mm-313 === Subject: : Non-uniqueness, polynomial factorsIt's easy to demonstrate that polynomials can be factored in aninfinite number of ways--that is, there isn't a unique factorization.I've started showing that by giving you the other side of afactorization I've used for a while by showing(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where f_1(0) = f_2(0) = f_3(0) = 0.That expression represents one way to show a *family* offactorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.Now I'll do some simple algebraic manipulations with that expression.Ok, let g_3(x) = f_3(x) + 3, so I have(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22and now, I multiply both sides by 7, which gives(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),which gives(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)which should look familiar to some of you.Now then, what is required for the a's to be algebraic integer functions?James HarrisMy math discoveries, found for profithttp://mathforprofit.blogspot.com/ === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.Once again you fail to restrain yourself in your definitions and thenproceed to attack other mathematicians for restraining themselves whenthey do things. When proper legitimate mathematicians talk aboutunique factorization of polynomials over a field, they mean that forthis unique factorization, they are restricting themselves to a_fixed_ unique factorization domain R and a _fixed_ polynomial ringR[x_i], where i ranges over some _fixed_ index set T.That ring is a unique factorization domain. When a mathematician wantsto do something like adding a new variable to the variable set or bytaking the ring modulo some funky set of relations, (s)he willexplicitly declare that this is happening instead of automaticallydoing it all the time mentally and therefore claiming that the uniquefactorization theorem I mentioned isn't true. This is what you seem todo. Or do I misunderstand you, given the way you cite the algebraicclosure of C (which makes this unique factorization theorem especiallyeasy to prove for the one-variable polynomial ring over that field)which you probably don't understand how to prove yourself?---- David === Subject: : Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0.You have not shown that that is possible with algebraic integer functions.So we can only assume they are algebraic functions. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions?7 f1(x) and 7 f2(x) are algebraic integer functions. I see no otherrequirement at this moment.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: Non-uniqueness, polynomial factors Adjunct Assistant Professor at the University of Montana. [.snip.]>You have not shown that that is possible with algebraic integer functions.>So we can only assume they are algebraic functions.I would suggest algebraic number-valued functions, since algebraicfunctions has a different meaning (functions obtained frompolynomial functions by use of quotients, positive fractionalexponents, addition, and multiplication...) ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Non-uniqueness, polynomial factors> [.snip.]>You have not shown that that is possible with algebraic integer functions.>So we can only assume they are algebraic functions.> I would suggest algebraic number-valued functions, since algebraic> functions has a different meaning (functions obtained from> polynomial functions by use of quotients, positive fractional> exponents, addition, and multiplication...)Wouldn't you also need functional inversion to handle something likethe inverse function of x^5 + x (as a function from the reals to thereals)? Radicals don't get you that one.---- David === Subject: : Re: Non-uniqueness, polynomial factors Adjunct Assistant Professor at the University of Montana.> [.snip.]>You have not shown that that is possible with algebraic integer functions.>>So we can only assume they are algebraic functions.> I would suggest algebraic number-valued functions, since algebraic>> functions has a different meaning (functions obtained from>> polynomial functions by use of quotients, positive fractional>> exponents, addition, and multiplication...)>Wouldn't you also need functional inversion to handle something like>the inverse function of x^5 + x (as a function from the reals to the>reals)? Radicals don't get you that one.Probably. The point being that algebraic function already has adifferent meaning, so what Dik is refering to is an algebraicnumber-valued function.= =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Non-uniqueness, polynomial factors >>You have not shown that that is possible with algebraic integer functions. >>So we can only assume they are algebraic functions. > I would suggest algebraic number-valued functions, since algebraic >> functions has a different meaning (functions obtained from >> polynomial functions by use of quotients, positive fractional >> exponents, addition, and multiplication...) >Wouldn't you also need functional inversion to handle something like >the inverse function of x^5 + x (as a function from the reals to the >reals)? Radicals don't get you that one. > Probably. The point being that algebraic function already has a > different meaning, so what Dik is refering to is an algebraic > number-valued function.Indeed. But when answering to James I get sloppy on occasion.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integer functions?You mean, for example, a_1(x) = 0, a_2(x) = 0 {both for all x},and a_3(x) = 60025*x^3 - 3675*x^2 - 72*x + 3,perhaps ?Andrzej > James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/ === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integer functions?> James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be takenas *conclusive* proof that you cannot answer.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.----http://www.crbond.com === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integer functions?> James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/> You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken> as *conclusive* proof that you cannot answer.Hey, it's not my fault you don't know enough mathematics to handle thequestion.I notice that the usual suspects are quiet.Where is Dik Winter, Nora Baron, or Rick Decker now?The question is a simple one, what is required for the a's to bealgebraic integer functions?James HarrisMy math discoveries, found for profithttp://mathforprofit.blogspot.com/ === Subject: : Re: Non-uniqueness, polynomial factorsJames HarrisC. Bond> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing[same old ]> Now then, what is required for the a's to be algebraic integerfunctions?> You tell us, Mr. Magnificent Smart Ass. Your failure to respond will betaken> as *conclusive* proof that you cannot answer.> Hey, it's not my fault you don't know enough mathematics to handle the> question.What is an algebraic integer, Harris? === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integer functions?> James Harris> My math discoveries, found for profit> http://mathforprofit.blogspot.com/> You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be > taken> as *conclusive* proof that you cannot answer.> Hey, it's not my fault you don't know enough mathematics to handle the> question.> I notice that the usual suspects are quiet.> Where is Dik Winter, Nora Baron, or Rick Decker now?> The question is a simple one, what is required for the a's to be> algebraic integer functions?> James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/Since Harris does not offer any incentives for coming up with an answers to his questions, I suggest that they be ignored.I am certain that everyone can find better things to do than to pander to Harris' already oversized ego. === Subject: : Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 and now, I multiply both sides by 7, which gives (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. > I notice that the usual suspects are quiet. > Where is Dik Winter, Nora Baron, or Rick Decker now?Sorry, but I am not continuously on-line reading news. > The question is a simple one, what is required for the a's to be > algebraic integer functions?a1, a2 and a3 are algebraic integer functions when 7 f1, 7 f2 and f3are. I do not think there are other requirements. But as you havenot shown yet that 7 f1, 7 f2 and f3 *are* algebraic integer functions,we are in limbo...-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: Non-uniqueness, polynomial factors> You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken> as *conclusive* proof that you cannot answer.> Hey, it's not my fault you don't know enough mathematics to handle the> question.It's your fault if *you* don't know enough mathematics to handle the question. Note that youstill didn't answer. The ball's in *your* court.> I notice that the usual suspects are quiet.> Where is Dik Winter, Nora Baron, or Rick Decker now?> The question is a simple one, what is required for the a's to be> algebraic integer functions?Your failure to answer is conclusive proof that you cannot answer.--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: : Re: Non-uniqueness, polynomial factors> It's easy to demonstrate that polynomials can be factored in an> infinite number of ways--that is, there isn't a unique factorization.> I've started showing that by giving you the other side of a> factorization I've used for a while by showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integerfunctions?> James Harris> My math discoveries, found for profit> http://mathforprofit.blogspot.com/> You tell us, Mr. Magnificent Smart Ass. Your failure to respond will betaken> as *conclusive* proof that you cannot answer.> Hey, it's not my fault you don't know enough mathematics to handle the> question.I'm sure C. Bond can answer his own question, he's just testing you. You arenot smart enough to realize that.David Moran> I notice that the usual suspects are quiet.> Where is Dik Winter, Nora Baron, or Rick Decker now?> The question is a simple one, what is required for the a's to be> algebraic integer functions?> James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/ === Subject: : Re: Non-uniqueness, polynomial factors> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> which should look familiar to some of you.> Now then, what is required for the a's to be algebraic integer functions?> James HarrisA fool can ask questions that wise men cannot answer. It is not clear that the fool asking this question has any idea of the answer to his own question.If past history is anything to go by, he is asking, rather than stating, because he doesn't know the answer, and will steal the result from anyone foolish enough to give him the answer. === Subject: : [JSH] Just Stupid Harrassment of Mathematicians: Re: Non-uniqueness, polynomial factors> Now then, what is required for the a's to be algebraic integer functions?Once again, the top number theorist in the universe realizes that all of hisarguments are fatally flawed and now turns the table once again!What happened, again? Have you finally realized, after a year of chestpounding, that you have no clue of what you speak?You are such a nasty person and I would not like to know you personally asyou have no class and do not know how to have a reasonable discussion on atopic.Month upon month of objections has once again brought you to a point ofbegging for others to show you a proof of something that you cannot possiblydo yourself.You are so sad, but many of us love the ABSOLUTE NONSENSE that youcontinue to spew out! I think you are up to eight years now of wasting yourlife. Perhaps more alcohol is in order!Please take some algebra courses! === Subject: : Re: [JSH] Just Stupid Harrassment of Mathematicians: Re: Non-uniqueness, polynomial factors> Now then, what is required for the a's to be algebraic integerfunctions?> Once again, the top number theorist in the universe realizes that all ofhis> arguments are fatally flawed and now turns the table once again!> What happened, again? Have you finally realized, after a year of chest> pounding, that you have no clue of what you speak?> You are such a nasty person and I would not like to know you personally as> you have no class and do not know how to have a reasonable discussion on a> topic.> Month upon month of objections has once again brought you to a point of> begging for others to show you a proof of something that you cannotpossibly> do yourself.> You are so sad, but many of us love the ABSOLUTE NONSENSE that you> continue to spew out! I think you are up to eight years now of wastingyour> life. Perhaps more alcohol is in order!> Please take some algebra courses!James claims that he uses PDE's, but he has shown repeatedly that he has apoor grasp on algebra and calculus, much less DE. He really needs to takesome math classes and maybe he'd see we're not the liars he makes us out tobe.David Moran === Subject: : Re: Online Calculus III course (Multi-variable)>> Everyone;>> I'm trying to find sources for reasonably priced, good math courses>> online. I'm specifically looking for Calculus III and beyond. I'm not>> interested in any of the pseudo schools that want your money to sell>> you a degree, I want the information.. I'm limited from pursuingthis>> in a formal school environment due to disabilty although over the>> years I did manage to struggle through the first of the Physics>> sequence and Calculus II.> Is there an instructor out there that might consider a self-studywith>> me? I am slow so a self-paced program would work best. A directed>> self-study would probably be the very best.>> Thank you,>> Jo>Many Colleges/Universities offer Extension courses or Distance>Education courses (or some similar name). What is offered and how the>course is administered will vary. I suggest you pick out some schools>near you and check their web sites.>-->Paul Sperry>Columbia, SC (USA)> Probably not likely that such a course is commonly available as shown in> an> extension catalog, but do not ignore the possiblility of openuniversity> method of course registration. If anything online exists for Calculus> III,> student must be very very self motivated to participate this way as> student.> Consider reveiwing the coursework on your own for semester I and IIfirst> before studying semester III.> That is probably not true any more. Tennessee Tech (or is it Belmont> University?) offers a Distance MBA. Well, Business Week Online> (http://www.businessweek.com/bschools/02/distance.htm) says there are 60> distance MBA programs. This list doesn't include at least two other> programs that allow an awful lot of coursework to be done online. UIUC> offers at least many finance courses online -- as real online courses(with> discussions), not just electronic correspondence courses. I've heard that> MIT offers all its courses (even graduate courses?) online. As you cansee,> I haven't verified all these assertions. But visit the schools' websites.> See what they offer.> University of Utah offers Calc I, II and III online.> wunnerful?) They require in person tests, but I'm sure somearrangements> can be made for that (especially if you say the words ADA).> I'm going to bet that you can find a lot. What I don't expect is that you> can find it for a lower price than traditional classroom courses.> Jon MillerI'm currently taking math 2057 from LSU independent study program.The course is Multidimensional Calculus - a good solid 3rd semestercalculus course.Very thorough with a focus on applications rather than formal proofs.Covering areas such as: functions of several variables, diferentiation &integration, vector calculus,line, surface, volume integrals, Stoke's, Greens, Divergence theorems, div,curl, LaGrange, etc.Very affordable. Take a look at http://www.is.lsu.eduI spent a lot of time searching for the best course of this type, and thiswas among the most complete + cheapest. === Subject: : complex contour integralNeed some help with this one:evaluate, if possible, the following contour integral around the simpleclosed contour C,where C is the counter-clockwise circle |z| = 1. Repeat for |z| = 2.(a) Integral[ sin[z]/z dz ]let f(z) = sin[z]/zso if the function f(z) = sin[z]/z is analytic on & inside the smooth closedcontour C we could useCauchy's thm or Cauchy-Goursat and Integral[ f(z) dz ] = 0.But on first inspection it looks like we have a singularity/discontinuity atz = 0, the origin. so we've got a problem.So I try Cauchy's integral formula, this time letting f(z) = sin[z] and thusIntegral[ f(z)/z-a dz ] = 2*Pi*i*f(a), where in this case a = 0Integral[ sin[z]/z dz ] = 2*Pi*i*f(0) = 0.Looking at f(z) = sin[z]/z if we expand in Maclaurin series we get1/z(z - z^3/3! + z^5/t! + ...) = 1 - z^2/3! + z^4/5! + ...So there really isn't a singularity??? And in this case could we really haveused Cauchy-Goursat? The result is the same = 0...Confused, === Subject: : Re: complex contour integralin message :> Need some help with this one:> evaluate, if possible, the following contour integral around the simple> closed contour C,> where C is the counter-clockwise circle |z| = 1. Repeat for |z| = 2.> (a) Integral[ sin[z]/z dz ]> let f(z) = sin[z]/z> so if the function f(z) = sin[z]/z is analytic on & inside the smooth> closed contour C we could use> Cauchy's thm or Cauchy-Goursat and Integral[ f(z) dz ] = 0.> But on first inspection it looks like we have a singularity/discontinuity> at z = 0, the origin. so we've got a problem.> So I try Cauchy's integral formula, this time letting f(z) = sin[z] and> thus Integral[ f(z)/z-a dz ] = 2*Pi*i*f(a), where in this case a = 0> Integral[ sin[z]/z dz ] = 2*Pi*i*f(0) = 0.> Looking at f(z) = sin[z]/z if we expand in Maclaurin series we get> 1/z(z - z^3/3! + z^5/t! + ...) = 1 - z^2/3! + z^4/5! + ...> So there really isn't a singularity??? And in this case could we really> have used Cauchy-Goursat? The result is the same = 0...> Confused,Look up removable singularity in your textbook.-- Jim Heckman === Subject: : Re: complex contour integral complette solution in your e-mail box--thanks for having the decencyto provide the address.>Need some help with this one:>evaluate, if possible, the following contour integral around the simple>closed contour C,>where C is the counter-clockwise circle |z| = 1. Repeat for |z| = 2.>(a) Integral[ sin[z]/z dz ]>let f(z) = sin[z]/z>so if the function f(z) = sin[z]/z is analytic on & inside the smooth closed>contour C we could use>Cauchy's thm or Cauchy-Goursat and Integral[ f(z) dz ] = 0.>But on first inspection it looks like we have a singularity/discontinuity at>z = 0, the origin. so we've got a problem.>So I try Cauchy's integral formula, this time letting f(z) = sin[z] and thus>Integral[ f(z)/z-a dz ] = 2*Pi*i*f(a), where in this case a = 0>Integral[ sin[z]/z dz ] = 2*Pi*i*f(0) = 0.>Looking at f(z) = sin[z]/z if we expand in Maclaurin series we get>1/z(z - z^3/3! + z^5/t! + ...) = 1 - z^2/3! + z^4/5! + ...>So there really isn't a singularity??? And in this case could we really have>used Cauchy-Goursat? The result is the same = 0...>Confused, === Subject: : Factorizations: Connecting the argumentsLet S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22Now consider (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where f_1(0) = f_2(0) = f_3(0) = 0.That expression represents one way to show a *family* offactorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.Now I'll do some simple algebraic manipulations with that expression.Ok, let g_3(x) = f_3(x) + 3, so I have(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22and now, I multiply both sides by 7, which gives(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),which gives(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078,where x is in the ring of algebraic integers.Some smart regrouping of terms gives P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where you should note that using v = -1 + 49x, givesP(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3and now I let the a's be roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice that if I let x=0, I haveP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's at x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) anda_2(0) to equal 0, which leaves a_3(0) with a value of 3.P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7)P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22).Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x)which means that(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22)has a factor of 49.However, the constant term of P(x)/49 is 22, which is verified byagain setting x=0, which gives P(0)/49 = 22.But for two of the factors of P(x), the constant terms is 7, which isNOT a factor of 22. Therefore, *none* of the constant terms ofP(x)/49 as they multiply to give 22 can have 7 as a factor.Given that the constant terms are independent of x's value, it must bethe case that dividing P(x) by 49 divides the two constant terms equalto 7, by 7.But to divide 7 from those constant terms requires dividing throughtwo of the factors, so(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22from reverse use of the distributive property, which gives constantterms that don't have 7 as a factor, as required.Which is(5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22with which I started.James HarrisMy math discoveries, found for profithttp://mathforprofit.blogspot.com/ === Subject: : Re: Factorizations: Connecting the arguments> Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22> Now consider> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.[ Some boring stuff snipped here ]> (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> with which I started.> James Harris Which merely proves that James Steven Harris goes around in circles. === Subject: : Re: Factorizations: Connecting the arguments [snip a boring argument that get us to a point we have been many times before and that nobody disputes]> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22)> has a factor of 49.James then make a correct argument for the behaviour at x=0> However, the constant term of P(x)/49 is 22, which is verified by> again setting x=0, which gives P(0)/49 = 22.> But for two of the factors of P(x), the constant terms is 7, which is> NOT a factor of 22. Therefore, *none* of the constant terms of> P(x)/49 as they multiply to give 22 can have 7 as a factor.> Given that the constant terms are independent of x's value, it must be> the case that dividing P(x) by 49 divides the two constant terms equal> to 7, by 7.James then incorrectly tries to extent this result to values ofx not equal to 0, The factors are not polynomial factors sothis does not work. Just because we divide the first two factorsby 7 at x=0 does not mean we have to divide the first two factorsby 7 at x not equal to 0. > But to divide 7 from those constant terms requires dividing through> two of the factors, so> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> from reverse use of the distributive property, which gives constant> terms that don't have 7 as a factor, as required.> Which is> (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> with which I started.Note f_1(x) and f_2(x) are not algebraic integers for all values ofx. -William Hughes === Subject: : Re: Factorizations: Connecting the arguments > Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 > Now consider > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.Note that you do *not* claim here that the f's are algebraic integerfunctions. If you with they are, please show such. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, > where x is in the ring of algebraic integers. > Some smart regrouping of terms gives > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3v is irrelevant at this point. > and now I let the a's be roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Can you show that the a's that you derived with multiplying the f's by7, 7 and 1 *must* be roots of this cubic? Let's try another regrouping: P(x) = 7^2(2401 x^3 - 147 x^2 + 2x)(5^3) - 3(-1 + 44 x)(5)(7^2) + 7^3and I let the a's be roots of: a^3 + 3(-1 + 44x)a^2 - 49(2401 x^3 - 147 x^2 + 2x). > Notice that if I let x=0, I have > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3.My cubic defining the a's has exactly the same properties. > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22). > Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x) > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) > has a factor of 49. > However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7.Yup, right upto this point. > But to divide 7 from those constant terms requires dividing through > two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22And this is a wrong conclusion. Again, the same wrong conclusion.Dividing off by three functions w1(x), w2(x) and w3(x) such thatw1(0) = w2(0) = 7 and w3(0) = 1 gives exactly the same. And aslong as w1(x), w2(x) and w3(x) multiply together to get 49, thisis also a valid division. Note that the constant term of (5 a1(x) + 7) / w1(x) = (5 a1(0) + 7) / w1(0) = 7/7 = 1(using the definition *you* gave for constant term).So each triple of functions w with the above definitions satisfyyour requirement. And I may also note that this is an infinitefamily of triples. > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required.Reverse use of the distributive property *is not valid*. E.g. ina ring, whenever (a + b)/c is in that ring, (a + b)/c = a/c + b/cis *only* true when all three are in that ring. So while it is truefor (5 a1(x) + 7)/w1(x)for the w's I have previously shown, it is *not* true for those w's for (5 a3(x) + 22)/w3(x)Because neither 5 a3(x)/w3(x), nor 22/w3(x) are algebraic integers. Buttheir *sum* is (with the w's I provided before). Note: w3(x) = gcd(5 a3(x) + 22, 7)which is not necessarily 1... Nor are w1(x) and w2(x) constant. > Which is > (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > with which I started.Yup, but you have not yet shown that f1 and f2 are algebraic integerfunctions.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: Factorizations: Connecting the arguments> Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22> Now consider> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078,> where x is in the ring of algebraic integers.> Some smart regrouping of terms gives > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> where you should note that using v = -1 + 49x, gives> P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3> and now I let the a's be roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Since you say above that a_1(x) = 7*f_1(x), wemust have that 7*f_1(x) is a root of the above polynomialalso. This implies 7^3*[f_1(x)]^3 + 3*(-1 + 49*x)*7^2*[f_1(x)]^2 -49*(2401*x^3 - 147*x^2 + 3*x) = 0, or, cancelling out 49, 7*[f_1(x)]^3 + 3*(-1 + 49*x)*[f_1(x)]^2 - (2401*x^3 - 147*x^2 + 3*x) = 0.Now specializing to x = 1 and letting w = f_1(1) gives 7*w^3 + 144*w^2 - 2257 = 0.The polynomial in w on the left has integer coefficients,and is primitive, irreducible, and *non-monic*. Thereforeits roots cannot be algebraic integers. Therefore f_1(1) cannot be an algebraic integer. Thereforef_1(x) cannot be an algebraic integer function. This contradictsyour implicit assumption at the beginning. Why do you keep doing this ???> Notice that if I let x=0, I have> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic defining the a's at x=0 is> a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and> a_2(0) to equal 0, which leaves a_3(0) with a value of 3.> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7)> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22).> Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x)> which means that> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22)> has a factor of 49.> However, the constant term of P(x)/49 is 22, which is verified by> again setting x=0, which gives P(0)/49 = 22.> But for two of the factors of P(x), the constant terms is 7, which is> NOT a factor of 22. Therefore, *none* of the constant terms of> P(x)/49 as they multiply to give 22 can have 7 as a factor.> Given that the constant terms are independent of x's value, it must be> the case that dividing P(x) by 49 divides the two constant terms equal> to 7, by 7. There is an unspoken assumption here: that the productof the constant terms is the constant term of the product.No problem with that; it is a true fact! Moreover the product of the constant terms, after division by 49, is an integer: 22.Another true fact. However it is NOT true that each of the constant terms, after the division has occurred, is necessarily an algebraic integer. Two of them are (those corresponding to the 7's, but the third is not (the one corresponding to 22). Note well, *this is where you are making your mistake*. Youthink that whatever you divide 22 by, the result must be analgebraic integer. Not true! There is no reason to assume this!> But to divide 7 from those constant terms requires dividing through> two of the factors, so> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> from reverse use of the distributive property, which gives constant> terms that don't have 7 as a factor, as required. What you have overlooked here is that there is *another*way to divide through by 49, and it works perfectly wellwith respect to the constant terms. There are algebraicintegers r, s, and t such that 49 = r*s*t, and a_1(x)/r, 7/r, a_2(x)/s, 7/s, and (5*f_3(x) + 22)/tare all algebraic integers. Here r, s, and t are, likea_1(x), etc., dependent on x: one could write r = r(x),s = s(x), and t = t(x). Also note that the constant terms multiply out exactlyas they should: (7/r)*(7/s)*(22/t) = (49/(r*s*t))*22 = 1*22 = 22.And again: 22/t is ***NOT*** an algebraic integer; moreover,contrary to what you are assuming, *** there is no reasonit needs to be ***.> Which is> (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> with which I started. Yes, this is what you started with. You have successfullygone in a circle. The problem is, what you started withhas no solutions in algebraic integer functions, if in additionyou require, *** as you have said above *** that 7*f_1(x) is a root of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0. To recap yet again: your claims here lead to acontradiction; they must therefore be false. Moreover,I have specified exactly where you are making an error: essentially in assuming that 22/t must be analgebraic integer. Your claims are false and I havepinpointed your error. Time to concede. Nora B.> James HarrisMy math discoveries, found for profit> http://mathforprofit.blogspot.com/ === Subject: : Re: Factorizations: Connecting the arguments> Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22> Now consider> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0.> That expression represents one way to show a *family* of> factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> Now I'll do some simple algebraic manipulations with that expression.> Ok, let g_3(x) = f_3(x) + 3, so I have> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) => 300125 x^3 - 18375 x^2 - 360 x + 22> and now, I multiply both sides by 7, which gives> (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x),> which gives> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078,> where x is in the ring of algebraic integers.> Some smart regrouping of terms gives> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> where you should note that using v = -1 + 49x, gives> P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3> and now I let the a's be roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice that if I let x=0, I have> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic defining the a's at x=0 is> a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and> a_2(0) to equal 0, which leaves a_3(0) with a value of 3.> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7)> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22).> Now P(x) has a factor of 49 as> P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x)> which means that> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22)> has a factor of 49.> However, the constant term of P(x)/49 is 22, which is verified by> again setting x=0, which gives P(0)/49 = 22.P(0) evaluates the polynomial at x=0. P(0)/49 = 22P(1) evaluates the polynomial at x=1. P(1)/49 = 281412P(2) evaluates the polynomial at x=2. P(2)/49 = 2326802--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: : JSH: 22 versus 7I just gave a post which *can* clear up the main issues if you followit all the way through, but it's long, the issues are advanced, and Idon't think many of you care what the mathematical truth is.The primary point that settles things is that 22 is coprime to 7 inthe ring of algebraic integers. Posters using various primitivearguments, and mostly just *claiming* one position that they cannotprove, are stuck with their position forcing them to need 22 if youlook one way, and 7 if you believe them, and look another.But you see 22 is not 7.By showing(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where f_1(0) = f_2(0) = f_3(0) = 0, I'm pushing the underlyingpolynomial to the fore, and for *that* polynomial the constant term is22.Make no mistake, what I do is use special tools of analysis on thepolynomial 300125 x^3 - 18375 x^2 - 360 x + 22.It doesn't take much to see that if the f's are linear functionsmaking the factorization into polynomial factors that multiplyingf_1(x) and f_2(x) by 22 will give you algebraic integer functions.But I've also used a more complicated solution for the f's wheremultiplying them times 7 gives algebraic integer functions.What gives?That's what I'm not going to explain in detail to you, except to saythat the simplest explanation is that the definition for the ring ofalgebraic integers is screwed up enough to let you do some wackystuff.Now then, consider the *demonstrated* apparent contradiction with(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where f_1(0) = f_2(0) = f_3(0) = 0, as you look for 22 times f_1(x)and f_2(x) in one case, while in another you look at 7 times them withthe guarantee that you get algebraic integer functions.If you have any sense at all, you might be now wondering, what does 7have to do with anything?James Harris === Subject: : Re: JSH: 22 versus 7> I just gave a post which *can* clear up the main issues if you follow> it all the way through, but it's long, the issues are advanced, and I> don't think many of you care what the mathematical truth is.> The primary point that settles things is that 22 is coprime to 7 in> the ring of algebraic integers. Posters using various primitive> arguments, and mostly just *claiming* one position that they cannot> prove, are stuck with their position forcing them to need 22 if you> look one way, and 7 if you believe them, and look another.> But you see 22 is not 7.> By showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0, I'm pushing the underlying> polynomial to the fore, and for *that* polynomial the constant term is> 22.> Make no mistake, what I do is use special tools of analysis on the> polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.> It doesn't take much to see that if the f's are linear functions> making the factorization into polynomial factors that multiplying> f_1(x) and f_2(x) by 22 will give you algebraic integer functions.> But I've also used a more complicated solution for the f's where> multiplying them times 7 gives algebraic integer functions.> What gives?> That's what I'm not going to explain in detail to you, except to say> that the simplest explanation is that the definition for the ring of> algebraic integers is screwed up enough to let you do some wacky> stuff.> Now then, consider the *demonstrated* apparent contradiction with> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0, as you look for 22 times f_1(x)> and f_2(x) in one case, while in another you look at 7 times them with> the guarantee that you get algebraic integer functions.What contradiction? The two cases have no connection. Nor would you expect them to.You have noted that there are an infinite number of solutions to (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22Indeed, let f_1(x) be *any* function such that f_1(v) is not equal to -1/5for any v not a root of (300125 x^3 - 18375 x^2 - 360 x + 22). Thenf_1(x) is part of a triplet that solves the equation. Two possibilitiesare the constants, f_1(x) = 3/2 and f_1(x)=5/3. In the first casewhen we multiply by 2 we get an integer, in the second case when wemultiply by 3 we get an integer. The integers 2 and 3 are coprime.Do you think this is a contradiction? -William Hughes === Subject: : Re: JSH: 22 versus 7> But you see 22 is not 7.I'm learning so much from this posting, my head is about to explode ...'22 is not 7'. I need a time to digest this notion. === Subject: : [JSH] Nonsense Continues: Re: 22 versus 7> If you have any sense at all, you might be now wondering, what does 7> have to do with anything?Lets see:It is at least the number of years that you have been spewing nonsense invarious newsgroups?Yeah, that probably it! === Subject: : Re: [JSH] Nonsense Continues: Re: 22 versus 7 > If you have any sense at all, you might be now wondering, what does 7 > have to do with anything? > Lets see: > It is at least the number of years that you have been spewing nonsense in > various newsgroups? > Yeah, that probably it!No. Multiplied by 6 we get that answer to everything.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: JSH: 22 versus 7> I just gave a post which *can* clear up the main issues if you follow> it all the way through, but it's long, the issues are advanced, and I> don't think many of you care what the mathematical truth is.You don't think -- period.> The primary point that settles things is that 22 is coprime to 7 in> the ring of algebraic integers. Posters using various primitive> arguments, and mostly just *claiming* one position that they cannot> prove, are stuck with their position forcing them to need 22 if you> look one way, and 7 if you believe them, and look another.> But you see 22 is not 7.May I quote you? But you see 22 is not 7. -- James Harris> By showing> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0, I'm pushing the underlying> polynomial to the fore, and for *that* polynomial the constant term is> 22.> Make no mistake, what I do is use special tools of analysis on the> polynomial 300125 x^3 - 18375 x^2 - 360 x + 22.You made the mistake, unless you'd care to cite which tools of analysisyou are claiming to use here.> It doesn't take much to see that if the f's are linear functions> making the factorization into polynomial factors that multiplying> f_1(x) and f_2(x) by 22 will give you algebraic integer functions.It would help if you gave the values for f_1(x) and f_2(x).> But I've also used a more complicated solution for the f's where> multiplying them times 7 gives algebraic integer functions.> What gives?> That's what I'm not going to explain in detail to you, except to say> that the simplest explanation is that the definition for the ring of> algebraic integers is screwed up enough to let you do some wacky> stuff.Wrong. The simplest explanation is that you are an ignorant blowhard withthe intelligence of an imbecile and the cognitive faculties of a hockypuck.> Now then, consider the *demonstrated* apparent contradiction with> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where f_1(0) = f_2(0) = f_3(0) = 0, as you look for 22 times f_1(x)> and f_2(x) in one case, while in another you look at 7 times them with> the guarantee that you get algebraic integer functions.> If you have any sense at all, you might be now wondering, what does 7> have to do with anything?I'm more interested in whether you might have eaten too much lead paint asa child.--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: : Re: Algebra> Prove that the 2D quadratic from aX^2 +bXY+ cY^2 is a positive> definite iff a>0 and 4ac - b^2 > 0> I can see that it must come from the quadratic equation but cannot get> any further, can anyone give me any help? (The proof of your result _does_ look a little bit like the manipulations that one goes through in deriving the quadratic formula ... ) First, note that aX^2 + bXY + cY^2 = a(X^2 + (b/a)XY) + cY^2 = a(X^2 + 2(b/2a)XY) + cY^2 and completing the square inside those parentheses gives aX^2 + bXY + cY^2 = a(X + (b/2a)Y)^2 + (4ac - b^2)/4a Y^2 . If a > 0 and 4ac - b^2 > 0, then aX^2 + bXY + cY^2 >= 0 for all X, Y and it is = 0 iff X + (b/2a)Y = Y = 0 iff X = Y = 0. So your form _is_ positive definite in this case. For the other direction, if the form is positive definite, evaluate it for X nonzero and Y = 0 to see that a > 0. Then evaluate it for some nonzero Y and X = -(b/2a)Y to see that (4ac - b^2)/4a > 0 (and, hence, that 4ac - b^2 > 0, as you already know that a > 0). === Subject: : Re: Algebra> Prove that the 2D quadratic from aX^2 +bXY+ cY^2 is a positive> definite iff a>0 and 4ac - b^2 > 0> I can see that it must come from the quadratic equation but cannot get> any further, can anyone give me any help?First, lemma:2D quadratic form aX^2 + bY^2 is positive definite iff a>0 and b>0If a>0 and b>0, clearly aX^2 + bY^2>0 when x or y is not zero. In the other direction, assume a<0; then the form is <0 when y = 0, and likewise for b<0.Second, transform the form you are given into that form:a X^2 + b XY + c Y^2 =a (X^2 + b/a XY + c/a Y^2) =a (X^2 + 2 * b/2a XY + c/a Y^2) =a (X^2 + 2 * X * b/2a Y + (b/2a Y)^2 - (b/2a Y)^2 + c/a Y^2) =a ( (X + b/2a Y)^2 + (c/a - b^2 / 4a^2) Y^2) =a ( (X + b/2a Y)^2 + (4ac - b^2)/4a^2 * Y^2)a * (x + b/2a Y)^2 + (4ac - b^2)/4a * Y^2By the lemma, this is positive definite if a>0 and (4ac - b^2)/a > 0.hthmeeroh-- If this message helped you, consider buying an itemfrom my wish list: Prove that the 2D quadratic from aX^2 +bXY+ cY^2 is >positive definite iff a>0 and 4ac - b^2 > 0==> set y = 0, x > 0; ax^2 > 0; a > 0; c > 0 likewise set x = 1/sqr a, y = +-1/sqr c ax^2 + bxy + cy^2 = 2 +- b/(sqr a)(sqr c) > 0 2 > |b/(sqr a)(sqr c)|; 4 > b^2 / ac<==if ax^2 + bxy + cy^2 = d; ax^2 + bxy + cy^2 - d = 0 x = (-by +- sqr(b^2 y^2 - 4a(cy^2 - d))/2a; x real 0 <= b^2 y^2 - 4a(cy^2 - d) = y^2 (b^2 - 4ac) + 4ad 0 <= y^2 (4ac - b^2) <= 4ad; 0 <= d if d = 0: y = 0; x = 0---- === Subject: : JSH: I WILL GET MY MONEYIf you ing morons think that I will let you get away with notgiving me credit for my ing math discoveries then you have anothering thing coming.What the ??!!!Where's pure math now, huh? Where's loving math for the ingbeauty of it now you ing s??!!!LOOK AT IT!!!Here is the partial difference equation and instructions forintegrating.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.And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dSfrom dS(x,2) to dS(x,y).http://mathforprofit.blogspot.com/You ing s. I will get credit for my discovery and get ingpaid, and you best believe that I will not ing let you stupids get away with your ing stupid bull--pure math myass--without me coming at you ers with some ing PURE ingPURE AS math that you stupid s have been ting on for overa ing YEAR!!!You goddamn S!!!What the is wrong with you s??!!! Don't you even believe inyour own stupid ? Where's pure math now?Where is it?James Harris === Subject: : Re: I WILL GET MY MONEY I will get credit for my discovery and get> paid,Money is one of two things mathmeticians do not get much of And this would be the other ... James Harris === Subject: : Re: Thank you, James!Thanks, James! For the first time, I have actually learned somethingfrom one of your posts. Unfortunately, it's not mathematics, butgrammar usage. The lively discussion between another thing andanother think has taught me the actual use of the term (I alwaysheard it as thing, cpompletely unaware of the think version).Oh yeah, the word also seems to be used quite a bit in yourpost too. Not really sure I have learned much from that.I can't remember anything else about your post. I think I got themost important parts out of it though.Jonathan hoyle === Subject: : Re: Thank you, James!> Thanks, James! For the first time, I have actually learned something> from one of your posts. Unfortunately, it's not mathematics, but> grammar usage. The lively discussion between another thing andanother think has taught me the actual use of the term (I always> heard it as thing, cpompletely unaware of the think version).> Oh yeah, the word also seems to be used quite a bit in your> post too. Not really sure I have learned much from that.Remember, the FCC recently said it's ok to drop F-bombs on network television, as long as it's not used to refer to anything sexual. So JSH's usage does in fact conform to current FCC policy. === Subject: : Re: JSH: I WILL GET MY MONEYFirst, do you have a f*ck fetish? Your use of f*ck is extreme. Ihave never seen anybody knowledgealbe in math use f*ck nearly as muchas you have here.> If you f*cking morons think that I will let you get away with not> giving me credit for my f*cking math discoveries then you have another> f*cking thing coming.What discovery? You can have your fatally flawed factorization. Asfor your PDE, it looks like a rehash of Legendre's Formula. See:http://mathworld.wolfram.com/LegendresFormula.html> What the f*ck??!!!> Where's pure math now, huh? Where's loving math for the f*cking> beauty of it now you f*cking s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You f*cking s. I will get credit for my discovery and get f*cking> paid, and you best believe that I will not f*cking let you stupidChange 'paid' to 'laid', it would be more consistent with your 'f*ck'rampage.> f*cks get away with your f*cking stupid bull--pure math my> ass--without me coming at you f*ckers with some f*cking PURE f*cking> PURE AS F*CK math that you stupid s have been ting on for over> a f*cking YEAR!!!> You goddamn F*CKS!!!> What the f*ck is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James Harris === Subject: : Re: I WILL GET MY MONEYHi James. I checked your math - it's wrong. I willbe cancelling all of your posts later this eveningso that no one will have to read them. In thefuture, kindly do not post any text to these newsgroups.Cheers! === Subject: : Re: JSH: I WILL GET MY MONEY>If you ing morons think that I will let you get away with not>giving me credit for my ing math discoveries then you have another>ing thing coming.>What the ??!!!>Where's pure math now, huh? Where's loving math for the ing>beauty of it now you ing s??!!!You got Tourette's syndrome now, Jimmy? That's a pretty serious issue.Doug === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.snip...Forgive me for posting to the original (I can read usenet from mynewsreader, but not write -- so I post on Google) as this is reallyaddressed to the far off-topic (but far more interesting) sub-threadabout ... another thing/k coming. I heard a phrase on /. the otherday that went something like If I ever see a spammer in person THEDEVIL'S GOING TO NEED A WIG. Has anyone heard that phrase before? Forsome reason, that cracked me up. I interpret it to mean that the guyis going to do something so wicked that the devil will need to wear adisguise out of shame. Can anyone clarify?> James Harris- Brett === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with > not giving me credit for my ing math discoveries then you > have another ing thing coming.> snip...> Forgive me for posting to the original (I can read usenet from my> newsreader, but not write -- so I post on Google) as this is > really addressed to the far off-topic (but far more interesting) > sub-thread about ... another thing/k coming. I heard a phrase > on /. the other day that went something like If I ever see a > spammer in person THE DEVIL'S GOING TO NEED A WIG. Has anyone > heard that phrase before? For some reason, that cracked me up. I> interpret it to mean that the guy is going to do something so > wicked that the devil will need to wear a disguise out of shame. > Can anyone clarify?I've never heard that phrase before, so you'll have to take thiswith a grain of salt, but, irregardless, I would guess thatthe devil refers to the spammer rather than The Devil,and needing a wig is a variation on a (somewhat) mild threatthat I _have_ heard, something along the lines of I'm goingto snatch (you, him, whoever) bald-headed. I think snatching bald-headed is intended to symbolize a seriousbut not life-threatening punishment, such as a mother mightoffer to her child (well, me, really ).Hopefully, I'm right, but, well, whatever, you know?Jim Burns === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James Harris ____ /(( )) ( )6 6( ) (_) l (_) DR. QUINN, DISPOSABLE COCKSUCKER <> ) ____) (_____ ( ____/ ) ) ( )( ) ( / / / / / / )==( / / / / / '/ //' '|` '|` / / ) ( / / / / / / `-...., ,..-' `-..-' === Subject: : Re: JSH: I WILL GET MY MONEYfuffy> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!> Where's pure math now, huh? Where's loving math for the ing> beauty of it now you ing s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James Harris === Subject: : Re: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!> Where's pure math now, huh? Where's loving math for the ing> beauty of it now you ing s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?It's behiiiind yooou - over THERE....![pure math exits stage left, runs round back of stage and appears stageright...]> Where is it?No, OVER THERE...! IT'S BEHIND YOUUU!!!!> James Harris === Subject: : Re: I WILL GET MY MONEYMike Terry was kind enoughto quote the latest JSH eruption in its entirety so some of us couldsee the poetic phrasing to which Mr Harris has descended:>> You ing s. I will get credit for my discovery and get ing>> paid,Um, one needs to get into an older profession than mathematician for that.(Who, exactly, does he think pays for mathematical discoveries?) === Subject: : Re: I WILL GET MY MONEY> Mike Terry was kind enough> to quote the latest JSH eruption in its entirety so some of us could> see the poetic phrasing to which Mr Harris has descended:>> You ing s. I will get credit for my discovery and get ing>> paid,> Um, one needs to get into an older profession than mathematician forthat.> (Who, exactly, does he think pays for mathematical discoveries?)The math fairy. See you leave your proofs under the pillow before you go tobed, and in the morning you'll find in place of said proof a direct depositreceipt.HTHMB === Subject: : Re: I WILL GET MY MONEY>Mike Terry was kind enough>to quote the latest JSH eruption in its entirety so some of us could>see the poetic phrasing to which Mr Harris has descended:> You ing s. I will get credit for my discovery and get ing> paid,>Um, one needs to get into an older profession than mathematician for that.>(Who, exactly, does he think pays for mathematical discoveries?)People have been wondering that ever since he first said he was inthis for the money.Of course the fact that none of us has ever been paid for discoveringa bit of mathematics doesn't really prove anything, since nobody inhistory has ever made the _sort_ of mathematical discovery he has.Presumably there are plenty of people out there with millions ofdollars, just waiting to hear about it when someone finally provessomething non-trivial. === Subject: : Re: JSH: I WILL GET MY MONEYContent-transfer-encoding: 8bit> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!! edy .Okay, now we've both got that out of our systems, at least for awhile...Who has made a commitment to pay you? === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.That should be: '... you have another THINK coming.' How appropriate.> What the ??!!!> Where's pure math now, huh? Where's loving math for the ing> beauty of it now you ing s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You ing s. I will get credit for my discovery and get ing> paid, Who is going to pay you? Hmmm? Surely no one here. Well..., someonemight pay you to shut up.> and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James HarrisYou should include this message with your journal submissions. Surelythe editors wouldn't resist your eloquence, despite all the phonecalls and e-mails they receive from your detractors here at sci.math. === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.Dear James,the correct expression is If you ing morons think X, you haveanother ing *think* coming, not another ing *thing* coming,which makes no sense.I hope this calms you down a bit. Cordially --Jeremy === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!> Where's pure math now, huh? Where's loving math for the ing> beauty of it now you ing s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James HarrisWith all due contempt, James, you have now flunked out of Sandbox 101.Your inexcusable social behavior and immature language are eloquenttestimony that you must return to the Playpen for a course in remedialcounting.If you fail again, you will be sent to crash dummy training school.--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: : Re: JSH: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!That's a lot of f-bombs dude, what's up? === Subject: : Re: I WILL GET MY MONEY> If you ing morons think that I will let you get away with not> giving me credit for my ing math discoveries then you have another> ing thing coming.> What the ??!!!> Where's pure math now, huh? Where's loving math for the ing> beauty of it now you ing s??!!!> LOOK AT IT!!!> Here is the partial difference equation and instructions for> integrating.> 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.> 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).> http://mathforprofit.blogspot.com/> You ing s. I will get credit for my discovery and get ing> paid, and you best believe that I will not ing let you stupid> s get away with your ing stupid bull--pure math my> ass--without me coming at you ers with some ing PURE ing> PURE AS math that you stupid s have been ting on for over> a ing YEAR!!!> You goddamn S!!!> What the is wrong with you s??!!! Don't you even believe in> your own stupid ? Where's pure math now?> Where is it?> James HarrisMAYBE IF YOU LEARNED SOME MATH, YOU'D REALIZE THAT WE WERE TELLING THETRUTH. You've been drinking again, haven't you? You are way too immature forbeing an adult.David Moran === Subject: : Re: JSH: The reverse factorization argument Adjunct Assistant Professor at the University of Montana. [.snip.]> I hereby invite Arturo back into the fray, to respond at least to >posts of others, if not to your own.I am reading the threads, but I do not want to reply even indirectlyto any of James's arguments. == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: JSH: The reverse factorization argumenthttp://mathforprofit.blogspot.com/Was this really profitable, and did you claim it as income ?Has anyone here ever thought of Math as a martial art ? === Subject: : Re: JSH: All the dumb crap in journals> Sci.math has nothing to do with what gets published in math journals,> for the simple reason that most of the 'prestigious' journals are> owned by, and run in the interest of, a handful of large corporations,> or professional guilds like the ASL, which have no reason whatsoever> to publish anything which departs from prevailing mathematical norms;> whereas sci.logic and sci.math provide forums (in principle at least),> for work in progress which flaunts these norms, and in so doink> challenges the credibility and authority of gate-keepers...> Some people think that the mathematics I do indeed belongs on the> usenet. -Dar S. Kabatoff> Name one, other than your alter egos.> Your god isn't powerful enough to...> a) give you a Book of instruction> b) provide you with your name> c) know how to countYou did NOT answer my question: Name one person (other than one ofyour alter egos) who thinks that the mathematics you do belongs on theusenet. === Subject: : Re: JSH: All the dumb crap in journals> ...> ...> Sci.math has nothing to do with what gets published in math journals, > for the simple reason that most of the 'prestigious' journals are> owned by, and run in the interest of, a handful of large corporations,> or professional guilds like the ASL, which have no reason whatsoever> to publish anything which departs from prevailing mathematical norms;> whereas sci.logic and sci.math provide forums (in principle at least),> for work in progress which flaunts these norms, and in so doing> challenges the credibility and authority of gate-keepers like yourself.> Even if what you say about a the manipulating powers behind journals> were true, flaunting the norms of mathematics would still be no better> method for advancing mathematics. You use the term prevailing as if> the norms of mathematics could be something else, but they just are> period, regardles of anybody's personal, political or whatsoever> interests.Mathematics is like art and fashion, it is defined by its producersand consumers and it has its trends and staples. Everything changesin mathematics, some faster than others.--Vaughan Pratt, FOM> The conflict between you and JSH (among others) is brought about> by factors that are evident in Budd Schulberg's contrast of the> businessman with the artist.> and the artist:> A good businessman...aims to please as many people as possible while> minimizing risk and standardizing production. The aim of the good> artist, on the other hand, is exactly the opposite: he turns his> back on every formula, keeps breaking new ground, risks everything,> and whether he succeeds or fails, prepares to risk again.> (Budd Schulberg, Movies in America: After Fifty Years, _The> Atlantic Monthly_, November 1947; cited in Leo Gurko, _Highbrows> and the Popular Mind_, Charter Books, 1953, p. 165)> This is quite ridiculous. A good businessman aims to maximize his> profit and the aim of an artist is not definable in a few lines.Businessman maximize profit by controlling markets and stiffingcompetitors. What is more, nothing is more foreign to the businessmentality than JSH's 'go-for-broke' approach to FLT.> However it is pretty clear that rebellious, revolutionary> avant-gardism, is only one of a multitude of artistic attitudes which> becomes fashionable from time to time, and is in itself no guarantee> for artistic creation.> The implication of your posting is, that the mere act of rebellion> against norms has in itself a positive effect towards whatever purpose> you are intent on attaining, but anyone can see that this is not the> case most of the time.> You wouldn't want to fly in a plane piloted by someone whose only> credentials as pilot are that he refuses to see any benefit in> actually learning to fly.> A pilot who revolutionizes aviation will first have to learn theprevailing norms or he won't have a chance of contributing anything> new to his craft (unless he finds a way to do so in afterlife).> And by the way, JSH openly states that he is into revolutionizing math> for profit, as a businessman in this regard, he has failed both> pleasing many people, as your Budd Schulberg requires, and making> any profit.More relevant to the sci.math treatment of JSH (who has neverexpressed an interest in 'revolutionizing aviation'), ishow society treats inventors. Writing in 1955, Gurko says:Inventors are the real professionals of a mechanical society; onewould expect them, of all people, to be regarded with respect,if not actual reverence. Yet few groups are more scorned. Until,that is, their inventions succeed, at which time they becomegarlanded heroes instantly escorted to the hall of fame, wherethey are set up as exemplars for future generations of boys toenvy and emulate. Before an invention is commercialized, however,the inventor must be prepared for calumny. He is a crank, crackpot,dabbler, nut; he is a fool slightly tilted toward the bughouse; heis an irresponsible loafer probably neglecting his wife and children;he just isn't right in the head. But the minute he hits the jackpot,these accusations are washed away and forgotten in a flood of acclaim.His eccentricity has been known all along to be a sign of genius. Hisimpracticality has become vision; his neglect of loved ones a desireto win for them a greater security; his indifference to criticismis no longer a mark of footlessness but of courage. So long asthe inventor has only his brains, he will be labeled with contemptuousepithets, from harmless crank to raving lunatic. Only the successfulmarketing of his product can save him... Leo Gurko, _Heroes,Highbrows and the Popular Mind_ (New York: Charter Books, 1953),pp. 54-55) === Subject: : Re: JSH: All the dumb crap in journalsExcuse me John, can you, aside from quoting others, please give ananswer to what I said or is this conversation pointless? I gave yousome thoughts on the fact that you are wrong assuming that asuming amaverick attitude is a positive step in itself towards attaining apurpose. You come to me with ...JSH (who has never expressed aninterest in 'revolutionizing aviation')..., have you never heard ofmetaphores?Then you are implying that JSH is receiving the treatment of inventorsas described by Gurko. Well, that description does not include thetreatment he has been given by many here, who have in the past andstill keep even now, taken the trouble of reading his posts andpointing out his mistakes on the mere ground of human cordiality. Youknow, math is a field where you are either right or wrong but theresnothing in between. The nice thing is you can prove you're right. Thereal trouble with JSH or your defending him, is the use of wrongreasons. His postulates are not going to become right by arguing abouthow he is treated, this is not politics.> ...> ...> Sci.math has nothing to do with what gets published in math journals, > for the simple reason that most of the 'prestigious' journals are> owned by, and run in the interest of, a handful of large corporations,> or professional guilds like the ASL, which have no reason whatsoever> to publish anything which departs from prevailing mathematical norms;> whereas sci.logic and sci.math provide forums (in principle at least),> for work in progress which flaunts these norms, and in so doing> challenges the credibility and authority of gate-keepers like yourself.> Even if what you say about a the manipulating powers behind journals> were true, flaunting the norms of mathematics would still be no better> method for advancing mathematics. You use the term prevailing as if> the norms of mathematics could be something else, but they just are> period, regardles of anybody's personal, political or whatsoever> interests.Mathematics is like art and fashion, it is defined by its producers> and consumers and it has its trends and staples. Everything changes> in mathematics, some faster than others.> --Vaughan Pratt, FOM> The conflict between you and JSH (among others) is brought about> by factors that are evident in Budd Schulberg's contrast of the businessman with the artist.> and the artist:> A good businessman...aims to please as many people as possible while minimizing risk and standardizing production. The aim of the good> artist, on the other hand, is exactly the opposite: he turns his> back on every formula, keeps breaking new ground, risks everything,> and whether he succeeds or fails, prepares to risk again.> (Budd Schulberg, Movies in America: After Fifty Years, _The> Atlantic Monthly_, November 1947; cited in Leo Gurko, _Highbrows> and the Popular Mind_, Charter Books, 1953, p. 165)> This is quite ridiculous. A good businessman aims to maximize his> profit and the aim of an artist is not definable in a few lines.> Businessman maximize profit by controlling markets and stiffing> competitors. What is more, nothing is more foreign to the business> mentality than JSH's 'go-for-broke' approach to FLT.> However it is pretty clear that rebellious, revolutionary> avant-gardism, is only one of a multitude of artistic attitudes which> becomes fashionable from time to time, and is in itself no guarantee> for artistic creation.> The implication of your posting is, that the mere act of rebellion> against norms has in itself a positive effect towards whatever purpose> you are intent on attaining, but anyone can see that this is not the> case most of the time.> You wouldn't want to fly in a plane piloted by someone whose only> credentials as pilot are that he refuses to see any benefit in> actually learning to fly.> A pilot who revolutionizes aviation will first have to learn theprevailing norms or he won't have a chance of contributing anything> new to his craft (unless he finds a way to do so in afterlife).> And by the way, JSH openly states that he is into revolutionizing math> for profit, as a businessman in this regard, he has failed both> pleasing many people, as your Budd Schulberg requires, and making> any profit.> More relevant to the sci.math treatment of JSH (who has never> expressed an interest in 'revolutionizing aviation'), is> how society treats inventors. Writing in 1955, Gurko says:Inventors are the real professionals of a mechanical society; one> would expect them, of all people, to be regarded with respect,> if not actual reverence. Yet few groups are more scorned. Until,> that is, their inventions succeed, at which time they become> garlanded heroes instantly escorted to the hall of fame, where> they are set up as exemplars for future generations of boys to> envy and emulate. Before an invention is commercialized, however,> the inventor must be prepared for calumny. He is a crank, crackpot,> dabbler, nut; he is a fool slightly tilted toward the bughouse; he> is an irresponsible loafer probably neglecting his wife and children;> he just isn't right in the head. But the minute he hits the jackpot,> these accusations are washed away and forgotten in a flood of acclaim.> His eccentricity has been known all along to be a sign of genius. His> impracticality has become vision; his neglect of loved ones a desire> to win for them a greater security; his indifference to criticism> is no longer a mark of footlessness but of courage. So long as> the inventor has only his brains, he will be labeled with contemptuous> epithets, from harmless crank to raving lunatic. Only the successful> marketing of his product can save him... Leo Gurko, _Heroes,> Highbrows and the Popular Mind_ (New York: Charter Books, 1953),> pp. 54-55) === Subject: : Re: JSH: All the dumb crap in journalsYes, visionaries are sometimes considered cranks. But far more often cranksare considered cranks. The reason we don't grab every nutty doofus thatcomes along and follow them around as though they were the next coming ofjesus is that most are in fact cranks. So, human experience is vindicated.Every once and awhile some lunatic will also be a genius (or occasionallylucky) and revolutionalize a field. The vast majority, however, will justgive their neighbors something to gawk at. There are alot of inept peopleall over the place. Alot of them make outrageous claims. For example, bagladies, bums, religious wackos, etc. We don't listen, because they arenuts.To sum it up for you John, JSH is not a revolutionary. He is doesn't knowanything about the math he claims to use.One more point. Most people who have changed the world through theirinventions weren't considered cranks. Most quietly laboured away foryears. The crank-genius is an exception, of course their stories are farmore interesting than the rest. (Euler, Fourier, Gauss, Hilbert, Legendre,Leibnitz, Mandelbrot, Noether, Poincare, Riemann, Stokes, Taylor, Weyl,Abel, Bessel, Boole, Des Cartes, Dirichlet, Kronecker, Lie, Lobachevsky,Markov, Mobius, von Neumann, Newton, Maxwell, Einstein, Lagrange, Laplace,Hamilton, Faraday, Planck, Bohr, Fermi, Schrodinger, Heisenberg, Pauli,Dirac. This is just a short list of those that weren't considered duringcranks. What was your point again? Which revolutionaries did I miss?Apparently there must be some, since that was the point of your post. Oh,btw, Galois was never considered a crank, so let's not try to use him. Alotof the people in this list were quirky, but they were all recognized basedon their merits without reference to their quirks. The fact is,mathematicians don't care how crazy you are, they only are about yourmathematics.)Justin Van Winkle> ...> ...> Sci.math has nothing to do with what gets published in math journals,> for the simple reason that most of the 'prestigious' journals are> owned by, and run in the interest of, a handful of large corporations,> or professional guilds like the ASL, which have no reason whatsoever> to publish anything which departs from prevailing mathematical norms;> whereas sci.logic and sci.math provide forums (in principle at least),> for work in progress which flaunts these norms, and in so doing> challenges the credibility and authority of gate-keepers likeyourself.> Even if what you say about a the manipulating powers behind journals> were true, flaunting the norms of mathematics would still be no better> method for advancing mathematics. You use the term prevailing as if> the norms of mathematics could be something else, but they just are> period, regardles of anybody's personal, political or whatsoever> interests.Mathematics is like art and fashion, it is defined by its producers> and consumers and it has its trends and staples. Everything changes> in mathematics, some faster than others.> --Vaughan Pratt, FOM> The conflict between you and JSH (among others) is brought about> by factors that are evident in Budd Schulberg's contrast of the businessman with the artist.> and the artist:> A good businessman...aims to please as many people as possiblewhile minimizing risk and standardizing production. The aim of the good> artist, on the other hand, is exactly the opposite: he turns his> back on every formula, keeps breaking new ground, risks everything,> and whether he succeeds or fails, prepares to risk again.> (Budd Schulberg, Movies in America: After Fifty Years, _The> Atlantic Monthly_, November 1947; cited in Leo Gurko, _Highbrows> and the Popular Mind_, Charter Books, 1953, p. 165)> This is quite ridiculous. A good businessman aims to maximize his> profit and the aim of an artist is not definable in a few lines.> Businessman maximize profit by controlling markets and stiffing> competitors. What is more, nothing is more foreign to the business> mentality than JSH's 'go-for-broke' approach to FLT.> However it is pretty clear that rebellious, revolutionary> avant-gardism, is only one of a multitude of artistic attitudes which> becomes fashionable from time to time, and is in itself no guarantee> for artistic creation.> The implication of your posting is, that the mere act of rebellion> against norms has in itself a positive effect towards whatever purpose> you are intent on attaining, but anyone can see that this is not the> case most of the time.> You wouldn't want to fly in a plane piloted by someone whose only> credentials as pilot are that he refuses to see any benefit in> actually learning to fly.> A pilot who revolutionizes aviation will first have to learn theprevailing norms or he won't have a chance of contributing anything> new to his craft (unless he finds a way to do so in afterlife).> And by the way, JSH openly states that he is into revolutionizing math> for profit, as a businessman in this regard, he has failed both> pleasing many people, as your Budd Schulberg requires, and making> any profit.> More relevant to the sci.math treatment of JSH (who has never> expressed an interest in 'revolutionizing aviation'), is> how society treats inventors. Writing in 1955, Gurko says:Inventors are the real professionals of a mechanical society; one> would expect them, of all people, to be regarded with respect,> if not actual reverence. Yet few groups are more scorned. Until,> that is, their inventions succeed, at which time they become> garlanded heroes instantly escorted to the hall of fame, where> they are set up as exemplars for future generations of boys to> envy and emulate. Before an invention is commercialized, however,> the inventor must be prepared for calumny. He is a crank, crackpot,> dabbler, nut; he is a fool slightly tilted toward the bughouse; he> is an irresponsible loafer probably neglecting his wife and children;> he just isn't right in the head. But the minute he hits the jackpot,> these accusations are washed away and forgotten in a flood of acclaim.> His eccentricity has been known all along to be a sign of genius. His> impracticality has become vision; his neglect of loved ones a desire> to win for them a greater security; his indifference to criticism> is no longer a mark of footlessness but of courage. So long as> the inventor has only his brains, he will be labeled with contemptuous> epithets, from harmless crank to raving lunatic. Only the successful> marketing of his product can save him... Leo Gurko, _Heroes,> Highbrows and the Popular Mind_ (New York: Charter Books, 1953),> pp. 54-55) === Subject: : Re: JSH: All the dumb crap in journals> Yes, visionaries are sometimes considered cranks. But far more often cranks> are considered cranks. The reason we don't grab every nutty doofus that> comes along and follow them around as though they were the next coming of> jesus is that most are in fact cranks. > So, human experience is vindicated.Human experience is vindicated?Oh, dahling! Oh, oh, o-o-oh!> Every once and awhileEvery once and awhile?Oh, dahling! Oh, oh, o-o-oh!> some lunatic will also be a genius (or occasionally> lucky) and revolutionalize Oh, dahling! Oh, oh, o-o-oh!> a field. The vast majority, however, will just> give their neighbors something to gawk at. There are alot of inept people> all over the place. Alot Right. Inept. Alot.> of them make outrageous claims. For example, bag> ladies, bums, religious wackos, etc. We don't listen, > because they are nuts.If YOU get listened to, who shouldn't be?> To sum it up for you John, JSH is not a revolutionary. He is doesn't know> anything about the math he claims to use.'He sure is doesn't!> One more point. Can I pass?> Most people who have changed the world through theirinventions weren't considered cranks. > Most quietly laboured away for> years.Oh, dahling! Oh, oh, o-o-oh!> The crank-genius is an exception, > of course their stories are far> more interesting than the rest. (Euler, Fourier, Gauss, Hilbert, Legendre,> Leibnitz, Mandelbrot, Noether, Poincare, Riemann, Stokes, Taylor, Weyl,> Abel, Bessel, Boole, Des Cartes, Dirichlet, Kronecker, Lie, Lobachevsky,> Markov, Mobius, von Neumann, Newton, Maxwell, Einstein, Lagrange, Laplace,> Hamilton, Faraday, Planck, Bohr, Fermi, Schrodinger, Heisenberg, Pauli,> Dirac. > This is just a short list of those that weren't considered duringcranks. No, dahling! Not during! Oh, oh, o-o-oh!> What was your point again? > Which revolutionaries did I miss?Which ones did you hit on?> Apparently there must be some, > since that was the point of your post.> btw, Galois was never considered a crank, so let's not try to use him.Not Galois. Please. Not Galois. Please, no.> Alot> of the people in this list were quirky,Quirky. In this list. Alot,> but they were all recognized based> on their merits without reference to their quirks.> The fact is,> mathematicians don't care how crazy you are, they only are about your> mathematics.)> Justin Van Winkle === Subject: : Re: James Harris - Challenge problem> Somehow I doubt it. Even if your proof were to be found to be entirely> correct, nobody would care. Why? Because the thing you assert is> completely irrelevent. Why don't you know this? Because you don't know any> mathematics. You don't know any mathematics. Seriously, think about this.> You don't know any of the mathematics you are trying to do. Generally,> proofs, at first glance, either seem right or not. You lack this intuition.Mathies are creatures of habit, who--because they combine hermeticallysealed minds with Bush-esque egos--are frightfully resentful of anythingthat is not hand-me-down, cookie-cutter mathematics.> You don't know any of the mathematics you are trying nto do. You could> study, but you seem to think that that is a rediculous idea. So, you will> continue to be an internet crank. By the time we are in a position to be> anywhere in mathematics you will still not know any mathematics, and you'll> spend your time posting to newsgroups as a crank. Have a nice life.Go back to your hand-me-down, cookie-cutter mathematics and leave JSH alone.