mm-443 Subject: Re: JSH: All the dumb crap in journals> In fact, I don't believe JSH has ever used a sock puppet in the past.>When he set up his live online chat room (The Algebra of>Factorizations), he logged in under the psuedonym Flatrings (this>was when he still didn't grasp what a ring was, and he needed his>rings to be flat in order to make his FLT proof work).>Everybody knew it was JSH, so I can't really accuse him of using a>sock puppet, bur he does have lint under his fingernails.> No lint at all. There was obviously no intent to deceive. Not only> everyone knew who Flatrings was, but JSH never pretended otherwise.> There is a big difference between an online nickname and a sock> puppet. You'd think that someone going by Mensanator would know> this.Absolutely. And when you see posts from Rotanasnem orMensanator666 it should be obvious that there is no intention todeceive. However, the nickname _is_ a sock puppet.>Unfortunately, the discussion was not to his liking and he>dismantled The Algebra of Factorizations after only one session (a>transscript was posted to sci.math), so that was the end of>Mr. Flatrings.> His use of the Flatrings moniker adds no evidence at all that JSH has> ever used sock puppets I did not say it was evidence that he has used sock puppets.> or would be willing to.That's where I disagree. I see it just the opposite. Someone with ahistory ofusing multiple nicknames (lint) would be more likely to use a sockpuppet, in my opinion.Now it may be that a moral person like Jesse Hughes would never use asock puppet, but JSH has done things that pale in comparison, so whywould I thinkhe would be unwilling to do so? === Subject: Re: JSH: All the dumb crap in journals Discussion, linux)> There is a big difference between an online nickname and a sock> puppet. You'd think that someone going by Mensanator would know> this.> Absolutely. And when you see posts from Rotanasnem orMensanator666 it should be obvious that there is no intention to> deceive. However, the nickname> _is_ a sock puppet.Not sure who refers to.> That's where I disagree. I see it just the opposite. Someone with a> history of using multiple nicknames (lint) would be more likely to> use a sock puppet, in my opinion.If so, only marginally so.> Now it may be that a moral person like Jesse Hughes would never use> a sock puppet, but JSH has done things that pale in comparison, so> why would I think he would be unwilling to do so?The fact that JSH has performed acts which you regard as immoral seemsa very poor predictor of whether he will use sock puppets. Unless Isee better evidence that JSH has at least considered sock puppetry inthe past, I won't guess that he's likely to do it in the future. Thewhole exercise is simply much too speculative.In fact, since we are now reduced to arguing whether JSH *would* usesock puppets rather than whether he *has* used them, perhaps we shouldjust drop it.Certainly he who can digest a second or third fluxion neednot, methinks, be squeamish about any point in divinity. George Berkeley, 1734 === Subject: Re: JSH: All the dumb crap in journals> With all the dumb crap in math journals, you people know that my find> of a way to count prime numbers by integrating a partial difference> equation is worthy of publication somewhere.Write it up and submit it to one of the journals then. The worst that canhappen is that someone finds out what you did was actually done a centuryago.> But you sit by as if there's nothing sinister going on, when all these> people are fighting such a choice result.> Yet in your ENTIRE CAREERS most of you will never publish anything> that's even close in neatness, but you'll fill up journals anyway.I have one result published -- a one-page proof of a result originally donein 1979 (15 pages) and redone in 1990 (six pages), which also provides alinear-time (very quick) algorithm, and which has a nice characterizationwhen equality holds in a certain inequality. Another one is a problem whichhad been open since 1976, and is currently being refereed. Both papers canbe found on my website: http://math.asu.edu/~checkman/#papers . This is allI've published (so far).And not every result is pretty. A lot of mathematics (especially specializatedmath) involves looking for the truth on a theoretical level (for truth's sake).> That's the math world.> If there's anything I really like here it's proving the lie to all of> you of pure math, as sure, you may convince others that my work> isn't important, but those of you desperate to find something worth> publication in your publish or perish world know the reality.Pure math is a fraud, just something you say, when you don't believe> in it.Pure math is a consequence of specialization, which only came about inthe 20th century. (Poincare was supposedly the last mathematician who knewall mathematics of his time.)> It's just a way to get by and pay your s. How do you get _your_ money? Sit at home and poke your belly button? Doingwork of any kind is just a way to get by and pay your s. Doing workyou don't have to qualifies as pure math.> Yes, to me you are> losers, too weak to handle the truth, so you think you can crap on me,> when what you do anyway is fill up journals with junk.Hasty overgeneralization.> I'm better than you, and you know it.There is no way for me to evaluate the first part of that statement, sinceI don't know who you are. -- Christopher Heckman === 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 instructionb) provide you with your namec) know how to count === Subject: Re: JSH: The reverse factorization argument> In sci.math, Nora Baron> 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 the f's are algebraic integer functions, and f_1(0) = f_2(0) = f_3(0) = 0.>Multiplying both sides by 49, I have>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>which is>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 5(3) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>and using g_3(x) = f_3(x) + 3, I have>(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)>where as shown in previous posts 7 f_1(x), 7 f_2(x), and g_3(x) are the roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).That is incorrect, as in fact, it's *one* possibility. The mistake isthe claim that they *are* the roots of the cubic that followed.There are an infinity of other ones.OOPS! My mistake.Now then, a math expert, i.e. a mathematician, should know that thereare an infinity of possibilities, and just in case you don't believeme, here's one that doesn't fit witha^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x),which is f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,which should look familiar if you've looked at the other thread.Notice that Nora Baron fell for the trap, and then yet anotherposter came in and cheered that poster!!!> Your premise is wrong. There are no algebraic integer>functions f_1(x_), f_2(x), and f_3(x) which are such that>7*f_1(x), etc., are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Proof: Let x = 1. Assume that 7*f_1(1) satisfies>the polynomial just given. Then after simplifying >a bit, you get> 7*[f_1(1)]^3 + 144*[f_1(1)]^2 - 2257 = 0.> The polynomial just given is primitive, irreducible,>and *non-monic*. Therefore f_1(1) cannot be an algebraic>integer. QED.> Short, simple, devastating.> I like it. :-)I want readers to pay attention to the *social forces* here as in factneither poster knows what they're doing!However, the second poster knows that by cheering Nora Baron he canpush the idea that they do!!!It's a technique that's simply devastating, to borrow from thesidekick poster.> [rest snipped]The reality is that given(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the f's are algebraic integer functions, and f_1(0) = f_2(0) =f_3(0), there are an *infinity* of possible f's.I'm showing the other side of the factorization, where for a whileI've emphasized *one* factorization out of infinity both to show theness of the mathematics, and to show you why posters like NoraBaron are, well, incompetent. === Subject: Re: JSH: The reverse factorization argument>In sci.math, Nora Baron>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 the f's are algebraic integer functions, and f_1(0) = f_2(0) = f_3(0) = 0.>Multiplying both sides by 49, I have>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>which is>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 5(3) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>and using g_3(x) = f_3(x) + 3, I have>(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)>where as shown in previous posts 7 f_1(x), 7 f_2(x), and g_3(x) are the roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> That is incorrect, as in fact, it's *one* possibility. The mistake is> the claim that they *are* the roots of the cubic that followed.> There are an infinity of other ones.> OOPS! My mistake.> Now then, a math expert, i.e. a mathematician, should know that there> are an infinity of possibilities, and just in case you don't believe> me, here's one that doesn't fit with> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x),> which is > f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,> which should look familiar if you've looked at the other thread. Might as well reply to this aspect of things also. The claimhere is that there are algebraic integers g1, g2, and g3 suchthat (5*g1*x + 1)*(5*g2*x + 1)*(5*g3*x + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22. This means that -1/(5*g1) is a root of the equation 300125 x^3 - 18375 x^2 - 360 x + 22 = 0. This simplifies to 22*g1^3 + 72*g1^2 - 5*147*g1 - 2401 = 0,and of course this polynomial has integer coefficients,is primitive, non-monic, and irreducible. Therefore g1cannot be an algebraic integer. Therefore your present claim is just as false as what you said before. Another dead end! Nora B. === Subject: Re: JSH: The reverse factorization argument>In sci.math, Nora Baron>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 the f's are algebraic integer functions, and f_1(0) = f_2(0) = f_3(0) = 0.>Multiplying both sides by 49, I have>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>which is>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 5(3) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>and using g_3(x) = f_3(x) + 3, I have>(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)>where as shown in previous posts 7 f_1(x), 7 f_2(x), and g_3(x) are the roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> That is incorrect, as in fact, it's *one* possibility. The mistake is> the claim that they *are* the roots of the cubic that followed.> There are an infinity of other ones.> OOPS! My mistake.> Now then, a math expert, i.e. a mathematician, should know that there> are an infinity of possibilities, and just in case you don't believe> me, here's one that doesn't fit with> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x),> which is > f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,> which should look familiar if you've looked at the other thread.> Notice that Nora Baron fell for the trap, and then yet another> poster came in and cheered that poster!!!> Your premise is wrong. There are no algebraic integer>functions f_1(x_), f_2(x), and f_3(x) which are such that>7*f_1(x), etc., are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Proof: Let x = 1. Assume that 7*f_1(1) satisfies>the polynomial just given. Then after simplifying >a bit, you get> 7*[f_1(1)]^3 + 144*[f_1(1)]^2 - 2257 = 0.> The polynomial just given is primitive, irreducible,>and *non-monic*. Therefore f_1(1) cannot be an algebraic>integer. QED.>Short, simple, devastating.>I like it. :-)> I want readers to pay attention to the *social forces* here as in fact> neither poster knows what they're doing!> However, the second poster knows that by cheering Nora Baron he can> push the idea that they do!!!> It's a technique that's simply devastating, to borrow from the> sidekick poster.>[rest snipped]> The reality is that given> (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 the f's are algebraic integer functions, and f_1(0) = f_2(0) => f_3(0), there are an *infinity* of possible f's.> I'm showing the other side of the factorization, where for a while> I've emphasized *one* factorization out of infinity both to show the> ness of the mathematics, and to show you why posters like Nora> Baron are, well, incompetent. Look, wienie. I am going to explain this in simplest possibleterms: 1. *You* said that a_1(x) is a root of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 2. *You* said that a_1(x) = 7*f_1(x), where f_1(x) is an algebraic integer function, which clearly means that it takes its values in the algebraic integers. 3. I showed that this implies that if x = 1, the two preceding statements imply that 7*[f_1(1)]^3 + 144*[f_1(1)]^2 - 2257 = 0. 4. I then noted that the polynomial in this equation has integer coefficients, and is primitive, irreducible, and *non-monic*. Therefore f_1(1) cannot be an algebraic integer. 5. I conclude that your statement that f_1(x) is an algebraic integer function cannot be true. So far as I can tell, you have no disagreements with this logic. It was too bitter, perhaps, to read that The Ghost in the Machineagreed heartily with what I said. It *was* devastating. You had to respond. Now you are saying [above] regarding your *own original claim*: That is incorrect, as in fact, it's *one* possibility. The mistake is the claim that they *are* the roots of the cubic that followed. There are an infinity of other ones. OOPS! My mistake.and then you claim that it is *** I *** who fell for the trap (but you, of course, did not!). You give no explanation or justificationfor the claim that There are an infinity of other ones. You then try to pretend that is I and the Ghost who have been stupid about this, not you. But it was YOU who made the false claim; YOU who made the mistake (OOPS!), and YOU who have been proven wrong.And now YOU who claim there is an infinity of possible f's, without producing a single one! Face it. You're an incompetent dolt with an incredible instinctfor wrong mathematics. Here you are trying to put up an arrogant,superior know-it-all false front, pretending that you have some secret understanding that has eluded the rest of us poor saps: big claimswhich you have no intention of backing up with proofs. Claims of superiority, like profits in a business, have to beearned the old-fashioned way. The currency here is rigorous proof, not braggadocio. You have nothing to show. It is you who has been proven wrong here, you who has been forced to say OOPS!! My mistake, not me. In truth, your argument at this point, rotten to the core from Day One, is now a total shambles. It's impossible to guess now what you actually think; you know that if you make a definite falsifiable statement, one of us will instantly prove it wrong. It's dishonest. You're not fooling anyone any more, if you everdid, I think not even yourself. If you actually were doing math for profit by now you would be completely bankrupt and out of business. I would love to hear Arturo's comments on this. You rudely accepted his invitation to quit posting because, obviously, you couldn't refutewhat he said any more than you can refute me, Dik Winter, or othershere. You called him a liar a thousand times over with never anyjustification, and you used obscenities in your replies to him. And now, here, *** proven wrong ***, you dissemble and posture and swaggerlike you're King Tut and pretend everyone else is stupid. Truly unbelievable behavior by someone who thinks he is the rightful heir to Gauss or Dedekind. I hereby invite Arturo back into the fray, to respond at least to posts of others, if not to your own. Nora B.> === Subject: Re: JSH: The reverse factorization argument>In sci.math, Nora Baron>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 the f's are algebraic integer functions, and f_1(0) = f_2(0) =f_3(0) = 0.>Multiplying both sides by 49, I have>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>which is>(5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 f_3(x) + 5(3) + 7) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)>and using g_3(x) = f_3(x) + 3, I have>(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)>where as shown in previous posts 7 f_1(x), 7 f_2(x), and g_3(x) arethe roots of>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>That is incorrect, as in fact, it's *one* possibility. The mistake is>the claim that they *are* the roots of the cubic that followed.>There are an infinity of other ones.>OOPS! My mistake.>Now then, a math expert, i.e. a mathematician, should know that there>are an infinity of possibilities, and just in case you don't believe>me, here's one that doesn't fit with>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x),>which is>f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,>which should look familiar if you've looked at the other thread.>Notice that Nora Baron fell for the trap, and then yet another>poster came in and cheered that poster!!!> Your premise is wrong. There are no algebraic integer>> functions f_1(x_), f_2(x), and f_3(x) which are such that>> 7*f_1(x), etc., are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>> Proof: Let x = 1. Assume that 7*f_1(1) satisfies>> the polynomial just given. Then after simplifying>> a bit, you get>> 7*[f_1(1)]^3 + 144*[f_1(1)]^2 - 2257 = 0.> The polynomial just given is primitive, irreducible,>> and *non-monic*. Therefore f_1(1) cannot be an algebraic>> integer. QED.Short, simple, devastating.I like it. :-)>I want readers to pay attention to the *social forces* here as in fact>neither poster knows what they're doing!>However, the second poster knows that by cheering Nora Baron he can>push the idea that they do!!!>It's a technique that's simply devastating, to borrow from the>sidekick poster.>[rest snipped]>The reality is that given>(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 the f's are algebraic integer functions, and f_1(0) = f_2(0) =>f_3(0), there are an *infinity* of possible f's.>I'm showing the other side of the factorization, where for a while>I've emphasized *one* factorization out of infinity both to show the>ness of the mathematics, and to show you why posters like Nora>Baron are, well, incompetent.> Look, wienie. I am going to explain this in simplest possible> terms:> 1. *You* said that a_1(x) is a root of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> 2. *You* said that a_1(x) = 7*f_1(x), where f_1(x) is an> algebraic integer function, which clearly means that> it takes its values in the algebraic integers.> 3. I showed that this implies that if x = 1, the two preceding> statements imply that> 7*[f_1(1)]^3 + 144*[f_1(1)]^2 - 2257 = 0.> 4. I then noted that the polynomial in this equation has> integer coefficients, and is primitive, irreducible, and> *non-monic*. Therefore f_1(1) cannot be an algebraic> integer.> 5. I conclude that your statement that f_1(x) is an algebraic> integer function cannot be true.> So far as I can tell, you have no disagreements with this logic.> It was too bitter, perhaps, to read that The Ghost in the Machine> agreed heartily with what I said. It *was* devastating. You had> to respond.> Now you are saying [above] regarding your *own original claim*:> That is incorrect, as in fact, it's *one* possibility.> The mistake is the claim that they *are* the roots of> the cubic that followed.> There are an infinity of other ones.> OOPS! My mistake.> and then you claim that it is *** I *** who fell for the trap (but> you, of course, did not!). You give no explanation or justification> for the claim that There are an infinity of other ones.> You then try to pretend that is I and the Ghost who have been stupid> about this, not you. But it was YOU who made the false claim; YOU> who made the mistake (OOPS!), and YOU who have been proven wrong.> And now YOU who claim there is an infinity of possible f's,> without producing a single one!> Face it. You're an incompetent dolt with an incredible instinct> for wrong mathematics. Here you are trying to put up an arrogant,> superior know-it-all false front, pretending that you have some secret> understanding that has eluded the rest of us poor saps: big claims> which you have no intention of backing up with proofs.> Claims of superiority, like profits in a business, have to be> earned the old-fashioned way. The currency here is rigorous proof,> not braggadocio. You have nothing to show. It is you who has been> proven wrong here, you who has been forced to say OOPS!! My mistake,> not me.> In truth, your argument at this point, rotten to the core from Day One,> is now a total shambles. It's impossible to guess now what you> actually think; you know that if you make a definite falsifiable> statement, one of us will instantly prove it wrong.> It's dishonest. You're not fooling anyone any more, if you ever> did, I think not even yourself. If you actually were doing math for> profit by now you would be completely bankrupt and out of business.> I would love to hear Arturo's comments on this. You rudely accepted> his invitation to quit posting because, obviously, you couldn't refute> what he said any more than you can refute me, Dik Winter, or others> here. You called him a liar a thousand times over with never any> justification, and you used obscenities in your replies to him. And> now, here, *** proven wrong ***, you dissemble and posture and swagger> like you're King Tut and pretend everyone else is stupid. Truly> unbelievable behavior by someone who thinks he is the rightful heir> to Gauss or Dedekind.> I hereby invite Arturo back into the fray, to respond at least to> posts of others, if not to your own.> Nora B.>I agree wholeheartedly with Nora. James has become increasingly arrogantlately, for no good reason. It is clear to me that he does not have a goodgrasp on mathematics because he showed in a previous post he can't integratecorrectly. Also, I showed him an example of where you add and subtract thesame thing via demonstrating how to write a function of the form y=ax^2+bx+cin the form y=a(x-h)^2+k and he called me an F***ing Liar. That's himrefuting basic algebra, the very thing he accuses everyone else of. Hisrefutation of my example further shows his incompetence in algebra because Ilearned this in my algebra 1 class my freshman year of high school. His namecalling is extremely childish and distasteful, something a GROWN MANshouldn't have to resort to. It's sad that he has to resort to such tactics.David Moran === Subject: Re: JSH: The reverse factorization argument... > f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,Which won't fit either. > The reality is that given > (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 the f's are algebraic integer functions, and f_1(0) = f_2(0) = > f_3(0), there are an *infinity* of possible f's.Pray show that that is true. Show even *one* possible set of f's.home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: The reverse factorization argumentIn sci.math, Dik T. Winter:> ...>f_1(x) = g_1 x, f_2(x) = g_2 x, f_3(x) = g_3 x,> Which won't fit either.>The reality is that given>(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 the f's are algebraic integer functions, and f_1(0) = f_2(0) =>f_3(0), there are an *infinity* of possible f's.> Pray show that that is true. Show even *one* possible set of f's.I'll admit I have no idea what an algebraic integer function is,unless one assumes that, when fed an algebraic integer, it yieldsone.As it is, things get interesting. One can, for instance,divide the original equation by 125:(f_1(x)+ 1/5)*(f_2(x) + 1/5)*(g_3(x) + 22/5) = 2401*x^3 - 147*x^2 - 72/25*x + 22/125and now one can pick and choose amongst many possibilities.For example,f_1(x) = 2401*x^3 - 147*x^2 - 72/25*x + 22/125f_2(x) = 4/5f_3(x) = -17/5works perfectly fine.Of course these aren't algebraic integer functions.If one substitutes g_i(x) = 1/(5*f_i(x)) and multipliesby g_1(x)*g_2(x)*g_3(x), one gets the following:(1 + g_1(x)) * (1 + g_2(x)) * (1 + 22*g_3(x)) = (300125*x^3 - 18375*x^2 - 360*x + 22) * (g_1(x)*g_2(x)*g_3(x))James is probably making the naive assumption that g_1(x)*g_2(x)*g_3(x)= 1 / (125 * f_1(x) * f_2(x) * f_3(x)) = 1 / (125 * 300125 / 125)= 1 / 300125, but this is obviously false; my functions abovefor example yield a triple product off_1(x) * f_2(x) * f_3(x) = -163268/25*x^3 + 9996/25*x^2 + 4896/625*x - 1496/3125Clearly this isn't even close to 300125/125 = 2401.If one requires the f_i(0) to be equal, however, there are someinteresting issues. Unfortunately, it's not clear that theyare alegebraically integral ones.We know thatF(0) = (5*f_1(0)+ 1)(5*f_2(0) + 1)(5*f_3(0) + 22) = 22.(assuming F(x) = 300125*x^3 - 18375*x^2 - 360*x + 22).If we set all f_i(0) to be equal to a value z, we of course get(5*z + 1)*(5*z + 1)*(5*z+22) - 22 = 0125*z^3 + 600*z^2 + 225*z = 05*z^3 + 24*z^2 + 9*z = 0after dividing by 25 in the final step.One of the roots is obviously z = 0. The other two arenot algebraic integers, and therefore the f_i() are notalgebraic integer functions as I've defined them above,for the nonzero roots.Are they algebraic integer functions when f_i(0) = 0?Good question.If I substitute F(1) = 281412 = (5*f_1(1)+ 1)*(5*f_2(1) + 1)*(5*f_3(1) + 22)I can come up with all sorts of combinations.Note that 281412 = 2^2 * 3^2 * 7817 -- an interesting number butwithout 5, 7, or 11 as factors.One could assume f_1(1) = 1, f_2(1) = 1, and f_3(1) = 1559.Not bad, if a bit ad hoc.F(-1) = -318118 = (5*f_1(-1)+ 1)(5*f_2(-1) + 1)(5*f_3(-1) + 22).If we assume the f_i() are linear and that f_i(0) = 0,we get (5*(-1) + 1)(5*(-1) + 1)(5*(-1559) + 22) = -124368.Whoops! Guess that ad hoc assumption of mine was a bit futzed...If we remove the ad hoc assumption but still assume the f_i() arelinear with f_i(0) = 0 and a nonzero leading coefficient,then by necessity f_i(x) = h_i * x, and that therefore(h_1 * x + 1) * (h_2 * x + 1) * (h_3 * x/22 + 1) = F(x)/22.Since F(x)/22 doesn't have integral coefficients (it's equalto (300125/22*x^3 - 18375/22*x^2 - 180/11*x + 1) h_1, h_2, and h_3/22aren't algebraic integers. I can't tell regarding h_3 at themoment.If we assume that at least one of the f_i(x) are constant (whichis possible), we get into a number of permutations and combinationswhich would probably take too long for me to delve into right now,as it's nearly midnight. :-)If we allow radical forms things get even more, well, radical.My brain's beginning to hurt. :-)#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: JSH: The reverse factorization argument Well I see you're trying to be a smart-ass Dik Winter, but you see, I just showed *one* of an infinity of possibilities for the the f's, which, if you were a math expert, you should have realized.> I'll make it clear to you now Dik Winter that what I'm doing is using a simple example to show that you're not competent for these discussions. Now then, here's the test again. (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 the question is, can the f's be algebraic integer functions, where f_1(0) = f_2(0) = f_3(0) = 0?> The answer is *no*, as I have shown in a part you snipped....I looked back because you said that and found that the statement isuntrue.You didn't prove anything but tried to rely on *one* factorization ofthe possible infinity, which *I* gave you.You're not looking good here Dik Winter.At least don't try the old trick of claiming you've already answered aquestion that you haven't. I've pointed out that you can't just pick some values that *I* have used before Dik Winter as there's no uniqueness of factorization to help you. And I'm emphasizing to other readers that I'm showing the lack of competence of certain posters, like Dik Winter. Now then Dik Winter, can you answer the test correctly?> Can you answer the test correctly? I have shown what polynomial f1(x)> and f2(x) are roots of. And also shown that that polynomial is not> monic and irrecudible when x != 0. So how *can* they be algebraic> integer functions?And I've pointed out that there are an *infinity* of possible f's, andin fact, I'll give a *different* one from the what you tried to usejust to show you up, as consider(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22with f_1(x) = g_1 x, f_2(x) = g_2 x, and f_3(x) = g_3 x,which some readers will note was already given in another thread, sothey might wonder why I have bring it up again.The answer is, Dik Winter doesn't know what he's doing, but pretendsvery well.I'm running two threads so you can see some bizarre stuff like meusing the *same* example of a factorization to counter Dik Wintermaking the *same* mistake in two different threads.The reality is that you can't give a *single* polynomial that willhandle all the possible f's, which is a mathematical truth that DikWinter seems willing to deny.If you trust him, look back to check his claim of having already giventhat polynomial, and notice how primitive his attempt is, as in fact,he's just trying to use work that I'd given.But you see, that's my point, Dik Winter is a hack. He follows alongwhere I've spent effort to be understood and then he confuseseverything but acts as if he's an expert.Here now though, I'm not helping him play his dark game, as I'm notexplaining everything so he can then step in after the real work hasbeen done to muddy the waters. === Subject: Re: JSH: The reverse factorization argumentfuffy> Well I see you're trying to be a smart-ass Dik Winter, but you see, I> just showed *one* of an infinity of possibilities for the the f's,> which, if you were a math expert, you should have realized.I'll make it clear to you now Dik Winter that what I'm doing is using> a simple example to show that you're not competent for these> discussions.> Now then, here's the test again.> (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 the question is, can the f's be algebraic integer functions,> where f_1(0) = f_2(0) = f_3(0) = 0?>The answer is *no*, as I have shown in a part you snipped....> I looked back because you said that and found that the statement is> untrue.> You didn't prove anything but tried to rely on *one* factorization of> the possible infinity, which *I* gave you.> You're not looking good here Dik Winter.> At least don't try the old trick of claiming you've already answered a> question that you haven't.> > I've pointed out that you can't just pick some values that *I* have> used before Dik Winter as there's no uniqueness of factorization to> help you.> And I'm emphasizing to other readers that I'm showing the lack of> competence of certain posters, like Dik Winter.> Now then Dik Winter, can you answer the test correctly?>Can you answer the test correctly? I have shown what polynomial f1(x)>and f2(x) are roots of. And also shown that that polynomial is not>monic and irrecudible when x != 0. So how *can* they be algebraic>integer functions?> And I've pointed out that there are an *infinity* of possible f's, and> in fact, I'll give a *different* one from the what you tried to use> just to show you up, as 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> > with f_1(x) = g_1 x, f_2(x) = g_2 x, and f_3(x) = g_3 x,> which some readers will note was already given in another thread, so> they might wonder why I have bring it up again.> The answer is, Dik Winter doesn't know what he's doing, but pretends> very well.> I'm running two threads so you can see some bizarre stuff like me> using the *same* example of a factorization to counter Dik Winter> making the *same* mistake in two different threads.> The reality is that you can't give a *single* polynomial that will> handle all the possible f's, which is a mathematical truth that Dik> Winter seems willing to deny.> If you trust him, look back to check his claim of having already given> that polynomial, and notice how primitive his attempt is, as in fact,> he's just trying to use work that I'd given.> But you see, that's my point, Dik Winter is a hack. He follows along> where I've spent effort to be understood and then he confuses> everything but acts as if he's an expert.> Here now though, I'm not helping him play his dark game, as I'm not> explaining everything so he can then step in after the real work has> been done to muddy the waters.> === Subject: Re: JSH: The reverse factorization argument> Well I see you're trying to be a smart-ass Dik Winter, but you see, I> just showed *one* of an infinity of possibilities for the the f's,> which, if you were a math expert, you should have realized. > > I'll make it clear to you now Dik Winter that what I'm doing is using> a simple example to show that you're not competent for these> discussions.> Now then, here's the test again.> (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 the question is, can the f's be algebraic integer functions,> where f_1(0) = f_2(0) = f_3(0) = 0? > The answer is *no*, as I have shown in a part you snipped.... > I looked back because you said that and found that the statement is > untrue.Yes, it was false indeed. > You didn't prove anything but tried to rely on *one* factorization of > the possible infinity, which *I* gave you.Oh, well... > You're not looking good here Dik Winter. > At least don't try the old trick of claiming you've already answered a > question that you haven't.> I've pointed out that you can't just pick some values that *I* have> used before Dik Winter as there's no uniqueness of factorization to> help you.> And I'm emphasizing to other readers that I'm showing the lack of> competence of certain posters, like Dik Winter.> Now then Dik Winter, can you answer the test correctly? > Can you answer the test correctly? I have shown what polynomial f1(x) > and f2(x) are roots of. And also shown that that polynomial is not > monic and irrecudible when x != 0. So how *can* they be algebraic > integer functions? > And I've pointed out that there are an *infinity* of possible f's, and > in fact, I'll give a *different* one from the what you tried to use > just to show you up, as 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 > with f_1(x) = g_1 x, f_2(x) = g_2 x, and f_3(x) = g_3 x, > which some readers will note was already given in another thread, so > they might wonder why I have bring it up again.In that other thread I have explicitly shown (with mathematical proof)that g1 and g2 are *not* algebraic integers. g3 is and 22 g1 and 22 g2are also algebraic integers. > The answer is, Dik Winter doesn't know what he's doing, but pretends > very well.I think the shoe fits another foot. > I'm running two threads so you can see some bizarre stuff like me > using the *same* example of a factorization to counter Dik Winter > making the *same* mistake in two different threads.Not the same. You claim that there are algebraic integer functionsf1 to f3 that satisfy the factorisation. It is *to you* to show thatthat is indeed possible. It is not *to us* to show that it is notpossible.But even when you do that you get (when you multiply the polynomialby 49, and two of the factors by 7) *not* the a1 to a3 that are rootsof your favoured monic irreducible polynomial. Because it has beenshown (in absurdo) that two of those are *not* divisible by 7 in thealgebraic integers. > The reality is that you can't give a *single* polynomial that will > handle all the possible f's, which is a mathematical truth that Dik > Winter seems willing to deny.You do not even give the beginning of a proof that such algebraic integerfunctions indeed do exist.home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Non-uniqueness, polynomial factosIt'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.My math discoveries, found for profithttp://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factos>It's easy to demonstrate that polynomials can be factored in an>infinite number of ways--that is, there isn't a unique factorization.Well, duh - nobody's ever disputed this (since we're talkingabout non-polynomial factors, which is not what is usuallymeant here.) (No, when someone says something can be done in many waysit does not follow that he's unaware that it can be donein infinitely many ways...)>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.>My math discoveries, found for profit>http://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factos> 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.> My math discoveries, found for profit> http://mathforprofit.blogspot.com/What's your point? Anyone can take an equation involving several variablesand substitute other combinations of other variables into it. So what? Youstated in another thread that there were an *infinite* number of solutionsfor the 'f's, but did not provide even a *single* one of them. Here youcompletely avoid the subject and simply replace the 'f's with expressionsinvolving other symbols. But substituting other combinations of variablesand constants for the 'f's does *not* change the values of the 'f's orincrease the number of valid solutions for them.Why not give us one of the sets of values for the 'f's from the *infinite*number of values available? (No fair substituting arbitrary expressionsfor the 'f's and claiming that this has any impact on the values of the'f's -- it doesn't!)--I think some village is missing its idiot.--.--http://www.crbond.com === Subject: AlgebraProve that the 2D quadratic from aX^2 +bXY+ cY^2 is a positivedefinite iff a>0 and 4ac - b^2 > 0I can see that it must come from the quadratic equation but cannot getany further, can anyone give me any help? === 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?aX^2 + bXY + cY^2 is positive definite iff the corresponding 2x2 matrix(a b/2)(b/2 c)has strictly positive eigenvalues; an equivalent condition (check this) isthat both a and the determinant of the matrix are strictly positive.P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. === Subject: Re: Is zero in the domain of functions ?>Suppose that F:D_1--->R , G:D_2 --->R> F(x)=sqrt(x^2(x-1)*(x+1)) , G(x)= |x|*sqrt{(x-1)*(x+1)}>where D_1 ,D_2 are the domains of definitions , for instance> D_1 := {x in R ; F(x) is real} .> This sort of question, about implied domains, has been asked before in> math newsgroups.I was slightly surprised at the shortness of this thread. I had thought itmight spark some debate.But I was even more surprised when, looking through this November's issueto this very topic.