponding covering map of X. But the deck transformations are homeomorphisms ... === Subject: Problem from Herstein 4 For those who have the book, it is I.N.Herstein, Topics in Algebra ( 2nd edition), chapter 2 ( Groups), section 2.9 ( Homomorphisms), problem 19 ( I include problem 18 for reference purposes) 18. Let G be the group of all real 2-by-2 matrices (a b) (c d), with ad - bc non-zero, under matrix multiplication, and let N be the subgroup consisting of those elements of G with ad - bc = 1. Prove that N contains G', the commutator subgroup of G. 19. In problem 18 show, in fact, that N = G'. No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has determinant 1, and thus so do all products of elements of this form. Thus G' is contained in N. To show that G' actually equals N, I first thought of working out the form of a general element of G' and solving the resulting equations, but these turned out to be rather horrible. The next problem in the book is a simpler one of the same form, and I was able to show you could actually get away with quite simple A and B to get the general element. My attempts to do the same thing with this one have been fruitless so far. Any suggestions? === Subject: Re: Problem from Herstein 4 > For those who have the book, it is I.N.Herstein, Topics in Algebra ( > 2nd edition), chapter 2 ( Groups), section 2.9 ( Homomorphisms), > problem 19 ( I include problem 18 for reference purposes) > 18. Let G be the group of all real 2-by-2 matrices > (a b) > (c d), with ad - bc non-zero, under matrix multiplication, and let N > be the subgroup consisting of those elements of G with ad - bc = 1. > Prove that N contains G', the commutator subgroup of G. > 19. In problem 18 show, in fact, that N = G'. > No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has > determinant 1, and thus so do all products of elements of this form. > Thus G' is contained in N. To show that G' actually equals N, I first > thought of working out the form of a general element of G' and solving > the resulting equations, but these turned out to be rather horrible. > The next problem in the book is a simpler one of the same form, and I > was able to show you could actually get away with quite simple A and B > to get the general element. My attempts to do the same thing with this > one have been fruitless so far. > Any suggestions? I believe problem 19 is a difficult one. Your last suggestion of finding simple A and B's to get the general element is the way to go. That is, you want to find generators of N and show that each generator is in G'. Finding suitable generators of N is not easy. Artin, in his book Geometric Algebra, gives a geometric way of arriving at them. But, even there, it is not easy to see how to arrive at them. One hint would be to think of row-echelon reduction of matrices that you used to solve linear equations. But, even after you found the generators, it is not easy to show that they are in G'. A hint here is to consider the homomorphism f: G -> G/G' and show that f(n) = 1 mod G' for each generator n of N. Make use of the fact that the image f(G) is abeian, being a subgroup of the abelian group G/G'. I think the time spent on working on this difficult problem will be worht it. -- Bill Hale === Subject: Re: Problem from Herstein 4 > For those who have the book, it is I.N.Herstein, Topics in Algebra ( > 2nd edition), chapter 2 ( Groups), section 2.9 ( Homomorphisms), > problem 19 ( I include problem 18 for reference purposes) > 18. Let G be the group of all real 2-by-2 matrices > (a b) > (c d), with ad - bc non-zero, under matrix multiplication, and let N > be the subgroup consisting of those elements of G with ad - bc = 1. > Prove that N contains G', the commutator subgroup of G. > 19. In problem 18 show, in fact, that N = G'. > No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has > determinant 1, and thus so do all products of elements of this form. > Thus G' is contained in N. To show that G' actually equals N, I first > thought of working out the form of a general element of G' and solving > the resulting equations, but these turned out to be rather horrible. > The next problem in the book is a simpler one of the same form, and I > was able to show you could actually get away with quite simple A and B > to get the general element. My attempts to do the same thing with this > one have been fruitless so far. > Any suggestions? I assume, that your matrices have their coefficients in some commutative field k. If k* are the nonzero elements of k (with multiplication), you can verify that (a b) (c d) --> ad-bc does indeed give a surjective homomorphism det: G ---> k*. This will exhibit N = ker(det) and G/N = k* commutative. This line of reasoning still works over any commutative ring; you would start with those matrices with _invertible_ determinant instead of nonzero determinant, and you would use the invertible elements U(R) of R in in place of k*. Marc Marc === Subject: Re: Problem from Herstein 4 :> 18. Let G be the group of all real 2-by-2 matrices :> (a b) :> (c d), with ad - bc non-zero, under matrix multiplication, and let N :> be the subgroup consisting of those elements of G with ad - bc = 1. :> Prove that N contains G', the commutator subgroup of G. :> 19. In problem 18 show, in fact, that N = G'. : I assume, that your matrices have their coefficients in some commutative : field k. If k* are the nonzero elements of k (with multiplication), : you can verify that : (a b) : (c d) --> ad-bc : does indeed give a surjective homomorphism det: G ---> k*. : This will exhibit N = ker(det) and G/N = k* commutative. All true, but this only shows that N contains G'; the hard part is the other direction (problem 19, show that N = G'), which is not coming to me right now. Ted === Subject: Re: Mathematical Integrity |However, lately, I surfed through several math forums and I found many |unpleasant posts. This isn't just a property of math forums; unmoderated usenet groups typically have many unpleasant posts. Remember that what you see isn't a random cross-section of the participants; you see more postings from the people who are all excited by something, whether cheerfully or because they're annoyed. Keith Ramsay For those who have the book, it is W. Rudin, Principles of Mathematical Analysis, chapter 2 ( Basic Topology), problem 18: [A set E is perfect if E is closed and every point of E is a limit point of E] Is there a nonempty perfect set in R which contains no rational number? This one has really had me stumped. My first guess was that the answer was no, but I didn't get anywhere trying to prove it. I then tried to construct such a set: since the set of rationals is countable, we can write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of length 1/2^n centred at r_n, let I be the union of the I_n, and let X be the complement of I in R. Since the total length of the intervals is at most 1, X is nonempty. As the complement of an open set, X is closed. But I can't show that X is perfect. So: Is the answer yes or no? If yes, does my set work as an example? Any hints would be most welcome. John Harrison === Subject: Re: details of joke wanted by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACDdUn10207; >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. Perhaps this one: A group of scientists were doing an investigation into problem-solving techniques, and constructed an experiment involving a physicist, an engineer, and a mathematician. The experimental apparatus consisted of a water spigot and two identical pails, one of which was fastened to the ground ten feet from the spigot. Each of the subjects was given the second pail, empty, and told to fill the pail on the ground. The physicist was the first subject: he carried his pail to the spigot, filled it there, carried it full of water to the pail on the ground, and poured the water into it. Standing back, he declared, There: I have solved the problem. The engineer and the mathematician each approached the problem similarly. Upon finishing, the engineer noted that the solution was exact, since the volumes of the pails were equal. The mathematician merely noted that he had proven that a solution exists. Now, the experimenters altered the parameters of the task a bit: the pail on the ground was still empty, but the subjects were presented with a pail that was already half-filled with water. The physicist immediately carried his pail over to the one on the ground, emptied the water into it, went back to the spigot, *filled* the pail, and finally emptied the entire contents into the pail on the ground, overflowing it and spilling some of the water. Upon finishing, he commented that the problem should have been better stated. The engineer, in turn, thought for some time before going into action. He then took his half-filled pail to the spigot, filled it to the brim, and filled the pail on the ground from it. Again he noted that the problem had an exact solution, which of course he had found. The mathematician thought for a long time before stirring. At last he stood up, emptied his pail onto the ground, and declared, The problem has been reduced to one already solved. === Subject: Re: Fastest factorial algorithm by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACFAmI17280; >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? ln (n!)= (ln(n)-1)*n+(1/2)*ln((2*pi*n)) + 1/(12*n) - 1/(360*n^3) more terms can be added when the n is smaller hjs. a people that don't remember their past are doomed to repeat it. === Subject: Re: Gaussian Continued Fractions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACFB0L17294; >> ... >> The traditional continued fraction series can be related to a >> regular tiling of the hyperbolic plane with triangles which have >> all three angles equal to zero. These triangles can be paired to >> produce squares, or a triangle can be grouped with its 3 neighbors >> to produce a regular hexagon. The groups need to be placed >> symmetrically as if the joining edge is a mirror to make a square >> or a hexagonal tiling pattern. >> The Gaussian version is related to a 3-dimensional hyperbolic >> tiling with octahedra which have 90 degree angles between faces >> and vertex angles equal to zero. >> The Eisenstein CF expansion is related to a tiling with tetrahedra >> which have 60 degree angles between faces and vertex angles equal >> to zero. By surrounding one of these tetrahedra with four others, >> a cube is formed, thus making a cube tiling pattern. >Where can I find more information about this? >Mike Here is a web page about the first: http://mathworld.wolfram.com/ModularGroupLambda.html The gray region is a square which is divided by the imaginary axis into two triangles. (This is shown in the half plane projection.) The vertices of one of the triangles can be labeled 0/1, 1/1 and 1/0. Farey fractions are the labels of the vertices of the triangles. For one vertex of each triangle, the numerator and denominator are the sums of the numerators and denominators of the other two vertices. The vertices of an octahedron of the second are 0, 1, i, 1+i, (1+i)/2 and infinity. The faces extend to meet the complex plane at Re(z)=0, Re(z)=1, Im(z)=0, Im(z)=1, |z-.5|=.5, |z-.5i|=.5, |z-1-.5i|=.5 and |z-.5-i|=.5. The vertices of a tetrahedron of the third are 0, 1, -w^2 and infinity. The faces extend to meet the complex plane at Im(z)=0, Im(z w)=0, Im((z-1)w^2)=0, and |3z-1+w^2|=sqrt(3). In the half plane projection, a horocycle is a circle tangent to the projection boundary or a straight line parallel to the projection boundary. If s is distance along a horocycle, and d is the chord length, then s = 2 sinh(d/2). This is a special case of the half plane projection distance formula 2 sinh(d/2)=|z-w|/sqrt(Im(z)Im(w)). The terms of the traditional continued fraction series can be viewed as hyperbolic distance traveled along horocycles, alternately curving left and right for each term. The starting point is one unit above the origin and the boundary headed straight right. (This is equivalent to curving to the left in the original hyperbolic plane.) A similar thing can be done with horospheres. The locus of possible trajectories consists of a plane one unit above the boundary and a packing of unit diameter spheres between the plane and the boundary which are tangent at the Gaussian or Eisenstein integer coordinates, and more smaller spheres below those spheres. The intrinsic geometry of a horosphere is Euclidean. Fuchsian Groups by Svetlana Katok is about hyperbolic symmetries including the modular group: http://www.math.psu.edu/katok_s/books.html Stephen === Subject: Re: JSH: Deprogramming needed? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACFAqg17284; >> (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 >1 for all a,b>0 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACIVpL31583; How to proof this? === Subject: Re: a^b+b^a>1 for all a,b>0 > How to proof this? Let b=xa. Wolog 0= (a^a)^x + x (a^a) >= u^x + x u > 1 where u is the minimum value of a^a. You should be able to fill in the missing details. === Subject: Turing Machine by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACIVjK31570; Hey all, can anybody help me with this question: M is a polynomially time bounded non deterministic TM; K(M) is the language it recognizes. Prove or disprove: If L reduces to K(M) then L is in NP. Zuliya === Subject: Re: Something Stupid? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACIVuR31612; >
>>  10  10  10
>>       950
>>How would you get these three 10s to 950 using only one straight line 
(any
>>length)?
>What in the hell are you talking about?
>Doug
>
Put a straight line over the second 10 to make a TO. === Subject: Re: rearranging equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hACLX1k13655; >>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!) >>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. >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu I would not assume that t1 and t2 are integers. The equation looks like one that would occur in first year circuit analysis for an E.E. course. In general it must be solved numerically (or graphically, if you got your degrees in the 1960s). Phil === Subject: SUM-PART-PART-PRODUCT by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADD8SM13422; My 9 year old son keeps coming home with these papers called sum-part-part product. The sum is filled in as is the product and he must fill in the part boxes. I can't, for the life of me, figure out a mathmatical way to calculate the parts....there must be a way. Can anyone help? ------------------------------- SUM | 20 | 40 | 16 | 30 | 14 | 60 | ------------------------------- PRT | | | | | | | ------------------------------- PRT | | | | | | | ------------------------------- === Subject: Re: SUM-PART-PART-PRODUCT > My 9 year old son keeps coming home with these papers called sum-part-part product. The sum is filled in as is the product and he must fill in the part boxes. I can't, for the life of me, figure out a mathmatical way to calculate the parts....there must be a way. Can anyone help? > ------------------------------- > SUM | 20 | 40 | 16 | 30 | 14 | 60 | > ------------------------------- > PRT | | | | | | | > ------------------------------- > PRT | | | | | | | > ------------------------------- Well, it appears that we're looking for two parts--I'll call them A and B--so that the sum of A and B and the product of A and B are as specified. In the first column, for example, we could fill in the parts A = 10, B = 10, since A + B = 20 and A * B = 100. In this particular example, there's a simple pattern: the two parts are equal and each is equal to half the sum (in the fifth colums, for example, the two parts are 7 and 7). Presumably, some of the exercises don't fit this pattern, in which case a 9-year-old might approach this by looking at the ways the sum can be made and seeing which one gives the desired product. Suppose, for example, that SUM = 18 and 18 = 0 + 18 Product = 0 (nope, that doesn't work) 18 = 1 + 17 Product = 17 (nope) 18 = 2 + 16 Product = 32 (nope, but we're getting closer) 18 = 3 + 15 Product = 45 (nope, but we're even closer) 18 = 4 + 14 Product = 56 (hooray! We've found an answer!) This will always work if there is a solution in positive integers, but it's a bit time-consuming if SUM is large. There is a quicker way that always works, but it's likely not something your son has seen yet. Still, all is not lost--if you try some more examples, you might observe that if you haven't found an answer when the first term is less than or equal to SUM / 2, you won't find one at all. Another help is to look at numbers that divide PRD. In our example, for instance, we know that 56 can be divided evenly by 4, so maybe 4 is one of the parts. It is, in this case, but it won't always be. There's a pattern here, too--try to find one. Hope this helps. Rick === Subject: [no subject] by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADFBl122015; Could anyine graph y= 3 cos x step by step? === Subject: Re: [no subject] > Could anyine graph y= 3 cos x step by step? Step 1: What are the zeros? Step 2: Between each zero, is the graph positive or negative? Step 3: How does the graph compare to the graph of y = cos x? === Subject: Re: compact definition.... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADFC0E22047; >book-topology >def) >A collection B of subsets of a space X is said to cover X, >or to be a covering of X, if the union of the elements of B is equal to X. >it is called an open covering of X if its elements are open subsets of X. >def) >A space X is said to be compact >if every open covering B of X contains a finite subcollection that also >covers X >-------------------------------- >i think that meaning of compact is X = U(G_i_n) by upper definition. >but in the other book, >i saw that definition of compact is X C U(G_i_n) {C : inclusion sign} >which of definiton is right?? >advice ...please.....thank you...... The second expression is the correct one for compact set. Every singleton set is compact but generally is not EQUAL to a union of open sets. Did you notice that the Munkres definition you give is for a compact SPACE? If X is a topological space, as opposed to a proper subset of a topological space, then it can't be properly contained in any set! === Subject: Re: Recommendations on Complex Analysis books? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADFBti22043; >The subject says it -- however, I'd like to clarify >one detail: I'm looking for a book on *analysis*, as >opposed to Calculus (i.e., that covers rigorously >the concepts and proofs on Complex numbers and >Complex variables functions). >However, I'm just a hobbyist, so I'm not looking for >the ultimate, advanced reference book (i.e., I'm not >a mathematician or even a student in Mathematics; I'm >an engineer, who already knows (at least *knew* very >well :-)) about complex numbers, but I'm beginning >to appreciate and enjoy the rigorous side of maths, >and I find that complex numbers do have a great >appeal. >Carlos One of my favorites is Invitation to Complex Analysis by Ralph Boas, published by Random House. === Subject: Re: origin of homotopy and homology by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADFER322310; >hi, >Does anyone know how Poincare came up with the idea of homotopy and homology >and how he defined them originally? See the paper A History of Homological Algebra at Charles Weibel's Home Page http://www.math.rutgers.edu/~weibel/ HTH D.L. === 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 hADHUtC32614; >> How do you prove that the circumference of a circle is proportional to >> its radius? .... and how did the Greeks prove it? > You've asked about history. You shall have history! >on the 8th June 2001, so here's a copy of that. >> Is there a straightforward way to prove that the ratio of >> circumference to the diameter of any arbitrary circle is always a >> constant? It feels like this should be easy but I have not seen such a >> proof in any book so far. >> That's good mathematical thinking! You've seen that rough ideas >> about similarity aren't enough. It was the ancient Greeks who realized >> that it's not straightforward at all, and invented the first technique of >> limits to handle just such problems. You've already had limit arguments >> outlined by DWIII and Peter M. Jack, so I'll just mention some of the >> history. >> The area was tackled before the circumference, perhaps because it's >> easier. Euclid XII.2 (probably due to Eudoxus) proved that Circles are >> to one another as the squares on the diameters. A modern statement of >> that might say >> (area of first circle)/(area of second circle) >> = (diameter of first circle)^2/(diameter of second circle)^2. >> This shows that our modern area formula (pi)(r^2) or (pi)(d^2)/4 Ken , Is this really a modern formula for circle's area? If You see it is equal to the product of [pi*d/4]*[d]. pi*d/4=quarter circle circumference . Thus the area comes after one has completed the required details and relationships of circumference and diameter. One has to start finding circumference [quadrature] relationships with square sides [Perimeter]. This is at least a logical procedure. [ Further analysis ,may be found below: http://www.stefanides.gr/quadcirc.htm http://www.stefanides.gr/piquad.htm ] I, believe that the Greek Philosophers started first with the circumference relationships,before going to the area ones. Panagiotis Stefanides httm://www.stefanides.gr >> So your unease about the constancy of pi was well justified. The >> question needed some very great mathematical minds to handle. >> Ken Pledger. === Subject: Re: Largest number ever written down? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADHV3R32645; Look up the word power tower. It looks like an exponent, but it is on the left hand side of a number. Example. 85 = 5^(5^(5^(5^(5^(5^(5^5)))))) The power tower exponet tells you the number of times to write the number down as an exponet to itself. Another example. 53 = 3^(3^(3^(3^3))) So I will claim the prize before someone adds 1 to my number === Subject: Re: Proof of Shannon's Theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hADLubR20531; >
>Well, I've worked out the issues.
>> I've been wanting to read a proof of Shannon's Theorem (existence of
>> good codes).  The proof provided in J.H. van Lint's book Introduction
>> to Coding Theory is nice and concise.  But maybe a bit too concise for
>> me since I have some questions.
>> In the middle of p.28, the expression for P_i in terms of g_i(y) puzzles
>> me unless each g_i(y) happens to be 0 or 1 or unless the expression is
>> supposed to be an inequality (<=) instead of an equality.
>Here I'm convinced it should be an inequality, which is fine since an
>upperbound is being established.
>> At the bottom of p.28, the expected value of P_C over all possible codes
>> C has expectation apparently carrying through all the summations and
>> then multiplicatively acting on the two terms.  Why is this?  Also, it
>> is unclear to me how the final equality that this then leads to is
>> reached.  Perhaps if I saw the steps written out in more detail I would
>> better understand.  After this part of the proof, I'm think I'm in the
>> clear.
>Each of the codewords x1, x2, ..., xM are independently chosen from the 
set
>{0,1}^n.  This means repetition is allowed, so surprisingly obviously
>suboptimal codes are included in the average.  I was under the assumption
>that repetition would be excluded when randomly choosing a code.
>At first this seems like generous overbounding, but as it turns out the
>finally inequality (as n->inf) is tight.  That is, once R >= 1 + plogp +
>qlogq, then arbitrarily good codes can no longer be found, though this is
>not proved.
>> I'll be quite happy when I fully understand a proof of this exciting
>> theorem.
>I'm happy.  However, now I'm interested in the case when R >= 1 + plogp +
>qlopq.  Also, I may look into a proof of Chebychev's inequality since it 
is
>used in the proof of Shannon's theorem.
>> Oscar
>
=== Subject: Re: Need advice on letters of recommendation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAE2cAR08252; >>letters of recommendation are about the most ludicrous mechanism in >>place in academia - it's a shame. institutional admissions exams would >>be far superior. >Letters of recommendation may well be ludicrous. I have no reason >to believe, and much reason to disbelieve, that institutional >admissions exams would be any better at deciding who would be >a good graduate student. perhaps you do not believe that a solid undergraduate preparation is key component in graduate studies success. is that it? >Go into detail on how you'd make such an exam, if you don't mind. i do not need to go into any detail to ascertain that a resounding reason for the lack of institutional exams in the united states is correlated with the economic and logistics required to put such systems in place. >I, by the way, am sorely lacking in the experience of administering >admissions evaluations of any sort whatever (but I do have all too >much experience at setting exams, hence my dubiety that they would >be any damned good for this purpose). While you're describing your >proposed exam, you might just run over your own experience in >administering admissions. >Lee Rudolph === Subject: Re: About big numbers... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAE2c9f08236; >What's the biggest number that can be made with 'three' nines? Refer to the other post on Big numbers. I would go with the power tower answer. 999 Power tower means 35 = 5^(5^5) 56 = 6^(6^(6^(6^6))) So 999 = 9^9^9^9^9 ... ^ 9 There will be 99 's in that list, and this number itself is HUGE === Subject: Re: The solution to off-topic posts and pests of Usenet; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAE2c7r08203; >
>>  I have been talking with a Kiewit employee as to controlling the
>>expiration of posts to my newsgroup. This gentleman's name is Stephen
>>Campbell and I find him very exasperating to ever communicate with. I
>>do not know if he is the ken Hreikin or the ken phrea, but I do know
>>this. He controls the expiration of my newsgroup.
>> That will come as a big surprise to every system administrator on the
>> planet.
>> On the contrary, each news admin controls expiration on their own
>> group at all.
>> You can't control expiration for a.s.p.p.  It isn't yours, or
>> Darthmouth's or Kiewit's (whoever they are) or any other's to
>> control.
>> -- 
>> Chris Lewis: _Una confibula non sat est_
>  The facts do not concur with your statements. 
>  (1) the Dartmouth newsfeed for sci.physics has 1823 posts as of this
>moment of writing, and simultaneously the Web newsfeed on sci.physics,
>the same newsgroup has 555 posts. It has none of the 1999 expiring
>posts, and none of my posts of 8 Aug.
>  You Chris Lewis claim that there is no standard to Internet, but
>obviously there is a standard for the Internet and it is the Web.
>Sci.physics from the Web is less apt to be biased or prejudicial as
>what the local feed can be and is here at Dartmouth at the moment.
>  (2) The newsgroup a.s.p.p was set up for the purpose of --to my
>colleagues-- my teachings (and to my critics -- my babble). If Stephen
>Campbell uses the control switch of expiration in 1995 to his whim and
>the same in 1996 and I have no say in that switch, then obviously the
>newsgroup a.s.p.p was established more for the use of Campbell than to
>Plutonium, and it should be renamed alt.sci.physics.campbell.  This
>hacking around over the word my newsgroup is superflous in the
>argument. It is my newsgroup in the context of the creation of the
>newsgroup and therefore I should have the control of expiration within
>reasonable use, not some outsider to the newsgroup. And since
>Dartmouth is the Homebase for that newsgroup, if I expire all posts
>with 1 day and 1 day only lifetime, then on the Web, all web viewings
>will show only the 1 day.
>  It sounds to me Chris that you are motivated by hate because your
>above statements just do not jive with the facts of the Web being a
>universal standard. Local feeds can deviate from what the Web
>newsgroups have, but still the Web is a standard to which the Local
>feeds can be compared. It is the fact that sci.physics from Dartmouth
>has 1823 posts and the Web sci.physics has 555 leads me to suspect that
>this is yet another directed attack on me by persons frustrated in
>their failing to kick me off of the Net.
>of mine concerning forging my name to subscription lists. None of those
>posts contained science. But should you go berserk because what I
>posts, posts with science content that I am arguing. I will become very
>take sides where you unrightfully go too far. 
>  This latest form of attack is pathetic because it is a burden on
>Dartmouth community readers of sci. newsgroups. And I feel that if
>someone over there at Kiewit is grilled over this cheap-shot-attack,
>they will just fobb it off by saying Oh, I forgot to turn the
>expiration switch, or that machine is blah blah...   When really it is
>a deliberate and planned attack on me.
>that if he were in charge, I would have been removed from the Net years
>ago. I think Randy tends to get flamboyant. Apparently he forgot the
>concepts of Academic Freedom and that the president of this College
>believes and fully practices in Academic Freedom. Don't get me wrong,
>I like Randy for he wears cowboy western hats just like I do. And he
>even has heard of Quantum Mechanics which I do not expect from many
>computer people.
>   I just cannot say this thought enough. That so many people get bent
>out of shape, not because of what I post or where I post or how
>frequently I post or what I say in my posts or anything, but the PU
>theory. People do not like to hear for the first time that God = Atom.
>And some people who hear that for the first time get so bent out of
>shape that they just cannot shake it but that it consumes them. They
>burned Hypatia at the stake because the Atomic theory meant more to her
>than did all of that religion winds coming out of the Middle East. They
>shut down Galileo's science lab because he trespassed too much into the
>religion coffers and chapels.   Imagine what they would like to do to
>the man who said God = ATOM of 231PU.  People, like Chris, like
>Stephen, like Randy who behave exemplary in all types of situations
>seem to lose it when they  are confronted by a revolutionary scientist
>who has written on him God = ATOM
>  In summary, contrary to Chris Lewis, there is a international
>expiration standard for newsgroups and that standard is the Web. Local
>newsfeeds can vary from the Web newsgroups and if they vary, then the
>question by the local readers should be why are we varying, and for
>what reason, in what the Web newsgroup is showing ?
>
=== Subject: Re: limit of modulus continuity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAE2c8s08220; >im using the following definition of uniform modulus of continuity of >f:X->Y, where d_x and d_y are the metrics in X and Y respectively: >omega_f (delta) = sup { d_y(f(x), f(a)) : a,x in X and d_x(a,x)<=delta >problem: if limit of omega_f(delta)/delta = 0 as delta goes to zero, >then f is constant. i have already proved that f is constant iff >omega_f(delta)=0 for all delta>=0. i also proved that >omega_f(t/n) >= 1/n omega_f(t) for all n in N and t in R >which seemed useful at first, but now i can't get anywhere. >can someone give me a hint? Here is a hint for R: omega_f(delta) signifies how much f can change in intervals of length delta. Suppose f changes by some quantity in [0,1]. What can you say about how much it changes in [0,1/2] and [1/2,1]? (use triangle inequality). In general metric spaces the theorem is incorrect. For instance in any discrete space omega_f(delta) is zero for all f's, given a sufficiently small delta. === Subject: Re: JSH: Deprogramming needed? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAE2cCN08261; >(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 indeed? >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. I'm not sure what people you are referring to. Surely some people are just in it for the gags, this is Usenet after all. But I do think several smart and learned people have tried to help you with your proofs over the years. For instance Arturo Magidin has been reading and correcting your posts for years. I have never seen him write anything provoking (which I can't say for David Ullrich, for instance). Yet you have repeatedly referred to him as an evil dude (among other more unpleasant terms). If you wish to be taken seriously, I think you should strongly consider apologizing or at least stop using such terms. *This is getting you nowhere*. >Aren't there any *other* people who suppose they are rational, who can >follow a logical argument, who might comment? This may seem strange to you, but reasonable people and even competent mathematicians may find it very hard to review your work. The reason being simply that you are not accostumed to formal proof writing and standard math terminology. That doesn't make your ideas wrong, even great mathematical geniuses like Ramanujan who weren't trained in writing formal proofs often produced hard-to-decipher material. However, this is exactly why you should be greatful to people like Arturo Magidin who have spent years on trying to understand your work. Most people simply get lost in the bulk of unquantified theorems, lemmas whose statement makes up their proof body, unannounced functions and variables, non-standard terminology, etc. >Why is it always the same people? Or people imitating them in hostile >and mocking displays of animosity or anger? The reason it is the same people is that they are the ones who understand your argument. I for one know enogth algebra to be able to read your works, but simply can't make any sense of it. The ones that criticize you just for the sake of criticism, with no actual math-oriented criticism, are indeed a distraction. You should consider, however, that you are attracting them to you by making weird and sometimes borderline psychotic posts. I often think you are making these posts just to arouse the crowd, as I doubt you really ever intended all that math powers that be are going to kill you instead of me stuff. But making such posts does make you a target of non-math-related criticism and mockery. >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? Objectiveness is a sliding scale. I don't think most people on Usenet are trying to make you look stupid. But like I've said, most people can't really read your work without some level of translation. >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. I don't think the objections raised by Arturo Magidin and Nora Baron for example are irrational. Of course, like I said, I must rely on them and other posters to understand what it is that you're saying, because I don't quite understand your terminology myself. Anyway this whole marketing book thing is just a distraction. I think you have gotten to used to use distractions in order to avoid direct discussion of the real issue, which brought you and your proof to deadlocks in the past. If you really want an honest discussion only post math, the rest is just fluff and fuels exactly those criticisms you complain about. >Think about it. I'll come back to the subject later. >James Harris >http://mathforprofit.blogspot.com / Thomas Billups === Subject: Does this ideal contain 1? Hi I have a problem on commutative algebra that I'm stuck on. Or, perhaps, it's really just a calculation I want to do: Let k be the complex field, r an even number (not too small, greater than 12, say) and suppose c,f,r,t,u,v,d are elements in k satisfying the five relations: c = -ut, v=ur, t=-fr, u-d+fvc^((r/4)-1)=0, r-f-ftc^((r/4)-1)=0, . Then I want to know if the ideal generated by the seven polynomials C, F, Y_1, Y_2, T_1, T_2, D in k[x,y] are the whole of the ring: C := x^2y^2-c F := xy-f(x^(r/2) - y^(r/2)) Y_1 := x^2y-rx^((r/2)+1) + ty^(r-1) Y_2 := -xy^2 - ry^((r/2)+1) -tx^(r-1) T_1 := x^(r+1) - ux - vy^((r/2)-1) T_2 := y^(r+1) - uy + vx^((r/2)-1) D := x^r + y^r -d I've tried using Magma to compute a couple of examples, and the answer is negative. However, of course, I would like to know if it is true in general. If anyone has suggestions on alternative approaches, I'd be very grateful! Jason === 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 hAEE3FP20059; >> 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? >> /david >Very simple. It's enough to accept that equal cords substends equal >arcs. How does one accept this ,though ! Panagiotis Stefanides http://www.stefanides.gr >If you have a circumference and an inscrit polygon, its n sides >(cords) are proportional to radius. If you now have a bigger polygon, >the new n arcs must increase proportional to radius. Otherwise the new >cords cannot substend equal arcs. L.Rodriguez === Subject: Re: Inversion? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEE3oS20129; >When I did A level maths at school, about 30+ years ago, in my >analytical geometry course I was taught a subject called 'inversion'. >As I recall, one defined a point O (centre of inversion?) and used >this to transform a curve. Any point P on the curve was transformed >to P', such that O, P, & P' were on a straight line and OP*PP' = k^2 >(k = radius of inversion?). >The one result I can remember is that a circle can be transformed into >a straight line, and vice versa. >I have never come across this again, and the subject seems to heve >disappeared from text books. It seemed a bit of an odd topic to me at >the time. Does anyone know how/when this concept came about? Did it >ever have any practical application? You are talking about inversion in a circle that maps each point inside a given circle to a point outside the circle. Points on the circle are mapped into themselves and the center of the circle is mapped to the point at infinity. I don't know how practical an application you would consider it but it is used in the Poincare disk and plane models for non-Euclidean geometry. Since they model straight lines by circles, reflection about a line becomes inversion in the circle. The Dutch artist, M. C. Escher, did a number of paintings based on the Poincere model and inversion. >Yours in curiosity, >Jon === Subject: Re: IRRATIONALITY AND TRANSCENDENCE OF e AND Pi (Part B) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEH8Uw32582; Re 6.1, Fourier published his proof of the irrationality of e in 1815 in his Cours d'analyse, according to Feldman & Nesterenko's Number Theory IV, p. 79. Here are some typos for Part B. In SUM(k=1 to infinity) of [k^(N!)]^(-1) which appears twice, to get a Liouville number the sum should be on N, not on k. (As it stands, the sum equals zeta(N!).) Similarly, in the example of a transcendental number, N=10 should be k=10. In If x is an algebraic number of degree n, where n is at least two, then there exists 0 < C less= 1 such that | x - p/q | > C*q^(-n) for all rational numbers p/q (p and q > 0 are integers). the exponent -n should be -n-1. For example, for any algebraic number x of degree n=2 (in fact, for any irrational number x at all) there are infinitely many p,q>0 with | x - p/q | < q^(-2). Also, it was the last < inequality that Dirichlet proved, rather the former > inequality. === Subject: Re: logical programmation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEIWZ806060; You can take a look on the Maple website, and perhaps on the LISP ' Head WebSite, and if you don't find,you can also send your request by mail to Pr. Silklossy... Au fait, le p'tit con qui arrivait toujours en retard .88 mes cours c'.8etait pas toi, par hasard??? B.Morlaye (N'est ce pas) >Hello ! >I'm looking for documentations about logical programmation tools like >prolog. >Does others langages exists ? and what are they ? perhaps using others >logics ? >Do you have informations for me (or good links) ? >Olivier === Subject: another quaternion question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEKErq13504; What role does the Jacobson radical aÁb = a+b - ab play in the quaternion algebra (lets say, over the reals), if any? ...anyone know if it has come up in any applications to physics? Good day, C. Dement === Subject: Re: Indian firsts in Maths by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEMoIM24432; >INVENTIONS & DISCOVERIES >INVENTION OF NUMERALS >Numerals are found in the inscriptions of Ashoka The Great in the 3rd >Century BC. This knowledge traveled from there to Europe and West. In >Arab countries even now numerals are known as HINDSE: from India. La >time, It is India that gave us the ingenious method of expressing all >numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's >Journal >INVENTION OF ZERO >Brahmagupta was the first mathematician to treat ZERO (0) as a number >and showed its mathematical operations >INVENTION OF ARITHMETIC >Arithmetic was discovered by Indians in about 2nd Century BC. >Bhaskaracharya's book Lilavathi is regarded as the first book on >modern arithmetic. The Arabs learnt and adopted it from India and >spreaded it to Europe. In 499 AD Aryabhatta finished his work >Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) >INVENTION OF ALGEBRA >In Western Europe the knowledge of Algebra was borrowed, not from >Greece but from Arabs, who acquired this from India. Algebra is the >only Arabic name for Bijaganitha. Aryabhatta was one of the first to >use Algebra (Encyclopedia Britannica) >INVENTION OF GEOMETRY AND TRIGNOMETRY >The brick work of Harappa and Mohenjodaro excavations show that people >of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta >formulated the rules for finding the area of a ëtriangle', which led >to the origin of Trignometry. >DISCOVERY OF ASTRONOMY >The knowledge of the motion of heavenly bodies was discovered by >Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for >calculating the timing of eclipses. In Surya Sidhanta' Latadeva, >talked about the earth's axis and called it SUMERU. That the earth is >a sphere and it rotates on its own axis, was known to Varahamihira >and other Indian astronomers much before Copernicus published this >theory. (Jewish Encyclopedia) >INVENTION OF CALENDAR MAKING >Discovery of measurement of time and discovery of nomenclature of >days, month and years and invention of calendar making was made in >India. In his book ëSurya Sidhanta' Latadeva (505 AD) divided the year >into 12 months. Seven planets of the solar system effect the earth's >atmosphere and their names were added to the seven days of the week, >which was accepted all over the world. >DISCOVERY OF THEORY OF GRAVITATION >In his book ëSidhanta Shiromani' Bhaskaracharya mentions about force >of attraction resembling gravity, discovered centuries later by >Newton. (Jewish Encyclopedia) >INVENTION OF IRON PRODUCTS IN 3000 BC >The word AYAS occurs in the four Vedas which denotes iron. Ashoka >pillar at Mehrauli, New Delhi and another iron pillar in Karnataka >stand proof of India's metallurgical heritage (A study published in >the magazine ëThe Current Science'). >INVENTION OF COPPER, BRONZE AND ZINC >The copper and bronze artifacts dates back to Indus Valley >Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was >made in around 50 BC at Zawar in Rajasthan (India). >INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS >Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone >to prepare the sulphate of copper, zinc and iron and carbonates of >lead and iron. RASAVIDYA or Indian alchemy made its appearance around >5th Century AD (National Science Centre, New Delhi) Yes, but what has India done for Mathematics lately? Phil === Subject: Re: Indian firsts in Maths > Yes, but what has India done for Mathematics lately? > Phil I know the above is Python, but my hat is off to Agrawal, Kayal,and Saxena for the long awaited proof that PRIMES is in P. For me, it's the single most exciting discovery in maths for the last few decades, and the crown for the achievement lies squarely on three Indian mathematicians' heads. Credit where credit is due. Country, nationality, race, religion and all that non-mathematical stuff is irrelevant. It's either good maths or it's not good maths, that's all that matters. Phil -- Unpatched IE vulnerability: Timed history injection Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/BackMyParent2/BackMyParent2-Content.HTM Exploit: http://www.safecenter.net/liudieyu/BackMyParent2/BackMyParent2-MyPage.HTM === Subject: Re: Fastest factorial algorithm by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAEMoJP24436; Just to compare. and show that 10000! is not that big On my Mac: 10000! took .05 seconds to run 100000! took 1.88 seconds 1000000! took 57.32 seconds === Subject: Re: transfinite series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAF0O7E31187; >> > Take the infinite series expansion for e and put it into the infinite > series expansion of e to the power of x and multiply the terms and you > end up with a series of aleph 1 terms >> >> This is nonsense. Even when you multiply everything out, there are >> still only countably many terms. >> Yes, compare countability of the rational numbers. >> > that sum to a finite result e to > the power of e. If you repete this process for e to the power of e to > the power of e do you end up with aleph 2 terms? >> For series with an arbitrary number of terms, look up the term summable >> family. But a necessary condition for convergence is that at most >> countably many terms are nonzero, this follows from the archimedean axiom >> of the real numbers. >If aleph 1 is the set of all subsets of aleph 0 No, aleph1 is first uncountable ordinal. You must mean bet1, which is usually written c. Whether or not aleph1=c is called the Continuum Hypothesis, and is undecidable in ZFC. Anyway, your argument is incorrect for c as well. > then take the counting >numbers the first subset is the empty set next come the sets with only >one member that is 1,2,3 etc next is the sets with two members these >can be arranged in a two dimensional array 1 2,1 3,1 4 etc 2 3,2 4,2 >5 etc 3 4,3 5,3 6 etc etc all the way up to those with an infinite >number of terms which can be arranged in an array with an infinite >number of dimensions. Now if you want each one of these sets has a >complement however I think when you get to the sets with an infinite >number of terms then each subset is duplicated. which subset did I >miss? You missed infinite subsets. Your counting scheme simply won't work for the infinite number of dimensions case. You did not describe the counting scheme for that case. >As for e to the power of e the first term is 1 next comes 1 + x >+ (x^2)/2 + then comes 1/2 + x/2 + (x^2)/4 + , x/2 + (x^2)/2 + (x^3)/4 >+ , (x^2)/4 + (x^3)/4 + (x^4)/8 + etc arranged in a two dimensional >array all the way up to those with an infinite number of terms that >can be arranged an array with an infinite number of dimensions. you >have a one to one correspondence or at least a one to two >correspondence. Cron === Subject: Re: Leonidas Alaoglu by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAFDHJl14417; I know this is quite a late reply, but since this message turns up on a Google search for Alaoglu... >
>[...]
>                           [in Alaoglu]
>>      Now I know that oglu is an ending for Turkish names and means son
>> of.  I supposed he was Turkish as a result of learning this, so I am
>> surprised that his first name was Leonidas, a Greek name from the famous
>> battle of Thermopylae (or however that is spelled).
>He certainly was Greek!
>Some Greek surnames' prefixes / suffixes are Turkish.
>-oglu is a suffix example.
>A prefix one is Cara- in Caratheodory
>(Kara- = Black, in Turkish)
Yes, I have noticed that, for Greeks, Kara or Cara is a popular 
surname prefix, and oglu or oglou is a popular surname suffix.  Some 
Greek surnames even seem completely Turkish (including root-word).  Also, 
many Armenian surnames have Turkish root-words with ian endings.  The 
phenomena is not surprising.
That being said: However, if you search Alaoglu Greek on Google, you 
don't get any pages referring to Greeks aside from, possibly, Leonidas 
Alaoglu.  On the other hand, if you search Alaoglu Turkish on Google, you 
get several pages referring to present-day Turks with that surname.  
Moreover, the word ala also has a meaning in Turkish (i.e. very good, 
excellent).
Yet, clearly, Leonidas is a Greek name and not a Turkish name.
So, here's what I'm wondering: Is it possible that the mathematician 
Leonidas Alaoglu had a Greek mother and a Turkish father?
Of course, it doesn't matter too much either way, but I just think that it 
would be an interesting possibility (especially given his particular time 
period).
>Antreas (whose the prefix of his surname, namely Hatzi- = pilgrim, is 
Arabic)
Interesting. :)  Even my real name, first and last, (Aynur is meaningful 
pseudonym) has no Arabic influence (and I'm Turkish).
Aynur
>
=== Subject: Re: Leonidas Alaoglu >That being said: However, if you search Alaoglu Greek on Google, you don't get any pages referring to Greeks aside from, possibly, Leonidas Alaoglu. On the other hand, if you search Alaoglu Turkish on Google, you get several pages referring to present-day Turks with that surname. Moreover, the word ala also has a meaning in Turkish (i.e. very good, excellent). >Yet, clearly, Leonidas is a Greek name and not a Turkish name. >So, here's what I'm wondering: Is it possible that the mathematician >Leonidas Alaoglu had a Greek mother and a Turkish father? Unlikely -- marriages between Greeks (read Orthodox Christians) and Turks (read Muslims) in the Ottoman Empire were very uncommon, moreover a Turkish man marrying a Christian woman back then (if not today as well) would have no incentive at all (to put it mildly) to give a non-Muslim/Turkish name to his son ... while a possibility you left out (Greek father, Turkish mother) would be even less likely for reasons I am asking you to guess :-) [Now look at me: Greek national, father's last name Turkish, father's first name Greek (Christos), father's parents both Greeks (Georgios, Antonia)...] baloglouAToswego.edu === Subject: Re: Vector notation for polynomials If x^3 + ax^2 + bx + c = 0 then Xv = xv where v is the vector (1, x, x^2) and X is the matrix with rows (0,1,0), (0,0,1) and (-c, -b, -a). In this case X^3 + aX^2 + bX + c = 0, which is a matrix equation. Now define w = Pv where P is any matrix with an inverse, and Y = PXP^(-1) Then Yw = xw (the roots of the x equation are also eigenvalues of Y, not only of X) and Y^3 + aY^2 + bY + c = 0 Not sure if this is what you're looking for, but you should at least find it interesting. The same logic applies to any polynomial. === Subject: Re: Vector notation for polynomials Steve: >I've been looking at a vector based representation of a second order polynomial. >I wondered if it is possible to represent higher order polynomials in the same way? >I cant see a straight-forward extension to 3rd order for example, if there is one. >This is the seond order representation: >f(x) = x'Ax+b'x+c Since no one else has answered, as a guess, will tensor notation help? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Argument with professor >> I am not a mathematician by trade, but I was talking to a maths >> professor and he absolutely refused to acknowledge the concept of a >> cross product for two vectors that are not 3 dimensional. >> >> For me, >> >> Let A be vector in N space >> Let B be vector in N space >> >> A x B = C >> >> Now, I can prove that C has the property >> >> C.A / |C||A| = C.B / |C||B| >> >> and that C exist and not be a 3-vector. >> We already have this concept. We call it vector >> orthogonal to A and B. There are infinitely many vectors >> C with the property that >> C.A / |C||A| = C.B / |C||B| = 0 >> But that's not what's generally meant by cross product. >> Even in 3-space, there are infinitely many such vectors. >> But only one is A x B. >> So, one could define the cross product between two n-vectors as an >> n-vector with the property >> >> C.A / |C||A| = C.B / |C||B| >> >> where C = A x B >> >> But, am I wrong? Or is the professor wrong? >> You've focused on one property that's easy to generate. >> You haven't stated a general way to calculate this >> thing, nor have you shown a way to do so that gives >> the other properties associated with cross products, >> such as anti-symmetry. >> However, I've heard (in this newsgroup) that a cross >> product can be defined for 7 dimensions. >What is it that makes something a cross product? Is it sufficient to >be a Lie product (anti-symmetric and Jacobi identity)? > A cross product is a special case of the Clifford wedge product. The wedge product of two vectors is called a bi-vector. It just so happens that in R^3 bi-vectors are dual to vectors, so we can interpret the three dimensional bi-vector as corresponding to a special type of vector (usually referred to as an axial vector). This correspondence breaks down in higher dimensions, since the ranks of dual objects must add up to the number of dimensions. Hence, in R^4, bi-vectors are dual to themselves and cannot be interpreted as vectors at all. In even higher dimensions, the situation becomes all the more hopeless. === Subject: Re: Argument with professor > [...] > tell the professor that dimensions 1, 3 and 7 have cross products. > References supplies if you want them. > Does that also work for 11, 13, 17 dimensions? I'm sure it should work for > 11 dimensions. No, sorry. References supplies if you want them. -- G.C. === Subject: Re: Argument with professor >> However, I've heard (in this newsgroup) that a cross >> product can be defined for 7 dimensions. > Nope. The combinatorics do not work out. Yes! Sort of ... There is a bilinear map from R^7 x R^7 -> R^7 (i'll write this as x) where R^7 has the usual dot product satisfying |a x b|^2 = |a|^2 |b^2| - |a.b|^2 (this implies (exercise) that x is antisymmetric). Alas this x is not equivariant under SO(7) (the usual cross product in 3 dimensions is: Ma x Mb = M(a x b) whenever M is an orthogonal transformation of determinant 1). In this case the matrices M satisfying this form a proper subgroup of SO(7). This subgroup is a 14-dimensional compact Lie group: the compact Lie group with root system G_2. An excellent reference for all this is _On Quaternions and Octonions_ by John H. Conway and Derek A. Smith. -- 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: Argument with professor > Ask yourself: if you can come up with something as utterly trivial as > this and someone a whole lot smarter than you who's answered similar > questions from students for years disagrees -- doesn't it strike you > as a lot more plausible to ask obviously I'm missing something here, > could someone fill me in what that is? or some such? What is there was no professor? Don't you look stupid, believer. === Subject: Re: Argument with professor >What is there was no professor? Don't you look stupid, believer. When I call someone stupid, I try to make a habit of making the sentence actually parse. Just a hint for next time. Doug === Subject: Re: Argument with professor >> Ask yourself: if you can come up with something as utterly trivial as >> this and someone a whole lot smarter than you who's answered similar >> questions from students for years disagrees -- doesn't it strike you >> as a lot more plausible to ask obviously I'm missing something here, >> could someone fill me in what that is? or some such? > What is there was no professor? Don't you look stupid, believer. Nice troll. Hope for you you will not have any serious question to ask someday. === Subject: Re: Argument with professor > > I am not a mathematician by trade, but I was talking to a maths > professor and he absolutely refused to acknowledge the concept of a > cross product for two vectors that are not 3 dimensional. > He is correct. > For me, > > Let A be vector in N space > Let B be vector in N space > > A x B = C > > Now, I can prove that C has the property > > C.A / |C||A| = C.B / |C||B| > > and that C exist and not be a 3-vector. > > So, one could define the cross product between two n-vectors as an > n-vector with the property > > C.A / |C||A| = C.B / |C||B| > > where C = A x B > > But, am I wrong? Or is the professor wrong? > You need better math under your belt. If in more than 3-D, what is > the unique direction of the resultant crossproduct vector? There > isn't any. Right-handed and left-handed coordinate systems > interconvert without reflection in 4-D, do they not? Here's some slightly better math from a quick thread from 10 years ago where John Baez and I were tweaking some sap who couldn't even get the 3-d case. -Bruce bbowen@pacbell.nnnnnnnnnnnnnet === Subject: Re: importance of Riemann hypothesis > I'm guessing a complete answer is too much to put in > one post, so what I'd really like to see are > subject, or any good books which discuss some of the > fruitful implications the hypothesis has. Any > suggestions? > Tyler The importance of Riemann's Hypothesis is based in the the consequences it will brings to the future of Mathematics if it results inassailable. How is that a proposition pertaining to complex analysis cannot be handled with all the mathematical paraphernalia? The best web site to consult is: http://www.maths.ex.ac.uk/~mwatkins/zeta/riemannhyp.htm === Subject: Re: importance of Riemann hypothesis >> The Riemann hypothesis obviously gets a lot of good >> publicity (see, e.g., the recent nytimes science >> section). But even as a (beginning) graduate student, >> I don't quite understand why it is so important. > I think the short answer is that there are a lot of important mathematical > proofs that somewhere within them have a step that includes the phrase > now, assuming that Riemann's hypothesis is correct. If Riemann > ultimately turns out to be incorrect, all these other proofs will have to > be reworked not to use Riemann's result. I don't think that is likely. I can imagine people saying, If the Riemann hypothesis holds then ... For example, I believe the Miller-Rabin test for primality can be shown to complete in polynomial time if the generalised Riemann hypothesis holds. (Now of course there is an alternative algorithm proved to complete in polynomial time.) But I can't imagine the result being assumed in a proof, as you suggest. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Big Number Game >> times. > *named after former Senator Roman Hruska, who once gave a speech > extolling the virtues of mediocre people. I think Nixon once nominated some judge for the supreme court and the general (legal) opinion was that he was a mediocre judge. Nixon (or was it Hruska?--this could have been around the same time) responded something to the effect that mediocre people need representation too. I think the judge's name was Carswell. === Subject: Re: billion (was: Big Number Game | A quote from: | The New Fowler's Modern English Usage, Third Edition, Edited by R. W. | Burchfield, The acknowledged authority on English usage | [all of that from the front of the dust jacket...] | | Under the topic billion: | It is best now to work on the assumption that the word means 'a | thousand millions' in all English-speaking areas... The system of number naming with million=10^6, billion=10^12 and so on was described by the Parisian Nicolas Chuquet in an unpublished manuscript in 1484. It was adopted in various European countries before the French (in the 16th century?) began using the system with billion=10^9. Americans adopted the then-French system (along with driving on the right) early on. The British and Germans and others kept the original system. Then in the 20th century France switched back to the original system, and as has been said already the British began using the newer one. Keith Ramsay === Subject: Re: Big Number Game > In message , Richard Heathfield > A trillion seems a little high. (I hear that billion isn't a word > in the UK? Does that mean that trillion isn't either?) >>In the UK, these words do exist, but with different meanings than those to >>which you might be accustomed. Billion means 10^12 (a million squared), >>trillion means 10^18 (a million cubed), quadrillion means 10^24 (a >>million to the power 4), etc etc (except in the hands of ignorant people >>such as politicians, businessmen, and the media). > I think Richard is referring to people born since 1940. No. The people who introduced and enforced the American usage of the word billion were certainly born before 1940. It was introcuced by politicians and the financial press in the 1960s for purely infantile reasons: the wish to talk about billions of pounds in the same way as the big boys (the Americans) were talking about billions of dollars. -- 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: billion (was: Big Number Game > A quote from: > The New Fowler's Modern English Usage, Third Edition, Edited by R. W. > Burchfield, The acknowledged authority on English usage > [all of that from the front of the dust jacket...] > Under the topic billion: > It is best now to work on the assumption that the word means 'a > thousand millions' in all English-speaking areas... But that is a false assumption. My home is an English-speaking area in which the word does not mean that. If the book suggests that it is best to work on false assumptions, then perhaps it isn't that good a book. -- Richard Heathfield : binary@eton.powernet.co.uk Usenet is a strange place. - Dennis M Ritchie, 29 July 1999. C FAQ: http://www.eskimo.com/~scs/C-faq/top.html K&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: billion (was: Big Number Game <4al8rvk1fk87q049sp17kpk615bjqk022o@4ax.com> In message , Richard Heathfield >> A quote from: >> The New Fowler's Modern English Usage, Third Edition, Edited by R. W. >> Burchfield, The acknowledged authority on English usage >> [all of that from the front of the dust jacket...] >> Under the topic billion: >> It is best now to work on the assumption that the word means 'a >> thousand millions' in all English-speaking areas... >But that is a false assumption. My home is an English-speaking area in which >the word does not mean that. >If the book suggests that it is best to work on false assumptions, then >perhaps it isn't that good a book. at least a county, rather than an individual dwelling. Nick -- Nick Wedd nick@maproom.co.uk === Subject: Re: billion (was: Big Number Game > In message , Richard Heathfield > A quote from: > The New Fowler's Modern English Usage, Third Edition, Edited by R. W. > Burchfield, The acknowledged authority on English usage > [all of that from the front of the dust jacket...] > Under the topic billion: > It is best now to work on the assumption that the word means 'a > thousand millions' in all English-speaking areas... >>But that is a false assumption. My home is an English-speaking area in >>which the word does not mean that. >>If the book suggests that it is best to work on false assumptions, then >>perhaps it isn't that good a book. > at least a county, rather than an individual dwelling. According to my dictionary, area can certainly mean that, but it has numerous other meanings as well, including part of a building. If he does not wish to be misinterpreted, perhaps he should make his meaning clearer. -- Richard Heathfield : binary@eton.powernet.co.uk Usenet is a strange place. - Dennis M Ritchie, 29 July 1999. C FAQ: http://www.eskimo.com/~scs/C-faq/top.html K&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: Big Number Game >> The metric system is not American and still everyone know what a kilo >> is. :-) > The British don't. >> Yes, it's 2.204 lb. > Where's Gene? > (i.e. there's a numerical error in that, he normally spots > such things...) Well, the exact value is 1/0.45359237 pounds, which is closer to 2.205 than 2.204. Now, how about the definition of a liquid ounce? (I know in the USA, it's 231/128 x (2.54 cubed) cc, and I think in the UK it's the volume of one ounce (mass) of water at 62 degrees F.) -- Don === Subject: Bredon corrections. I made the (possibly controversial) step of teaching myself homotopy theory from GB's Geometry and Topology book, and noticed a shed load of typos. Does anyone know if they've all been catalogued and whether I could access a list of them somewhere? === Subject: Re: transfinite series >Take the infinite series expansion for e and put it into the infinite >series expansion of e to the power of x and multiply the terms and you >end up with a series of aleph 1 terms that sum to a finite result e to >the power of e. If you repete this process for e to the power of e to >the power of e do you end up with aleph 2 terms? > > You still only get aleph 0 terms. > I think the OP is making a very common error: mistaking the set of > all subsets of N with the set of all *finite* subsets of N Working in binary after the decimal point first you have .000... and its compliment .111... next comes those with one one .1 .01 .001 etc and their compliment .0111... .00111... .000111... then those with two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 .001001 etc etc and their compliments these can be arranged in two two dimensional arrays all the way up to those with an infinite number of ones that can be arranged in two infinite dimensional arrays. Which real number did I miss? So now we have a one to two correspondence between my original construction and the reals between zero and one therefore a series with a continium of nonzero terms sum to a finite result. Mr fritz says this is in impossible. So does e^(e^e) have c members or 2^c members === Subject: Re: transfinite series In sci.math, charles ramsey > Take the infinite series expansion for e and put it into the infinite > series expansion of e to the power of x and multiply the terms and you > end up with a series of aleph 1 terms that sum to a finite result e to > the power of e. If you repete this process for e to the power of e to > the power of e do you end up with aleph 2 terms? No, you end up with aleph 0 terms. If f(x) = a_0 + a_1*x + a_2*x^2 + ... + a_n*x^n and g(x) = b_0 + b_1*x + b_2*x^2 + ... + b_n*x^n then one can do things like f(x)^2 = a_0^2 + 2*a_1*a_0 * x + (2*a_2*a_0 + a_1^2)*x^2 + (2*a_3*a_0 + 2*a_2*a_1)*x^3 + ... + c_n*x^n + ... where c_n is an arithmetic expression in terms of the a_i. (There's probably a name for it but I don't know it offhand.) e^(f(x)) ultimately resolves to something with aleph 0 terms, although actually writing out the n'th term might take quite a bit of doing... :-) The issue is similar to counting the rationals; a positive rational number r = p/q can be associated with (p,q) in (N x N), and that gives one aleph 0 terms as (p,q) can be associated with J by simply sorting the terms using (p+q,p) as the sort key (or, in more conventional form, by threading a diagonal wire through the infinite square array): (1,1) (1,2),(2,1) (1,3),(2,2),(3,1) (1,4),(2,3),(3,2),(4,1) ... or 1 <-> (1,1) 2 <-> (1,2) 3 <-> (2,1) 4 <-> (1,3) 5 <-> (2,2) 6 <-> (3,1) etc. So the rationals don't have aleph 1 elements, either. (Note that the association of r with (p,q) is not one-to-one. It doesn't matter.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Prime conjecture - puzzle The arithmetic mean of two consecutive primes minus the geometric mean of the same two consecutive primes equals less than 1/2. What's the name of the original conjecture? Tapio === Subject: five vectors Hallo, Ich have given 5 vectors and have to find a base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth vector through this base. I think the general approach is to try every possible linear combination: 1.) a*v1+b*v2+c*v3+d*v4 = v5 or 2.) a*v1+b*v2+c*v3+d*v5 = v4 or 3.) a*v1+b*v2+c*v5+d*v4 = v3 or 4.) a*v1+b*v5+c*v3+d*v4 = v2 or 5.) a*v5+b*v2+c*v3+d*v4 = v1 The vectors are: ^^^^^^^^^^^^^^^^^^ v1 = (1,-2,0,3); v2 = (2,-5,-3,6); v3 = (0,1,3,0); v4 = (2,-1,4,-7) v5 = (5,-8,1,2) I got the following results: 1.) a + 2c = 1 and b - c = 1 and d = 1 2.) a + 2b = -3 and c - b = 1 and d = 1 3.) a - d = 2 and b - d = -1 and c + d = 0 4.) a - 3d = 2 and b + d = 0 and c + d = -1 5.) a + d = 0 and 2b - 3d - 1 = 0 and 2c - d - 1 = 0 I have expected to get simple numerical values for a,b,c and d (like: a = 2,...) but not this. Can you tell me what my mistake was? Karl. === Subject: Re: five vectors > Hallo, > Ich have given 5 vectors and have to find a > base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth > vector through this base. > I think the general approach is to try every possible linear > combination: > 1.) a*v1+b*v2+c*v3+d*v4 = v5 > or > 2.) a*v1+b*v2+c*v3+d*v5 = v4 > or > 3.) a*v1+b*v2+c*v5+d*v4 = v3 > or > 4.) a*v1+b*v5+c*v3+d*v4 = v2 > or > 5.) a*v5+b*v2+c*v3+d*v4 = v1 > The vectors are: > ^^^^^^^^^^^^^^^^^^ > v1 = (1,-2,0,3); v2 = (2,-5,-3,6); > v3 = (0,1,3,0); v4 = (2,-1,4,-7) > v5 = (5,-8,1,2) Inspection shows that v1 + v2 + v4 = v5. Further inspection shows that 2v1 - v2 = v3. -- 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: Reality of response to my work > ... > > But are not prime numbers important? > > Interesting, barely important. > > > Shouldn't a method for counting > > them that is a first in recorded human history be of some interest? > > Well, if it was a first, perhaps, yes. It has been shown that your > method is not so very original. > > > Why should people get away with attacking a discoverer who is telling > > the truth? > > We do not know whether you are talking the truth. You have a partial > difference equation and create from that a partial differential > equation and claim they lead to the same results. It is to you to > *prove* that that is so, because it is not necessarily true. Examples > have been given that show that it is indeed not true in a number of > cases. > > Moreover, you state it is easy to program, you have said earlier that > you are a programmer, and now you ask others to provide numerical > results. Why do you not do the programming yourself? > If this guy really has found a way to count prime numbers or to easily > determine if a number is prime or even easily sieve out numbers that > are either prime or not prime, then it is a major discovery. If it is > original, he should get credit regardless of his ego or the egos of > anybody else here. If it is real, he should look into intellectual > property aspects of the work. The market will determine its > usefulness, not the egos of posters here who can only call names. There is no question of the legitimacy of my method for counting prime numbers by integrating a partial difference equation. There is also no question that I'm the first person in recorded history to ever present such a method. There's also no question that partial difference equations are analogs to partial differential equations. I *have* talked to mathematicians about my work, but they seem intent on pursuing a dog-eat-dog academics strategy, where since my find doesn't appear to benefit any of them personally, they refuse to acknowledge its importance. I, of course, find that behavior problematic. Consider also that I found an over one hundred year old error from a definition, which mathematicians just seem intent on denying. So before I can make any money, I have to get past mathematicians, the dark gatekeepers testing their ability to deny knowledge from the general public. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Reality of response to my work > ... > > But are not prime numbers important? > > Interesting, barely important. > > > Shouldn't a method for counting > > them that is a first in recorded human history be of some interest? > > Well, if it was a first, perhaps, yes. It has been shown that your > method is not so very original. > > > Why should people get away with attacking a discoverer who is telling > > the truth? > > We do not know whether you are talking the truth. You have a partial > difference equation and create from that a partial differential > equation and claim they lead to the same results. It is to you to > *prove* that that is so, because it is not necessarily true. Examples > have been given that show that it is indeed not true in a number of > cases. > > Moreover, you state it is easy to program, you have said earlier that > you are a programmer, and now you ask others to provide numerical > results. Why do you not do the programming yourself? > If this guy really has found a way to count prime numbers or to easily > determine if a number is prime or even easily sieve out numbers that > are either prime or not prime, then it is a major discovery. If it is > original, he should get credit regardless of his ego or the egos of > anybody else here. If it is real, he should look into intellectual > property aspects of the work. The market will determine its > usefulness, not the egos of posters here who can only call names. > general consensus here is that it is > 1) not original > 2) slower than current established methods to to the same thing If he files for IP protection, the USPTO will determine originality. If it is slower than established methods, the market (reality) will determine it. I suggest that rather than discussing it here, he should be talking to an IP attorney. He is correct that good original work can be done outside the academic community. For most of recorded history, mathematics was the province of amateurs and there is probably still a lot of room for such. If you are concerned about his ego, then you must not be too familiar with academia where ego is ALL important. Name calling says more about the one calling names than about the person being maligned. === Subject: Re: Reality of response to my work I think you must have come into JSH's life recently. His algorithm was encoded in Java along with existing algorithms, and all were optimised by many people in this ng. His algorithm was no better for any tytpes of tests than existing tests. The underlying theory was shown to be identical to a method from (I think) the mid nineteenth century. That's not actually too bad for an amateur, but nobody likes him because of his revolting attitude. You can killfile JSH, but you can't killfile JSH threads. Lets leave him be. Peter Webb > ... > > But are not prime numbers important? > > Interesting, barely important. > > > Shouldn't a method for counting > > them that is a first in recorded human history be of some interest? > > Well, if it was a first, perhaps, yes. It has been shown that your > method is not so very original. > > > Why should people get away with attacking a discoverer who is telling > > the truth? > > We do not know whether you are talking the truth. You have a partial > difference equation and create from that a partial differential > equation and claim they lead to the same results. It is to you to > *prove* that that is so, because it is not necessarily true. Examples > have been given that show that it is indeed not true in a number of > cases. > > Moreover, you state it is easy to program, you have said earlier that > you are a programmer, and now you ask others to provide numerical > results. Why do you not do the programming yourself? > > If this guy really has found a way to count prime numbers or to easily > determine if a number is prime or even easily sieve out numbers that > are either prime or not prime, then it is a major discovery. If it is > original, he should get credit regardless of his ego or the egos of > anybody else here. If it is real, he should look into intellectual > property aspects of the work. The market will determine its > usefulness, not the egos of posters here who can only call names. > general consensus here is that it is > 1) not original > 2) slower than current established methods to to the same thing > If he files for IP protection, the USPTO will determine originality. > If it is slower than established methods, the market (reality) will > determine it. I suggest that rather than discussing it here, he > should be talking to an IP attorney. > He is correct that good original work can be done outside the academic > community. For most of recorded history, mathematics was the province > of amateurs and there is probably still a lot of room for such. If > you are concerned about his ego, then you must not be too familiar > with academia where ego is ALL important. Name calling says more > about the one calling names than about the person being maligned. === Subject: Re: Reality of response to my work Discussion, linux) > If he files for IP protection, the USPTO will determine > originality. I don't believe mathematical proofs are patentable. Sadly, software algorithms *are* patentable in the United States, and so it is unclear where to draw the distinction. Also sadly, the USPTO is inept at determining originality of software patents, but that would be a point in JSH's favor. I guess to be fair, James never did offer a proof but only a description of how to count primes, as far as I can recall. Who knows? Maybe the patent office in its infinite wisdom will believe that such an algorithm ought to be patentable, centuries of previous mathematical research practices notwithstanding. > If it is slower than established methods, the market (reality) will > determine it. I suggest that rather than discussing it here, he > should be talking to an IP attorney. Dik Winter's recent discussion on primality proving will perhaps shed light on whether James's algorithm. > He is correct that good original work can be done outside the > academic community. On this, I don't think anyone disagrees. Your claim that ego is all-important in academics is churlish and false in my experience, but the above statement is undoubtedly true. Of course, amateurs have a harder time contributing than in past centuries, but this is true of all sciences as well. It is a natural feature of building large bodies of knowledge. The number of questions still open to interested amateurs shrinks as the time required to master prerequisite material grows. Nonetheless, there are still many areas, I assume, in which an amateur might contribute. Prime counting might even be one, but I wouldn't know. -- Even if [...] a communistic regime should come [to China], the old tradition [...] will break Communism and change it beyond recognition, rather than Communism [...] break the old tradition. It must be so. -- Lin Yutang on Socialism with Chinese characteristics in 1935 === Subject: Re: Reality of response to my work > Name calling says more > about the one calling names than about the person being maligned. Now if you could only convince JSH of this, perhaps we could all get back to just mathematics. === Subject: Re: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around Bill, I think you meant /factors of large prime numbers/ ? ;-) Phil -- Unpatched IE vulnerability: Security zone transfer Description: Automatically opening IE + Executing attachments Published: March 22nd 2002 Reference: http://security.greymagic.com/adv/gm002-ie/ === Subject: Re: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around > Bill, I think you meant /factors of large prime numbers/ ? Large prime numbers don't have proper factors. === Subject: Re: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around > > Bill, I think you meant /factors of large prime numbers/ ? > Large prime numbers don't have proper factors. And Dik's name isn't Bill. Sheesh, some people... Phil -- Unpatched IE vulnerability: Notepad popups Description: Opening popup windows without scripting Reference: http://computerbytesman.com/security/notepadpopups.htm Followup: http://msgs.securepoint.com/cgi-bin/get/bugtraq0308/55.html === Subject: Re: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around > Bill, I think you meant /factors of large prime numbers/ ? Yup, something like that. 2^n-1. Factors of large prime numbers would be boring. The group I have been working with has been specified in the papers as a group that was working on factoring large primes... So much about newspaper competence. -- 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: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around > > Bill, I think you meant /factors of large prime numbers/ ? >Yup, something like that. 2^n-1. Factors of large prime numbers would >be boring. >The group I have been working with has been specified in the papers as a >group that was working on factoring large primes... So much about >newspaper competence. There was once a brief note in _Scientific American_ about someone's work on factoring large primes... >-- >dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 >home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ David C. Ullrich === Subject: Re: Reality of response to my work >There was once a brief note in _Scientific American_ about someone's >work on factoring large primes... I assume the person had already managed to factor small primes and had thus moved on to greater challenges. === Subject: Re: Reality of response to my work You know, you're a failure. You're no better at persuading people to buy your flimsy arguments than the people who criticize you are at convincing you their wrong. I guess the posters and the postee deserve each other. Ciao, MB > Some of you might have noticed that I took a break from talking about > the key proof with which I can show an over one hundred year old error > to once again bring up my find of a way to count prime numbers by > integrating a partial difference equation. > Here you can see the truth, clearly presented, and how posters react > to it. > My point is that these posters have an agenda, they're not logical, > and they will simply attack my work no matter how correct it is. > They are apparently *very* angry people, and if you look over my posts > on that partial difference equation, you should be hard-pressed to > find a good reason for their responses. > Maybe you think I'm using hyperbole in my posts by saying that I'm the > only one in recorded human history to find a way to count prime > numbers by integrating a partial difference equation. > But that is a fact. It's not hyperbole to simply state a fact. > However, let's imagine that you're a jealous and angry person, who > finds it INFURIATES you that there's this person daring to post about > their discovery, which makes you look bad, you think. > Now then, if you believe that you can smear this person, and > negatively impact by posting something against them, why not? > My fear is that some of you are attributing traits to posters because > you believe that mathematicians as a group *always* have certain > traits of intelligence and rationality. > But are not prime numbers important? Shouldn't a method for counting > them that is a first in recorded human history be of some interest? > Why should people get away with attacking a discoverer who is telling > the truth? > Why do *you* let them? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Reality of response to my work > You know, you're a failure. You're no better at persuading people to buy > your flimsy arguments than the people who criticize you are at convincing > you their wrong. I guess the posters and the postee deserve each other. > Ciao, time to bring out the hammer! About $1000 a year James to go ahead with a patent if you can see industry relying on the techniques within 20 years. Atleast with usenet you got the nearly machine readable version which would be patent quality. 0.00001 % will follow. Unless you branch into education which is where 98% of money in mathematics is, you must give James credit as an able teacher at times. Marketing step 1 ~ find what product your audience needs. Try getting an email list of maths graduates be your best bet I've pleaded for years for assistance to escape the pervert Truman Company who literally ridicule me the second I wake up to the second I fall asleep, Hey Trueman... what up, my uncle told me you post here... he's did some audio on your secrete show and gave me some tapes. love the show. I'm from Townsville and YOU ARE the Truman I was in Townsville over the weekend, and I heard him. Very spooky! I'm in Townsville. We're sick of you You rule Truman. >Do you know if the truman is living in Townsville? I've been hearing stuff, yeah. http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 http://tinyurl.com/rv5f http://tinyurl.com/v1yf Herc === Subject: Re: Reality of response to my work > You know, you're a failure. You're no better at persuading people to buy > your flimsy arguments than the people who criticize you are at convincing > you their wrong. I guess the posters and the postee deserve each other. > Ciao, > time to bring out the hammer! > About $1000 a year James to go ahead with a patent if you can see industry > relying on the techniques within 20 years. Atleast with usenet you got the > nearly machine readable version which would be patent quality. > 0.00001 % will follow. Unless you branch into education which is where 98% of > money in mathematics is, you must give James credit as an able teacher at times. > Marketing step 1 ~ find what product your audience needs. > Try getting an email list of maths graduates be your best bet > I've pleaded for years for assistance to escape the pervert Truman Company who > literally ridicule me the second I wake up to the second I fall asleep, > Hey Trueman... > what up, my uncle told me you post here... he's did some audio on your > secrete show and gave me some tapes. love the show. > I'm from Townsville and YOU ARE the Truman > I was in Townsville over the weekend, and I heard him. > Very spooky! > I'm in Townsville. We're sick of you > You rule Truman. >Do you know if the truman is living in Townsville? > I've been hearing stuff, yeah. > http://tinyurl.com/iky5 > http://tinyurl.com/iky8 > http://tinyurl.com/iky9 > http://tinyurl.com/iky4 > http://tinyurl.com/rv5f > http://tinyurl.com/v1yf > Herc My experience with patents is that the provisional application which but I advise him to contact a patent attorney to write it. This costs me about $400 for this provisional application. The provisional application protects him for 1 year till he files the complete application. This also gives him time to shop the idea around to companies. After a year he files the complete application. My experience is that this could cost from $6000-10000 when using a patent attorney. I wish him luck as the world needs new ideas. === Subject: Re: Reality of response to my work > You know, you're a failure. You're no better at persuading people to buy > your flimsy arguments than the people who criticize you are at convincing > you their wrong. I guess the posters and the postee deserve each other. > Ciao, > time to bring out the hammer! > About $1000 a year James to go ahead with a patent if you can see industry > relying on the techniques within 20 years. Atleast with usenet you got the > nearly machine readable version which would be patent quality. > 0.00001 % will follow. Unless you branch into education which is where 98% of > money in mathematics is, you must give James credit as an able teacher at times. > Marketing step 1 ~ find what product your audience needs. > Try getting an email list of maths graduates be your best bet > I've pleaded for years for assistance to escape the pervert Truman Company who > literally ridicule me the second I wake up to the second I fall asleep, > Hey Trueman... > what up, my uncle told me you post here... he's did some audio on your > secrete show and gave me some tapes. love the show. > I'm from Townsville and YOU ARE the Truman > I was in Townsville over the weekend, and I heard him. > Very spooky! > I'm in Townsville. We're sick of you > You rule Truman. >Do you know if the truman is living in Townsville? > I've been hearing stuff, yeah. > http://tinyurl.com/iky5 > http://tinyurl.com/iky8 > http://tinyurl.com/iky9 > http://tinyurl.com/iky4 > http://tinyurl.com/rv5f > http://tinyurl.com/v1yf > Herc Although I am sure I could understand the math if I tried, I dont have the time. However, what I do see is many ppl who have allowed their egos get in the way of rationality in this discussion. This guy is correct that much valueable math can be done by non-academics. It has been my experience (physicist techno-entrepreneur) that academics almost never have any knowledge of marketability or practicality of a technology or method in the real world regardless of their level of expertise. I strongly advise this guy to not listen to any of the arguments here and instead apply for some type of intellectual property protection. If it is original, this will be determined by that process. If it is useful, the market will determine that. Even if it is not as useful as he thinks it is, it may be more useful when combined with some other method so he could profit from it. In the US, I think he can still apply for IP protection for 1 year after publication. === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) > Nor have > you provided a relation between energy and mass if you don't > accept relativity. > Socks For a relation between mass and energy using Galilean relativity, see Evidence Against Emission Theories J. G. Fox, Amer. J. Phys. 33, 1 (1965). Androcles === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >HenriWilson skrev i melding >> >> Of course you are funny. >> If electric and magnetic fields acted instantaneously, light would travel at >> infinite speed. >I note with interest that your statement is so ambiguously put >that it may be right as well as wrong. >But I take for granted that your statement is meant to be >relevant to my claim, which was: > as it enters a static electric field. >So assuming that acting instantly is to be understood >in this sense, an unambiguous version of your statement >becomes: > travel at infinite speed. >Was this what you meant to say, Henry? >If not, what DID you mean to say, and what is the relevancy >of what you meant to say to your action time of the static >accelerating field in an accelerator? >Paul, finding the acrobatic show a little monotonous > Remember the old vacuum tube triodes. > Acccording to you, a signal on the grid would be instantly felt at the anode. > If that's not instantaneous communication, what is? > Henri Wilson. > See the Stupidity of Relativity. > www.users.bigpond.com/hewn/index.htm I think Paul's acrobatics are no match for yours, H. His plea of monotony is his way of giving up. Paul wants a static field at the grid. That's ok, until the electron reaches the grid, then that static charge has to reverse sign, and is no longer static. Leaving aside the slew rate of the amplifier driving the grid (which in any real experiment we could not do), if the grid reversed from positive (attracting the electron) to negative as it passed through (propelling the electron on), it would still take a finite amount of time for the field beyond the grid where the electron is heading to change, because c is finite. Hence at the instant the electron reaches the grid, even if the grid potential is zero, there is a negative field ahead of the electron to slow it down. Hence the electron cannot attain c, by this method. However, a positron coming down the other way would have a closing velocity with the electron that was greater than c. That would cause Paul to point to the relativist method of composition of velocities, but of course that requires a universal frame against which those velocities are computed. In this case, that frame would be the frame of the tube, stationary in the universal frame. Move the tube in the universal frame, and that wouldn't affect the closing velocities of the electron and positron. Androcles === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Remember the old vacuum tube triodes. >> Acccording to you, a signal on the grid would be instantly felt at the >anode. >> If that's not instantaneous communication, what is? >> Henri Wilson. >I think Paul's acrobatics are no match for yours, H. His plea of monotony is >his way of giving up. Paul wants a static field at the grid. That's ok, >until the electron reaches the grid, then that static charge has to reverse >sign, and is no longer static. Leaving aside the slew rate of the amplifier >driving the grid (which in any real experiment we could not do), if the grid >reversed from positive (attracting the electron) to negative as it passed >through (propelling the electron on), it would still take a finite amount of >time for the field beyond the grid where the electron is heading to change, >because c is finite. Hence at the instant the electron reaches the grid, >even if the grid potential is zero, there is a negative field ahead of the >electron to slow it down. Hence the electron cannot attain c, by this >method. However, a positron coming down the other way would have a closing >velocity with the electron that was greater than c. Precisely. Paul Andsernon doesn't know what he's talking about. >That would cause Paul to point to the relativist method of composition of >velocities, but of course that requires a universal frame against which >those velocities are computed. In this case, that frame would be the frame >of the tube, stationary in the universal frame. Move the tube in the >universal frame, and that wouldn't affect the closing velocities of the >electron and positron. The velocity addition equation is a very neat mathematical trick that allows SR to prove itself by circularity. >Androcles Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> >> Of course you are funny. >> If electric and magnetic fields acted instantaneously, light would travel at >> infinite speed. >I note with interest that your statement is so ambiguously put >that it may be right as well as wrong. >But I take for granted that your statement is meant to be >relevant to my claim, which was: > as it enters a static electric field. >So assuming that acting instantly is to be understood >in this sense, an unambiguous version of your statement >becomes: > travel at infinite speed. >Was this what you meant to say, Henry? >If not, what DID you mean to say, and what is the relevancy >of what you meant to say to your action time of the static >accelerating field in an accelerator? >Paul, finding the acrobatic show a little monotonous > Remember the old vacuum tube triodes. > Acccording to you, a signal on the grid would be instantly felt at the anode. So if: as it enters a static electric field. it follows: A signal on the grid would be instantly felt at the anode Explain why, please. > If that's not instantaneous communication, what is? Henry, you know of course that you are babbling nonsense. There simply isn't possible to be as stupid as you pretend. Paul === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> > > Of course you are funny. > If electric and magnetic fields acted instantaneously, light would travel at > infinite speed. >> >>I note with interest that your statement is so ambiguously put >>that it may be right as well as wrong. >> >>But I take for granted that your statement is meant to be >>relevant to my claim, which was: >> as it enters a static electric field. >> >>So assuming that acting instantly is to be understood >>in this sense, an unambiguous version of your statement >>becomes: >> travel at infinite speed. >> >>Was this what you meant to say, Henry? >>If not, what DID you mean to say, and what is the relevancy >>of what you meant to say to your action time of the static >>accelerating field in an accelerator? >> >>Paul, finding the acrobatic show a little monotonous >> >> >> Remember the old vacuum tube triodes. >> Acccording to you, a signal on the grid would be instantly felt at the anode. >So if: > as it enters a static electric field. >it follows: >A signal on the grid would be instantly felt at the anode >Explain why, please. >> If that's not instantaneous communication, what is? >Henry, you know of course that you are babbling nonsense. >There simply isn't possible to be as stupid as you pretend. >Paul Let's get this straight. You say that the force on a charge due to an electric field acts instantaneously. Correct? Let us consider a charged sphere somewhere in the universe. It exerts a force on every other charge. If we can arrange for it to lose that charge somehow, you are claiming that all those forces disappear INSTANTLY. I will continue when I receive your answer (if one is forthcoming). Henri Wilson. See the Stupidity of Relativity. www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >HenriWilson skrev i melding >> >> >>HenriWilson skrev i melding >> > > Of course you are funny. > If electric and magnetic fields acted instantaneously, light would travel at > infinite speed. >> >>I note with interest that your statement is so ambiguously put >>that it may be right as well as wrong. >> >>But I take for granted that your statement is meant to be >>relevant to my claim, which was: >> as it enters a static electric field. >> >>So assuming that acting instantly is to be understood >>in this sense, an unambiguous version of your statement >>becomes: >> travel at infinite speed. >> >>Was this what you meant to say, Henry? >>If not, what DID you mean to say, and what is the relevancy >>of what you meant to say to your action time of the static >>accelerating field in an accelerator? >> >>Paul, finding the acrobatic show a little monotonous >> >> >> >> Remember the old vacuum tube triodes. >> >> Acccording to you, a signal on the grid would be instantly felt at the anode. >So if: > as it enters a static electric field. >it follows: >A signal on the grid would be instantly felt at the anode >Explain why, please. >> If that's not instantaneous communication, what is? >Henry, you know of course that you are babbling nonsense. >There simply isn't possible to be as stupid as you pretend. >Paul > Let's get this straight. > You say that the force on a charge due to an electric field acts > instantaneously. Correct? > Let us consider a charged sphere somewhere in the universe. It exerts a force > on every other charge. If we can arrange for it to lose that charge somehow, > you are claiming that all those forces disappear INSTANTLY. > I will continue when I receive your answer (if one is forthcoming). Interesting question, Henry. I've used it myself, on the discussion of gravity propagation. If a star were to convert *all* it mass to radiation in one super-supernova, how long would it take for us to detect its loss of gravity? I don't mean watch its planet suddenly fly away, that would be a distant observation and would reach us at c. I mean detect the gravity loss right here. This would be the biggest gravity pulse (negative going) that we could possibly detect. But in order to detect the loss, where would have to be aware of it in the first place. As it turns out, the nearest star to us (other than the sun) is 3.9 light-years away, and that is just too far to detect it's gravity directly. So I fail to understand why anyone would attempt an experiment to search for gravity waves, other than to give themselves some funding on a futile attempt. There's a lot of money to be made out of relativity, and very few people are altruistic. Androcles === Subject: Re: How many continuous functions are there? Martin === Subject: Re: Homeomorphisms and compactness > I have a small question which I have been unable to solve on my own. I > know that if f: M->N is a homeomorphism, then > K compact in M <=> f(K) compact in N > and I'd like to know if the opposite way holds, that is, if f is > bijective and > K compact in M <=> f(K) compact in N > is f then a homeomorphism? I have proven this to be true for locally > compact spaces, but I would like to know whether the result holds > generally. > Any help? > -Alexander It's true for compactly generated topological spaces (aka k-spaces), which include locally compact and metric spaces. See Dugundji's book. If X is not compactly generated and X' is X equipped with the compactly generated topology induced from the original topology, then X' --> X is a counterexample. === Subject: plotting elliptic curves for LaTeX I am not sure if I have to ask this in the tex newsgroup, but I think the math people know more about this type of problem. I wanted to know if there is a package I can use to plot elliptic curves in LaTeX, can one do it in gnuplot? I may just be able to use plot command and square root of x^3+ax+b and then the negative of it, then stick them together.. but that's messy, probably theres a better way (I mean.. what if I have an algebraic curve defined by f(x,y)=0 and I can't express y in terms of x very simply? is there no way to do it with gnuplot? or maybe other free softwares, or latex packages). As a last resort, I might just travel to university just to use their for any suggestions. Jose Capco === Subject: Start Starting with asking if this is a good forum to ask for math help? I am a college calculus student who is already having trouble, and the semester has barely begun. Thought I would ask if it was alright to post a question before I did so. KL === Subject: Re: Factoring problem, solved > I thought I'd bring this thread back to the subject at hand, as some > posters have pushed it off into a different subject area. > > Remember the problem I'm talking about is the factoring problem. Is it? I thought it was an example of why routine head-exams is a good idea for some people. > > More than likely you used some software today that depends on the > supposed difficulty of factoring very large numbers. supposed difficulty... A few days ago you claimed to have solved this easy problem. Then your algorithm turned out to not work. Then you were puzzled. then it was a half solution now it's gotta worth the theory out... Well clearly you don't have a solution. So isn't it fair to say it IS a difficult problem to solve? Tom === Subject: Re: Factoring problem, solved > I thought I'd bring this thread back to the subject at hand, as some > posters have pushed it off into a different subject area. > > Remember the problem I'm talking about is the factoring problem. > > Is it? I thought it was an example of why routine head-exams is a good > idea for some people. > > > More than likely you used some software today that depends on the > supposed difficulty of factoring very large numbers. > > supposed difficulty... > > A few days ago you claimed to have solved this easy problem. Then > your algorithm turned out to not work. Then you were puzzled. then it > was a half solution now it's gotta worth the theory out... > > Well clearly you don't have a solution. So isn't it fair to say it IS > a difficult problem to solve? > > Tom Stupid, stupid, stupid, your post is so damn stupid considering how often I've explained this now. I have solved the factoring problem. The algorithm I found theoretically should work well over 90% of the time. However, I have a program implementing the algorithm that is heavily recursive, so that it starts factoring at a worse level. Some other program used to fully factor T, the surrogate, with the current algorithm should give a factor over 90% of the time. There is theory, and there is implementation. Now you people are not very bright, so you talk here in reply to me as if this is just some usual not important stuff that typically ends up on Usenet, but if what I'm saying is true, and I prove it mathematically, then someone smarter than you can NOW use the information to factor VERY large numbers, and your chatter will not change a thing. And, being stupid does not protect you from the fall-out, if this thing goes badly. Some of you may still be calmly telling yourselves that I can't be right, even as you become penniless. Your stupidity is not a protection against someone using information that you choose not to believe. James Harris === Subject: Re: Factoring problem, solved > Stupid, stupid, stupid, your post is so damn stupid considering how > often I've explained this now. > > I have solved the factoring problem. Ok. You said it's polynomial time and now you claimed to have solved it. If both of these are true you should be able to [provided the polynomial isn't too unfavourable] factor the RSA composites and go on vacation! > The algorithm I found theoretically should work well over 90% of the > time. Oh. I thought you had a solution. Now one sentence later it's 90%. > However, I have a program implementing the algorithm that is heavily > recursive, so that it starts factoring at a worse level. That shouldn't matter. If you execute a polynomial number of polynomial time steps it will still be polynomial. E.g. O(n^{k+m}) is still polynomial in n. > Some other program used to fully factor T, the surrogate, with the > current algorithm should give a factor over 90% of the time. Wait... So you're saying you require some algorithm [call it A] to factor your surrogate [we can call that a congruence right? Why not stick to REAL math terminology?] What is the running time A? > Now you people are not very bright, so you talk here in reply to me as > if this is just some usual not important stuff that typically ends up > on Usenet, but if what I'm saying is true, and I prove it > mathematically, then someone smarter than you can NOW use the > information to factor VERY large numbers, and your chatter will not > change a thing. Character attacks don't change the nature of your posts, theory, implementation or results. You might as well not include them in your posts. > And, being stupid does not protect you from the fall-out, if this thing > goes badly. Typical paranoid dillusion... > Some of you may still be calmly telling yourselves that I can't be > right, even as you become penniless. > > Your stupidity is not a protection against someone using information > that you choose not to believe. There is a difference between believing you and thinking the worst. Will factoring become polynomial time in the future? It's possible. Is your algorithm the way to get there? Hell no. Oh but please reply and show me how wrong I am. Tom === Subject: Re: Factoring problem, solved >More than likely the correct number is greater than 90% factoring, and >the heavy recursion of my program hides how well the algorithm works. > > A couple of days ago you were asserting that the > 50% number came from Quadratic Residues, and that > you'd proved it. Well, either you're wrong now, or > your proof was wrong then. We can't tell, because > Bull. I started the argument, but didn't finish it in my post, but what I started was quite correct, and not gibberish. It turns out that quadratic residues do play a role, but actually indicate it works better than I realized at first. So I'm figuring things out and clarifying as I go along. I'll go ahead and post the full argument that goes into quadratic residues so that people can see just how crazy Usenet posters who reply to me really are. Oh, to understand the argument, you HAVE to understand things like mod and, of course, even know what a quadratic residue actually is. James Harris === Subject: Re: Factoring problem, solved >More than likely the correct number is greater than 90% factoring, > and >the heavy recursion of my program hides how well the algorithm > works. > > A couple of days ago you were asserting that the > 50% number came from Quadratic Residues, and that > you'd proved it. Well, either you're wrong now, or > your proof was wrong then. We can't tell, because > > > Bull. Correction noed. === Subject: Re: Factoring problem, solved You're saying that Greg doesn't understand elementary number theory? .... Dude, do some research before you perform character attacks. He's a long term cryptographer with involvement in the field for the last [several?] decades. He organized various USENIX conferences and Crypto'03. He's written several published papers and is actively involved in GSM and CDMA groups. He also happens to be a VP of QUALCOMM. Saying Greg doesn't know the math is about the most untrue thing you've said yet in this thread to date. Tom === Subject: How I operate, surrogate factoring Ok, let me explain the purpose of my postings and how I operate as I'm seeing posts from people indicating you don't understand what's going on. I talk out problems I'm working on. You don't need to reply at all, or you can reply, and I may or not respond depending on my mood, as since most of you replying don't bother with mathematics, I typically end up replying when I wish to let off steam by dumping on one of you. I hypothesize in posts, and try to figure out difficult areas where I'm stuck. I have been stuck for a while, which is why you see this explosion of posts. The theory indicated that my method should work a lot better than I thought I was seeing as I expected 100% factoring and was getting less, much less as I used bigger numbers, which I can now explain as being a result of the heavy recursion. On its own based on the mathematics I've just worked out in the last couple of days, the method factors well above 90%, but less than perfect, if you understand recursion at all, means that as my implementation recurses it pulls down even a starting high percentage to one that drops rapidly to well below 50%. That's not a big deal to me now, as it kind of pissed me off for a while, and I just know there must be a way to get 100% factoring, but 90+ percent is good enough to still claim the factoring problem solved. Now I'm focusing more on the underlying theory, and trying to get a handle on some quirks. Besides the fundamental question remains: how do you pick the surrogates? Great fun. As I solve problems and feel less of a need to talk things out, the postings will drop, and I may delete some out as well. Eventually, I'll stop. You need not reply to me. If no one replies at all, it won't make much a difference. I just like talking problems I'm working on out. If you wish to reply, that's your business. If you say something I can seize on when I'm in a bad mood--frustrating work figuring these things out--then I may happily do my best to jump all over you. That's how I operate. Been doing it for years now. Haven't changed much in all that time. James Harris === Subject: Re: How I operate, surrogate factoring Ok, you've solved factoring. I'll give you that. Now give us the running time of your algorithm [or as precise estimate you can]. Tom === Subject: software for plotting equations not arranged into y = f(x) format I'm looking for some plotting sofware that can handle plotting 2d graphs based on equations that are not arranged in the form y = f(x). For example, with gnuplot, I can easily plot a half-circle by typing: plot sqrt(1-x**2) In effect, it is plotting y = sqrt(1-x^2). However, I would like to be able to plot graphs from an equation such as: 1 = x**2 + y**2 but I think gnuplot can't do such a thing. (I know that I can easily rearrange this to be in terms of y, as in the first plot command in this post, but that is beside the point.) What software would be able to handle this? mathcad? maple? Anything free would be nice, but I'll consider commercial software too.. alex === Subject: Re: Basically a sieve method, relation to quantum > The original algorithm in my program, will, my current analysis > shows, > factor about 50% of the time, which is astounding. > > I'm not sure why that's astounding. There are lots of algorithms > that > factor numbers 100% of the time. > > > Yeah but my algorithm does it in polynomial time. Have you proven that point?