mm-234 > This is not correct. Note that a1/f = b1, or a1 = f*b1. > Putting this into the equation above that the a's > satisfy and simplifying, one obtains f*b1^3 + 3*(-1 + m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m). Clearly b2 is a root of the same polynomial. Since > b3 = a3, b3 satisfies a different equation: b3^3 + 3*(-1+mf^2)b3^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0. >> >>Um, Nora Baron do you accept that there exists a SINGLE cubic which >>has the b's as it's roots? >> >>That is, ONE cubic with the roots b_1, b_2, and b_3? Um, , since b3=a3, but b2 is not equal to a2 and b1 is >> not equal to a1, they cannot be roots of the same cubic UNLESS the >> cubic that defines the a's is reducible. > >It's not true that a_3 = b_3 in general, which I no in a follow-up >post. I was mistaken here as a_3 = b_3 in general. > And unless you are in a specific family of cases, you must have > a_3=b_3 in every other case. I've no my mistake in saying that a_3 does not equal b_3 in general. >What is true is that at m=0, a_3 = b_3 = 3. > >Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 >as its roots, and in fact, at m=0, I can give that single cubic as Sigh. Given any three COMPLEX numbethere exists a cubic that has > those three complex numbers as roots. Namely, if you have the complex > numbers a, b, and c, then the single cubic that has those three > complex numbers as roots is (x-a)(x-b)(x-c). That's basic precalculus > stuff. Readers should note the Sigh. and consider what its purpose was. Now then, I've pushed that there's a *single* cubic because the properties of that single cubic refutes attempts at hiding the correctness of the argument that shows their is a definition problem at the heart of mathematics. Now notice how he tried to maneuver, first with the Sigh. and then by trying to downplay the assertion as if it were childish. > The point, however, is that any old cubic is not good enough. If you > want to conclude stuff about the integrality of b_1, b_2, and b_3 > (e.g., whether or not they are algebraic integers) then you need the > cubic to be monic and to have algebraic integer (or better yet, > integer) coefficients. Now creates his own requirements by mentioning something that is basically true, but out of context. That is, readers might now assume that I've claimed that the cubic for the b's has algebraic integer coefficients, which I did, early on (see my original post in this thread) but I had to post corrections as that is wrong. Since then I've been using b^3+ (...)b^2 + (...)b+...- (m^3 f^4 - 3m^2 f^2 + 3m) so there's no reason to believe that I'm looking for algebraic integeand definitely not for integers as coefficients. will create a requirement that he makes up, so that he can then claim that I'm wrong when that requirement isn't fulfilled. > In any case when b_3=a_3 (which is every case in which a_1x is not > equal to -uf and a_2x is not equal to -uf), you have b_3=a_3 but b_2 > different from a_1 and a_2, and b_1 different from a_1 and > a_2 (assuming f is not a unit, which you do). IF the polynomial that > defines the a's is irreducible over Q, then ANY polynomial with > integer coefficients that has a_3 as a root must be a multiple of the > polynomial that defines the a's. And therefore, it is either of degree > strictly larger than 3, or else it is a scalar multiple of the > polynomial that defines the a's and has exactly a_1, a_2, and a_3 as > roots. Therefore, in all such cases, it is impossible for there to be > a single cubic WITH INTEGER COEFFICIENTS which has b_1, b_2, and b_3 > as roots. Now notice how this poster has used *his* requirement, with emphasis, as if he's made some important point. The naive reader might assume at this point that I need integer coefficients for my argument, when in fact the proper cubic is b^3+ (...)b^2 + (...)b+...- (m^3 f^4 - 3m^2 f^2 + 3m) and there's no indication that the coefficients are integers. I'm showing you how this poster has crea an illusion of competence while attacking valid mathematics. He basically uses typical reader patterns against readers to convince them of things that are not true. > This is simple, basic, precalculus algebra of polynomials. Yet another rather obvious maneuver. >The SINGLE cubic is important as it can't be non-monic. > >If it were non-monic, then how would it give a monic at m=0? Let f(x) = (2m+1)x^3 - 3x^2 +1. It's a single cubic. It gives a monic cubic when m=0. Are you claiming > that it is ALWAYS monic? [.rest dele.] Now having the poster goes for the finish where the attempt is to finally completely fool the reader with a trivial example, while refusing to handle the full situation. Here the full situation is that in cases where this poster claims the cubic for the b's is not monic, it's necessary that the leading coefficient have f as a factor. Some more information is pertinent: R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) It gets worse. As provably the cubic defining the b's has algebraic integer coefficients, and is also monic when yet another cubic a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) is reducible to linear terms over Q, then the poster further needs that the leading coefficient of the b's varies to 1 for any value where it does, and then switches to having a factor of f, when that cubic is irreducible over Q. As that position is illogical and against mathematics, you can see why the poster needs tactics to fool readers. As an example of the cubic for the b's with algebraic integer coefficients, consider b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 which is the correct cubic, when m=1, f=sqrt(2), and notice it is monic. === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. >> This is not correct. Note that a1/f = b1, or a1 = f*b1. >> Putting this into the equation above that the a's >> satisfy and simplifying, one obtains f*b1^3 + 3*(-1 + m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m). Clearly b2 is a root of the same polynomial. Since >> b3 = a3, b3 satisfies a different equation: b3^3 + 3*(-1+mf^2)b3^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0. > >Um, Nora Baron do you accept that there exists a SINGLE cubic which >has the b's as it's roots? > >That is, ONE cubic with the roots b_1, b_2, and b_3? Um, , since b3=a3, but b2 is not equal to a2 and b1 is > not equal to a1, they cannot be roots of the same cubic UNLESS the > cubic that defines the a's is reducible. >> >>It's not true that a_3 = b_3 in general, which I no in a follow-up >>post. > >I was mistaken here as a_3 = b_3 in general. > And unless you are in a specific family of cases, you must have >> a_3=b_3 in every other case. > >I've no my mistake in saying that a_3 does not equal b_3 in >general. And my post was made before you went back and no that mistake. So stop yer whining. >>What is true is that at m=0, a_3 = b_3 = 3. >> >>Now then, there MUST exist a *single* cubic that has b_1, b_2 and b_3 >>as its roots, and in fact, at m=0, I can give that single cubic as Sigh. Given any three COMPLEX numbethere exists a cubic that has >> those three complex numbers as roots. Namely, if you have the complex >> numbers a, b, and c, then the single cubic that has those three >> complex numbers as roots is (x-a)(x-b)(x-c). That's basic precalculus >> stuff. > >Readers should note the Sigh. and consider what its purpose was. > >Now then, I've pushed that there's a *single* cubic because the >properties of that single cubic refutes attempts at hiding the >correctness of the argument that shows their is a definition problem >at the heart of mathematics. > >Now notice how he tried to maneuver, first with the Sigh. and then >by trying to downplay the assertion as if it were childish. You complain about length. Here you whine for 7 lines about a ->single<- word, not even once noting that what I said was ->correct<-. >> The point, however, is that any old cubic is not good enough. If you >> want to conclude stuff about the integrality of b_1, b_2, and b_3 >> (e.g., whether or not they are algebraic integers) then you need the >> cubic to be monic and to have algebraic integer (or better yet, >> integer) coefficients. > >Now creates his own requirements by mentioning >something that is basically true, but out of context. It's not out of context. When I talked about cubics for the b's, ->I<- always talked about their coefficients, and I said that when b3=a3, and the polynomial that defines the a's is irreducible, there is no cubic with integer coefficients that defines the b's. Instead of understanding that, you have launched yet another of your libelous personal attacks and appeals to the gallery, replete with smelly, stinking, red herrings. >That is, readers might now assume that I've claimed that the cubic for >the b's has algebraic integer coefficients, which I did, early on (see >my original post in this thread) but I had to post corrections as that >is wrong. So... You ->had<- made the claim I addressed. So it was not out of context for me to address that claim. But somehow, it is out of context for me to address it, because you had changed your mind? Quit your whining. [.snip.] > will create a requirement that he makes up, so that he >can then claim that I'm wrong when that requirement isn't fulfilled. That's a lie. I sta a property. YOU took issue with that property. To disprove my statement about a cubic with the b's as roots, you produced a polynomial that (a) failed to meet the hypothesis of my statement; and (b) agreed with the conclusion of my statement. Thus wasting time and space. I did not say you are wrong, until you star claiming that what ->I<- had said was wrong. In ->that<-, you were wrong. Your representation above is a lie. [.snip.] >> In any case when b_3=a_3 (which is every case in which a_1x is not >> equal to -uf and a_2x is not equal to -uf), you have b_3=a_3 but b_2 >> different from a_1 and a_2, and b_1 different from a_1 and >> a_2 (assuming f is not a unit, which you do). IF the polynomial that >> defines the a's is irreducible over Q, then ANY polynomial with >> integer coefficients that has a_3 as a root must be a multiple of the >> polynomial that defines the a's. And therefore, it is either of degree >> strictly larger than 3, or else it is a scalar multiple of the >> polynomial that defines the a's and has exactly a_1, a_2, and a_3 as >> roots. Therefore, in all such cases, it is impossible for there to be >> a single cubic WITH INTEGER COEFFICIENTS which has b_1, b_2, and b_3 >> as roots. > >Now notice how this poster has used *his* requirement, with emphasis, >as if he's made some important point. I did: the point was that it was part of my original statement, WHICH YOU IGNORED. [.snip.] >> This is simple, basic, precalculus algebra of polynomials. > >Yet another rather obvious maneuver. Yet another waste of space. [.snip.] >>The SINGLE cubic is important as it can't be non-monic. >> >>If it were non-monic, then how would it give a monic at m=0? Let f(x) = (2m+1)x^3 - 3x^2 +1. It's a single cubic. It gives a monic cubic when m=0. Are you claiming >> that it is ALWAYS monic? >Now having the poster goes for the finish where the attempt is to >finally completely fool the reader with a trivial example, while >refusing to handle the full situation. No. I took YOUR statement: if it were non-monic, then how would it give a monic at m=0? and I addressed ->THAT<- statement. This is a favorite tactic or yours: you invoke a false general principle to justify your claims. When you are given an example that the general principle does not hold, you complain that the example is not the situation you had. But you invoked it as a ->general<- principle. So you are engaging in a bait-n-switch. [.rest of bait-n-switch dele.] == Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticise. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b^3 - 3b^2, which is clearly > monic, at m=0. > > f_0(m).b^3 + f_1(m).b^2 + f_2(m).b + f_3(m), > > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? > > Which would require that > f_3(m)/f_0(m) = -(m^3 f^4 - 3m^2 f^2 + 3m) > so then, does anyone else besides this poster accept that possibility? > > Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) > algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), > obviously this is a monic cubic with constant term -b1.b2.b3 = > -(m^3.f^4 - 3m^2.f^2 + 3m). However we do not know whether the > coefficients of b^2 and b are algebraic integer for all m. But for each > m they are algebraic numbers. So choose an arbitrary m: m0. Say the > coefficient for b^2 is c2 and the coefficient for b is c1. We can write > c2 and c1 as the quotient of an algebraic integer and a normal integer. > Take r0 the lowest common multiple of the two normal integers. Multiply > the polynomial by r0, then > r0.(b-b1)(b-b2)(b-b3) > has algebraic integer coefficients for m=m0. Define the f_i(m0) > accordingly. Do that for each integer m and you get the given functions > f_i(m). Note that the r's depend on m! > > Well readedid anyone see what's wrong with this poster's position? > > He suggests > > f_0(m).b^3 + f_1(m).b^2 + f_2(m).b + f_3(m), > > where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? > > But the poster fails to point out that his position requires that for > certain values of m and another important variable I call f, that his > f_0(m) *must* have f as a factor, so it's basically a step function. > > Is that the right word step function? I think not. I have no idea what you mean with that term. Can you explain what you mean? > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > Here reducibility over rationals of the cubic defining the a's > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > has been given as an issue, and the posters position requires that his > f_0(m) vary such that it's 1 if that cubic is reducible over > rationals, and f, if it's irreducible over rationals. See my other post. It is a fallacy. With m=1 and f=2, the polynomial in a is reducible. The polynomial I come up with is: 2b^3 + (25 + sqrt(105))/2.b^2 + (-7 + sqrt(105))/2.b - 14. > O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? > > I'll direct a question to the poster now. > > Well, then, what is f_0(m) if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > is irreducible over Q? Depends on the roots a1 to a3, which two you divide by f, how the resulting b's can be written as a quotient of an algebraic integer and a normal integer (in lowest form), and perhaps a bit more. It is *not* possible to give a general formula. But I have given the algorithm through which you can calculate that number for each m. > Readethis post is a checkmate post. > > O. Why? This is not sufficient to show that the b's are algebraic > integers. To show that you must also have that f_0(m) is for each m > a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that > you must have that the coefficients of (b-b1)(b-b2)(b-b3) are > algebraic integeand to show that you must have that b1, b2 and b3 > are algebraic integers. Looks a bit circular to me. > > The poster has no room to run and has walked into the checkmate. > > All that's left is for the poster to concede (or run away) that his > position requires that f_0(m) behave as described. I have shown that that requirement is not there. But *even when that requirement was there*, my paragraph above is *still* true. The requirement is not even important. > Checkmate. Yup. -- === Subject: Re: Finishing argument, core error proven > ... > Now Nora Baron needs that SINGLE cubic to be non-monic for some > values of m, but that leads to a problem, there's no mathematical way > for the cubic to be non-monic, and have b^3 - 3b^2, which is clearly > monic, at m=0. f_0(m).b^3 + f_1(m).b^2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? Which would require that > f_3(m)/f_0(m) = -(m^3 f^4 - 3m^2 f^2 + 3m) > so then, does anyone else besides this poster accept that possibility? Oh, yes, for each integer m. But are f_1(m)/f_3(m) and f_2(m)/f_3(m) > algebraic integers for all m? Consider the cubic (x-b1)(x-b2)(x-b3), > obviously this is a monic cubic with constant term -b1.b2.b3 = > -(m^3.f^4 - 3m^2.f^2 + 3m). However we do not know whether the > coefficients of b^2 and b are algebraic integer for all m. But for each > m they are algebraic numbers. So choose an arbitrary m: m0. Say the > coefficient for b^2 is c2 and the coefficient for b is c1. We can write > c2 and c1 as the quotient of an algebraic integer and a normal integer. > Take r0 the lowest common multiple of the two normal integers. Multiply > the polynomial by r0, then > r0.(b-b1)(b-b2)(b-b3) > has algebraic integer coefficients for m=m0. Define the f_i(m0) > accordingly. Do that for each integer m and you get the given functions > f_i(m). Note that the r's depend on m! Well readedid anyone see what's wrong with this poster's position? He suggests f_0(m).b^3 + f_1(m).b^2 + f_2(m).b + f_3(m), where the f_i(m) are algebraic integer functions of m and f_0(0) = 1, > f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? But the poster fails to point out that his position requires that for > certain values of m and another important variable I call f, that his > f_0(m) *must* have f as a factor, so it's basically a step function. Is that the right word step function? I think not. I have no idea what you mean with that term. Can you explain > what you mean? Your position requires that f_0(m) either be a unit or have f as its only non-unit factor, depending on certain conditions as m varies from 0 to positive infinity. The condition is that if a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) is irreducible over Q, by your position, f_0(m) must have f as its only non-unit factor. > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > Here reducibility over rationals of the cubic defining the a's > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > has been given as an issue, and the posters position requires that his > f_0(m) vary such that it's 1 if that cubic is reducible over > rationals, and f, if it's irreducible over rationals. See my other post. It is a fallacy. With m=1 and f=2, the polynomial > in a is reducible. The polynomial I come up with is: > 2b^3 + (25 + sqrt(105))/2.b^2 + (-7 + sqrt(105))/2.b - 14. That looks ok as one of the roots is 1 as required, as one of the b's is 1. If it is the correct polynomial, then *each* of the coefficients has 2 as a factor, but NOT in the ring of algebraic integeas the middle coefficients are arbitrarily excluded by the definition of algebraic integers from having 2 as a factor. Those who doubt that assessment can instead believe in the step function, which somehow varies the leading coefficient from f to 1 based on whether or not a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) is reducible over Q. Note that for the example above, it reduces to a linear and a quadratic. Now there's no *mathematical* way to have a function f_0(m) which will vary in such an extreme manner depending on whether or not some cubic is reducible over Q or not, that is consistent with the other mathematical facts here. > O, yes for each integer m, f_0(m) is a factor of f_3(m). So what? I'll direct a question to the poster now. Well, then, what is f_0(m) if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > is irreducible over Q? Depends on the roots a1 to a3, which two you divide by f, how the resulting > b's can be written as a quotient of an algebraic integer and a normal > integer (in lowest form), and perhaps a bit more. It is *not* possible to > give a general formula. But I have given the algorithm through which you > can calculate that number for each m. That's nonsense. It's not necessary to give the expression explicitly as given that only f is being divided off, it follows that only f can be the non-unit factor, if there is one. Remember, this poster is trying to cast doubt on the factorization (b_1 x + u)(b_2 x + u)(b_3 x + uf) which comes from the factorization (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and the a's are given by a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) which is a monic with integer coefficients. So *only* f is the only non-unit available for the leading coefficient, if the poster's assertions are correct. Oh yeah, I can give an example myself, where you can see the cubic for the b's as b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 which works when m=1, f=sqrt(2), so if you wish to believe this poster you need to believe in that odd function which varies from being a unit to having f as a factor dependent on the reducibility of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > Readethis post is a checkmate post. O. Why? This is not sufficient to show that the b's are algebraic > integers. To show that you must also have that f_0(m) is for each m > a factor (in the algebraic integers) of f_1(m) and f_2(m). To show that > you must have that the coefficients of (b-b1)(b-b2)(b-b3) are > algebraic integeand to show that you must have that b1, b2 and b3 > are algebraic integers. Looks a bit circular to me. The poster has no room to run and has walked into the checkmate. All that's left is for the poster to concede (or run away) that his > position requires that f_0(m) behave as described. I have shown that that requirement is not there. But *even when that > requirement was there*, my paragraph above is *still* true. The > requirement is not even important. Part of the problem I face with certain posters is that they may not be smart enough to realize when they've been shown to be wrong. This poster either is unwilling to concede to the math, or is incapable of realizing the flaw in his position. The reality is that a function with the behavior he needs doesn't exist in this context, as his f_0(m) can't vary in the way he needs as m goes from 0 to positive infinity, where it is a unit for certain values, but switches to having f as its only non-unit factor for others. What I'm showing here is a outlandish consequence of the posters position, since this poster and others have continued to argue against the mathematical argument which shows that the leading coefficient is always monic as in fact two of the a's have a factor that is f, as shown by their terms independent of m, for all m. You see, m=0 is NOT a special case. I'll further note that this poster and others have gotten away with an especially odd position by trying to convince others that given terms independent of m--so m's value doesn't matter--that m=0 is a special case, as if those independent terms vary with m, which would mean they are dependent on m. > Checkmate. Yup. My assessment is that this poster probably doesn't quite realize the mathematical requirements of his claim. But then again, if his mathematical abilities were greater, he'd never have challenged using m=0 to clear out m to get to terms independent of m in the first place. After all, the technique isn't all that complica or hard to understand. === Subject: Re: Finishing argument, core error proven ... [ Note, it is about 3 a's of which exactly 2 should be divisible by f. The b';s are defined as follows: b1 = a1/f, b2 = a2/f, b3 = a3.] > The question is to the solution for the b's, just like earlier in the > post a solution for the a's is given as they are roots of the cubic > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > Readers should note the need for me to put *back* in math that is > needed to properly evaluate various claims, as this poster left them > out. I'll tackle again. > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) Yup. > The reason is that posters like this one have worked to convince > readers of mathematically false things, and I can show that their > assertions would force the cubic that has the b's for roots to be > *selectively* non-monic, where if they are correct, the cubic has to > be non-monic if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > is irreducible over Q, and monic if it's not. Lessee whether that is true. Let's take m = 1, f = 2. The polynomial becomes: a^3 + 9.a^2 - 28, this one is reducible: (a + 2)(a^2 + 7a - 14). Nevertheless, every polynomial with integer coefficients that defines the b's is not monic, because the b's are not all algebraic integers. So your claim is false. > That is, you'd need a function of m that would equal 1, when > > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > > is reducible over Q, and f, when it is not, and that function would be > the leading coefficient of the cubic defining the b's. And so the conclusion is false. > Pray show. (I think you mean a cubic polynomial?) > > > The cubic cannot, in general, be given without using ..., as then it > is > > b^3 + (...)b^2 + (...)b - (m^3 f^4 - 3m^2 f^2 + 3m). > > However, for certain values of m and f, the terms are algebraic > integeand the cubic is easily displayed. For instance, for m=1, > f=sqrt(2), it is I thought that f should be integer? But why do you not try the same with m=1 and f=2? Do you not dare because for those values the terms are *not* algebraic integers? > b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 > and note that here > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m), gives > a^3 + 3a^2 - 2, > which IS reducible, but intriguingly, isn't even reducible to linear > terms over Q. That is a red herring. > The answer is that posters arguing against the cubic defining the b's > being in general monic, are just wrong, and reducibility over Q just > determines whether or not you can easily *see* that they are wrong. No. See m=1, f=2. > At m=0 it is b^2.(b-3). > > Do you concede it Dik T. Winter? > > So, yup, no problem. > > What's important about that readers is it forces the poster to either > accept that the cubic is monic in general, which means accepting the > core problem, or claim that there's a *function* of m, which varies > in some odd way. And that is indeed the whole point. There's a *function* of m that varies in some odd way. > There is: (x - b1)(x - b2)(x - b3). But, does it have integer coefficients, > or rational coefficients? Not when f = 5 and u = 1, as shown above. > > But is it monic? > > As written, yes it is monic. No problem. But it does not necessarily > have integer coefficients, so what do you conclude from it being monic? > > Well, no, it doesn't necessarily have integer coefficients, like I > showed before with f=sqrt(2), m=1, where > b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 > but readers can see just how misleading Nora Baron was in posts > showing TWO cubics, where one must exist. A red herring. Nora Baron has shown that b1 and b2 are roots of a primitive irreducible non-monic polynomial with integer coefficients. So b1 and b2 are *not* algebraic integers. That is actually easy to prove. Here a similar proof: The a's are roots of (a^3 + 3(-1+mf^2).a^2 - f^2.(m^3.f^4 - 3m^2.f^2 + 3m). b1 = a1/f and b2 = a2/f are roots of the same polynomial with a replaced by b.f: f^3.b^3 + 3.f^2.(-1 + mf^2).b^2 - f^2.(m^3.f^4 - 3m^2.f^2 + 3m). We can divide by f^2 (as f is nonzero) and get: f.b^3 + 3.(-1 + mf^2).b^2 - (m^3.f^4 - 3m^2.f^2 + 3m). So two of the b's are roots of this polynomial, which is in general non-monic. > Now for certain values, the cubic has algebraic integer coefficients, > while for otheit does not because the ring of algebraic integers > doesn't include certain numbers that it should, which leads to a > problem in core as you can have the appearance of two proofs > contradicting each other. Eh? See below. > That's bogus, as I can show with > b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1. A red herring. > Now this poste may *want* a function as the leading coefficient of the > general b cubic such that it magically equals f when > > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > > is irreducible over Q, and switches to 1, when it is reducible, but > there's no logical basis for that position. False and false. > It's simply some person *wishing* something they cannot prove. See the proof above. > In fact, the coefficients provably cannot be algebraic integers for > all values of m and f, like for instance, with m=1, f=sqrt(5), they > are not. Eh? What are the coefficients with m=1 and f=sqrt(5)? And why can they not become algebraic integers when you multiply the cubic by an integer? >Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, > > Here's the second sneak, as yes, the b's are in fact algebraic > numbeeven when they're not algebraic integebut the poster has > moved to a field. So what? You also want to move to something different from the algebraic integers. But what I am showing here is that *given three algebraic numbers b, there is a cubic polynomial with algebraic integer coefficients that has the b's as root*. You just claimed above that that was false with the b's resulting for some m and f. > so all coefficients are algebraic numbers. Each algebraic number can be > written as the quotient of an algebraic integer and a normal integer. > > Here's the next point of the attempt at snowing the reader, as the > poster is pushing the position that the b's can be written like, say, > x/f, where x is some algebraic integer that doesn't have f as a > factor, but possibly shares non-unit factors with f. Yes. Do you want a proof? Take the defining polynomial for an algebraic number b (note: I say defining polynomial, so it is primitive and irreducible): a_n.x^n + a_[n-1].x^(n-1) + ... + a1.x + a0 where the a_n are normal integers. Multiply the polynomial by a_n^(n-1), we get: a_n^n.x^n + a_[n-1].a_n^(n-1).x^(n-1) + ... + a1.a_n^(n-1)x + a0.a_n^[n-1], rewrite as a polynomial in (a_n.x): (a_n.x)^n + a_[n-1].(a_n.x)^(n-1) +...+ a1.a_n^(n-2).(a_n.x) + a0.a_n^[n-1], note that this is a monic polynomial with integer coefficients, so its roots are algebraic integers. One of the roots is a_n.b, so b = (a_n.b)/a_n where a_n.b is an algebraic integer and a_n is integer. All pretty basic. > Do > so. Multiply by the LCM of the denominators and you have a polynomial with > algebraic integer coefficients of which the three b's are roots. The question > is: is that polynomial monic? The answer is: in general, no. > > Now notice, given all of that, you'd think that there'd be some > *logical* argument, but instead you have the poster simply asser > the desired conclusion. Is it not clear enough? The leading coefficient is the LCM of all the denominators. Even if we express the original coefficients in lowest term as quotient of an algebraic integer and an integer, the LCM of the integers is in general *not* 1. It is only 1 when all coefficients *are* already algebraic integers. > It's this type of poster that gives me fits as Nora Baron and Arturo > Magidin both engage in this behavior. Because you do not understand very basic math? > Later, the poster comes back with the same techniques, never actually > managing to refute a single point of my argument mathematically. What parts of the refutations above do you not understand? -- === Subject: Magidin is too many (was Re: Finishing argument, core error proven) == > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A man's capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made[1] at least > ten publications, full of figures[2] few readers can critize[3]. A great > many people are staggered[4] to this extend[5], that they imagine there > must be the[6] indefinite something in the mysterious all this[7]. > They are brought to the point of suspicion that[8] the mathematicians[9] > ought not to treat all this with such undisguised contempt, > at least[10]. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan [1] 'make' publications: 'make' instead of 'do'. [2] 'figures': wrong word [3] 'critize' publications; a quasi-verb, formed by attaching an English suffix to a Spanish root (e.g. 'criticar' - 'icar' = 'crit'; 'crit' + 'ize' = 'critize') [4] many people are 'staggered' to this extend: inappropriate use of informal register [5] many people are staggered to this 'extend': participle error [7] there must be the indefinite something in the mysterious *all this*: Google translations are worth what they cost. [8] they are 'brought' to the 'point' of 'suspicion' that: Google translations are worth what they cost. [10] with such undisguished contempt, 'at least': Google translations are worth what they cost. --John > = > magidin@math.berkeley.edu === Subject: Re: Magidin is too many (was Re: Finishing argument, core error proven) == >> Why do you take so much trouble to expose such a reasoner as >> Mr. Smith? I answer as a deceased friend of mine used to answer >> on like occasions - A man's capacity is no measure of his power >> to do mischief. Mr. Smith has untiring energy, which does >> something; self-evident honesty of conviction, which does more; >> and a long purse, which does most of all. He has made[1] at least >> ten publications, full of figures[2] few readers can critize[3]. A great >> many people are staggered[4] to this extend[5], that they imagine there >> must be the[6] indefinite something in the mysterious all this[7]. >> They are brought to the point of suspicion that[8] the mathematicians[9] >> ought not to treat all this with such undisguised contempt, >> at least[10]. >> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan It's very hard to see what your point is. Regardless, you seem unable to recognize obvious typos, unaware of the fact that accep English usage was different when de Morgan was alive (1806-1871), and also unaware of various constructions that are perfectly standard even today. Then there's the question of why you assume this is a translation, by Google or otherwise - de Morgan was British. Amounts to a standard spelling flame, except an unusually In detail: >[1] 'make' publications: 'make' instead of 'do'. Huh? One does a publication? I don't think so. >[2] 'figures': wrong word Huh? How did you deduce that the writer was not referring to figures? As in pictures, diiagrams, etc. >[3] 'critize' publications; a quasi-verb, formed by attaching an >English suffix to a Spanish root (e.g. 'criticar' - 'icar' = 'crit'; >'crit' + 'ize' = 'critize') Huh? This seems to be simply either a typo for criticise or a typo for an old (and/or British) spelling of criticise. >[4] many people are 'staggered' to this extend: inappropriate use of >informal register >[5] many people are staggered to this 'extend': participle error Huh? Again, seems clearly a typo for staggered to this extent (or again, possibly an archaic usage). something was meant - I don't see why you assume that, the indefinite something makes perfect sense. >[7] there must be the indefinite something in the mysterious *all >this*: Google translations are worth what they cost. >[8] they are 'brought' to the 'point' of 'suspicion' that: Google >translations are worth what they cost. Uh, right. Huh? >[10] with such undisguished contempt, 'at least': Google >translations >are worth what they cost. Or a typo for undisguised. >--John > >> = >> magidin@math.berkeley.edu ************************ === Subject: Re: Magidin is too many linux) == >> Why do you take so much trouble to expose such a reasoner as >> Mr. Smith? I answer as a deceased friend of mine used to answer >> on like occasions - A man's capacity is no measure of his power >> to do mischief. Mr. Smith has untiring energy, which does >> something; self-evident honesty of conviction, which does more; >> and a long purse, which does most of all. He has made[1] at least >> ten publications, full of figures[2] few readers can critize[3]. A great >> many people are staggered[4] to this extend[5], that they imagine there >> must be the[6] indefinite something in the mysterious all this[7]. >> They are brought to the point of suspicion that[8] the mathematicians[9] >> ought not to treat all this with such undisguised contempt, >> at least[10]. >> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan [1] 'make' publications: 'make' instead of 'do'. > [2] 'figures': wrong word > [3] 'critize' publications; a quasi-verb, formed by attaching an > English suffix to a Spanish root (e.g. 'criticar' - 'icar' = 'crit'; > 'crit' + 'ize' = 'critize') > [4] many people are 'staggered' to this extend: inappropriate use of > informal register > [5] many people are staggered to this 'extend': participle error > [7] there must be the indefinite something in the mysterious *all > this*: Google translations are worth what they cost. > [8] they are 'brought' to the 'point' of 'suspicion' that: Google > translations are worth what they cost. > [10] with such undisguished contempt, 'at least': Google > translations > are worth what they cost. What a shock. John Correy is being a presumptuous moron. Arturo has already said that his copy of de Morgan's work is by Dover books, if I recall correctly. Hence, it is plausible that it is an English translation of the text, so John's stupid criticisms ought to be direc to the translator and not Arturo (aside from the typos in the words criticize and extent, presumably). In any case, John's criticism of the use of figures is bumfuzzling. What word is preferable there? Also, what is wrong with [1]? Surely one does not *do* publications, but makes them? I presume [4], [6] and [7] are faithful translations of de Morgan's tone and intent, so John's complaint is with the author, not Arturo. The same goes for the majority of the remainder. Usenet criticisms of translations of historic texts are worth what they cost, I suppose. -- The sole cause of all human misery is the inability of people to sit quietly in their rooms. -- Blaise Pascal === Subject: Re: Magidin is too many > == >> Why do you take so much trouble to expose such a reasoner as >> Mr. Smith? I answer as a deceased friend of mine used to answer >> on like occasions - A man's capacity is no measure of his power >> to do mischief. Mr. Smith has untiring energy, which does >> something; self-evident honesty of conviction, which does more; >> and a long purse, which does most of all. He has made[1] at least >> ten publications, full of figures[2] few readers can critize[3]. A great >> many people are staggered[4] to this extend[5], that they imagine there >> must be the[6] indefinite something in the mysterious all this[7]. >> They are brought to the point of suspicion that[8] the mathematicians[9] >> ought not to treat all this with such undisguised contempt, >> at least[10]. >> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan [1] 'make' publications: 'make' instead of 'do'. > [2] 'figures': wrong word > [3] 'critize' publications; a quasi-verb, formed by attaching an > English suffix to a Spanish root (e.g. 'criticar' - 'icar' = 'crit'; > 'crit' + 'ize' = 'critize') > [4] many people are 'staggered' to this extend: inappropriate use of > informal register > [5] many people are staggered to this 'extend': participle error > [7] there must be the indefinite something in the mysterious *all > this*: Google translations are worth what they cost. > [8] they are 'brought' to the 'point' of 'suspicion' that: Google > translations are worth what they cost. > [10] with such undisguished contempt, 'at least': Google > translations > are worth what they cost. What a shock. John Correy is being a presumptuous moron. No, is being a presumptuous moron. I've noticed a consistent double standard from posters. > Arturo has already said that his copy of de Morgan's work is by Dover > books, if I recall correctly. Hence, it is plausible that it is an > English translation of the text, so John's stupid criticisms ought to > be direc to the translator and not Arturo (aside from the typos in > the words criticize and extent, presumably). Here apparently *wishes* to defend , and has decided to attack John Corry, apparently out of anger. > In any case, John's criticism of the use of figures is bumfuzzling. > What word is preferable there? Also, what is wrong with [1]? Surely > one does not *do* publications, but makes them? I presume [4], [6] > and [7] are faithful translations of de Morgan's tone and intent, so > John's complaint is with the author, not Arturo. The same goes for > the majority of the remainder. Usenet criticisms of translations of historic texts are worth what > they cost, I suppose. The way I see it, was being a smartass with a translation, and was unaware of problems with his usage of a text, and the meaning as John Correy sees it from reading it in its original language. Rather than just nod at what's not necessarily a big deal--another screw-up from --the poster decides to create another thread to attack John Correy, and THEN at the end belabors Usenet criticism of translations!!! If anything is true in general about Usenet, it's that people can go on and on about just about anything. === Subject: Re: Magidin is too many > == >> Why do you take so much trouble to expose such a reasoner as >> Mr. Smith? I answer as a deceased friend of mine used to answer >> on like occasions - A man's capacity is no measure of his power >> to do mischief. Mr. Smith has untiring energy, which does >> something; self-evident honesty of conviction, which does more; >> and a long purse, which does most of all. He has made[1] at least >> ten publications, full of figures[2] few readers can critize[3]. A >> great >> many people are staggered[4] to this extend[5], that they imagine there >> must be the[6] indefinite something in the mysterious all this[7]. >> They are brought to the point of suspicion that[8] the >> mathematicians[9] >> ought not to treat all this with such undisguised contempt, >> at least[10]. >> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan [1] 'make' publications: 'make' instead of 'do'. > [2] 'figures': wrong word > [3] 'critize' publications; a quasi-verb, formed by attaching an > English suffix to a Spanish root (e.g. 'criticar' - 'icar' = 'crit'; > 'crit' + 'ize' = 'critize') > [4] many people are 'staggered' to this extend: inappropriate use of > informal register > [5] many people are staggered to this 'extend': participle error > [7] there must be the indefinite something in the mysterious *all > this*: Google translations are worth what they cost. > [8] they are 'brought' to the 'point' of 'suspicion' that: Google > translations are worth what they cost. > [10] with such undisguished contempt, 'at least': Google > translations > are worth what they cost. What a shock. John Correy is being a presumptuous moron. No, is being a presumptuous moron. I've noticed a consistent double standard from posters. JSH's standards of judging mathematics are at least double, one low standard for his own work and total rejection of anyone else's criticism of it. > > Arturo has already said that his copy of de Morgan's work is by Dover > books, if I recall correctly. Hence, it is plausible that it is an > English translation of the text, so John's stupid criticisms ought to > be direc to the translator and not Arturo (aside from the typos in > the words criticize and extent, presumably). Here apparently *wishes* to defend , and > has decided to attack John Corry, apparently out of anger. Here JSH demonstrates his belief in his own abilities to read minds. > > In any case, John's criticism of the use of figures is bumfuzzling. > What word is preferable there? Also, what is wrong with [1]? Surely > one does not *do* publications, but makes them? I presume [4], [6] > and [7] are faithful translations of de Morgan's tone and intent, so > John's complaint is with the author, not Arturo. The same goes for > the majority of the remainder. Usenet criticisms of translations of historic texts are worth what > they cost, I suppose. The way I see it, was being a smartass with a > translation, and was unaware of problems with his usage of a text, and > the meaning as John Correy sees it from reading it in its original > language. That just shows how ceyed JSH's vision is, since English, and 19th century English, is the native language of English speaking Irishman mathematician and logician Augustus de Morgan, and is the original language of A Budget of Paradoxes. Rather than just nod at what's not necessarily a big deal--another > screw-up from --the poster decides to > create another thread to attack John Correy, and THEN at the end > belabors Usenet criticism of translations!!! Magidin did not screw up here, and on the rare ocasions when he does, is the first to acknowledge it. But JSh has screwed up again in trying to attack his betters.. If anything is true in general about Usenet, it's that people can go > on and on about just about anything. As is evidenced by JSH's idiot postings. > === Subject: Re: Magidin is too many <8765iqu9lr.fsf@phiwumbda.org> linux) What a shock. John Correy is being a presumptuous moron. No, is being a presumptuous moron. I've noticed a consistent double standard from posters. No double standard here. John Correy attemp to criticize Arturo's translation of a quotation by de Morgan. His criticism was particularly stupid since he didn't realize that de Morgan was British and the quotation was untransla (a fact I didn't realize either, but then I didn't criticize the translation). >> Arturo has already said that his copy of de Morgan's work is by Dover >> books, if I recall correctly. Hence, it is plausible that it is an >> English translation of the text, so John's stupid criticisms ought to >> be direc to the translator and not Arturo (aside from the typos in >> the words criticize and extent, presumably). Here apparently *wishes* to defend , and > has decided to attack John Corry, apparently out of anger. No, no, no, I did not wish to defend Magidin. I simply chose once again to call John Correy an offensive and stupid bastard. It's not as if this is uncharacteristic of me. On the contrary, I regularly call John Correy an offensive and stupid bastard. Mostly, I do this when John Correy is being an offensive and stupid bastard (hence the regularity). I'm not particularly angry at him. Why would I be? Arturo can defend himself, obviously, but no one else can call John an offensive and stupid bastard for me. Some things I just feel I gotta do for my damn self. >> In any case, John's criticism of the use of figures is bumfuzzling. >> What word is preferable there? Also, what is wrong with [1]? Surely >> one does not *do* publications, but makes them? I presume [4], [6] >> and [7] are faithful translations of de Morgan's tone and intent, so >> John's complaint is with the author, not Arturo. The same goes for >> the majority of the remainder. Usenet criticisms of translations of historic texts are worth what >> they cost, I suppose. The way I see it, was being a smartass with a > translation, and was unaware of problems with his usage of a text, and > the meaning as John Correy sees it from reading it in its original > language. Yeah, you'd think that. But, I guess, that isn't so plausible given that it's untransla. > Rather than just nod at what's not necessarily a big deal--another > screw-up from --the poster decides to > create another thread to attack John Correy, and THEN at the end > belabors Usenet criticism of translations!!! I didn't create another thread, you silly boy. My newsreader asked me whether to omit the (Was: ....) part of the subject line, which is standard for replies to new subjects. Blame Google for failing to thread according to references or failing to recognize that Subject 1 is part of the same thread as Subject 1 (Was: Subject 2). Or blame yourself for general stupidity. Why would I defend Arturo by starting a thread with a subject line Magidin is too many? > If anything is true in general about Usenet, it's that people can go > on and on about just about anything. -- If anything is true in general about Usenet, it's that people can go on and on about just about anything. -- speaks the truth. === Subject: Re: Magidin is too many Visiting Assistant Professor at the University of Montana. >The way I see it, was being a smartass with a >translation, and was unaware of problems with his usage of a text, and >the meaning as John Correy sees it from reading it in its original >language. The original language ->is<- English; Augustus de Morgan was an English mathematician, one time president of the Royal Astronomical Society, and _A Budget of Paradoxes_ was written in English. I did have two typos, critize for criticize, and extend for extent. They have been correc. The way I see it, you don't know anything, but you simply assume that I must be doing something wrong... So much for your alleged concern for truth. >Rather than just nod at what's not necessarily a big deal--another >screw-up from -- Do please list the many screw-ups of mine. Because, although you spend enormous amounts of time and space whining about my screw-ups, except for the current arguments, each and every time you have eventually admit that I was correct and the only screwing up was you. So, when you talk about another screw-up, you're just lying. Again. == Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticise. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: Re: Magidin is too many >I did have two typos, critize for criticize, Not criticise? Just wondering... -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: Magidin is too many Visiting Assistant Professor at the University of Montana. > >>I did have two typos, critize for criticize, > >Not criticise? Just wondering... I'll check my copy at home tonight. It's possible my fingers got ahead of me there. The Open Court edition, which I own, was re-edi and published in Chicago, with additional notes by David Eugene Smith. It's possible that some of the spelling may have been altered to reflect american standards along the way... == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) == === Subject: Re: Magidin is too many >I did have two typos, critize for criticize, >>Not criticise? Just wondering... >I'll check my copy at home tonight. It's possible my fingers got ahead >of me there. >The Open Court edition, which I own, was re-edi and published in >Chicago, with additional notes by David Eugene Smith. It's possible >that some of the spelling may have been altered to reflect american >standards along the way... According to the OED, criticize is the preferred spelling. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Magidin is too many linux) According to the OED, criticize is the preferred spelling. What does the OED say about, say, generalize/generalise? -- So, at this time, I'd like to assure you that I am not interes in making sure mathematicians worldwide get fired. I've rethought my desire to go to Congress and try to get funding for mathematicians cut. -- is a reasonable man. Whew! === Subject: Re: Magidin is too many >I did have two typos, critize for criticize, Not criticise? Just wondering... >I'll check my copy at home tonight. It's possible my fingers got ahead >>of me there. >The Open Court edition, which I own, was re-edi and published in >>Chicago, with additional notes by David Eugene Smith. It's possible >>that some of the spelling may have been altered to reflect american >>standards along the way... According to the OED, criticize is the preferred spelling. This is what Fowler prefers too. Many mistakenly regard the -ize ending as an Americanisms and so avoid it, however the -ise they use is something of a Gallicism But there are exceptions: televise --- a back formation from television for one. (Half Greek and half Latin -- nothing good will ever come from it) -- === Subject: Re: Magidin is too many > This is what Fowler prefers too. Many mistakenly regard the -ize > ending as an Americanisms and so avoid it, however the -ise they > use is something of a Gallicism But there are exceptions: > televise --- a back formation from television for one. > (Half Greek and half Latin -- nothing good will ever come from it) Indeed a source of confusion, particularly for international speakers. Of course, being forced to use a sed up language to begin with, (for international communication, that is) doesn't help, either. Now, there's a good chap...:*) > -- > -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: Re: Magidin is too many > What a shock. John Correy is being a presumptuous moron. I've been pleased to see that almost nobody has been rising to Correy's bait lately. Obviously the troll is getting lonely and looking for new tactics. === Subject: Re: Magidin is too many > What a shock. John Correy is being a presumptuous moron. I've been pleased to see that almost nobody has been rising > to Correy's bait lately. Obviously the troll is getting lonely > and looking for new tactics. I find that killfiling Correy works pretty well -- still occasionally people quote his garbage (sometimes containing personal attacks). -- === Subject: Re: Magidin is too many Visiting Assistant Professor at the University of Montana. >Arturo has already said that his copy of de Morgan's work is by Dover >books, if I recall correctly. Actually, I own a copy of the 1915 Open Court Press edition. >Hence, it is plausible that it is an >English translation of the text, so John's stupid criticisms ought to >be direc to the translator and not Arturo (aside from the typos in >the words criticize and extent, presumably). Yes; I've correc those typos. at all. >In any case, John's criticism of the use of figures is bumfuzzling. de Morgan meant figures; the individual in question was attempting to square the circle. >I presume [4], [6] >and [7] are faithful translations of de Morgan's tone and intent, so >John's complaint is with the author, not Arturo. [4], [6], and [7] (modulo typos) are de Morgan's ->words<-, as written by him in the first place. == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) == === Subject: Re: Magidin is too many <8765iqu9lr.fsf@phiwumbda.org> linux) > at all. Ha! That's funny! I had simply assumed that John was correct that it was a translation. Well, I guess you simply should have quo a better writer, or something. -- What I've learned is that [mathematicians are] the gatekeepeand seem to have almost absolute power when it comes to mathematics. -- , on All I Really Ever Needed to Know I Learned in /Ghostbusters/. === Subject: Re: Magidin is too many >[...]I had simply assumed that John was correct Wow. Gonna take you some time to live that down... ************************ === Subject: Re: Finishing argument, core error proven [cut] >Let's take (x-b1)(x-b2)(x-b3); all b's are algebraic numbers, Here's the second sneak, as yes, the b's are in fact algebraic > numbeeven when they're not algebraic integebut the poster has > moved to a field. There is nothing wrong with moving to a field (the field of algebraic numbers). Are you saying that this is an invalid step in his reasoning? Why would it be invalid? Hale === Subject: JSH rants again (was): Finishing argument, core error proven posting-host=adsl-66-126-135-58.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1066173344 http://mygate.mailgate.org/mynews/sci/sci.math/ 9f71621e2e85daf0d7225103900419 d0.48257%40mygate.mailgate.org > Hey, that's correct, which means I had it right the first > time, but later second guessed myself about a_3 and b_3. And there in a nutshell you have pretty much wrapped up the reasons you will always be a buffoon in sci.math, James. Math is not _about_ guessing, it is about _proving_, and you don't have and won't be bothered to learn the skills to craft correct proofs. Math is not _about_ picking a goal and finding all the lame argumentation you can assemble, worthy or not, to support that goal, math is about going where the rock solid evidence leads you; it is exploration, not construction, and is absolutely not a debate class. Math is not played by the rules of politics, so all your ranting about other poster's tactics are, all your jockeying for someone to publish your incorrect results as if simply by being published that would make them less false or less of a waste of everyones time is, an indication of insanity and entirely beside the point. Math is played by the rules of Math and by no otheand you just don't have what it takes to be a player because you cannot bring yourself to play by the rules of the game and you won't be taken seriously unless you do. I understand from hints elsewhere that you are a member of a minority and think that is somehow involved in your reception. Trust me, it is your membership in the majority, who are math illiterates like you, that accounts for your lack of finding warmth and fellowship here, among those who have spent large portions of their lives becoming conversant with humankind's most splendid achievement(*), not your religion, race, place of origin, or whatever triviality it is that seems to you to be the issue. I know it will be hard for you to accept that your path to fame and glory is not to be found in the field of math, but as many years as you have was chasing that ephemera should by now constitute a clue capable of penetrating even a brick like the one you use to decorate the top of your neck. Find a life. xanthian, reading sci.math like a kid with his nose pressed against the candy store window, wishing and wanting, but without the proper coin of the realm; in a sense, I can really share James' pain, but not admire its outcome. (*) Which the acceptance of the nonsense you espouse would only tend to tear down, degrade, and destroy. -- Pos via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Re: JSH rants again (was): Finishing argument, core error proven <9f71621e2e85daf0d7225103900419d0.48257@mygate.mailgate.org>, > Math is not played by the rules of politics, Which is just as well. If we used the rules of politics, then we would all pretend that Harris claimed weapons of math destruction actually exist and blast him out of existence. Then in a year or so, we would start admitting that no proof for any of his theorems has so far been found, but we are quite sure that the proofs will be found. In ten years or so, Harris would be forgotten. === Subject: Re: Finishing argument, core error proven > For me there have been two perspectives as I work to figure out how to > explain the definition problem in mathematics with LOTS of opposition, > and I wonder about mathematicians so dedica to attacking an > argument that is clearly correct. I remind of that as I present what should finish their ability to > distract, as I've seen a strange and dedica effort to ignore the > actual math, and simply toss up just about anything rather than face > the truth. All variables are in the ring of algebraic integers unless otherwise > sta. Let P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and let R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f so P(m) = f^2 R(m). Now consider P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the a's are given by the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). Then it must be true that the following factorization exists R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where the b's are given by the following cubic: b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), should be b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m) where a_3 = b_3, and at m=0, b_3 = 3. It's not true in general that a_3 = b_3. Um, here's one spot where I got it wrong, when I was right the first > time, as poin out in this thread on sci.math by the poster Arturo > Magidin. And in fact, in general, a_3 = b_3. The cubic I gave was wrong though, again, as the correct cubic is b^3+ (...)b^2 + (...)b+...- (m^3 f^4 - 3m^2 f^2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. I am new to this, but a fresh look can always help. Am I wrong? If so, perhaps you could provide a few more numerical examples or a function in terms of f and m. > === Subject: Re: Finishing argument, core error proven The cubic I gave was wrong though, again, as the correct cubic is b^3+ (...)b^2 + (...)b+...- (m^3 f^4 - 3m^2 f^2 + 3m) which may seem strange, but consider m=1, f=sqrt(2), gives b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1 where you can see what's hidden in general. > I am new to this, but a fresh look can always help. For that specific case where m=1, f=sqrt(2), but you can't generalize the way you're trying, like, for m=0, without regard to the value of f, you have b^3 - 3b^2. Remember R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) and the cubic, in general, that has the b's for roots has inexpressible terms, which is why I use b^3+ (...)b^2 + (...)b+...- (m^3 f^4 - 3m^2 f^2 + 3m). > Am I wrong? If so, perhaps you could provide a few more numerical examples > or a function in terms of f and m. My point in using the example is to show that certain posters are full of it, as they *have* been attacking rather basic algebra, especially with one particularly annoying tactic. It's not such a complica thing in analysis when you have terms independent of a particular variable to set that variable to 0 to clear it out, but these posters attacked that basic technique and claimed that m=0 was a special case!!! The insinuation being that terms independent of m were in fact dependent on m, but when challenged, they simply talked around that reality, and continued to attack algebra. Now back to your question, in terms of figuring out particular cubics for the b's, like how I did for m=1, f=sqrt(2), it's a matter of working it out and I haven't worked out any more examples at this time. === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >My point in using the example is to show that certain posters are full >of it, as they *have* been attacking rather basic algebra, especially >with one particularly annoying tactic. > >It's not such a complica thing in analysis when you have terms >independent of a particular variable to set that variable to 0 to >clear it out, but these posters attacked that basic technique and >claimed that m=0 was a special case!!! Your basic technique seems to be a mess. Whenever people ask you what the term is, instead of answering you just start new threads. You claim something changes, when people ask you what and how, you don't answer, you just start new threads and repeat the same claim. I went to the Hong Kong forum you were recommending. I no that as soon as a rather basic question about your basic technique was asked, you ran away like a frightened little kid. No wonder you haven't been posting there any more. Let me quote it here, maybe you'll have the courage to actually answer As there are continued postings it might help to consider the alternative. So, consider again $P(m)=g_1 g_2 g_3$, $g_1 = a_1 x + uf$ where it's established that $uf$ is independent of $m$ as determined by setting $m=0$, note $P(0)=u^2f^2(3x + uf)$. Now suppose that instead of $f$ as a factor $g_1$ had $sqrt{f}$ when $m=5$, then you'd have $g_1/sqrt{f} = a_1/sqrt{f} x + usqrt{f}$ so the independent term has changed. You were asked which function had an independent term that changed, what it changed from, and what it changed to. Rather than answering that simple question (just three things to say: what had changed, from what, and to what), you replied: Let $P(m)=f^2 Q(m)$ where I introduce $Q(m)$ as I fear using $P(m)/f^2$ may be confusing to Random. Now then, there exists factors I'll call $h_1$, $h_2$, $h_3$ such that $Q(m)=h_1 h_2 h_3$ and $h_1$ is a factor of $g_1$, $h_2$ is a factor of $g_2$, and $h_3$ is a factor of $g_3$. So I can get the terms *independent* of m, with the h's, in the same way that I did for the g's, by setting $m=0$, which gives $h_1=u$, $h_2=u$, and $h_3=3x+uf$, which is a conclusion that may be deba. Now then, I note that if the value of $m$ affects what factor of $f$ it is that $g_1$ has then the *independent* terms of the h's will vary with it. For instance, if $g_1$ has a maximum factor of $f$ that is $sqrt{f}$ then it would force the *independent* term of $h_1$ to have $sqrt{f}$ as a factor, which violates two things: 1. The term independent of $m$ would vary with $m$. 2. The independent term of $Q(0)$ is $u^2(3x + uf)$ which is coprime to $f$. You were never clear on something. Are h_1(m) = g_1(m)/f, h_2(m) = g_2(m)/f, and h_3(m) = g_3(m), or is the definition h_1(m) = g_1(m)/gcd(g_1(m),f) h_2(m) = g_2(m)/gcd(g_2(m),f), h_3(m)=g_3(m)/gcd(g_3(m)/f) ? Simple question, should have a simple answer. Another simple question: Independent term means what you get when you evaluate at m=0. What function has the independent term that would vary with m, if g_1(5) were a multiple of sqrt(f)? Is it P(m), Q(m), g_1(m), g_2, g_3, h_1, h_2, h_3 (and with what definition)? What was the independent term before, and what did it become after when m=5? Make sure you explain how you can tell it changed. Simple, no? But no, instead of answering that simple question, you make statements like: >The insinuation being that terms independent of m were in fact >dependent on m, but when challenged, they simply talked around that >reality, and continued to attack algebra. == Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticise. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: Re: Finishing argument, core error proven [.snip.] >My point in using the example is to show that certain posters are full >of it, as they *have* been attacking rather basic algebra, especially >with one particularly annoying tactic. > >It's not such a complica thing in analysis when you have terms >independent of a particular variable to set that variable to 0 to >clear it out, but these posters attacked that basic technique and >claimed that m=0 was a special case!!! Your basic technique seems to be a mess. Whenever people ask you > what the term is, instead of answering you just start new > threads. You claim something changes, when people ask you what and > how, you don't answer, you just start new threads and repeat the same > claim. That's a rather interesting falsehood. A key expression has long been g_1 = a_1 x + uf, where, though not seen, certain variables are dependent on m, such that some posters want me to write g_1(m) = a_1(m) x + uf, and at m=0, g_1 = uf, as uf is the independent term. So it is not true that I don't answer what the independent term is. I went to the Hong Kong forum you were recommending. I no that as > soon as a rather basic question about your basic technique was > asked, you ran away like a frightened little kid. No wonder you > haven't been posting there any more. I last pos there a few days ago, which doesn't mean I won't post there again. The poster doesn't give a link so I'll give one for readers who wish to see the full discussion on the Hong Kong website, where there's an additional advantage as they allow use of LaTeX. See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 > Let me quote it here, maybe you'll have the courage to actually answer As there are continued postings it might help to consider the > alternative. So, consider again $P(m)=g_1 g_2 g_3$, $g_1 = a_1 x + uf$ where it's > established that $uf$ is independent of $m$ as determined by setting > $m=0$, note $P(0)=u^2f^2(3x + uf)$. Now suppose that instead of $f$ as a factor $g_1$ had $sqrt{f}$ when > $m=5$, then you'd have $g_1/sqrt{f} = a_1/sqrt{f} x + usqrt{f}$ so the independent term has changed. You were asked which function had an independent term that changed, > what it changed from, and what it changed to. Oh, I remember that, as some poster kept going on and on asking the same question after I answered. > Rather than answering that simple question (just three things to say: > what had changed, from what, and to what), you replied: Let $P(m)=f^2 Q(m)$ where I introduce $Q(m)$ as I fear using > $P(m)/f^2$ may be confusing to Random. > > Now then, there exists factors I'll call $h_1$, $h_2$, $h_3$ such that > $Q(m)=h_1 h_2 h_3$ and $h_1$ is a factor of $g_1$, $h_2$ is a factor > of $g_2$, and $h_3$ is a factor of $g_3$. So I can get the terms *independent* of m, with the h's, in the same > way that I did for the g's, by setting $m=0$, which gives $h_1=u$, > $h_2=u$, and $h_3=3x+uf$, which is a conclusion that may be deba. Here readers can see that I mention u as independent, and mention that I'd done the same for the g's where uf is independent of the value of m. And THAT is why I can use m=0 because an independent term is independent without regard to the value of m, so m=0 is as good as any other in one sense, and perfect in another, as it clears m out so that I can focus on the constant term. > Now then, I note that if the value of $m$ affects what factor of $f$ > it is that $g_1$ has then the *independent* terms of the h's will vary > with it. For instance, if $g_1$ has a maximum factor of $f$ that is > $sqrt{f}$ then it would force the *independent* term of $h_1$ to have > $sqrt{f}$ as a factor, which violates two things: 1. The term independent of $m$ would vary with $m$. 2. The independent term of $Q(0)$ is $u^2(3x + uf)$ which is coprime > to $f$. You were never clear on something. What I did was explain how refusing to accept the term as independent would cause a contradiction. Specifically I have a polynomial Q(m), which has a constant term that is u^2 (3x + uf), which is coprime to f since f is coprime to 3, x and u. Now the simple position is that at m=0, h_1 = u, showing u as the constant term, and in fact that's provable, but I no some might wish to debate it. Given that h_1 has a term independent of m that's u, while it is a factor of g_1 which has uf as a term independent of m, it follows that g_1 in general, without regard to the value of m, has a factor of f. That is, it must be true that g_1 = f h_1. > Are h_1(m) = g_1(m)/f, h_2(m) = g_2(m)/f, and h_3(m) = g_3(m), or > is the definition h_1(m) = g_1(m)/gcd(g_1(m),f) h_2(m) = g_2(m)/gcd(g_2(m),f), > h_3(m)=g_3(m)/gcd(g_3(m)/f) ? Focusing on the terms independent of m shows that it must be that g_1 = f h_1. > Simple question, should have a simple answer. Another simple question: Independent term means what you get when you evaluate at m=0. What function has the independent term that would vary with m, if > g_1(5) were a multiple of sqrt(f)? Is it P(m), Q(m), g_1(m), g_2, g_3, h_1, h_2, h_3 (and with what > definition)? What was the independent term before, and what did it become after > when m=5? Make sure you explain how you can tell it changed. Simple, no? Irrational. The term independent of m does not vary with m. It's a simple technique to find independent terms by setting m=0. > But no, instead of answering that simple question, you make statements > like: >The insinuation being that terms independent of m were in fact >dependent on m, but when challenged, they simply talked around that >reality, and continued to attack algebra. If this poster is willing to concede that terms independent of m are in fact independent of m, then there is no more room for argument about the definition error in core mathematics. Again, readers wishing to see the discussion on the Hong Kong math site should go to the following link: http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. [.snip.] >>My point in using the example is to show that certain posters are full >>of it, as they *have* been attacking rather basic algebra, especially >>with one particularly annoying tactic. >> >>It's not such a complica thing in analysis when you have terms >>independent of a particular variable to set that variable to 0 to >>clear it out, but these posters attacked that basic technique and >>claimed that m=0 was a special case!!! Your basic technique seems to be a mess. Whenever people ask you >> what the term is, instead of answering you just start new >> threads. You claim something changes, when people ask you what and >> how, you don't answer, you just start new threads and repeat the same >> claim. > >That's a rather interesting falsehood. Really? Let's see: >A key expression has long been g_1 = a_1 x + uf, where, though not >seen, certain variables are dependent on m, such that some posters >want me to write g_1(m) = a_1(m) x + uf, and at m=0, g_1 = uf, as uf >is the independent term. > >So it is not true that I don't answer what the independent term is. [changes], you just start new threads and repeat the same claim. What changed? >> I went to the Hong Kong forum you were recommending. I no that as >> soon as a rather basic question about your basic technique was >> asked, you ran away like a frightened little kid. No wonder you >> haven't been posting there any more. > >I last pos there a few days ago, which doesn't mean I won't post >there again. > >The poster doesn't give a link so I'll give one for readers who wish >to see the full discussion on the Hong Kong website, where there's an >additional advantage as they allow use of LaTeX. > >See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 > Let me quote it here, maybe you'll have the courage to actually answer As there are continued postings it might help to consider the >> alternative. So, consider again $P(m)=g_1 g_2 g_3$, $g_1 = a_1 x + uf$ where it's >> established that $uf$ is independent of $m$ as determined by setting >> $m=0$, note $P(0)=u^2f^2(3x + uf)$. Now suppose that instead of $f$ as a factor $g_1$ had $sqrt{f}$ when >> $m=5$, then you'd have $g_1/sqrt{f} = a_1/sqrt{f} x + usqrt{f}$ so the independent term has changed. You were asked which function had an independent term that changed, >> what it changed from, and what it changed to. > >Oh, I remember that, as some poster kept going on and on asking the >same question after I answered. Really? Where was the answer where you said WHAT changed, and from what to what? You replied, but you never ->answered<-. As usual. >> Rather than answering that simple question (just three things to say: >> what had changed, from what, and to what), you replied: Let $P(m)=f^2 Q(m)$ where I introduce $Q(m)$ as I fear using >> $P(m)/f^2$ may be confusing to Random. >> >> Now then, there exists factors I'll call $h_1$, $h_2$, $h_3$ such that >> $Q(m)=h_1 h_2 h_3$ and $h_1$ is a factor of $g_1$, $h_2$ is a factor >> of $g_2$, and $h_3$ is a factor of $g_3$. So I can get the terms *independent* of m, with the h's, in the same >> way that I did for the g's, by setting $m=0$, which gives $h_1=u$, >> $h_2=u$, and $h_3=3x+uf$, which is a conclusion that may be deba. > >Here readers can see that I mention u as independent, and mention that >I'd done the same for the g's where uf is independent of the value of >m. And readers can see that you did NOT say what changed from what to what. You did not say which independent term changed, you did not say what it was before, and you did not say what it was after. >And THAT is why I can use m=0 because an independent term is >independent without regard to the value of m, so m=0 is as good as any >other in one sense, and perfect in another, as it clears m out so that >I can focus on the constant term. Sophistry. >> Now then, I note that if the value of $m$ affects what factor of $f$ >> it is that $g_1$ has then the *independent* terms of the h's will vary >> with it. For instance, if $g_1$ has a maximum factor of $f$ that is >> $sqrt{f}$ then it would force the *independent* term of $h_1$ to have >> $sqrt{f}$ as a factor, which violates two things: 1. The term independent of $m$ would vary with $m$. 2. The independent term of $Q(0)$ is $u^2(3x + uf)$ which is coprime >> to $f$. You were never clear on something. > >What I did was explain how refusing to accept the term as independent >would cause a contradiction. The first thing you did here is to split my paragraph in two, something you ALWAYS whine about when people do to you. So nice to see your standards are still doubled. But you did NOT explain what term supposedly changed. You did NOT explain any contradiction. You just SAID the independent term would vary, but you did not SHOW it would. You just ->said<- it would. When people asked you to show it did, you changed the subject. >Specifically I have a polynomial Q(m), >which has a constant term that is u^2 (3x + uf), which is coprime to f >since f is coprime to 3, x and u. > >Now the simple position is that at m=0, h_1 = u, showing u as the >constant term, and in fact that's provable, but I no some might >wish to debate it. > >Given that h_1 has a term independent of m that's u, while it is a >factor of g_1 which has uf as a term independent of m, it follows that >g_1 in general, without regard to the value of m, has a factor of f. No, it does not follow. That's exactly where your mistake is. >That is, it must be true that g_1 = f h_1. That is, you would like it to be true, but you do not prove it. So, whose constant term supposedly changed when g_1(5) was divisible only by sqrt(5)? The constant term of h_1? How did it change? What was it before, and what is it after? >> Are h_1(m) = g_1(m)/f, h_2(m) = g_2(m)/f, and h_3(m) = g_3(m), or >> is the definition h_1(m) = g_1(m)/gcd(g_1(m),f) h_2(m) = g_2(m)/gcd(g_2(m),f), >> h_3(m)=g_3(m)/gcd(g_3(m)/f) ? > >Focusing on the terms independent of m shows that it must be that g_1 >= f h_1. Okay, so the answer is that h_1(m) is always defined as g_1(m)/f, and presumably h_2(m) is always defined as g_2(m)/f, and presumably h_3(m) is ALWAYS defined as g_3(m). >> Simple question, should have a simple answer. Another simple question: Independent term means what you get when you evaluate at m=0. What function has the independent term that would vary with m, if >> g_1(5) were a multiple of sqrt(f)? Is it P(m), Q(m), g_1(m), g_2, g_3, h_1, h_2, h_3 (and with what >> definition)? What was the independent term before, and what did it become after >> when m=5? Make sure you explain how you can tell it changed. Simple, no? > >Irrational. The term independent of m does not vary with m. Don't be an ass. YOU are the one who claimed that a term would change; that was the contradiction you claimed would arise from assuming that g_1(5) was a multiple of sqrt(f) but not of f. I asked you: let's assume, to see this contradiction, that g_1(5) is a multiple of sqrt(f) but not of f. You claim that this assumption will cause an independent term to vary. I want to know WHAT independent term (the independent term of h_1?) is supposed to vary, and how did it vary. Just answer that, James. >It's a simple technique to find independent terms by setting m=0. Yet you seem incapable of following up on YOUR OWN CLAIM. >> But no, instead of answering that simple question, you make statements >> like: >>The insinuation being that terms independent of m were in fact >>dependent on m, but when challenged, they simply talked around that >>reality, and continued to attack algebra. > >If this poster is willing to concede that terms independent of m are >in fact independent of m, then there is no more room for argument >about the definition error in core mathematics. This statement above is a big fat red herring. Nobody is saying that terms independent of m are not independent of m. The issue ->here<- is just what it is you think is independent of m, and what it is you think would NOT be indpendent of m is your claims of divisibility were false. As usualy, you do a bait-n-switch. As usual, you introduce red herrings. As usual, you lie. Good for you. At least you are consistent. == Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: ref. for infinite series I'm looking for a reference for the proof that the infinite series Matt Lafer === Subject: Re: ref. for infinite series > I'm looking for a reference for the proof that the infinite series Matt Lafer One of such proofs is a side result of the summation of Fourier series(for x^2 on [-pi,pi], if I am not mistaken). I found more fun and practice in going through the whole Fourier bit, rather than fishing for a straight, minimalist, probably obscure direct proof. Search for Fourier Series and take your pick. ZVK(Slavek). === Subject: Re: ref. for infinite series , > One of such proofs is a side result of the summation of Fourier > series(for x^2 on [-pi,pi], if I am not mistaken). I think it's x, not x^2. === Subject: Re: ref. for infinite series >, > >> One of such proofs is a side result of the summation of Fourier >> series(for x^2 on [-pi,pi], if I am not mistaken). > >I think it's x, not x^2. That's a different it. If you apply Parseval to the function x on [-pi,pi] you get the sum of 1/n^2. If you use the convergence of the Fourier series for x^2 you also get the sum of 1/n^2 coming out. ************************ === Subject: Re: ref. for infinite series > I'm looking for a reference for the proof that the infinite series Matt Lafer For a link to 14 proofs, see Evaluating zeta(2) on Robin Chapman's home page... === Subject: Conway about Fuller Here are some quotes of John Horton Conway about R. Buckminster Fuller's books Synergetics: an exploration in the geometry of thinking, and Synergetics 2 (see 1), from geometry-research (see 2). All of the quotes are completely out of context; you can't tell what he was replying to. I'll bet a lot of people won't read the Mathematica notebook SynergeticsApplication7 (see 3), or see anything new or worthwhile in the mathworld entry on synergetics coordinates (see 4) because of the answer to the question below. Do mathematicians follow JHC like he said some people follow RBF? (1) http://www.rwgrayprojects.com/synergetics/synergetics.html (2) http://www.mathforum.org/epigone/geometry-research (3) http://library.wolfram.com/infocenter/MathSource/600 (4) http://mathworld.wolfram.com/SynergeticsCoordinates.html I must say I find this synergetic language slightly irritating! Mathematicians have often found it convenient to parameterize n-space by n+1 coordinates adding to zero long before Bucky Fuller coined this word. I'm really disturbed to find a curriculum writer having such positive opinions about Fuller's synergetics. The state of geometry in the schools is bad enough as it is. A typical example of a meaningless statement from Fuller! But Bucky Fuller, though a wonderful architect, was close to crazy, as his books clearly show. He regularly used phrases (such as the fundamental structure of the plane is hexagonal) that sound wonderful but have no meaning. In fact it's a mere matter of convenience whether we use orthogonal or hexagonal coordinates - neither is intrinsically better than the other; it's just that some coordinate-systems are better sui to some problems than others. I've used dozens of different coordinate-systems in my life, as have most other professional mathematicians, and I prefer not to waste time by muttering meaningless mumbo-jumbo to show how one is somehow more moral than the others. I continued to the following paragraph because it fortuitously illustrates Bucky's megalomania. A pity he didn't tell the world those wonderful conceptual insights he was so confident of having discovered! ... The confusion swirls in Synergetics itself, as well as in its wake. Readers with a real appreciation for matters mathematical avoid it for that reason. ... It would be a disaster for writing such as this to be studied in our schools. That is why I was so concerned to find a Curriculum Developer so enamored of Fuller. I agree with you that Fuller was under no obligation to study earlier work. But if he had done so, his writings might have been much more useful. As it is, their effect is more negative than positive, and I am very much afraid that you will make it more negative still. I can only hope that few other teachers find this junk valuable. I don't know. But allow me to express my astonishment that despite the fact that any page of Fuller's writings displays this kind of megalomaniac loon-acy, he has plenty of followers. ...Fuller gives plenty of evidence of not knowing what a proof is, and single sentence. Moreover, the sentences Ive managed to understand have almost always been either crude self-promotion or utterly trivial. So to study his words is a waste of time. all. As I said, Fuller's writings on geometry are meaningful and probably usually correct. But let me assure you that you would have gained so much more by reading geometry from a more sensible source. --- Cliff Nelson === Subject: Re: Conway about Fuller > Here are some quotes of John Horton Conway about R. Buckminster Fuller's > books Synergetics: Most seem pretty commonsensical. Do mathematicians follow JHC like he said some people follow RBF? I doubt it. The followers of Fuller seem to admire his sayings however little they mean. On the other hand Conway is a mathematicians and other mathematicians treat him as a mathematician; they scrutinize his mathematical arguements for sense and insight (he is prolific in both) and dismiss his arguments when found to be wrong (fortunately a rare occurence). -- === Subject: Re: Simple curve question... If I have the following three data sets: > A B > -------------- > 1500 300 > 4000 500 > 6000 300 What is the best way to calculate the corresponding B for any given A? I > cannot assume any specific curve shape, meaning B could be 300 100 500, but > A will always be increasing. > You should have some sort of theoretical relationship. In economics, a new commercial generates increasing sales for a while, and then sale decay as could determine when they should introduce a new commercial in order to maximize profit to the client. Or at least an exercise for MBA students can be genera. Similarly, chemical reactions follow a sort of S-curve they follow and then plot the future of the reaction. For your data (whatever they are), there should be some sort of relationship. If the A's are wavelength and the B's are height, I think you've got a lot of trouble. But if they're something else, there should be a theoretical equation that relates A and B. If so, there are ways to find the curve of that form that best fits. Besides, why fit just any curve? The next data point is going to change it anyway. === Subject: Re: help me....my teacher....my problem is~~ > let |z-10i | =6 > let x : argument of z find max * min of 6cos(x) + 8sin(x) Carmody ha scritto nel messaggio > A circle radius 6 centred at 10i > 6 and 8, eh? sin and cos (at right angles) eh? > Combine those to get 10sin(x+a) for some a. hot-girl ha scritto nel messaggio > If (x+a) stays within (-Pi/2, Pi/2), then the the min > and max occur at extremal values of x, as 10sin(x+a) > is monotone increasing in that range. If however, (x+a) exceeds either or both those bounds then > the max and/or min will be +10 or -10 respectively. ------------------ > i am not understand. i think that interval of x is less than pi. told you about a circle radius 6 centered at 10i in the complex plane. Draw it and you will see that the argument x of the coplex number z can decrease from the angle of the left tangent from the origin to the circle to the right one. (i.e. from pi/2+arcsin(3/5) to arcsin(4/5) > addition explanation ...please I think you can get the min/max of the given function of x also deriving it. Paolo === Subject: Re: help me....my teacher....my problem is~~ > let |z-10i | =6 let x : argument of z find max * min of 6cos(x) + 8sin(x) -------------------------- i wait your hot-advice...please... thank in advance. |z-10i | =6 It is a circle with centre in 10i and radius 6. Let P the point on the circle, and Q = 10i. Then, when OP is tangent to the circle, OP _|_PQ and |OP| = sqrt(10^2 - 6^2) = 8. Let a = arcsin(3/5). Obviously, a < pi/4. pi/2 - a <= x <= pi/2 + a Dividing by 10, 10((3/5)cos(x) + (4/5)sin(x)) = 10(sen(a)cos(x) + cos(a)sin(x)) = 10sin(x+a) Then pi/2 <= x + a <= pi/2 + 2a The maximum of 10sin(x+a) is then 10, when x + a = pi/2. The minimum happen when x + a = pi/2 + 2a. Then sin(pi/2 + 2a) = cos(2a) = cos^2(a) - sin^2(a) = 16/25 - 9/25 = 7/25 and the minimum is 14/5. The product max*min is then 28, a perfect result ... I hope you cite it when you present your hot-work ... -- Saludos, Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosa@matematicas.net === Subject: Re: help me....my teacher....my problem is~~ > Ignacio Larrosa Ca?stro > A Coru? (Espa?) > ilarrosa@matematicas.net > > thank you very much....teacher... === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > > >>You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. > >I wonder if this is the reason for liberals to work on dumbing >down public schools? >> >>Might be, might be:-) > Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elec next year. > >>With a stress on appear. > >I'm pretty good at being able to identify functioning brains. >There is evidence that they can think cogently about baseball >and other things; but there's been massive brain damage out >there. I've heard/read these people take the fact that Event A >occured before Event B and reorder them because it would prove >that Person C is bad. Yes, I've encountered lots of this, recently. Kennedy has apparently forgotten that the >World Trade Center has ever exis. When he did his fraud >expultations, nobody seemed to notice that he sounded like a >drunk who whines and weeps after their first beer. > Aye, indeed. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >I'm pretty good at being able to identify functioning brains. >There is evidence that they can think cogently about baseball >and other things; but there's been massive brain damage out >there. I've heard/read these people take the fact that Event A >occured before Event B and reorder them because it would prove >that Person C is bad. Yes, I've encountered lots of this, recently. Really ... that's a new one on me. I thought you were leading up to the famous post hoc/propter hoc fallacy. But you're taking this a step further, and saying they lie as well as are stupid? Intriguing. > Kennedy has apparently forgotten that the >World Trade Center has ever exis. When he did his fraud >expultations, nobody seemed to notice that he sounded like a >drunk who whines and weeps after their first beer. Aye, indeed. I seem to have heard of this speech second hand twice now. The first was a letter prin in the local paper, telling the senator that fraud or perjury or whatever the term was, is what happens when a woman dies in your car and you try to get a friend to lie and say he was driving at the time. Though I don't actually know that this last happened ... I should be remember to be consistently skeptical even of allegations supporting my prejudices, even if the opposition don't have this standard. Speaking of the WTC, I had an eerie experience today. I heard some vague rumor on the subway that something happened. When I got off in Brooklyn, I heard seemingly every emergency vehicle in the world streaming south on the BQE -- an initial reponse only paralled once in my memory. However, it turned out that a Ferry had crashed into the pier in Staten Island during high winds, tearing open the side of the boat and killing maybe a dozen. Pretty spectacular and utterly unpreceden (until I read in the paper tomorrow and find out what the precedents were), but no terror attack. I think we will shortly learn that Staten Island ferries have no side thrusteand shouldn't have been operating without tugs ... which is the only way they ever operate ... in high winds. Since such things are known in nautical science, one can only imagine the edge of the safety envelope hadn't been tresspassed recently; obviously. Now, after operating in 50 knot gusts, we will be down to stopping service during 20 knot puffs. OTOH, initially it looks as if not only did he hit the pier, he hit the _wrong_ pier ... the right pier has sturdy wooden fendebut he put the sharp corner of a nearby concrete pier into the wooden boat. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > Claim: According to a study by the Lovenstein Institute, President > Bush has the lowest IQ of all presidents of past 50 years. > Status: False. Origins: No, > this isn't a real news report, nor does it describe a real study. > There isn't a Lovenstein Institute in Scranton, Pennsylvania (or > anywhere else in the USA), nor do any of the people quo in the > story exist.... http://www.snopes.com/inboxer/hoaxes/presiq.htm > http://www.museumofhoaxes.com/lovenstein.html Thx for sorting that out for me. I admit that it sounded a bit over the top, but it fit too well after the castrating comment to leave it out. Bush critics would like the idea in this special case, dubya sympathisers would see the impact on society of the whole measure. OK.. by now i guess i got myself in deep enough s to go on holiday.. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > >> >> >You mean, you still don't realize that the above is a fake and that >there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. >> >>I wonder if this is the reason for liberals to work on dumbing >>down public schools? > >'Scuse me? You're excused ;-). Note that I did do a quote job on the word. > >I happen to be a knee-jerk liberal, and I have no interest at all in >dumbing down schools. If so, then you can't possibly be a liberal of the Massachusetts flavor. > >> Even people who appear to have an ability >>to think use this rumor of Bush dumbness to get a Democrat >>elec next year. > >Every time I've ever seen a picture of him next to a senior advisor, I >get the impression of a 4-year-old on career day, visiting the grown >ups. Now, think about that. A really, really, _really_ good boss is one who stays in the background and doesn't attract attention. He doesn't have the ego that insists he be the center of attention all the time. IOW, he spends his efforts and his employees' efforts on problems rather than Hollywood image. Judging the performance of a guy based on whether he could star in a movie is stupid....IMO, of course. Shut the sound off the next time you see him on TV and really watch. //BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >> >> >You mean, you still don't realize that the above is a fake and that >there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. >> >>I wonder if this is the reason for liberals to work on dumbing >>down public schools? > >Might be, might be:-) > >> Even people who appear to have an ability >>to think use this rumor of Bush dumbness to get a Democrat >>elec next year. >> >With a stress on appear. I'm pretty good at being able to identify functioning brains. There is evidence that they can think cogently about baseball and other things; but there's been massive brain damage out there. I've heard/read these people take the fact that Event A occured before Event B and reorder them because it would prove that Person C is bad. Kennedy has apparently forgotten that the World Trade Center has ever exis. When he did his fraud expultations, nobody seemed to notice that he sounded like a drunk who whines and weeps after their first beer. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > You mean, you still don't realize that the above is a fake and that >there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. I wonder if this is the reason for liberals to work on dumbing > down public schools? Even people who appear to have an ability > to think use this rumor of Bush dumbness to get a Democrat > elec next year. Intelligence is nice, but most Bush-bashing on this score is probably from Mensans frustra that their ability to juggle puzzles doesn't make them leader of the free world. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >>You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. I wonder if this is the reason for liberals to work on dumbing >> down public schools? Even people who appear to have an ability >> to think use this rumor of Bush dumbness to get a Democrat >> elec next year. > >Intelligence is nice, but most Bush-bashing on this score is probably >from Mensans frustra that their ability to juggle puzzles doesn't >make them leader of the free world. Nope. If you recall, it got star due to snobbery. His style won't enhance your status if status is defined by nose altitude. I recall the reaction of the reporters on Washington Week in Review to a talk of his. They were stunned that the speech was boring and were greatly relieved--no more Hollywood productions. The guy talked about government business and not about himself. Ever since then, this lack of ego-display has been associa with stupidity. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > >> >> >You mean, you still don't realize that the above is a fake and that >there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. >> >>I wonder if this is the reason for liberals to work on dumbing >>down public schools? > >'Scuse me? > >I happen to be a knee-jerk liberal, and I have no interest at all in >dumbing down schools. > >> Even people who appear to have an ability >>to think use this rumor of Bush dumbness to get a Democrat >>elec next year. > >Every time I've ever seen a picture of him next to a senior advisor, I >get the impression of a 4-year-old on career day, visiting the grown >ups. > Really? Well, I must be really stupid then, since I see non of this. I just hear lots of apparent 4-year olds making statements like youabove. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > >Every time I've ever seen a picture of him next to a senior advisor, I >get the impression of a 4-year-old on career day, visiting the grown >ups. > >>Really? > >Are you telling me this is not my impression? > Oh, if you insist, I won't question it, of course. >> Well, I must be really stupid then, since I see non of this. >>I just hear lots of apparent 4-year olds making statements like >>youabove. > >So are you telling me I'm an idiot because I have this impression, or >that this is not really my opinion and I'm an idiot for not knowing my >own opinion? Oh, I just gave you my impression. As for what your opinion is and whether you know it, I don't have sufficient information to judge. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >>Every time I've ever seen a picture of him next to a senior advisor, I >>get the impression of a 4-year-old on career day, visiting the grown >>ups. >> >Really? Are you telling me this is not my impression? > Well, I must be really stupid then, since I see non of this. >I just hear lots of apparent 4-year olds making statements like >youabove. So are you telling me I'm an idiot because I have this impression, or that this is not really my opinion and I'm an idiot for not knowing my own opinion? === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > > >>You mean, you still don't realize that the above is a fake and that >>there is no such institution as The Lovenstein Institute of Scranton >>Pennsylvania. Even the Guardian which, at the time, jumped >>enthusiastically on this crap, had to retract and admit that they were >>taken for a ride. I suggest you contact them. > >I wonder if this is the reason for liberals to work on dumbing >down public schools? 'Scuse me? I happen to be a knee-jerk liberal, and I have no interest at all in > dumbing down schools. Even people who appear to have an ability >to think use this rumor of Bush dumbness to get a Democrat >elec next year. Every time I've ever seen a picture of him next to a senior advisor, I > get the impression of a 4-year-old on career day, visiting the grown > ups. Bush the Lesser is a hole that lucked into being surrounded by a donut. (Uncle Al is a right-wing fascist elitist.) -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes > Bush the Lesser is a hole that lucked into being surrounded by a > donut. >(Uncle Al is a right-wing fascist elitist.) > Uncle Al > Al, to complete the short version of your cv you should say: Uncle Al is a right-wing fascist elitist and also Uncle Al is a right-hand facilitator eliciting paid sperm donations ahahaha......ahahahanson === Subject: Re: Fun and Mental; Raisin for High; Achievements of Shoes >> >> >You mean, you still don't realize that the above is a fake and that >there is no such institution as The Lovenstein Institute of Scranton >Pennsylvania. Even the Guardian which, at the time, jumped >enthusiastically on this crap, had to retract and admit that they were >taken for a ride. I suggest you contact them. >> >>I wonder if this is the reason for liberals to work on dumbing >>down public schools? 'Scuse me? I happen to be a knee-jerk liberal, and I have no interest at all in >> dumbing down schools. Even people who appear to have an ability >>to think use this rumor of Bush dumbness to get a Democrat >>elec next year. Every time I've ever seen a picture of him next to a senior advisor, I >> get the impression of a 4-year-old on career day, visiting the grown >> ups. > >Bush the Lesser is a hole that lucked into being surrounded by a >donut. Maybe so. But you can't tell that by looking at his management style (which does not require him to always be center stage). > ..(Uncle Al is a right-wing fascist elitist.) I noticed. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Simulation of a sphereical pendulum === >Subject: Simulation of a spherical pendulum > [snip intro] >I wan to extend this >to a spherical pendulum (i.e. with two degrees of freedom - not a cone >pendulum), but found it significantly harder than I had anticipa. > >Is there a way of deriving the equations describing the angular >acceleration in the two directions (say, theta'' and phi'') using only >Newtonian mechanics? These could then be numerically integra to >provide a simulation of the chaotic behaviour. > >Movement outside of the plane is provided by either an initial angular >velocity (phi'), or perturbations provided by magnets moun below >the pendulum. If you are setting your pendulum in a uniform gravitational field you should find theta acceleration is proportional to sin theta and phi acceleration is zero. It's just like the reguluar old pendulum. pretty boring. If you are setting your pendulum on the surface of the earth you will find that the phi acceleration isn't zero and that the plane the pendulum swings back and forth in precesses slowly. If you are setting your pendulum above magnets (as you sta in your post) then I don't know what will happen. Where are the magnets exactly? What kind of magnets are they? You need to clear this up. adam === Subject: Re: Simulation of a sphereical pendulum oops. pos to the wrong newsgroup. Please ignore my previous post. === Subject: e=2,71. e=2,71...8281828459.... lim (1+1/x)^x, for x-->oo. (1+1/1000)^1000=2,71...6923932... i = 1/1000 ; Inf. = I=1000 ( 1+ i )^I = e -- Andrea Sorrentino, libero Ricercatore Web: http://digilander.libero.it/socratis === Subject: Re: e=2,71. > e=2,71...8281828459.... > lim (1+1/x)^x, for x-->oo. (1+1/1000)^1000=2,71...6923932... i = 1/1000 ; Inf. = I=1000 ( 1+ i )^I = e -- > Andrea Sorrentino, libero Ricercatore > Web: http://digilander.libero.it/socratis What is your point? You have demonstra a case that does not disprove convergence, but not that ( 1+ i )^I = e === Subject: Calculus - Graph problems... I have 2 quick questions... 1. Sketch the graph of the following function showing vertical and horizontal asymptotes and relative extrema. Label your x-axis in terms of a and your y-axis in terms of 1/a. f(x)=(a-x)/(x^2+a^2) where a is a constant, a greater than 0. I can find out the 2 stationary points, the HA, and that their are no VAs. I don't get how to find other points to make that graphs however, or inflection points etc. 2.Sketch the graph of the following: f(x) = (g(x^2 + 6)^(.5) ---------------- x+2 It's an odd function. The VA is x=-2. I can't figure out the relative extrema or IPs or HA for some reason though. Scott Eliason === Subject: Re: Calculus - Graph problems... > I have 2 quick questions... > 1. Sketch the graph of the following function showing vertical and > horizontal asymptotes and relative extrema. Label your > terms of a and your y-axis in terms of 1/a. > f(x)=(a-x)/(x^2+a^2) where a is a constant, a greater than 0. > I can find out the 2 stationary points, the HA, and that their are no > VAs. I don't get how to find other points to make that graphs > however, or inflection points etc. 2.Sketch the graph of the following: > f(x) = (g(x^2 + 6)^(.5) > ---------------- > x+2 > It's an odd function. The VA is x=-2. I can't figure out the relative > extrema or IPs or HA for some reason though. I presume you are supposed to know how to find derivatives of functions of these sorts. For such functions, with continous 2nd derivatives at all points of their domains, extrema can only occur when f'(x) = 0, though not all zeros of the derivative need be extrema, and inflections can only occur when f''(x) = 0, though not all zeros of the 2nd derivative need be inflections. === Subject: Re: Calculus - Graph problems... Scott Eliason escribi.97 en el > I have 2 quick questions... > 1. Sketch the graph of the following function showing vertical and > horizontal asymptotes and relative extrema. Label your > terms of a and your y-axis in terms of 1/a. > f(x)=(a-x)/(x^2+a^2) where a is a constant, a greater than 0. > I can find out the 2 stationary points, the HA, and that their are no > VAs. I don't get how to find other points to make that graphs > however, or inflection points etc. Study where the graph intersect axis and asymptotes (in this case the HA is the OX axis), and on what side of them is the graph. You can deduce easily that must be 4 inflection points: before the maximum, between maximum and minimum and after the minimum. But you can calculate them exactly f(x) = - 2(x^3 - 3x^2 - 3x + 1)/(x^2 + 1)^3 This cubic equation has the root x = -1. Dividing by (x + 1), you get x^2 - 4x + 1 = 0 with the roots 2 +/- sqrt(3). Sketch any date as soon as you know it and at the end you only need join them. > 2.Sketch the graph of the following: > f(x) = (g(x^2 + 6)^(.5) > ---------------- > x+2 > It's an odd function. The VA is x=-2. I can't figure out the relative > extrema or IPs or HA for some reason though. What is g? But no matter what is g, f(x) it isn't a odd function: f(-x) =/= -f(x). -- Best Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: How much longer must physics put up with F=ma? > The total force [F] used to move an object against friction or any other > restraint, is the total force exer on it minus the frictional - or other > restraining force; so that the _net force_ is what does the actual work, and > is [f = F-uw]; which leaves F = wa/g + uw; or f = wa/g! The coefficient of restraint [u] against moving something on a level surface > is usually a fraction of the object's weight [w], so that the product [uw] > is the measure of the restraining frictional force: On slopes the > coefficient of restraint is proportional to the sin of the angle of the > slope: For a 90 vertical slope the coefficient of restraint against driving > something poin directly into it will depend on the hardness of the > material: As is the case for driving stakes and such into the ground. > Yes, the total force needed to move an object would seem to include > the force necessary to overcome friction, and any other impeding > resistance, as well as the inertia of the object that results from > mass. But in physics, is force really defined to include the total > effort, or just that part that overcomes inertia to give acceleration? > If it isn't, it sure ought to be: The total force is the measure of the total effort; including the force necessary to overcome friction, and any other impeding resistance, as well as the inertia of the object.. > When pushing a rock ac the ground, frictional force usually seems > to arise quickly Frictional force doesn't _arise_, iI's already there, as 'static friction'; preventing the rock from moving on its own. In order to start the rock moving it's necessary to apply a force _greater_ than the static force: Immediately after the movement begins, the rocks begins to accelerate, and will continue to do so as long as the force is maintained. When the force is reduced the rock will slow its acceleration, and when the force ceases, the rock will be stopped: Brought to rest by 'kinetic' friction. to prevent the rock moving beyond a certain speed > without applying additional force. When force is lessened, the rock > slows immediately. It is easy to see how Aristotle came to think that > force was needed to keep an object moving at a constant speed. But > perhaps this is not what modern physics considers force. If you are pushing an object along the ground, there is also air > resistance, which is another form of friction. If you pound a stake > into the ground, you get ground resistance, which is really the forces > of atomic electric fields that you are running up against. > Oh horsefeathers: Pounding stakes into the ground simply requires If you hit a rock with a steel point, then there might be some electricity genera: Ever seen sparks and smelled brimstone, while working a pick and shovel to dig a hole or trench? > If I push against a brick wall, nothing will move. Am I really > exerting a force, or is this only tension, or stress? > If you push with a force beyond the wall's ability to withstand lt; you'll move that darn wall; in a quick fashon. > But when we consider an object in a frictionless state, there is > another question that arises. Newton said in his third law of motion: To every action there is always opposed an equal reaction; or the > mutual actions of two bodies upon each other are always equal, and > direc to contrary parts. Now when a force is applied to a mass in a frictionless state, the > inertial force opposes the motive force, which limits the acceleration > according to a=F/m. But if the inertial force is an equal and > opposite force to the motive force, as Newton's 3rd law seems to say, > then why do we have acceleration at all? matter, and/ or masses therof cannot simultaneously occupy or pass through the exact same place; they exert a mutual, equal and oppositely direc force on each other and cause each other to accelerate; by spinning slowing down or changing direction; in proportion to their elasticity and massiveness. Don't forget Newton and Einstein were only human, and they put their pants on one leg at a time, just like the rest of us. === Subject: Re: How much longer must physics put up with F=ma? > The total force [F] used to move an object against friction or any other > restraint, is the total force exer on it minus the frictional - or other > restraining force; so that the _net force_ is what does the actual work, and > is [f = F-uw]; which leaves F = wa/g + uw; or f = wa/g! The coefficient of restraint [u] against moving something on a level surface > is usually a fraction of the object's weight [w], so that the product [uw] > is the measure of the restraining frictional force: On slopes the > coefficient of restraint is proportional to the sin of the angle of the > slope: For a 90 vertical slope the coefficient of restraint against driving > something poin directly into it will depend on the hardness of the > material: As is the case for driving stakes and such into the ground. > Yes, the total force needed to move an object would seem to include > the force necessary to overcome friction, and any other impeding > resistance, as well as the inertia of the object that results from > mass. But in physics, is force really defined to include the total > effort, or just that part that overcomes inertia to give acceleration? If it isn't, it sure ought to be: The total force is the measure of > the total effort; including the force necessary to overcome friction, > and any other impeding resistance, as well as the inertia of the > object.. When pushing a rock ac the ground, frictional force usually seems > to arise quickly Frictional force doesn't _arise_, iI's already there, as 'static > friction'; preventing the rock from moving on its own. In order to > start the rock moving it's necessary to apply a force _greater_ than > the static force: Immediately after the movement begins, the rocks begins to accelerate, > and will continue to do so as long as the force is maintained. When the force is reduced the rock will slow its acceleration, and > when the force ceases, the rock will be stopped: Brought to rest by > 'kinetic' friction. > I agree with you that the frictional force must be there to begin with as static friction, but the frictional force must increase as a function of velocity, otherwise the rock wouldn't reach any terminal velocity, which experience shows us it usually does. > to prevent the rock moving beyond a certain speed > without applying additional force. When force is lessened, the rock > slows immediately. It is easy to see how Aristotle came to think that > force was needed to keep an object moving at a constant speed. But > perhaps this is not what modern physics considers force. If you are pushing an object along the ground, there is also air > resistance, which is another form of friction. If you pound a stake > into the ground, you get ground resistance, which is really the forces > of atomic electric fields that you are running up against. Oh horsefeathers: Pounding stakes into the ground simply requires resist being pushed apart by the stake? It's the electromagnetic forces of the atoms. If it were only gravity holding them together, it would be like pushing a stake through a pile of flour. > If you hit a rock with a steel point, then there might be some > electricity genera: Ever seen sparks and smelled brimstone, while > working a pick and shovel to dig a hole or trench? Not my line of work. If I push against a brick wall, nothing will move. Am I really > exerting a force, or is this only tension, or stress? If you push with a force beyond the wall's ability to withstand lt; > you'll move that darn wall; in a quick fashon. But when we consider an object in a frictionless state, there is > another question that arises. Newton said in his third law of motion: To every action there is always opposed an equal reaction; or the > mutual actions of two bodies upon each other are always equal, and > direc to contrary parts. Now when a force is applied to a mass in a frictionless state, the > inertial force opposes the motive force, which limits the acceleration > according to a=F/m. But if the inertial force is an equal and > opposite force to the motive force, as Newton's 3rd law seems to say, > then why do we have acceleration at all? matter, and/ or masses therof cannot simultaneously occupy or pass > through the exact same place; they exert a mutual, equal and > oppositely direc force on each other and cause each other to > accelerate; by spinning slowing down or changing direction; in > proportion to their elasticity and massiveness. Again, it is the electromagnetic fields of the atoms that cause this. But I still can't quite see how Newton's third law applies. Don't forget Newton and Einstein were only human, and they put their > pants on one leg at a time, just like the rest of us. even put his pants on in the morning! Double-A === Subject: Re: How much longer must physics put up with F=ma? pants on one leg at a time, just like the rest of us. > even put his pants on in the morning! Nor did Einstein. As I heard it, he met Queen Elizabeth II wearing a suit he had slept in, and once was escor home by a Princeton policeman who found him deep in thought on a park bench one Sunday morning, sans trousers. I prefer to believe these incidents were not accidental, and that he did these thing tongue in cheek because he knew that he, of all people in the world, could get away with it. With his exquisitely subtle sense of humor, he was having a private joke on all of US. Tom Davidson Richmond, VA === Subject: Re: The Passing of a Mathematician > In which particular way does the NSA keep alive the > value of democracy? By protecting the security of the Uni States, a major democratic > power, and its citizens. Unfortunately the only way it knows how to do this is by murdering liberty elsewhere. -- G.C. === Subject: Re: The Passing of a Mathematician > In which particular way does the NSA keep alive the > value of democracy? By protecting the security of the Uni States, a major democratic > power, and its citizens. Unfortunately the only way it knows how to do this is by murdering > liberty elsewhere. You might remember the old ad: Delta's ready when you are ...... === Subject: Re: The Passing of a Mathematician >In which particular way does the NSA keep alive the value of democracy? It helps protect the Uni States of America, a democratic nation, > from being invaded and conquered by foreign non-democratic nations. Which foreign non-democratic wish to invade and conquer the USA? It's the USA that does the invading. -- G.C. === Subject: Re: The Passing of a Mathematician ... > somewhat less justification, bewail the money spent on building bombs, > in the belief that it could be used to grow more food. (Since, as Mark There is no need for more food. People go hungry not because there is not enough food, but because they do not have enough money to buy the food of which there is plenty. > Twain no, land is something of which they're not making any more, > throwing more money at food production is likely only to drive up the > price. Of course, that could still feed the hungry - giving more money > to the right people, for example, could lead to less meat being eaten, > increasing the efficiency of food production.) John Savard > http://home.ecn.ab.ca/~jsavard/index.html -- G.C. === Subject: Re: The Passing of a Mathematician Come on, you don't seriously believe that growing more food while the > population remains unchecked is any kind of answer, do you? > What does the population have to do with it? There is more than enough food right now to feed every person on earth, as there probably has been for all of history. We have even, for at least 100 yeahad the means to physically transport it to where it's needed. Yet it still doesn't get there. Population remains unchecked? The population in the developed world is not replacing itself. Even in India, birthrates are falling. The population will check itself. In spite of Catholic prohibitions of such a thing. === Subject: Re: The Passing of a Mathematician Come on, you don't seriously believe that growing more food while the > population remains unchecked is any kind of answer, do you? What does the population have to do with it? There is more than enough food > right now to feed every person on earth, as there probably has been for all > of history. We have even, for at least 100 yeahad the means to > physically transport it to where it's needed. Yet it still doesn't get > there. Hey, I'm not the one who was wailing about scientists disappearing into the NSA instead of solving world hunger problems. Population remains unchecked? Well, if the more than enough food cannot be gotten to those who are starving, then the population is unchecked in that region. http://members.aol.com/owagiveaway/cony_slum_people_f.wav >The population in the developed world is not > replacing itself. Even in India, birthrates are falling. The population > will check itself. In spite of Catholic prohibitions of such a thing. === Subject: Re: The Passing of a Mathematician >Population remains unchecked? The population in the developed world is not >replacing itself. Even in India, birthrates are falling. The population >will check itself. In spite of Catholic prohibitions of such a thing. FWIW, the Roman Catholic church does not prohibit birth control; it prohibits artificial means of birth control. The church will happily advocate periodic abstinence, for example. The logic leading to this situation is somewhat tortuous. dave === Subject: Re: The Passing of a Mathematician >In which particular way does the NSA keep alive the value of democracy? It helps protect the Uni States of America, a democratic nation, > from being invaded and conquered by foreign non-democratic nations. Note, though, that the notion of democracy could be interpre (or practiced) differently. Some countries have even 'democratic' in their names (one example if the past is GDR.) The same with human rights or 'help' (donations) given to poor countries and sick people needing medicaments, etc. etc. M. K. Shen === Subject: Re: The Passing of a Mathematician part: >Come on, you don't seriously believe that growing more food while the >population remains unchecked is any kind of answer, do you? Limiting population growth, if possible, certainly is a real answer. For any given population, growing more food answers the current problem, and if we could grow *enough* more food, perhaps we could get the demographic transition some people hope for. Even if expecting nature to take its course is asking too much, the further ahead of the game we get, the easier it will be to also confront population. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Passing of a Mathematician > part: Come on, you don't seriously believe that growing more food while the >population remains unchecked is any kind of answer, do you? Limiting population growth, if possible, certainly is a real answer. For any given population, growing more food answers the current > problem, and if we could grow *enough* more food, perhaps we could get > the demographic transition some people hope for. Even if expecting nature to take its course is asking too much, the > further ahead of the game we get, the easier it will be to also > confront population. http://members.aol.com/owagiveaway/cony_slum_people_f.wav John Savard > http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Passing of a Mathematician John Savard scribbled the following: >>In which particular way does the NSA keep alive the value of democracy? > It helps protect the Uni States of America, a democratic nation, > from being invaded and conquered by foreign non-democratic nations. Care to list those nations? Do you view non-democratic as a proper or improper subset of foreign? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ You could take his life and... - Mirja Tolsa === Subject: Re: The Passing of a Mathematician >Care to list those nations? Do you view non-democratic as a proper or >improper subset of foreign? Proper. I live in Canada, another democratic nation, not equal to the Uni States. Finland is a democracy, although not long ago, it was menaced by the Soviet Union, and its people weren't able to freely choose their country's foreign policy. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Fundamental Reason for High Achievements of Jews > As can be seen from a few excerpts from his posts below, > Bob Kolker should remove the logs from his eyes. Bulls. I have identified the enemy. The enemy is Islam. The dreadful > happenings of 9/11 prove it. The way to defend against the evil your > enemies do is to kill your enemies. I don't agree with you. The principle is clear. Help and protect your friends. Kill your > enemies. It is as simple as that. And we are ALL in this fight. If the > U.S. government. would give me a nuke, I would gladly blow it and myself > up at at the next Haj in Meccah. I would love taking a hundred thousand > of the bastards with me. A few months ago I had dinner with several other couples and our kids. We all live in the same neighborhood, our kids play sports together and/or are involved in scouting together. Four families. Two christian, one jewish, one muslim. The muslem couple went to Mecca for the first time last year. It'd suck if they'd been blown up. Bob, you are a whack. Your comment about blowing people up shows that you are no better than the whacks that strap bombs to their chest and take out Jews at a mall somewhere in Israel, no better than the whacks that ran airplanes into the WTC, no better than the whacks that were sniping at people in the DC area last year, no better than whacks anywhere. === Subject: Re: Fundamental Reason for High Achievements of Jews > The principle is clear. Help and protect your friends. > Kill your enemies. It is as simple as that. > And we are ALL in this fight. If the U.S. government. > would give me a nuke, I would gladly blow it and myself > up at at the next Haj in Meccah. I would love taking a > hundred thousand of the bastards with me. - Kolker AHhahahhaha......Bob, are you having a bout of genocidal > upwellings again? As a Jew, you are supposed to be better > and more lovable then the Muslims...and 5.5 times smarter. > You said so yourself!. > Now, you've become like them, a Jewish Homicide bomber, > a Wailing-Wall-Martyr....a Sui-Holocauster........ > ahahahhha............ahahahahanson There is nothing wrong with killing people who have harmed you, > are will harm you eventually. We send our soldiers to die regularly. > It is business as usual. I just want a high body count. > Bob Kolker > Sheesh, Bob.......kill people who will harm you EVENTUALLY ....??? That was obviously the Nazi osophy. That is obviously the Al Qaeda osphhpy. Wow, and now is it also the Jewish osophy???......... So, the beat goes on, Bob, as it always did....with your tribe quarreling with everybody in sight?.......resulting in the reoccurring nightmare of having your collective ass either on the run or in the fire, periodically and epically. What sing prospects, Bob. But you guys must like that, you did so for the last 5761 years. Yet, somehow this doesn't ring to me like being 5.5 smarter..........hahahahaha...... ahahahha.......ahahanson PS: Potter is probably dancing in the streets now, Bob Kolker. You assis Tom Potter more than anyone else to prove his, Potters', vision, goal and point........ahahahahahaha..... === Subject: Re: Fundamental Reason for High Achievements of Jews >> Modern socialism is a balance of public and private >> enterprise with an adiquate social safety net. > >That means stealing from one and giving to other via taxes. >Taxation is Theft Stupidity alert! > Parki-pooh, you do not have to precede any of your intellectually empty posts with any alert!. Everyone knows that you are stupid. The poster is obviously a taxpayer, hence a contributor to your salary, since you are on the public payroll, as a college professor. Instead of denigrating the poster, who is a real contributor to your welfare, thank him for that and show/explain him that his portion is a very wise investment by pointing out the benefits he reaps from your public service accomplishments. That is what you should alert him to....not your stupidity. BTW, what are your accomplishments beside your green barking on the net? ahahahaha........ahahahanson > >(because it is enforced at gun point and is not voluntary like >charity). The adequate social safety is hardley ever adequate, is almost >never safe and is very unsocial to those who are taxed to pay for it. > >Bob Kolker > > === Subject: Re: Fundamental Reason for High Achievements of Jews > Modern socialism is a balance of public and private > enterprise with an adiquate social safety net. >> >>That means stealing from one and giving to other via taxes. >>Taxation is Theft >> >> Stupidity alert! >> >Parki-pooh, you do not have to precede any of your intellectually >empty posts with any alert!. Everyone knows that you are stupid. > >The poster is obviously a taxpayer, hence a contributor to your >salary, since you are on the public payroll, as a college professor. I see somebody is too dumb to know the difference between public universities and private ones. >Instead of denigrating the poster, who is a real contributor to your >welfare, thank him for that and show/explain him that his portion is >a very wise investment by pointing out the benefits he reaps from >your public service accomplishments. That is what you should alert >him to....not your stupidity. BTW, what are your accomplishments >beside your green barking on the net? >ahahahaha........ahahahanson > >> >>(because it is enforced at gun point and is not voluntary like >>charity). The adequate social safety is hardley ever adequate, is almost >>never safe and is very unsocial to those who are taxed to pay for it. >> >>Bob Kolker >> >> >