mm-1085 No. What you say about the value of the definite integral is correct - your idea that the integral from -infinity to infinity can be used in place of the antiderivative in an integration by parts shows amazing ignorance. >>Where do you get that from? >> Just explained that in a different post. >>The definite integral int -oo^+oo f(t).d(t-T) >>gives us f(T), but there is no information >>given to us about the indefinite integral >>nor of the anti-derivative. >>Indeed, the definite integral gives us f(T), which >>is a constant function having no aspect of t in its >>evaluation and which graphs as a horizontal >>line from -oo to +oo. This comes from the >>basic definitions of the Diracian. >> True, and of no relevance whatever to the question >> of what an antiderivative is. >> Hint: Youre _really_ showing over and over that >> you have no understanding of _basic_ calculus. >> You keep kindly informing me that Im making >> a laughingstock of myself, presumably youd >> want to know how utterly ignorant youre >> revealing yourself to be. >> PS: Look up the definition of ad hominem. >> Im not saying you must be wrong because youre >> stupid, Im saying youre looking very stupid >> because the things youre saying are wrong, >> at such a basic level. >>Your claimed anti-derivative has a strong >>dependence on t, and therefore has not >>come from anything formally defined. >Are you making this up as you go along? >Well, here is what David C. Ullrich posted.. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > An ad hominem attack from you of no value to the > discussion. > Shame on you. > You should no better. > I do not keep kindly informing you that youre making > a laughingstock of myself. I have suggested that once > and once only. > Please get your facts correct. > You say that the things quoted below from me are true > and then you say that Im utterly ignorant. > You are inconsistent. >>Where do you get that from? >> Just explained that in a different post. >>The definite integral int -oo^+oo f(t).d(t-T) >>gives us f(T), but there is no information >>given to us about the indefinite integral >>nor of the anti-derivative. >>Indeed, the definite integral gives us f(T), which >>is a constant function having no aspect of t in its >>evaluation and which graphs as a horizontal >>line from -oo to +oo. This comes from the >>basic definitions of the Diracian. >> True, and of no relevance whatever to the question >> of what an antiderivative is. >> Hint: Youre _really_ showing over and over that >> you have no understanding of _basic_ calculus. >> You keep kindly informing me that Im making >> a laughingstock of myself, presumably youd >> want to know how utterly ignorant youre >> revealing yourself to be. >> PS: Look up the definition of ad hominem. >> Im not saying you must be wrong because youre >> stupid, Im saying youre looking very stupid >> because the things youre saying are wrong, >> at such a basic level. >>Your claimed anti-derivative has a strong >>dependence on t, and therefore has not >>come from anything formally defined. >Are you making this up as you go along? >Well, here is what David C. Ullrich posted.. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. Hang on a moment... you say that I cannot use the properties of a definite integral.....and then you go on and do so yourself! If I cannot rely on the property of a definite integral that int -oo^+oo f(t).d(t-T) is f(T) then neither can you! Well, let us suppose that our f(t) is sin(wt)...how does your claimed anti-derivative then differentiate to become sin(wt).d(t-T)? Or, let us suppose that our f(t) is e^t...how does your claimed anti-derivative then differentiate to become e^t.d(t-T)? So, how does your anti-derivative differentiate to become two different products of functions? Remember, until we actually evaluate the definite integral that were dealing in whole functions over all t in symbolic maths. > Well, here is what David C. Ullrich posted > My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). === Subject: Re: Infantile authours degrading this NG. >Hang on a moment... you say that I cannot use the properties >of a definite integral.....and then you go on and do so yourself! >If I cannot rely on the property of a definite integral that >int -oo^+oo f(t).d(t-T) is f(T) then neither can you! I didnt say any such thing. I said that you need an antiderivative in the integration by parts formula. That formula you quote is _why_ the antiderivative is what I say it is. By definition an antiderivative for f(t).d(t-T) would be F, where F(x) = int_0^x f(t).d(t-T). If x < T then that gives F(x) = 0, while if x > T that gives F(x) = f(T), precisely because of that formula above. >Well, let us suppose that our f(t) is sin(wt)...how >does your claimed anti-derivative then differentiate >to become sin(wt).d(t-T)? Impossible to answer this question since you insist generalized functions have nothing to do with it - these antiderivatives are not differentiable in the classical sense, they have derivatives in the sense of distributions. >Or, let us suppose that our f(t) is e^t...how >does your claimed anti-derivative then differentiate >to become e^t.d(t-T)? >So, how does your anti-derivative differentiate to >become two different products of functions? >Remember, until we actually evaluate the definite >integral that were dealing in whole functions over >all t in symbolic maths. >> Well, here is what David C. Ullrich posted >> My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. Apart from throwing in the term generalised functions, how does the presented mathematics for your derivatives differ from those of the classical sense? Remember that any one function in your class of generalised functions represents the properties of the whole class. >Well, let us suppose that our f(t) is sin(wt)...how >does your claimed anti-derivative then differentiate >to become sin(wt).d(t-T)? > Impossible to answer this question since you insist > generalized functions have nothing to do with it - > these antiderivatives are not differentiable in > the classical sense, they have derivatives in the > sense of distributions. === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > Apart from throwing in the term generalised functions, > how does the presented mathematics for your > derivatives differ from those of the classical sense? > Remember that any one function in your class of > generalised functions represents the properties of the > whole class. >>Well, let us suppose that our f(t) is sin(wt)...how >>does your claimed anti-derivative then differentiate >>to become sin(wt).d(t-T)? >> Impossible to answer this question since you insist >> generalized functions have nothing to do with it - >> these antiderivatives are not differentiable in >> the classical sense, they have derivatives in the >> sense of distributions. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. >Apart from throwing in the term generalised functions, >how does the presented mathematics for your >derivatives differ from those of the classical sense? The antiderivative is not even continuous, much less differentiable in the classical sense. >Remember that any one function in your class of >generalised functions represents the properties of the >whole class. >>Well, let us suppose that our f(t) is sin(wt)...how >>does your claimed anti-derivative then differentiate >>to become sin(wt).d(t-T)? >> Impossible to answer this question since you insist >> generalized functions have nothing to do with it - >> these antiderivatives are not differentiable in >> the classical sense, they have derivatives in the >> sense of distributions. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. The anti-derivative of (sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2)) is pretty continuous in the classical sense, and indeed, in any sense you care to present. > The antiderivative is not even continuous, much less > differentiable in the classical sense. === Subject: Re: Infantile authours degrading this NG. >The anti-derivative of >(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2)) >is pretty continuous in the classical sense, and >indeed, in any sense you care to present. Uh, yes it is. If that function were a delta function youd have a point. >> The antiderivative is not even continuous, much less >> differentiable in the classical sense. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. Where does that assertion come from? Remember - you proscribe the definite integral int -oo^+oo f(t).d(t-T) > By definition an antiderivative for f(t).d(t-T) > would be F, where > F(x) = int_0^x f(t).d(t-T). > If x < T then that gives F(x) = 0, while if x > T > that gives F(x) = f(T), precisely because of that > formula above. === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > Where does that assertion come from? > Remember - you proscribe the definite > integral int -oo^+oo f(t).d(t-T) >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). >> If x < T then that gives F(x) = 0, while if x > T >> that gives F(x) = f(T), precisely because of that >> formula above. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. By what definition? - You have presented here a definite integral and not an anti-derivative which contradicts earlier assertions by you. Indeed, in the only related definition of which I am aware, the limits of integration should be +/- oo and not 0 to x. To which definition in your By definition do you refer? > By definition an antiderivative for f(t).d(t-T) > would be F, where > F(x) = int_0^x f(t).d(t-T). === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > By what definition? - You have presented here a > definite integral and not an anti-derivative which > contradicts earlier assertions by you. > Indeed, in the only related definition of which I > am aware, the limits of integration should be +/- oo > and not 0 to x. To which definition in your By definition > do you refer? >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. >By what definition? - You have presented here a >definite integral and not an anti-derivative which >contradicts earlier assertions by you. >Indeed, in the only related definition of which I >am aware, the limits of integration should be +/- oo >and not 0 to x. To which definition in your By definition >do you refer? You really need to find a calculus book somewhere. If F(x) = int_-infinty^x f(t) then F = f; this is a thing called the Fundamental Theorem of Calculus. (In the present context int_0^x is the same, since were assuming that f(t) = 0 for t < 0). >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. You really need to grow up. This is an international forum and not the staff room of the infantss school in which you apparently teach. Youre changing your tune, once again. The discussion was about int f(t).d(t-T), now your equivocating about int f(t) only. Integration and differentiation of products is not done by simple substitution. > You really need to find a calculus book somewhere. > If F(x) = int_-infinty^x f(t) then F = f; this > is a thing called the Fundamental Theorem of Calculus. > (In the present context int_0^x is the same, since > were assuming that f(t) = 0 for t < 0). === Subject: Re: Infantile authours degrading this NG. >You really need to grow up. >This is an international forum and not >the staff room of the infantss school in >which you apparently teach. >Youre changing your tune, once again. >The discussion was about int f(t).d(t-T), >now your equivocating about int f(t) only. Its hard to decide. Youre just trolling, or youre really unable to follow simple arguments. Hint: The f in the statement from me you quote below is not the same as the f that weve been talking about. If Id realized it was going to cause confusion I would have used a different letter... >Integration and differentiation of products >is not done by simple substitution. >> You really need to find a calculus book somewhere. >> If F(x) = int_-infinty^x f(t) then F = f; this >> is a thing called the Fundamental Theorem of Calculus. >> (In the present context int_0^x is the same, since >> were assuming that f(t) = 0 for t < 0). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > Youre changing your tune, once again. Quack? Honk? === Subject: Re: Infantile authours degrading this NG. Yes - I corrected that in a later post. However, the differentiation and integration was -s/-s giving unity. > <* is apparent the Ô- was taken out front. But since the beginning, dU should > have been Ô-e^(-st)s with the Ôs in the numerator, not denominator. I > suspect this error came from ARBs original attempt at the integration by === Subject: Re: Infantile authours degrading this NG. You may well have a point here, if my error is in not evaluating the definite integral over the whole term. However, other errors in what you propose nullify your argument, (see other posts) > Well, here is what David C. Ullrich posted >So that line should be > F(t).e^(-st) ...... === Subject: Re: Infantile authours degrading this NG. > You may well have a point here, if my error is in > not evaluating the definite integral over the whole > term. > However, other errors in what you propose nullify your > argument, (see other posts) The Ôother errors are on your part, not Davids. (see other replies) daestrom === Subject: Re: Infantile authours degrading this NG. Your argument falls down when T = 0, giving t -T as just t, for then F(0) = f(0), and not 0. > Well, here is what David C. Ullrich posted. >Now when we put in the limits t = 0 to t = infinity >the first term vanishes since F(0) = 0 === Subject: Re: Infantile authours degrading this NG. > Your argument falls down when T = 0, > giving t -T as just t, for then F(0) = f(0), > and not 0. Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) f(0). And the Dirac delta function evaluated at (0-T) is zero, regardless of the value of f(0). (assuming that T > 0) daestrom === Subject: Re: Infantile authours degrading this NG. if F(t) = f(t).d(t-T), then with T equal to zero, this comes down to f(t).d(t). This might be undefined because it is not expressed under an integral sign, but it is (probably) not 0. If it were to be expressed under an integral sign, then it would be f(0) as I stated. > Your argument falls down when T = 0, > giving t -T as just t, for then F(0) = f(0), > and not 0. > Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) f(0). > And the Dirac delta function evaluated at (0-T) is zero, regardless of the > value of f(0). (assuming that T > 0) === Subject: Re: Infantile authours degrading this NG. nope, youve got it all wrong bean. obviously youre incompetent. well we knew that from the start. need better than that if youre going to stay here. dr. x > if F(t) = f(t).d(t-T), then with T equal to zero, > this comes down to f(t).d(t). > This might be undefined because it is not expressed > under an integral sign, but it is (probably) not 0. > If it were to be expressed under an integral sign, then > it would be f(0) as I stated. >> Your argument falls down when T = 0, >> giving t -T as just t, for then F(0) = f(0), >> and not 0. >> Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) > f(0). >> And the Dirac delta function evaluated at (0-T) is zero, regardless of >> the >> value of f(0). (assuming that T > 0) --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. The evaluation yields f(T) which is a constant and is a graphed as a horizontal straight line having no expression of time in its evaluation. This is fundamental to the definition of the Diracian, where no other properties are specified. This property is used in other aspects of DSP and so it must be used consistently here. > Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. === Subject: Re: Infantile authours degrading this NG. > The evaluation yields f(T) which is a constant > and is a graphed as a horizontal straight line > having no expression of time in its evaluation. > This is fundamental to the definition of the Diracian, > where no other properties are specified. This > property is used in other aspects of DSP and so > it must be used consistently here. >> Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous function of t. Where t T it has a value of f(T) (note carefully the case of each statement). It has two possible values depending on whether you are evaluating it at tT. That is not the same thing as what youre saying, ... is a constant... daestrom === Subject: Re: Infantile authours degrading this NG. That does not agree with the definition with which I work, which is int -oo^+oo f(t).d(t-T) = f(T). In that evaluation, all dependence upon t disappears. What definition are you using to claim discontinuity? > The evaluation yields f(T) which is a constant > and is a graphed as a horizontal straight line > having no expression of time in its evaluation. > This is fundamental to the definition of the Diracian, > where no other properties are specified. This > property is used in other aspects of DSP and so > it must be used consistently here. >> Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. > No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous > function of t. === Subject: Re: Infantile authours degrading this NG. >That does not agree with the definition with which >I work, which is int -oo^+oo f(t).d(t-T) = f(T). >In that evaluation, all dependence upon t disappears. That evaluation of the _antiderivative_ is simply _wrong_. >What definition are you using to claim discontinuity? >> The evaluation yields f(T) which is a constant >> and is a graphed as a horizontal straight line >> having no expression of time in its evaluation. > This is fundamental to the definition of the Diracian, >> where no other properties are specified. This >> property is used in other aspects of DSP and so >> it must be used consistently here. >> Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. > No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous >> function of t. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. That evaluation is the one by which all electronic engineers are taught. >That does not agree with the definition with which >I work, which is int -oo^+oo f(t).d(t-T) = f(T). >In that evaluation, all dependence upon t disappears. > That evaluation of the _antiderivative_ is simply > _wrong_. === Subject: Re: Infantile authours degrading this NG. But the gives a different result if the integration is done in the opposite direction, and therefore cannot be valid. Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. === Subject: Re: Infantile authours degrading this NG. > But the gives a different result if the integration > is done in the opposite direction, and therefore > cannot be valid. > Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. Evaluating F(t) where tT you have a value of f(t). If evaluated in the reverse order (i.e. over the interval +inf to zero), you get -f(t) instead of f(t). This is exactly as it should be. daestrom === Subject: Re: Infantile authours degrading this NG. If you evaluate the anti-derivative given by Mr.Ullrich (without accepting it as a correct formula, BTW) then you would get the properties where F(t) = 0 for t > T and f(T) for t < T. > But the gives a different result if the integration > is done in the opposite direction, and therefore > cannot be valid. > Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. > Evaluating F(t) where tT you have a value of f(t). > If evaluated in the reverse order (i.e. over the interval +inf to zero), you > get -f(t) instead of f(t). This is exactly as it should be. > daestrom === Subject: Re: Infantile authours degrading this NG. >> But the gives a different result if the integration >> is done in the opposite direction, and therefore >> cannot be valid. >> Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. > Evaluating F(t) where tT you have a value of > f(t). Oops, where t> T you have F(t) = f(T), not f(t). My mistake. > If evaluated in the reverse order (i.e. over the interval +inf to zero), > you get -f(t) instead of f(t). This is exactly as it should be. That should be -f(T) instead of f(T), again, my mistake. daestrom === Subject: Re: Infantile authours degrading this NG. I disagree that it is wrong. The operation being undertaken is Integration By Parts and therefore Integration is required and not Anti-Differentiation. My f(T).e^(-st) is the result of evaluating U.int(V), and, as it is part of a sum, evaluting the integral between the limits of 0^oo to save further tedious writing. > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. >>Certainly, if you integrate by parts, you would choose >>int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >>I try this, I get 0!...... >>int(UV) = U.int(V) - int[dU.int(V)] giving..... >> int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >>with f(t).d(t-T) as V and e^(-sT) as U ..... >>f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >Your first step, > f(T).e^(-st) - int(e^(-st)/-s . f(T)) >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). === Subject: Re: Infantile authours degrading this NG. >I disagree that it is wrong. The operation being >undertaken is Integration By Parts and >therefore Integration is required and not >Anti-Differentiation. Hilarious. I think you might want to take Calc 101 again before lecturing us on all this stuff. >My f(T).e^(-st) is the result of evaluating >U.int(V), If int means int_0^infinty then yes it is, thats exactly whu its wrong. > and, as it is part of a sum, evaluting >the integral between the limits of 0^oo to save >further tedious writing. Making less and less sense - if youd done that evaluation already then there would be no t appearing at this point. Its really good of you to warn me about what a laughingstock Im making of myself, btw. >> Unfortunately, although he claimed to have posted such, it was >> masked by his over-riding urge to be insulting and was therefore >> not seen by me. >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >>Your first step, >> f(T).e^(-st) - int(e^(-st)/-s . f(T)) >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. >> I disagree that it is wrong. The operation being >> undertaken is Integration By Parts and >> therefore Integration is required and not >> Anti-Differentiation. > Hilarious. I think you might want to take > Calc 101 again before lecturing us on all > this stuff. >> My f(T).e^(-st) is the result of evaluating >> U.int(V), > If int means int_0^infinty then yes it is, thats > exactly why its wrong. i just realized now how far this is cross-posted. i suspect that David is hanging out on sci.math (but i dunno). David, he really doesnt get it. and he doesnt want to get it. if it were up to Beanie, we would rewrite all the calculus textbooks. lets not feed the troll. maybe if we stop feeding it, it will shrivel up and blow away. r b-j === Subject: Re: Infantile authours degrading this NG. > I disagree that it is wrong. The operation being > undertaken is Integration By Parts and > therefore Integration is required and not > Anti-Differentiation. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. > My f(T).e^(-st) is the result of evaluating > U.int(V), >> If int means int_0^infinty then yes it is, thats >> exactly why its wrong. >i just realized now how far this is cross-posted. i suspect that David is >hanging out on sci.math (but i dunno). Yes. > David, he really doesnt get it. >and he doesnt want to get it. up. Was actually trying to help the guy with the math in my first one or two replies, but since then its all been just for the fun of watching the way the guys technique. Hes pretty good, has things like being insulting and then whining when people reply insultingly, ignoring refutations and then claiming they dont exist, claiming that the fact that there have been a lot of replies proves he must be right, etc down pat. Weve got better here on sci.math but hes pretty good. I cant decide whether hes just trolling or really believes hes right... > if it were up to Beanie, we would rewrite >all the calculus textbooks. >lets not feed the troll. maybe if we stop feeding it, it will shrivel up >and blow away. What fun would that be? >r b-j ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. I see that you cannot resist the temptation to lower the tone. > Hilarious. I think you might want to take > Calc 101 again before lecturing us on all > this stuff. > Its really good of you to warn me about what a > laughingstock Im making of myself, btw. === Subject: Re: Infantile authours degrading this NG. >I see that you cannot resist the temptation to >lower the tone. If you had any sense youd stop making an idiot of yourself over the math, stop being so utterly hilarious complaining about the tone of peoples comments, and just take a calculus class. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. >> Its really good of you to warn me about what a >> laughingstock Im making of myself, btw. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. I see that you cannot resist the temptation to lower the tone. >I see that you cannot resist the temptation to >lower the tone. > If you had any sense youd stop making an idiot of > yourself over the math, stop being so utterly > hilarious complaining about the tone of peoples > comments, and just take a calculus class. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. >> Its really good of you to warn me about what a >> laughingstock Im making of myself, btw. === Subject: Re: Infantile authours degrading this NG. >I see that you cannot resist the temptation to > lower the tone. In an earlier post David pointed out legitimate problems with the math behind your argument. Rather than read, and learn from, that post, you accidentally deleted it. I suggest you find the post on Google and read it. You may learn something. Charles Perry P.E. === Subject: Re: Infantile authours degrading this NG. I suggest that you read what is written before you open your mouth and come out with untruths. You may learn something. I did not accidentally delete anything. In response to a barrage of quite unnecessary gratuitous and insulting remarks by Mr.Ullrich, I _DELIBERATELY_ deleted things. If Mr.Ullrich had anything valuable to offer, then it was lost in the fog of his childish outbursts. There is no concept of legitimacy. Law does not come into it. >I see that you cannot resist the temptation to > lower the tone. > In an earlier post David pointed out legitimate problems with the math > behind your argument. Rather than read, and learn from, that post, you > accidentally deleted it. I suggest you find the post on Google and read > it. You may learn something. === Subject: Re: Infantile authours degrading this NG. >I suggest that you read what is written before you open > your mouth and come out with untruths. > You may learn something. > I did not accidentally delete anything. In response to > a barrage of quite unnecessary gratuitous and insulting > remarks by Mr.Ullrich, I _DELIBERATELY_ deleted > things. One of the posts you _DELIBERATELY_ deleted is the one I quoted earlier, where David C. Ullrich, started out quite politely showing you where you made your mistakes. And in his more recent post, he states: If you take f(T) exp(-st), (assume s > 0), evaluate at t = infinity, t = 0 and subtract you get -f(t). If F is an antiderivative of f(t) delta(t-T), so F(t) = 0 for t < T, F(t) = f(T) for t > T, and you do the same evaluate-subtract with F(t)e^{-st} you get 0. This was clear from the difference in your calculation and my correct version of the same calculuation, btw. Not much Ôunnecessary gatuitous and insulting remarks in that unless you consider the last sentence such. And if you do, then you obviously have a much thinner Ôskin than the way you treat others. daestrom === Subject: Re: Infantile authours degrading this NG. lets ask crystal, shall we bean? dr. x >I suggest that you read what is written before you open > your mouth and come out with untruths. > You may learn something. > I did not accidentally delete anything. In response to > a barrage of quite unnecessary gratuitous and insulting > remarks by Mr.Ullrich, I _DELIBERATELY_ deleted > things. > If Mr.Ullrich had anything valuable to offer, then it was lost > in the fog of his childish outbursts. > There is no concept of legitimacy. Law does not come into it. >>I see that you cannot resist the temptation to >> lower the tone. >> In an earlier post David pointed out legitimate problems with the math >> behind your argument. Rather than read, and learn from, that post, you >> accidentally deleted it. I suggest you find the post on Google and >> read >> it. You may learn something. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. How does your Indefinite Integral or (Antiderivative) differ from a Definite Integral when the limits of integration are imposed? > Well, here is what David C. Ullrich posted >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. === Subject: Re: Infantile authours degrading this NG. >How does your Indefinite Integral or (Antiderivative) >differ from a Definite Integral when the limits of >integration are imposed? If you take f(T) exp(-st), (assume s > 0), evaluate at t = infinity, t = 0 and subtract you get -f(t). If F is an antiderivative of f(t) delta(t-T), so F(t) = 0 for t < T, F(t) = f(T) for t > T, and you do the same evaluate-subtract with F(t)e^{-st} you get 0. This was clear from the difference in your calculation and my correct version of the same calculuation, btw. >> Well, here is what David C. Ullrich posted >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. No you dont. You get -f(T). >How does your Indefinite Integral or (Antiderivative) >differ from a Definite Integral when the limits of >integration are imposed? > If you take f(T) exp(-st), (assume s > 0), evaluate > at t = infinity, t = 0 and subtract you get -f(t). === Subject: Re: Infantile authours degrading this NG. >> Unfortunately, although he claimed to have posted such, it was >> masked by his over-riding urge to be insulting and was therefore >> not seen by me. >Well, here is what David C. Ullrich posted.... (lines with >> are from Airy >R. Beans previous post, those preceded by Ô> are from David C. Ullrichs >pose, my comments are embedded with Ô<*<) >Show where it is wrong, or else resort >to infantile sneering. >>Ok. Took me a minute to find the place >>where you showed us the integration >>by parts: >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. >>Your first step, >> f(T).e^(-st) - int(e^(-st)/-s . f(T)) >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ><**>>Now when we put in the limits t = 0 to t = infinity >>the first term vanishes since F(0) = 0 and e^(-st) >>tends to 0 as t -> infinty (at least if s > 0; if s < 0 >>this integration by parts is not going to work.) >>So out Laplace transform becomes >> - int_0^infinity (e^(-st)/-s . F(t)) >>Since F(t) = 0 for t < T and f(T) for t > T >>this equals >> f(T) int_T^infinity (e^(-st)s ). ><*is apparent the Ô- was taken out front. But since the beginning, dU should >have been Ô-e^(-st)s with the Ôs in the numerator, not denominator. I >suspect this error came from ARBs original attempt at the integration by I pointed out explicitly, in lines you snipped, that Airy had written e^{-st}/s where it should have been e^{-st} s. I thought that Id fixed the error at that point, its possible that some erroneous cut&pasted lines were not edited properly. (Also pointed out at least twice so far that this is somewhat amusing. Before looking at it I knew that his integration by parts must be wrong, since it gave the wrong answer, but I assumed that the problem was with something subtle about delta functions. Turns out that in addition to not getting the antiderivative right his amazing conclusion is based in part on simply differentiating an exponential incorrectly! Its easy to see why this was the point at which he suddenly decided he needed to announce that he was not reading any more replies... Although announcing that hed deleted the replies containing math and then also complaining that the replies he was reading didnt contain any math was a very bold step.) >>Again, its easy to evaluate the last integral; >>int_T^infinity (e^(-st)s ) = e^(-sT), so we >>finally get f(T)e^(-sT) for the Laplace transform. ><**>daestrom ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. >>Well, here is what David C. Ullrich posted.... (lines with >> are from >>Airy >>R. Beans previous post, those preceded by Ô> are from David C. Ullrichs >>pose, my comments are embedded with Ô<*<) >>Show where it is wrong, or else resort >>to infantile sneering. >Ok. Took me a minute to find the place >where you showed us the integration >by parts: >>Certainly, if you integrate by parts, you would choose >>int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >>I try this, I get 0!...... >>int(UV) = U.int(V) - int[dU.int(V)] giving..... >> int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >>with f(t).d(t-T) as V and e^(-sT) as U ..... >>f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >>f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >>f(T).e^(-sT) - f(T).e^(-sT).... >>0. >Your first step, > f(T).e^(-st) - int(e^(-st)/-s . f(T)) >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). >><**>Now when we put in the limits t = 0 to t = infinity >the first term vanishes since F(0) = 0 and e^(-st) >tends to 0 as t -> infinty (at least if s > 0; if s < 0 >this integration by parts is not going to work.) >So out Laplace transform becomes > - int_0^infinity (e^(-st)/-s . F(t)) >Since F(t) = 0 for t < T and f(T) for t > T >this equals > f(T) int_T^infinity (e^(-st)s ). >><*>It >>is apparent the Ô- was taken out front. But since the beginning, dU >>should >>have been Ô-e^(-st)s with the Ôs in the numerator, not denominator. I >>suspect this error came from ARBs original attempt at the integration by > I pointed out explicitly, in lines you snipped, that Airy had > written e^{-st}/s where it should have been e^{-st} s. I thought > that Id fixed the error at that point, its possible that some > erroneous cut&pasted lines were not edited properly. > (Also pointed out at least twice so far that this is somewhat > amusing. Before looking at it I knew that his integration by > parts must be wrong, since it gave the wrong answer, but I > assumed that the problem was with something subtle about > delta functions. Turns out that in addition to not getting > the antiderivative right his amazing conclusion is based > in part on simply differentiating an exponential incorrectly! Yes, I was not trying to point the finger at you for the original error, just pointing out that at that particular point you corrected your explanation (for those that may have wondered how the Ôs term suddenly moved from denominator to numerator). And that, as I said, I suspect this error came from ARBs original... posting, not you per se. Interesting bit of calculus, I may have to go visit sci.math to see what other interesting things are bandyd about. daestrom === Subject: Re: Infantile authours degrading this NG. > Interesting bit of calculus, I may have to go visit sci.math to see what > other interesting things are bandyd about. > daestrom Take ARB with you. === Subject: Re: Infantile authours degrading this NG. The bold step that I took was to make a stand against infantile ad hominem attacks as is your wont. You continue with your motivation to insult in your posting below. If you have anything of value to offer, then offer it in a mature post. Shame on you. You should know better. (If you adopt the same arrogant stance to your pupils as you do here, and if any of those pupils are reading this NG, you can be sure that you are the laughing stock in your school (if not so already)) > Its easy to see why this was the point at which he suddenly > decided he needed to announce that he was not reading any > more replies... Although announcing that hed deleted the > replies containing math and then also complaining that > the replies he was reading didnt contain any math was > a very bold step.) === Subject: Re: Infantile authours degrading this NG. <310ft1F35b2uqU4@uni-berlin.de> <3132s3F30bqd5U1@uni-berlin.de> <315hu5F3491fhU4@individual.net> <315nt5F372fc6U1@individual.net> <316s92F37f4hjU1@individual.net> <38ttq0pn4ujphmaesmpmjfhj8m767apgos@4ax.com> <318bvsF35ggl4U15@individual.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The bold step that I took was to make a stand > against infantile ad hominem attacks as is your wont. > You continue with your motivation to insult in your > posting below. > If you have anything of value to offer, then offer it > in a mature post. > Shame on you. You should know better. The irony of it, the irony. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Infantile authours degrading this NG. >> The bold step that I took was to make a stand >> against infantile ad hominem attacks as is your wont. >> You continue with your motivation to insult in your >> posting below. >> If you have anything of value to offer, then offer it >> in a mature post. >> Shame on you. You should know better. >The irony of it, the irony. Indeed. But you left out the best bit, irony-wise: >(If you adopt the same arrogant stance to your >pupils as you do here, and if any of those pupils are >reading this NG, you can be sure that you are the >laughing stock in your school (if not so already)) ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > The bold step that I took was to make a stand > against infantile ad hominem attacks as is your wont. > > You continue with your motivation to insult in your > posting below. > > If you have anything of value to offer, then offer it > in a mature post. > > Shame on you. You should know better. >> The irony of it, the irony. > Indeed. But you left out the best bit, irony-wise: >> (If you adopt the same arrogant stance to your >> pupils as you do here, and if any of those pupils are >> reading this NG, you can be sure that you are the >> laughing stock in your school (if not so already)) Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie (real name Gareth) is not competent enough in the subject matter to be teaching it to anyone. despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002. shtml Even kids would be able to figure out quickly that they cannot learn (except by rote, decidedly a bad way to learn math) from Beanie because what he says makes no sense because it is false. even more so with adults. you cannot systemically learn underlying concepts from misconceptions. and you cannot learn math without learning conceptually. i imagine that Beanie is unemployed and (due to his character) unemployable. r b-j === Subject: Re: Infantile authours degrading this NG. > Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie > (real name Gareth) is not competent enough in the subject matter to be > teaching it to anyone. > despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002. shtml Wrong Gareth The Gareth that is giving you the run around, and that is exactly what he is doing, is in his 50s. He attended Essex University, although it appears there little evidence he graduated, and has an varied employment history. He has admitted being dismissed from on job. All the evidence is he screwed up big time. Other posts by him suggest that this was not an isolated case. He trolls varies newsgroups with his wind-ups on a range of topics, DSP is one his favorite topics- his theories on sampling first came to light several years ago. They have been refuted many times. Other topics have included a justification for the Soham murders (for US readers, two teenage girls were murdered in Soham a few years ago). He appears to be unemployable and becomes fixated on those who have won out arguements against him. Do some googling and you will find evidence of all the above. If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. === Subject: Re: Infantile authours degrading this NG. : He appears to be unemployable and becomes fixated on those who have won out : arguements against him. said NIMROD hiding in the shadows thowing stones from a safe distance : : If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. said NIMROD hiding in the shadows thowing stones from a safe distance === Subject: Re: Infantile authours degrading this NG. >> Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie >> (real name Gareth) is not competent enough in the subject matter to be >> teaching it to anyone. >> despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002. shtml > Wrong Gareth > The Gareth that is giving you the run around, and that is exactly what he is > doing, is in his 50s. i found a few old postings where *he* says hes class of Ô72 (not 2002). same middle name, same first name, same last name. education in mathematics. lotsa coincidences for someone on this side of the pond. im sure you UKers know him better. > He attended Essex University, although it appears there little evidence he > graduated, and has an varied employment history. He has admitted being > dismissed from on job. All the evidence is he screwed up big time. i picked that up. > Other posts by him suggest that this was not an isolated case. > He trolls varies newsgroups with his wind-ups on a range of topics, DSP is > one his favorite topics- his theories on sampling first came to light > several years ago. They have been refuted many times. i know that. another reason i thought it was the young guy in the photo is that his arguments are really neophyte (his integration-by-parts and his concepts surrounding bandlimited sampling are indicative of someone who hasnt really done it) and his means of argument are very immature. not reßective of someone in their 50s. > Other topics have included a justification for the Soham murders (for US > readers, two teenage girls were murdered in Soham a few years ago). > He appears to be unemployable and becomes fixated on those who have won out > arguements against him. > Do some googling and you will find evidence of all the above. its exactly what i did. > If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. i agree. r b-j === Subject: Re: Infantile authours degrading this NG. >same middle name, same first name, same last name. education in >mathematics. lotsa coincidences for someone on this side of the pond. im >sure you UKers know him better. There are some historical consistencies with the email addresses with which hes been associated that put him as an experienced ringer (a person who, with other folks, pulls the ropes to toll peals and carols from church bell towers), with email traffic to related lists back to at least 1998. He was upsetting some of those folks with his views, too, and his account of his historical geographic locations is consistent with where he is now. So I think the young guy in the website you found is unlikely to be Airy. The employment history issues also seem to indicate the same. He seems to be interested in and experienced with music, which is something he has in common with you and I, though. ;) Apologies to those of you whove dealt with this for way too long. Hes still a little new to us at comp.dsp, but weve had very similar experiences with other people. Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intels opinions. http://www.ericjacobsen.org === Subject: Re: Infantile authours degrading this NG. > So I think the young guy in the website you found is unlikely to be > Airy. The employment history issues also seem to indicate the same. ive come around to that conclusion. its just that with an identical first, middle, and last name and some formal study in electrical engineering and math and being in the UK, that just seemed to be too many coincidences for me initially. > Apologies to those of you whove dealt with this for way too long. > Hes still a little new to us at comp.dsp, but weve had very similar > experiences with other people. *very* similar? i think Beanie is in a league of his own. his trolling mastery *way* outstrips eBob. r b-j === Subject: Re: Infantile authours degrading this NG. Nimrod for a while caused the following to be typed: > If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE > TROLL. Best advice Ive heard on this subject so far. -- Radio glossary #21 Packet: An item which causes your XYL to not speak to you for five days when it arrives along with a Waters and Stanton invoice. === Subject: Re: Infantile authours degrading this NG. brießy between bottles, to write: >He appears to be unemployable and becomes fixated on those who have won out >arguements against him. Which lets be honest, is pretty much everybody. Nick. === Subject: Re: Infantile authours degrading this NG. >Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie >(real name Gareth) is not competent enough in the subject matter to be >teaching it to anyone. >despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002. shtml ... >i imagine that Beanie is unemployed and (due to his character) unemployable. >r b-j Im dont think thats the same Gareth Evans (way too young). Maybe he can enlighten us. It was evident to me yesterday that thats apparently not an uncommon name in the UK (or Australia). Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intels opinions. http://www.ericjacobsen.org === Subject: Re: Infantile authours degrading this NG. >Im dont think thats the same Gareth Evans (way too young). Maybe >he can enlighten us. It was evident to me yesterday that thats >apparently not an uncommon name in the UK (or Australia). Gareth is a pretty common Welsh name. Evans is very common Welsh name. Put them together and you get a fairly common, and very Welsh name :-) Steve === Subject: Re: Infantile authours degrading this NG. > Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie >(real name Gareth) is not competent enough in the subject matter to be >teaching it to anyone. >despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002. shtml >Even kids would be able to figure out quickly that they cannot learn (except >by rote, decidedly a bad way to learn math) from Beanie because what he says >makes no sense because it is false. even more so with adults. you cannot >systemically learn underlying concepts from misconceptions. and you cannot >learn math without learning conceptually. >i imagine that Beanie is unemployed and (due to his character) unemployable. >r b-j Not too sure about that logic. When I was doing A-level maths (A-levels are the UK exams at 18, before entering university) we had a really bad maths teacher. Everyone knew she was really bad, so we largely ignored her. We worked extra hard on our own from books, and did lots of exercising with the published past papers. The final result was we all achieved really excellent grades in which is a bit sad, but the bottom line is her incompetance actually achieved good results. :-) Steve === Subject: Re: Infantile authours degrading this NG. >> Beanie doesnt have pupils (except maybe two beneath his corneas). Beanie >> (real name Gareth) is not competent enough in the subject matter to be >> teaching it to anyone. ... >> Even kids would be able to figure out quickly that they cannot learn (except >> by rote, decidedly a bad way to learn math) from Beanie because what he says >> makes no sense because it is false. ... > Not too sure about that logic. > When I was doing A-level maths (A-levels are the UK exams at 18, before > entering university) we had a really bad maths teacher. Everyone knew > she was really bad, so we largely ignored her. makes sense. i was saying that they cannot learn ... *from* Beanie.... > We worked extra hard on > our own from books, and did lots of exercising with the published past > papers. The final result was we all achieved really excellent grades in > which is a bit sad, but the bottom line is her incompetance actually > achieved good results. :-) just because of two outcomes that might be correlated, it does not mean that one was the cause (ty teacher) and the other was the effect (good test results). you do not know how well yould have done with a really good teacher. perhaps it is that you did so well despite the poor teacher not because of the poor teacher. i am an advocate of good teaching. good, effective, teachers have to be, at the very least, competent in the subject and in related subjects that they are teaching. my $0.02 . r b-j === Subject: Re: Infantile authours degrading this NG. >You claim to be a mathematician of 20 years standing. >I suggest that you do not try to be a teacher of mathematics. >Mathematics is an abstract subject for which your students >must be relaxed in mind in order to be able to take on >abstractions. Your emotive and insulting posts >will disrupt such a state of mind. In my time as a part-time >tutor of mathematics to adults I came across this time and >time again - people who have been put off mathematics, not >by the subject per se, but by the aggression and personal >attacks addressed to them by their teachers. Guffaw. Every once in a while I get advice like this here on sci.math. Curiously, it always comes from someone whos totally wrong about some bit of mathematics, whos insisting hes right and refusing to accept simple explanations of why hes wrong. Been teaching math for 20 years, and no situation even remotely like that has ever come up in the classroom. >Your infantile tirade below illustrates well the persons proscribed >Shame on you. >(I notice that you do not address the mathematical points below) Guffaw again. The mathematical points have been addressed quite completely. Below I refer to a post where I do address a new mathematical point (that youre simply doing the basic calculus wrong), which post you _say_ that you deleted unread. Which is of course another classic crackpot technique: ignore substantive objections and then claim they dont exist (or repeatedly ignore substantive objections and then whine about the fact that theyre not repeated in every reply.) >> Classic crackpot. If nobody says youre wrong you must >> be right. But if a large number of people say youre wrong >> it follows you must be right as well. >> Things must be very pleasant in your little world. >> Btw, since you dont seem to have noticed my post >> where I reply to your request to show _where_ your >> integration by parts fails: When I said that >> integration by parts does not apply because delta >> is not a continuous I was assuming that you >> were making one sort of moderately subtle error. >> Hadnt looked at the actual argument you gave. >> In case you didnt see that post, in your >> integration by parts you simply do the basic >> calculus wrong. >>It is interesting the size of the amount of posts. >If I am wrong in what I say, it would be a simple >>matter to ignore me and what I posit. >That so many people, people who would seek to >>represent themselves as authorities, respond with >>emotional postings that do not address the >>points raised, seems to suggest that I am not wrong, and that >>they are embarrassed about a basic failing in their >>professed knowledge and so feel threatened. >the amount of posts, posts that are non-technical and >>of an undesirable ad hominem style has reached such a >>scale that I have binned them all this morning. >This leaves just one point that has not yet been addressed, and >>that is, that any mathematical analysis of a real system must, >>to be respectable, deal with measurements of that system. >There is no system of which I am aware that has sampling >>pulses whose measurements match those of the Diracian >>Impulse... >1. Their amplitudes do not approach infinity. >2. Their areas do not approach unity. >3. In any case, the operation of f(t).d(t - T) is not >>defined unless under an integral sign, and cannot be >>evaluated unless under that integral sign. >4. The evaluation of the spectrum of the Diracian relies >>on it being over all time, -oo^+oo. It is improper, therefore, >>to attempt to evaluate other operations using the Diracian >>with a reduced domain, and yet still rely on the spectrum >>derivation. > of posts that it has generated, it stands to reason that you are > unlikely to get a response suitable to you from them. >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. >Which is of course another classic crackpot technique: ignore >substantive objections and then claim they dont exist (or >repeatedly ignore substantive objections and then whine >about the fact that theyre not repeated in every reply.) At one time on uk.radio.amateur, Bean, in one of his previous sock-puppets, had requested help with a technical problem with an off-the-shelf amateur transceiver he had bought. He received a number of helpful, technically-based replies and suggestions. It came as little surprise that it was at that precise moment he suffered an ISP failure, and never saw the replies. He was the only one on the ng who used that particular ISP who reported such a failure..... FUs set to remove uk.radio.amateur. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. There is nothing amiss about deleting posts that are a thinly-disguised vehicles for delivering infantile abuse, such as yours below. Shame on you - you should know better if you claim 20 years experience as an expert. No expert needs to hind behind abusive and infantile tirades. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that youre simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. > Which is of course another classic crackpot technique: ignore > substantive objections and then claim they dont exist (or > repeatedly ignore substantive objections and then whine > about the fact that theyre not repeated in every reply.) === Subject: Re: Infantile authours degrading this NG. As with all escapees from the infants school playground, I invite you to repost any technically relevant posts, but without your childish and unnecessary infantile outpourings - then I will deal with the points raised. I assume that okstate.edu is some form of infants school and that you are copying the behaviour of your pupils? > There is nothing amiss about deleting posts that > are a thinly-disguised vehicles for delivering > infantile abuse, such as yours below. > Shame on you - you should know better if you > claim 20 years experience as an expert. > No expert needs to hind behind abusive and > infantile tirades. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that youre simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. > Which is of course another classic crackpot technique: ignore > substantive objections and then claim they dont exist (or > repeatedly ignore substantive objections and then whine > about the fact that theyre not repeated in every reply.) === Subject: Re: Infantile authours degrading this NG. I have deleted posts which are excuses for gratuitous abuse. No point reading them further. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that youre simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. === Subject: Re: Infantile authours degrading this NG. >Mathematics is an abstract subject for which your students >must be relaxed in mind in order to be able to take on >abstractions. >In my time as a part-time >tutor of mathematics to adults I came across this time and >time again - people who have been put off mathematics, not >by the subject per se, but by the aggression and personal >attacks addressed to them by their teachers. In my time I came across classes in which the range of mathematical abilities varied from the incompetent to the brilliant. I cant recall a case of anyone being put off by the teacher - but plenty that were put off by teaching methods. You may, of course, be referring to your own case - which is not necessarily generally true. FUs set to remove uk.radio.amateur. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. I suggest that if you wish to gain any respect and be seen as a potential mathematics professional that you drop your wont for making personal remarks and that you concentrate on the matters in hand. You seem to have no experience of acadaemic debate, and I can tell you that Argumentum Ad Hominem, which is your wont, has no place in reputable acadaemic institutions. Do us all a favour, Mr.Ullrich and grow up, or else leave this forum. > You claim to be a mathematician of 20 years standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. Your emotive and insulting posts > will disrupt such a state of mind. In my time as a part-time > tutor of mathematics to adults I came across this time and > time again - people who have been put off mathematics, not > by the subject per se, but by the aggression and personal > attacks addressed to them by their teachers. > Your infantile tirade below illustrates well the persons proscribed > Shame on you. > (I notice that you do not address the mathematical points below) > Classic crackpot. === Subject: Re: Infantile authours degrading this NG. I suggest that if you wish to gain anyones respect in this forum, you learn to quit top posting like a newbie, quit trying to impress everyone with your amazingly brilliant but trivial technobabble, and learn not to get all pissy the first time someone begs to differ with you. > I suggest that if you wish to gain any respect and be > seen as a potential mathematics professional that you > drop your wont for making personal remarks and > that you concentrate on the matters in hand. > You seem to have no experience of acadaemic debate, and > I can tell you that Argumentum Ad Hominem, which is your > wont, has no place in reputable acadaemic institutions. > Do us all a favour, Mr.Ullrich and grow up, or else leave > this forum. > You claim to be a mathematician of 20 years standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. Your emotive and insulting posts > will disrupt such a state of mind. In my time as a part-time > tutor of mathematics to adults I came across this time and > time again - people who have been put off mathematics, not > by the subject per se, but by the aggression and personal > attacks addressed to them by their teachers. > Your infantile tirade below illustrates well the persons proscribed > Shame on you. > (I notice that you do not address the mathematical points below) > Classic crackpot. -- Fairy A. Bean === Subject: Re: Infantile authours degrading this NG. >That so many people, people who would seek to >represent themselves as authorities, respond with >emotional postings that do not address the >points raised. But you yourself have deleted whole posts - some of which were long and detailed and which dealt with the points you raised - because you perceived something you defined as Ôchildish early on. If you believe that to be a mature approach to debate, then you are in a very small minority. You have been repeatedly told that the way to deal with (what you perceive as) Ôchildishness is to ignore it and deal with the substance. This (positive) behaviour reinforces the message that negativity is not the means to achieve a satisfactory outcome. Unfortunately, your approach merely reinforces negative behaviour and is ultimately self-defeating. FUs set to remove uk.radio.amateur, where Beans approach to discussion has been discussed ad nauseum. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. > I suggest that if you wish to gain any respect and be > seen as a potential mathematics professional that you > drop your wont for making personal remarks and > that you concentrate on the matters in hand. Alas, you are confused. And what is worse, given to top-posting. === Subject: Re: Infantile authours degrading this NG. authors. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Infantile authours degrading this NG. >It is interesting the size of the amount of posts. Please remove uk.radio.amateur from your FUs. TIA -- from Aero Spike === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? A > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. lets integrate by parts correctly: L[x(t).delta(t-T)] = int_0^infinity exp(-st).x(t).delta(t-T).dt using int_a^b udv = (uv)|_a^b - int_a^b v.du as our formulation of parts, choose: u = exp(-st) => du = -s.exp(-st)dt dv = x(t).delta(t-T) dt => v = int_0^t x(u) delta(u-T)du = 0 if t < T, or x(T) if t >= T so, L[x(t).delta(t-T)] = [exp(-st).v(t)]_0^inf - int_0^inf -s.exp(-st).v(t)dt = 0 + s.int_T^inf x(T)exp(-st)dt = [-x(T).exp(-st)]_T^inf = x(T).exp(-sT) hopefully that is satisfactory for you, I think the sloppiness is warranted. Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? hey Beanie, i have proposal for you. would you promise to shut up and go away (or, at least, just shut up) if i show you where your ßaw is in your integration by parts? its amazing why you dont just accept the simple proof (applying the definition of the dirac impulse function) that +inf integral{ x(t) d(t-T) e^(-st) dt} ( d(t) = dirac impulse function ) -inf +inf = integral{ [x(t) e^(-st)] d(t-T) dt} -inf +inf = integral{ x1(t) d(t-T) dt} ( where x1(t) = x(t) e^(-st) ) -inf +inf = integral{ x1(u+T) d(u) du} ( substitute u = t-T ) -inf +inf = integral{ x2(u) d(u) du} ( where x2(t) = x1(t+T) ) -inf = x2(0) ( by definition of the dirac impulse ) = x1(T) = x(T) e^(-sT) now thats how normal people without mental illness will evaluate that bilateral Laplace Transform integral. (people *with* mental illness may want to use a less direct and more difficult method so that they will more likely screw up and get the wrong answer.) for some reason Beanie wants to evaluate it by parts and he insists he gets a different answer (zero). setting aside the lesson of Occams Razor ( http://en.wikipedia.org/wiki/Occam%27s_razor the simplest explanation should be used), the answer *if* you integrate by parts *correctly* (which Beanie cant seem to do correctly), the integral turns out to be the same. Laplace{ x(t) d(t-T) } = x(T) e^(-sT) not zero as Beanie claims Beanie, if you promise to behave yourself and stop trolling this newsgroup, ill show your ßaw in your integration by parts. both integrating by parts and the straight forward integration comes out with the same answer (as if any of us suspected otherwise). but im gonna use the *textbook* definition of integrating by parts: b | b b integral{ u dv} = (uv)| - integral{ v du} a | a a (Beanie, youll need a mono-spaced font to read this ASCII math well.) Beanie wants (if he agrees to conventional notation) u = e^(-st) and dv = x(t) d(t-T) dt Beanie, if you agree to those (very reasonable) terms, i will show you (and everyone else reading it) the ßaw in your derivation that erroneously says the result is zero. r b-j > with as follows..... > You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to > determine your result. > You simply cannot say that f(t).d(t-T) yields f(T) unless > you do so under an integral. This arises from the fundamental > properties of the Diracian Delta function . > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... thats because your knowledge and insight is far less than you think and youre also quite mentally ill. (perhaps only the latter is true if youre the master troll Jerry makes you out to be, but both are true if youre simply the obnoxious crackpot and jerk that is more ostensible.) > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st), and claiming that f(t).d(t-T) equals > f(T) is not sound. We use the properties of the Diracian > Delta Function in so many other aspects of signal > processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? Because of your childish behaviour quoted below, something of value to contribute to the discussion, then please present it in a style more appropriate to an international forum. > hey Beanie, > i have proposal for you. would you promise to shut up and go away (or, at > least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? are you afraid to see your mistake explicitly exposed? cluck, cluck r b-j > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. >> hey Beanie, >> i have proposal for you. would you promise to shut up and go away (or, at >> least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? I am not afraid of anything, be that anything you, or the truth. It is simply that I refuse to deal with you when you indulge yourself in infantile outbursts. Shame on you. You should know better, including not indulging in silly yah-boo-sucks tirades as in your latest rant below. You do nothing for your reputation by so behaving. > are you afraid to see your mistake explicitly exposed? > cluck, cluck > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st) For what its worth, tongue firmly in cheek, and with apologies to David Ulrich, who did it properly: You seem to have a problem with the Dirac Delta definition. Anyway, presumably you are happy with the Laplace transform L[f(t).delta(t-T)] = int_0^infinity f(t).delta(t-T).exp(-st) dt Maybe the notion that the Delta function is, loosely speaking, zero nearly everywhere is sort-of vaguely familiar to you, so you might be convinced that something like this is true: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt And maybe considering f(t) and exp(-st) might get you ruminating on their continuity, and a daring insight like this may leap out: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt Forgive me if Im getting carried away, but isnt the Riemann Integral linear? I think it might be, so wouldnt that mean that L[f(t).delta(t-T)] = lim M->0 f(T).exp(-sT) int_(T-M)^(T+M) delta(t-T)dt I really think that I might be onto something here, but I need to work out how to integrate that pesky Dirac Delta ... any takers? === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? In your second equation, how come youve still got t in your expression of the Diracian, but youve replaced it by T in the other two terms? It would seem to me that if a principle is to be applied, then it must be applied everywhere, so that you must end up with d(T -T) = d(0). > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt > And maybe considering f(t) and exp(-st) might get you ruminating on > their continuity, and a daring insight like this may leap out: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> What I seek is a sound mathematical proof of the claim >> that f(t).d(t-T) gives rise to a spectrum contribution >> of f(T).e^-(st) ... > I really think that I might be onto something here, but I need to work > out how to integrate that pesky Dirac Delta ... any takers? its actually pretty straight-forward to do it right (its on the other post i just made). Beanie prefers to use integration-by-parts, and then makes a mistake (he actually doesnt do it right from the beginning) and comes out with zero so then he claims that all accumulated knowledge since Heaviside and Nyquist about it is in error. dont fall for his bull. he really doesnt know what hes typing about. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >>I really think that I might be onto something here, but I need to work >>out how to integrate that pesky Dirac Delta ... any takers? > its actually pretty straight-forward to do it right (its on the other post > i just made). Beanie prefers to use integration-by-parts, and then makes a > mistake (he actually doesnt do it right from the beginning) and comes out > with zero so then he claims that all accumulated knowledge since Heaviside > and Nyquist about it is in error. dont fall for his bull. he really > doesnt know what hes typing about. > r b-j Disclaimer, my previous post was not to be taken very seriously - the maths was intentionally hand-wavy and tongue-in-cheek. David Ullrich did it the standard way further up. I just integrated it by parts for Mr Bean, so he is hopefully now satisfied ... Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > I just integrated it by parts for Mr Bean, so he is hopefully now > satisfied ... but shucks! you gave it away! for free! i wanted to extract a promise (that the asshole would probably not keep) before telling him what he did wrong. oh well, its public knowledge anyway. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > For what its worth, tongue firmly in cheek, and with apologies to David > Ulrich, who did it properly: Even more apologies to David Ullrich for spelling his name wrong === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> For what its worth, tongue firmly in cheek, and with apologies to David >> Ulrich, who did it properly: >Even more apologies to David Ullrich for spelling his name wrong No problem - I assumed you were talking about someone else, wondered why I didnt see any posts from him... (the spelling with one l is much more common.) ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? Can you either stop intentially re-arranging the letters and spacings in the subject line of these posts ... or if its unintentional, fix your usenet posting application so it stops doing so? Its a little disorientating to have a thread breaking off into 5 or 6 different permutations within the group ( appears to be localized to your replies ). Its making a mess in Google as well. >with as follows..... >You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to >determine your result. >You simply cannot say that f(t).d(t-T) yields f(T) unless >you do so under an integral. This arises from the fundamental >properties of the Diracian Delta function . >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >What I seek is a sound mathematical proof of the claim >that f(t).d(t-T) gives rise to a spectrum contribution >of f(T).e^-(st), and claiming that f(t).d(t-T) equals >f(T) is not sound. We use the properties of the Diracian >Delta Function in so many other aspects of signal >processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform >like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. ( modify address for return mail ) www.numbersusa.com www.americanpatrol.com === Subject: Deviration Of The Spectrom Due To f(t).d(t - T)????? No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if its unintentional, fix > your usenet posting application so it stops doing so? Its a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). Its making a mess in Google as well. >with as follows..... >You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to >determine your result. >You simply cannot say that f(t).d(t-T) yields f(T) unless >you do so under an integral. This arises from the fundamental >properties of the Diracian Delta function . >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. >What I seek is a sound mathematical proof of the claim >that f(t).d(t-T) gives rise to a spectrum contribution >of f(T).e^-(st), and claiming that f(t).d(t-T) equals >f(T) is not sound. We use the properties of the Diracian >Delta Function in so many other aspects of signal >processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform >like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. > ( modify address for return mail ) > www.numbersusa.com > www.americanpatrol.com Author-Supplied-Address: Anonymous-Remailer See Comment Header === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Comments: This message did not originate from the above address. It was remailed by two or more anonymous mail services. Send complaints of REAL abuse ONLY to remaileradmin at optonline dot n Mail-To-News-Contact: abuse@dizum.com > No. Sorry. Good reason to do so. Your desperate need for attention is a good reason only in the mind of selfish arseholes like yourself. -=- This message was posted via two or more anonymous remailing services. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Stupid boy. I see that you dont have the strength of character Stupid cowardly boy. > No. Sorry. Good reason to do so. > Your desperate need for attention is a good reason only in the mind of > selfish arseholes like yourself. > -=- > This message was posted via two or more anonymous remailing services. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? And what would that good reason be? BTW, MS Outlook Express, despite its failings, does group things by thread even if the subject line has been changed. > No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if its unintentional, fix > your usenet posting application so it stops doing so? Its a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). Its making a mess in Google as well. === Subject: Re: Deviration Of Teh Spectrom Due To f(t).d(t - T)????? Confidential, sorry. > And what would that good reason be? BTW, MS Outlook Express, despite its > failings, does group things by thread even if the subject line has been changed. > No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if its unintentional, fix > your usenet posting application so it stops doing so? Its a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). Its making a mess in Google as well. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? > No. Sorry. Good reason to do so. >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. Your dU, ie dU/dt, ie d/dt e^(-sT) equals zero. Thus int[dU.int(V)] = int(0.f(T)) = 0. and U.int(V) = e^(-sT) . f(T), which is equal to the left handside. Wilbert === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Yes its a constant. Where did I cock up? Lets see..... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-sT) as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. Ah yes, typo, should read..... with f(t).d(t-T) as V and e^(-st) as U ..... also the next line is wrong, should read..... f(T).e^(-st) - int(-s.e^(-st) . f(T))..... However the effect of a consecutive differentiation followed by an integral is still to give -s/-s. Giving.... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-st) as U ..... f(T).e^(-st) - int(-s.e^(-st) . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. > No. Sorry. Good reason to do so. >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. > Your dU, ie dU/dt, ie d/dt e^(-sT) equals zero. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? > No. Sorry. Good reason to do so. Do what? === Subject: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Do what was complained of. > No. Sorry. Good reason to do so. > Do what? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Do what was complained of. Where? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? In a previous post. > Do what was complained of. > Where? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > In a previous post. Top-posting makes the whole thing impractical. === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Nothing wrong with top-posting. It is the preferred method, because you dont have to page down through much repeated and already-seen history. skipped over unread. > In a previous post. > Top-posting makes the whole thing impractical. === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Neener, and Neener that... > Nothing wrong with top-posting. Thats what clueless newbies always say. > It is the preferred method, because you dont have to > page down through much repeated and already-seen > history. > skipped over unread. > In a previous post. > Top-posting makes the whole thing impractical. -- Checkmate all rights reserved === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Nothing wrong with top-posting. Clueless! > It is the preferred method, because you dont have to Really? And what exactly is your supporting evidence? A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > A: Maybe because some people are too annoyed by top-posting. It is precisely because of this that he does it. His pathology requires that he annoy people. Bob -- Things should be described as simply as possible, but no simpler. A. Einstein === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Nothing wrong with top-posting. > It is the preferred method, because you dont have to > page down through much repeated and already-seen > history. More properly, It is preferred by some, because they dont have to page down through much repeated and already-seen history. It is not universally Ôpreferred by all use-net users. In fact, there are several use-nettiquite > skipped over unread. ... by those that dont want to scroll through the previous context. Others, that dont mind reading the postings in context dont mind at all. In fact, many prefer in-context posting as it is much clearer what the poster is replying to. In the end, in-context posting, with snippage of unrelated, repeatedly quoted material is Ôpreferred by many. You dont speak for the entire use-net community, so dont pretend that you do. daestrom === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? >> Nothing wrong with top-posting. >> It is the preferred method, because you dont have to >> page down through much repeated and already-seen >> history. > More properly, It is preferred by some, because they dont have to page > down through much repeated and already-seen history. It is not > universally > Ôpreferred by all use-net users. Id say it was preferred by wankers, but thats just my humble opinion. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Longest Thread Ever In de.sci.mathematik Thread: Hat Cantor doch geirrt? Actually 3095 postings === Subject: Re: Longest Thread Ever words ßoating about: >In >de.sci.mathematik >Thread: Hat Cantor doch geirrt? >Actually 3095 postings Dots nice, still no pun for discussion? === Subject: Re: Longest Thread Ever isnt > In > de.sci.mathematik > Thread: Hat Cantor doch geirrt? > Actually 3095 postings The Greatest Man Ever hit 20,000 postings last April. === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > > Just because the factors are divisible by 7 when x=0, it doesnt > follow that the factors are divisible by 7 for all values of x. > It follows from the distributive property. James Harris === Subject: Re: JSH:Understanding constant terms ... > Just because the factors are divisible by 7 when x=0, it doesnt > follow that the factors are divisible by 7 for all values of x. ... > It follows from the distributive property. Back on that again James? No, it does *not* follow from the distributive property. You have r_3(x) + 22 which is not divisible by any factor of 7 when x = 0. You claim that it is not divisible by any factor of 7 when x != 0 (and base that on the distributive property). The claim you made quite some time ago was that if for some x, and some factor f of 7, (r_3(x) + 22)/f was an algebraic integer that than (due to the distributive property) in r_3(x)/f + 22/f both terms should be algebaric integers. But that is *not* the distributive property, and it is false. Consider x + 1 in the algebraic integers. By your interpretation for all x (x + 1)/2 would not be an algebraic integer. On the other hand, you have r_1(x) + 7, which is divisible when x = 0, but is not necessarily divisible by 7 when x != 0. And I do not know how you intend to show such using the distributive property, because this is also obviously false. x + 2 is divisible by 2 when x = 0, but not for many other values of x. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH:Understanding constant terms > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > > > Just because the factors are divisible by 7 when x=0, it doesnt > follow that the factors are divisible by 7 for all values of x. > > It follows from the distributive property. No, you have two assertions: 1: (5 a_1(x) + 7) is divisible by 7 for all x 2: a_1(x) is divisible by 7 for all x If 1 is true then 2 follows from the distributive property. However, 1 does not follow from the distributive property (indeed 1 is false). -William Hughes > James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > > > Just because the factors are divisible by 7 when x=0, it doesnt > follow that the factors are divisible by 7 for all values of x. > > It follows from the distributive property. > James Harris How? === Subject: Re: JSH:Understanding constant terms >> which means that 7 is divided from the constant terms as well, and >> only two of the constant terms, c_1 and c_2 have 7 as a factor, so >> necessarily 7 is divided from the factors where the constant terms are >> 7. > Just because the factors are divisible by 7 when x=0, it doesnt >> follow that the factors are divisible by 7 for all values of x. >> It follows from the distributive property. In what way does it follow from the distributive property???? If something is not obvious, it would help for you to explain it. Otherwise, it is assumed to be just more crap from James. === Subject: Re: JSH:Understanding constant terms Discussion, linux) >> Just because the factors are divisible by 7 when x=0, it doesnt >> follow that the factors are divisible by 7 for all values of x. >> It follows from the distributive property. Care to step through that argument for us slowpokes? -- Jesse F. Hughes Love songs suck and losing you aint worth a damn. -- The poetry of Bad Livers === Subject: Re: JSH:Understanding constant terms >> >> Just because the factors are divisible by 7 when x=0, it doesnt >> follow that the factors are divisible by 7 for all values of x. >> >> It follows from the distributive property. > Care to step through that argument for us slowpokes? Sure. The factor g_1(x) has *two* parts, where one of them is the constant term, which is constant as it is in fact, 7, and 7 is constant. The other varies as x varies. Now the constant term goes from 7 to 1, which means that it is divided by 7. Understand? Well then, by the distributive property, the other term must be divided by 7 as well, as if you have two parts, then you cant get to one without going through the other. For example to help you understand the concept consider z = x + 7 versus z/7 = x/7 + 1 and notice that by the distributive property, you cant divide 7 from the second piece without going through the first! Thats how the distributive property applies. For other readers, note that I can give detail as necessary with the argument, and also notice the *level* of detail that posters wish to request! That makes the argument in my original post one of the most worked out in math history with a level of detail that few ever bother to try in todays math world. It is a perfect argument, in every detail, as a proof must be. James Harris === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you cant get to > one without going through the other. > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 Let me try, too. Let g(x) = 17 / (x+1). Whats the constant term? If I understand you correctly, it is g(0), i.e., 17. Right? Now, 17 divides 17. Right? So I should conclude (by the distributive property) that 17 divides g(x) for all x, right? In particular, I guess that 17 divides g(16). But g(16) = 1, so therefore, 17 divides 1, right? Where did I go wrong? -- Jesse F. Hughes I love Mathematics[...] She doesnt care about my feelings, or my pride or how I so wish to get out and travel. Truth is truth, and anything else is just plain wrong. -- James Harris, /A Love Story/ === Subject: Re: JSH:Understanding constant terms > Where did I go wrong? I can help here. You went wrong when you tried to explain some mathematics to JSH. This is a simple, but common, error. === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > > Just because the factors are divisible by 7 when x=0, it doesnt > follow that the factors are divisible by 7 for all values of x. > > It follows from the distributive property. > Care to step through that argument for us slowpokes? > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. Er, I guess. Trivially, we can write g_1(x) = (g_1(x) - g_1(0)) + g_1(0). ^^^^^^^^^^^^^^^^^ ^^^^^^ so-called constant bit ^^^^^^^^^^^^^^^^^ evidently the non-constant bit > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you cant get to > one without going through the other. What the distributive property says is (a + b)/c = a/c + b/c. You have g_1(0)/7 is an algebraic integer (or something). You want to conclude that for all x, g_1(x)/7 is an algebraic integer. All the distributive property seems to give you is: g_1(0) / 7 = (g_1(0) / 7 - g_1(0) / 7) + g_1(0) / 7 or, if you prefer, g_1(x) / 7 = (g_1(x) / 7 - g_1(0) / 7) + g_1(0) / 7 Neither of these equations proves that g_1(x) is divisible by 7. I dont see how I go from that to the claim that 7 divides g_1(x) - g_1(0) for all x. Maybe a little more detail? I still dont get it. -- I dont want to wine and dine and date you once or twice. I want to hold you now. I just want to spend the night. You tell me a better plan. Baby, Im not a patient man. -- Jimmy Lafave, the romantic troubadour. === Subject: Re: JSH:Understanding constant terms > >> >> Just because the factors are divisible by 7 when x=0, it doesnt >> follow that the factors are divisible by 7 for all values of x. >> >> > It follows from the distributive property. > > Care to step through that argument for us slowpokes? > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you cant get to > one without going through the other. > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you cant divide 7 from > the second piece without going through the first! > Thats how the distributive property applies. > For other readers, note that I can give detail as necessary with the > argument, and also notice the *level* of detail that posters wish to > request! > That makes the argument in my original post one of the most worked out > in math history with a level of detail that few ever bother to try in > todays math world. > It is a perfect argument, in every detail, as a proof must be. > James Harris You havent proved what you said you were going to prove. You said you were going to prove that the factor was divisible by 7 for all values of x. You havent proved that. === Subject: Re: JSH:Understanding constant terms > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you cant divide 7 from > the second piece without going through the first! Ok, youve shown that when z (which is better written z(x) since its a function of x) is divided by 7, both terms are divided by 7. Thats fine, and obviously true by the distributive law, as you claim. But what you were supposed to show was that if z(0) is divisible by 7, then z(x) is divisible by 7 for all x. Clearly thats NOT true, since in your example, z(0)=7 is divisible by 7, but z(1)=8 is not divisible by 7. --Mark === Subject: Re: JSH:Understanding constant terms > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you cant divide 7 from > the second piece without going through the first! > Ok, youve shown that when z (which is better written z(x) since its a > function of x) is divided by 7, both terms are divided by 7. Thats fine, > and obviously true by the distributive law, as you claim. But what you were > supposed to show was that if z(0) is divisible by 7, then z(x) is divisible > by 7 for all x. Clearly thats NOT true, since in your example, z(0)=7 is > divisible by 7, but z(1)=8 is not divisible by 7. > --Mark Ok, the actual expression from my post is g_1(x) = 5a_1(x) + 7 where 7 is a constant term as a_1(0) = 0. And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 as a multiple. Dividing P(x) by 49 gives a contant term of 22. That means that when P(x) is divided by 49, g_1(x) must be divided by 7, by the distributive property. Or do you disagree? James Harris === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? I dont see how youre appealing to the distributive property here. Can you be more explicit? The distributive property says a(b + c) = ab + ac. What is a, b and c? I guess that a is 1/y, b is 5 a_1(x) and c is 7, right? But so what? So g_1(x)/7 = 5 a_1(x) / 7 + 1. If I knew that g_1(x) was divisible by 7 (in what ring? algebraic integers?), then I could conclude that 5 a_1(x) is also divisible by 7. And if I knew that 5 a_1(x) was divisible by 7, then I would know that g_1(x) is, too. But I dont know either of these things, do I? -- I have to break the code of how [mere humans] work, and I have made a lot of progress over years of effort, and I feel like I am close to figuring out all the inner details of human wiring. -- James S. Harris on the extra problems of conveying his research === Subject: Re: JSH:Understanding constant terms >> Ok, youve shown that when z (which is better written z(x) since >> its a function of x) is divided by 7, both terms are divided by 7. >> Thats fine, and obviously true by the distributive law, as you >> claim. But what you were supposed to show was that if z(0) is >> divisible by 7, then z(x) is divisible by 7 for all x. Clearly >> thats NOT true, since in your example, z(0)=7 is divisible by 7, >> but z(1)=8 is not divisible by 7. > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? Before going back to your original posting, Id like to clear up this point first. Several people have made the claim that this statement, which Ill slightly generalize and call S, is false: S: If g(0) is divisible by k, then g(x) must be divisible by k for all x. You disputed these claims, seeming to be saying that you believe that S is true. Do you now agree that S is false? --Mark === Subject: Re: JSH:Understanding constant terms > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? Youre distributing different quantities when Ôx is not equal to 0. Your original decomposition into factors applied to the x = 0 case, not to the general case. Go ahead and substitute x = 1 into the equation, follow through with it and see what you get. (By the way, that information was given to you months ago, but you ignored it!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, > Up to this point you are fine. > so > necessarily 7 is divided from the factors where the constant terms are > 7. > > No. What you are saying here is that if 7 divides g(0) then > 7 must divide g(x) for all x. This is simply not true. It follows algebraically. If you need me to explain in further detail then I can, but simple denial is not going to work. > The rest of your post in nonsense. > - William Hughes I can explain the algebra in detail to you, but if you simply choose to claim its nonsense, then no progress can be made, right? Now then, if you will accept algebra, and point out an area of confusion then I can explain to you. But if you are simply going to just deny then theres little point. Which is it? Will you follow algebra, and give a mathematical area for me to explain, or are you just going to deny? James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, > > Up to this point you are fine. > > so > necessarily 7 is divided from the factors where the constant terms are > 7. > > > > No. What you are saying here is that if 7 divides g(0) then > 7 must divide g(x) for all x. This is simply not true. > It follows algebraically. If you need me to explain in further detail > then I can, but simple denial is not going to work. Actually it does not follow, algebraically or otherwise. Simple assertion is not going to work. I have explained in great detail, why you are wrong. You have ignored these posts. > > The rest of your post in nonsense. > > - William Hughes > I can explain the algebra in detail to you, but if you simply choose > to claim its nonsense, then no progress can be made, right? Good, start by explaining all the counterexamples I have provided. You have quite a backlog. As an easy first step explain why, with a(0) = 0 and w1(0) =1 the constant terms of (a(x) + 7) and (a(x)/w1(x) + 7/w1(x)) are different. > Now then, if you will accept algebra, and point out an area of > confusion then I can explain to you. But if you are simply going to > just deny then theres little point. > Which is it? Will you follow algebra, and give a mathematical area > for me to explain, or are you just going to deny? Which is it? Will you reply to the above and give mathematical arguments, or are you just going to ignore? > James Harris === Subject: Re: JSH:Understanding constant terms > Where am I going wron ? Ill explain. > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > OK, (x-1)(x-2) = x^2 -3x +2 > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > No idea what you mean by algebraic integr functions Well, just in case people didnt understand, I explained in my next paragraph. Ill skip past your next and go to that next. > Possibly: > A) (x-1)=0 and (x-2)=0 are both monic polys > wuth integer coefficients so their roots are > algebraic integers. > B) y = (x-1), Z = (x-2): if x is an algebraic integer > then y and z both are. > Which one of these do you mean or do you mean something else ? > If you mean that the product of the gs gives you P(x) > and that the root or roots of g_r(x) =0 are algebraic integers > then the gs must be monic and so must P(x). > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. And thats what I mean, as I went ahead and explained what I mean. An algebraic function f(x), will give an algebraic integer result for any algebraic integer x. Hopefully that clears things up. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > Yes this follows from Vietas theorem > Forn any poly x^n + B_1x^(n-1) + ... +B_n =0 > B_n = product of roots x(-1)^n > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > If and only if the gs are monic polys No. The gs are algebraic integer functions. Heres an example: f(x) = 2x is an algebraic integer function, as for any algebraic integer x, it gives an algebraic integer, but notice its not a monic polynomial. > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > r_1(x) = x-1 -1 > r_2(x) = x-2 -2 > => (x-1)(x-2) = x^2 -3x +2 > So you have gone round in a circle. Well, try it out on x^2 + 3x + 2 = (x+2)(x+1) with g_1(x) = x+2, and g_2(x) = x+1 where Im deliberately using polynomial factors, and youll find that c_1 = 2, r_1(x) = 2, and c_2 = 1, r_2(x) = x, so (r_1(x) + c_1)(r_2(x) + c_2) = (x+2)(x+1) and its trivial there, but that should give you the general idea. What Ive done is generalize from the basic to something a little more complicated where instead of polynomial factors I can use non-polynomial ones. > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > P(X) is not monic so it cannot be the P(X) > you describe above . Why not? Consider 2x^2 + 3x + 1 = (2x + 1)(x + 1) where here g_1(x) = 2x + 1 and its an algebraic integer function. > So Im confused - was your previous explabation irelevant > to what comes ? > Perhaps you should go through the steps as if you were > writing a computer program defining precisely > what each term means, how one step follows from the next. > Otherwise, you have no hope of communicating your valuable insights > to anyone. A computer would be easier, and I know as I can program computers. People are quirky, as they can choose to ignore information. You can lead a horse to water... > I am none the wiser as to what you are trying to prove. > Perhaps you should state the theorem you are trying to > prove clearly and then outline the strategy > of your proof with clear,precise definitions of all the terms > What criteria do the gs have to meet ? They are algebraic integer functions and factors of P(x). > Are they monic polys ? - presumably so if their roots > are to be albegraic integers. Perhaps you > are not sure yourself ? I am sure. And no they are not necessarily monic polynomials. > Are they irreducible or does this not matter ? It doesnt matter. > Does the factorisation have to be unique ? No. James Harris === Subject: Re: JSH:Understanding constant terms > Where am I going wron ? > Ill explain. Yes,I think I understand you now. To recap A g_i(x) is any function that produces an algebraic integer when x is an algebraic integer The roots r1,r2 of x^2 -2x +2 are akgebraic integers The roots, s1,s2,s3 of x^3 -7x +5 are likewise algebraic integers so g_1(x)= r1*x^(1/2) + s2*x + 2 g_2(x)= s2*x^(1/2) + r3 give algebaric integers when x is an algebraic integer. Their product P(x)is a polynomial in x^(1/2) and g_1(0)*g_2(0) = 2 r3 is an algebraic integer. So where does that get us ? === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). Dont the g_i, polys in x, also have to be algebraic integers whenever x is an algebraic integer ? So, say 6x^2 +(pi)x + 2 couldnt be a g_1 even though its constant term is a rational integer. === Subject: Re: JSH:Understanding constant terms >>If and only if the gs are monic polys >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? Im sorry. Was that English? You were assuming that the g_i(x) were polynomials, and that the product of the g_i(x) was equal to P(x). Then your statement if and only if the gs are monic polys was about the statement that each g_i(0) divides, in the algebraic integers, P(0). But that is false. For each g_i(0) to divide P(0) in the ring of all algebraic integers, it is necessary and sufficient for each of the values g_i(0) to be algebraic integers. For certainly we need each g_i(0) to be an algebraic integer. And if they all are, then clearly each of them divides P(0) in the ring of all algebraic integers. You dont need the g_i(0) to be monic; you seemed to be led astray by the fact that g_i(0) is plus or minus the product of the roots of g_i(x). Certainly if g_i(x) is a monic polynomial with algebraic integer coefficients, then g_i(0) will be an algebraic integer; but it is possible for g_i(0) to be an algebraic integer without all roots of g_i(x) to be algebraic integer. > So, say 6x^2 +(pi)x + 2 couldnt >be a g_1 even though its constant term >is a rational integer. Doesnt matter. Your assertion was about the g_i(0) dividing P(0) in the algebraic integers. For that statement, you dont need the g_i(x) to be monic (note that P(x) was not monic either). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH:Understanding constant terms >>If and only if the gs are monic polys >> >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? > Im sorry. Was that English? I cant say that Ive found your prose particularly pristine. Are you a non-native English speaker ? > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the gs are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). According to Harris, the g_i(x) are algebraic integers whenever x is ( his algebraic integer functions). Clearly g_i(0) has to be an algebraic integer and so the constant term of g_i(x)has to be an algerbaic integer. You say that the gi can be ANY polynominal. This is clearly not the case. For example pi*x +1. Are you saying that p1*1 +1 is an alegebraic integer ? > But that is false. For each g_i(0) to divide P(0) in the ring of all > algebraic integers, it is necessary and sufficient for each of the > values g_i(0) to be algebraic integers. Yes, > For certainly we need each > g_i(0) to be an algebraic integer. And if they all are, then clearly > each of them divides P(0) in the ring of all algebraic integers. > You dont need the g_i(0) to be monic; you seemed to be led astray by > the fact that g_i(0) is plus or minus the product of the roots of > g_i(x). No, that was not the problem. Certainly if g_i(x) is a monic polynomial with algebraic > integer coefficients, then g_i(0) will be an algebraic integer; but it > is possible for g_i(0) to be an algebraic integer without all roots of > g_i(x) to be algebraic integer. Yes. > So, say 6x^2 +(pi)x + 2 couldnt >be a g_1 even though its constant term >is a rational integer. > Doesnt matter. Im afraid it does matter if you are using the g_i(x) as defined by Harris. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. >If and only if the gs are monic polys > > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> Im sorry. Was that English? >I cant say that Ive found your >prose particularly pristine. I do try to avoid the use of cute abbreviations like polys, though. >Are you a non-native English speaker ? Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the gs are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. >You say that the gi can be ANY polynominal. Please note that I was not addressing Jamess argument. I was addressing your restriction to the g_i being polynomials, and your assertion, ->under that further restriction<-, that each g_i(0) would divide P(0) in the ring of all algebraic integers if and only if the g_i(x) were monic. ->That<- assertion is false. >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? No. I am saying that if g(x)=pi*x + 1, then it is entirely possible for g(0) to divide P(0) in the ring of all algebraic integers, EVEN THOUGH g(x) is not a monic polynomial. Hell: take g1(x) = 2x+1 and g2(x) = x-1. Then g1(x)*g2(x) = P(x) = 2x^2 - x - 1. g1(0) and g2(0) both divide P(0) in the ring of all algebraic integers, EVEN THOUGH g1(x) is not monic. You claimed the divisibility conclusion would hold if and only if the gs are monic poly[nomial]s. The if is true. The only if is false. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys > > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> >> Im sorry. Was that English? >I cant say that Ive found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. Cute ? Poly used to be a widespr. abb. for polytechnic. Ive never heard it reviled on the basis of cuteness before. But then you learn many strange things on sci.math. >Are you a non-native English speaker ? > Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the gs are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. >You say that the gi can be ANY polynominal. > Please note that I was not addressing Jamess argument. I was > addressing your restriction to the g_i being polynomials, and your > assertion, ->under that further restriction<-, that each g_i(0) would > divide P(0) in the ring of all algebraic integers if and only if the > g_i(x) were monic. ->That<- assertion is false. Yes, I agree >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? > No. I am saying that if g(x)=pi*x + 1, then it is entirely possible > for g(0) to divide P(0) in the ring of all algebraic integers, EVEN > THOUGH g(x) is not a monic polynomial. Yes, but we now know that the g_i(x) must meet another condition. What I said prev. was incorrect, but your statement that ANY polyno. would do is crassly passe in the light of new revelations. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. [.snip.] > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the gs are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >>whenever x is ( his algebraic integer functions). >>Clearly g_i(0) has to be an algebraic >>integer and so the constant term of g_i(x)has to be an >>algerbaic integer. >You say that the gi can be ANY polynominal. >> Please note that I was not addressing Jamess argument. I was >> addressing your restriction to the g_i being polynomials, and your >> assertion, ->under that further restriction<-, that each g_i(0) would >> divide P(0) in the ring of all algebraic integers if and only if the >> g_i(x) were monic. ->That<- assertion is false. >Yes, I agree >>This is clearly not the case. For example pi*x +1. >>Are you saying that p1*1 +1 is an alegebraic integer ? >> No. I am saying that if g(x)=pi*x + 1, then it is entirely possible >> for g(0) to divide P(0) in the ring of all algebraic integers, EVEN >> THOUGH g(x) is not a monic polynomial. >Yes, but we now know that the g_i(x) >must meet another condition. >What I said prev. was incorrect, >but your statement that ANY polyno. >would do is crassly passe in the light >of new revelations. So... I addressed your comment as written. Since that comment, you have now changed your mind about what the g_i(x) should be; based on that change of heart, you complain that my correction was wrong, because it failed to meet your newfound and not-previously asserted conditions on the g_i(x). How... interesting... -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH:Understanding constant terms > [.snip.] > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the gs are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >>whenever x is ( his algebraic integer functions). >>Clearly g_i(0) has to be an algebraic >>integer and so the constant term of g_i(x)has to be an >>algerbaic integer. >You say that the gi can be ANY polynominal. >> >> Please note that I was not addressing Jamess argument. I was >> addressing your restriction to the g_i being polynomials, and your >> assertion, ->under that further restriction<-, that each g_i(0) would >> divide P(0) in the ring of all algebraic integers if and only if the >> g_i(x) were monic. ->That<- assertion is false. >Yes, I agree >>This is clearly not the case. For example pi*x +1. >>Are you saying that p1*1 +1 is an alegebraic integer ? >> >> No. I am saying that if g(x)=pi*x + 1, then it is entirely possible >> for g(0) to divide P(0) in the ring of all algebraic integers, EVEN >> THOUGH g(x) is not a monic polynomial. >Yes, but we now know that the g_i(x) >must meet another condition. >What I said prev. was incorrect, >but your statement that ANY polyno. >would do is crassly passe in the light >of new revelations. > So... I addressed your comment as written. Since that comment, you > have now changed your mind about what the g_i(x) should be; based on > that change of heart, you complain that my correction was wrong, > because it failed to meet your newfound and not-previously asserted > conditions on the g_i(x). > How... interesting... The disingenuousness of your attempted exculpation is breathtaking. I am unversed in exegesising the pronouncements of the Supreme Coprolocutor. You, however, analyse His productions with monomaniacal intensity. Therefore, to claim that the conditions on the g_i(x), although new to me, were unknown to you is sophistry of the most egregious kind. A simple admission of your trivial error would bring the catharsis and ultimate peace of mind you yearn for. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. [.snip.] >> So... I addressed your comment as written. Since that comment, you >> have now changed your mind about what the g_i(x) should be; based on >> that change of heart, you complain that my correction was wrong, >> because it failed to meet your newfound and not-previously asserted >> conditions on the g_i(x). >> How... interesting... >The disingenuousness of your attempted >exculpation is breathtaking. >I am unversed in exegesising the >pronouncements of the Supreme Coprolocutor. >You, however, analyse His productions with >monomaniacal intensity. Therefore, >to claim that the conditions on the g_i(x), >although new to me, were unknown to you >is sophistry of the most egregious kind. Sigh. I addressed YOUR comment. I noted an error in YOUR comment, based on YOUR hypothesis. You acknowledge that error, but complain that I noted it. I did not address anything the original poster said. Presumably, you are either joking, or dense. I prefer to assume the former, though it seems hard to ignore the latter possibility. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH:Understanding constant terms > [.snip.] >> So... I addressed your comment as written. Since that comment, you >> have now changed your mind about what the g_i(x) should be; based on >> that change of heart, you complain that my correction was wrong, >> because it failed to meet your newfound and not-previously asserted >> conditions on the g_i(x). >> >> How... interesting... >The disingenuousness of your attempted >exculpation is breathtaking. >I am unversed in exegesising the >pronouncements of the Supreme Coprolocutor. >You, however, analyse His productions with >monomaniacal intensity. Therefore, >to claim that the conditions on the g_i(x), >although new to me, were unknown to you >is sophistry of the most egregious kind. > Sigh. I addressed YOUR comment. I noted an error in YOUR comment, > based on YOUR hypothesis. You acknowledge that error, but complain > that I noted it. I did not address anything the original poster said. Typical No, I wasnt complaining that you noted it at all. My soul was suffused with the light that you brought unto me. I wasnt saying that you explicitly addressed anything the OP said, but that you must have known implicitly what the OP said about the g_i(x) due to your regular enlightening intercourse with the OP. Perhaps a ßeeting fit of insanity overcame you as you pondered these deep matters causing you to cast aside your usual intellectual caution and make that rash and impetuous statement about ANY polynomial. > Presumably, you are either joking, or dense. Joking ? Gay persißage ? Badinage of an ironic nature ? Egad Sir,I would rather ritually disembowel myself than make light of your forensic quest for Truth. > I prefer to assume the former, though it seems hard to ignore the > latter possibility. Your sharpness of mind is matched only by your generosity of spirit. === Subject: Re: JSH:Understanding constant terms Discussion, linux) > No, I wasnt complaining that you noted it at all. > My soul was suffused with the light that you > brought unto me. Jesus, give it a rest and put away the thesaurus. Were already ing impressed. At least I know I am. -- Jesse F. Hughes To be honest, I dont have enough interest in math to spend the time it would take to clean up the mess that I believe has been created in the past 100 or so years. -- Curt Welch lets the world down. === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? > Im sorry. Was that English? >I cant say that Ive found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. > Cute ? Poly used to be a widespr. abb. Um Davidson, widespr. and abb. are not words in the English language. If you think they are acceptable abbreviations in the English language you are clearly not qualified to be asking others rudely if they are a non-native English speaker. > for polytechnic. Ive never heard So you were saying Dont the g_i, polytechnics in x, also have to be ...? KeithK > it reviled on the basis of cuteness before. > But then you learn many strange things on sci.math. >Are you a non-native English speaker ? > Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the gs are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. >You say that the gi can be ANY polynominal. > Please note that I was not addressing Jamess argument. I was > addressing your restriction to the g_i being polynomials, and your > assertion, ->under that further restriction<-, that each g_i(0) would > divide P(0) in the ring of all algebraic integers if and only if the > g_i(x) were monic. ->That<- assertion is false. > Yes, I agree >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? > No. I am saying that if g(x)=pi*x + 1, then it is entirely possible > for g(0) to divide P(0) in the ring of all algebraic integers, EVEN > THOUGH g(x) is not a monic polynomial. > Yes, but we now know that the g_i(x) > must meet another condition. > What I said prev. was incorrect, > but your statement that ANY polyno. > would do is crassly passe in the light > of new revelations. === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys > No. If and only if the values g_i(0) are algebraic integers. This > will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for > example, > if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? > Im sorry. Was that English? >I cant say that Ive found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. > Cute ? Poly used to be a widespr. abb. > Um Davidson, widespr. and abb. are not words in the English language. Yes, thats wherefore me chosed they. > If you think they are acceptable abbreviations in the English language you > are clearly not qualified to be asking others rudely if they are a > non-native English speaker. Me not it were which the one say first bad bad things about my spoke Brit-talk was. You cogitate big deep, you get it much quickly. === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys > No. If and only if the values g_i(0) are algebraic integers. This > will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for > example, > if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? > Im sorry. Was that English? >I cant say that Ive found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. > Cute ? Poly used to be a widespr. abb. > Um Davidson, widespr. and abb. are not words > in the English language. > Yes, thats wherefore me chosed they. > If you think they are acceptable abbreviations in the English language you > are clearly not qualified to be asking others rudely if they are a > non-native English speaker. > Me not it were which the one say > first bad bad things about my spoke Brit-talk was. > You cogitate big deep, you get it much quickly. Me humbled muchly your by delightf. reply. :) KeithK === Subject: Re: JSH:Understanding constant terms >If and only if the gs are monic polys >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? So, say 6x^2 +(pi)x + 2 couldnt >be a g_1 even though its constant term >is a rational integer. The g_i Ôs are not necessarily polynomials. In the specific case JSH considers later in his posting, the a_i(x) are the roots of a cubic whose coefficients are functions of x. They could be found explicitly using Cardanos formula: complex expressions involving cube- and square-roots of polynomials in x. (They are algebraic integers whenever x is, because the cubic involved is monic.) -- David Hartley === Subject: Re: JSH:Understanding constant terms >>If and only if the gs are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Dont the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? So, say 6x^2 +(pi)x + 2 couldnt >be a g_1 even though its constant term >is a rational integer. > The g_i Ôs are not necessarily polynomials. In the specific case JSH > considers later in his posting, the a_i(x) are the roots of a cubic > whose coefficients are functions of x. They could be found explicitly > using Cardanos formula: complex expressions involving cube- and > square-roots of polynomials in x. (They are algebraic integers whenever > x is, because the cubic involved is monic.) Apparently, all that is required of the g_i(x) is that when x is an algebraic integer g_1(x) is also an algebraic integer. So something of the form A*log(x) + pi + tan(x) would not do. As the sums, products and roots of algebraic integers are algebraic integers then, say, g_1(x) A x^(3/5) + Bx^(1/2) + C, where A,B,C are algebraic integers would be an algebaraic integer if x is an algebraic integer. But this is still a polynomial in fractional powers of x. Non-polynomial seems a bit misleading. === Subject: Re: JSH:Understanding constant terms ... > Apparently, all that is required of the g_i(x) > is that when x is an algebraic integer > g_1(x) is also an algebraic integer. > So something of the form A*log(x) + pi + tan(x) > would not do. > As the sums, products and roots of algebraic > integers are algebraic integers then, say, g_1(x) > A x^(3/5) + Bx^(1/2) + C, where A,B,C > are algebraic integers would be an algebaraic integer > if x is an algebraic integer. But this is > still a polynomial in fractional powers of x. > Non-polynomial seems a bit misleading. What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly call that non-polynomial. And what do you think of sqrt(x + 1)? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH:Understanding constant terms > ... > Apparently, all that is required of the g_i(x) > is that when x is an algebraic integer > g_1(x) is also an algebraic integer. > So something of the form A*log(x) + pi + tan(x) > would not do. > As the sums, products and roots of algebraic > integers are algebraic integers then, say, g_1(x) > A x^(3/5) + Bx^(1/2) + C, where A,B,C > are algebraic integers would be an algebaraic integer > if x is an algebraic integer. But this is > still a polynomial in fractional powers of x. > Non-polynomial seems a bit misleading. > What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly > call that non-polynomial. And what do you think of sqrt(x + 1)? I dont know what your definition of polynomial or non-polynomial might be. So I can only guess as to why you think your first example is a non-polynomial. A polynomial is finite sum of terms of the form Ax^k, where A is an algebraic integer, x a variable and k a positive rational number How are you defining polynomial and non-polynomial ? === Subject: Re: JSH:Understanding constant terms ... > Non-polynomial seems a bit misleading. > > What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly > call that non-polynomial. And what do you think of sqrt(x + 1)? > I dont know what your definition of polynomial or non-polynomial > might be. > So I can only guess as to why you think your first example > is a non-polynomial. In wat way is -x + sqrt(x^2 - 4) a polynomial? > A polynomial is finite sum of terms of the form Ax^k, where A > is an algebraic integer, x a variable and k a positive > rational number > How are you defining polynomial and non-polynomial ? The same. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the as are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the as gives > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > Algebraic integers are roots of monic polynomials with integer > coefficients. > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the as is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. Which is, of course, an out-and-out contradiction, right? So unless mathematics is inconsistent, there must be something wrong. What is it? It could be that algebraic number theory is wrong. Or possibly Galois theory. Or both. Or it could be that you have a logical error somewhere in what you have said above. You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly seems OK. I dont disagree with it. Then you noted that 49 = 7*7 must be factored out of the expression (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). Lets look in detail at that troublesome third term, r_3(x) + 22. The objective here is to examine whether I can divide a nonunit factor of 7 out of this expression, with the result after the factoring being an algebraic integer. Let w be a nonunit factor of 7. You note that 22 is coprime to 7. Therefore 22/w cannot be an algebraic integer. So that is potentially a problem. But recall that r_3(x) + 22 = 5 a_3(x) + 7. Suppose I divide the latter expression through by w: I get 5 a_3(x)/w + 7/w. The term 7/w is not a problem here because I was assuming that w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. Now, if such a w can be chosen so that a_3(x)/w is also an algebraic integer, I conclude that in the expression 5 a_3(x)/w + 7/w all the coefficients are algebraic integers, and the sum is an algebraic integer. Can such a w be defined ? If so, it is going to have to be a function of x: because when x = 0, w has to be 1. Whereas, for most other integers x, the polynomial that you mention above is irreducible, and w is a nonunit factor of 7. Here is how w_1(x), w_2(x), and w_3(x) need to be defined: w_1(x) = GCD(a_1(x), 7) w_2(x) = GCD(a_2(x), 7) w_3(x) = GCD(a_3(x), 7). Here GCD denotes the Ôgreatest common divisor function. It is defined in the ring of algebraic integers by a theorem of Dedekind. Obviously these definitions imply that all of (5 a_1(x) + 7)/w_1(x), (5 a_2(x) + 7)/w_2(x), and (5 a_3(x) + 7)/w_3(x) are algebraic integers. Now the crucial thing to show is that w_1(x) * w_2(x) * w_3(x) = 49. Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general not by any additional factors of 7. Why? Because 2401 x^3 - 147 x^2 + 3x is coprime to 7 [except when x itself is divisible by 7], and -49*(2401 x^3 - 147 x^2 + 3x) is the constant term of the polynomial that the as satisfy, as you note above. Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product must equal 49, as desired. What this shows is that it IS possible to divide 49 out of (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) in such a way that each factor is an algebraic integer. But what about the constant terms ? It is *not required* that the constant terms be respected. That is, even though it is true that (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, (as you can immediately check), there is NO REASON to require that each of these three terms be algebraic integers; and of course, 22/w_3(x) is not. Again, why? Isnt the product of the constant terms equal to the constant term of the product ??? Yes. But here is the last key fact. The constant term of (r_2(x) + 22)/w_3(x) is NOT what you think it is. You think it must be 22/w_3(x). It isnt. You need to remember YOUR OWN DEFINITION of constant term. It is instead 22/w_3(0). So then you note that w_3(0) = 1, and you have no contradiction. Everything hangs together. Again: you defined Ôconstant term in a perfectly reasonable way, and you should have stuck with your definition. Instead you got confused and assumed that whatever is in the POSITION of the constant term *is* the constant term. But here, 22/w_3(x) is not even constant, because by its definition, w_3(x) is not constant. Nora B. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > > Algebraic integers are roots of monic polynomials with integer > coefficients. > > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the as is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. > > Which is, of course, an out-and-out contradiction, right? There is the appearance of contradiction, which is readily resolved by not giving the ring of algebraic integers a special position. So then, just because the roots do not have 7 as a factor in the ring of algebraic integers, its not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? If you overrate the ring of algebraic integers, then you can make arguments that are wrong. Essentially, you have to understand that the requirement that a number be the root of a monic polynomial with integer coefficients is a meaningless technicality, with no real mathematical importance. There is no weight to the requirement mathematically that a number be the root of a monic polynomial with integer coefficients. Thats what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. It turns out to be a problem in algebraic number theory, which has lead to a mis-use of Galois Theory. Its not even really complicated, but there are reasons for people to get emotional over the issue, as its a mistake at the foundations of the discipline of mathematics. Its an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. You can continue with that assumption. I have no problem defending the math I outlined in my original post. Ultimately, it boils down to constants being constant. Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I dont disagree with it. Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Lets look in detail at that troublesome third term, Its not troublesome. Its asymmetrical with regard to the rest. The asymmetry is what blocks you out of the ring of algebraic integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. Now youre going to cheat, and Im going to explain how you cheat before you do it, so readers can see how it works. A number cannot be an algebraic integer if it is not the root of some monic polynomial with integer coefficients. You say nonunit above as if that applies globally, when youre going to rely on non-unit status in the ring of algebraic integers. So consider a unit function u(x), which is not a unit in the ring of algebraic integers. Now you can multiply with that unit, and then assert that you have a non-unit in the ring of algebraic integers, to claim that you can do that multiplication. But its only a non-unit in the ring of algebraic integers because its not the root of some monic polynomial with integer coefficients. Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. Notice that the poster didnt say, in the ring of algebraic integers. Its important that you note that the claim is in the ring of algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. The poster is apparently acknowledging that 22 properly is coprime to 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the Ôgreatest common divisor function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. Notice also that you have the implication that the factors do not have a constant term, which has to be made explicit in a bit. Essentially the ws the poster is trying to use are unit factors, but are not units in the ring of algebraic integers by a technicality that they are not roots of a monic polynomial with integer coefficients. Its a neat trick when you think about it that requires relying on the very problem that has been outlined to try and use the ring of algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the as satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. Well, yes, its possible. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That Notice the poster doesnt even try at this point, simply resorting to hand-waving by asserting that the constant terms dont need to be respected! The problem is that if you have a constant term that is 7, another that is 7, and one thats 22, then to get rid of the 7s, you have to divide out by 7, but thats algebra thats inconvenient to the poster! So instead the poster just TELLS you that the math doesnt care. Lets watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isnt the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isnt. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. Notice that the poster has asserted that the constant term is constant at x=0, but every where else its actually a function of x. So the full assertion is that the constant term is NOT CONSTANT, but is instead a function of x. Remember, when people try to fight mathematics they have to at some point rely on something that is just wacky. Here you can see that ultimately the poster is trying to get you to believe that a constant is in fact a function of x. But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = P(0), and notice, you dont have any functions of x, as x has been set to 0. So the poster is using unit factors, which by a technicality are not units in the ring of algebraic integers, unless you do wish to believe that mathematics is inconsistent, and so what? Your belief would be wrong. > Again: you defined Ôconstant term in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > Nora B. Well, I say youre using unit factors, which because they are not roots of a monic polynomial with integer coefficients are not units in the ring of algebraic integers. How do you answer? James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the as are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the as gives > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > Algebraic integers are roots of monic polynomials with integer > coefficients. > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the as is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. > Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, its not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. > Thats what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > Its not even really complicated, but there are reasons for people to > get emotional over the issue, as its a mistake at the foundations of > the discipline of mathematics. > Its an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I dont disagree with it. > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Lets look in detail at that troublesome third term, > Its not troublesome. Its asymmetrical with regard to the rest. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now youre going to cheat, and Im going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. > You say nonunit above as if that applies globally, when youre going > to rely on non-unit status in the ring of algebraic integers. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. > Now you can multiply with that unit, and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But its only a non-unit in the ring of algebraic integers because > its not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > Notice that the poster didnt say, in the ring of algebraic integers. > Its important that you note that the claim is in the ring of > algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > The poster is apparently acknowledging that 22 properly is coprime to > 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the Ôgreatest common divisor function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. > Essentially the ws the poster is trying to use are unit factors, but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. > Its a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the as satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. > Well, yes, its possible. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That > Notice the poster doesnt even try at this point, simply resorting to > hand-waving by asserting that the constant terms dont need to be > respected! > The problem is that if you have a constant term that is 7, another > that is 7, and one thats 22, then to get rid of the 7s, you have to > divide out by 7, but thats algebra thats inconvenient to the poster! > So instead the poster just TELLS you that the math doesnt care. > Lets watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > Again, why? Isnt the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isnt. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else its actually a function of x. What nonsense! Nora Baron said no such thing!! She said YOU fail to see that 22/w_3(x) is NOT the constant term of (r_2(x) + 22)/w_3(x). Why? because w_3(x) is a function of x. > So the full assertion is that the constant term is NOT CONSTANT, but > is instead a function of x. The full assertion is that 22/w_3(x) is NOT CONSTANT, but is instead a function of x. (DUH) KeithK > Remember, when people try to fight mathematics they have to at some > point rely on something that is just wacky. Here you can see that > ultimately the poster is trying to get you to believe that a constant > is in fact a function of x. > But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are > algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = > P(0), and notice, you dont have any functions of x, as x has been set > to 0. > So the poster is using unit factors, which by a technicality are not > units in the ring of algebraic integers, unless you do wish to believe > that mathematics is inconsistent, and so what? Your belief would be > wrong. > Again: you defined Ôconstant term in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > Nora B. > Well, I say youre using unit factors, which because they are not > roots of a monic polynomial with integer coefficients are not units in > the ring of algebraic integers. > How do you answer? > James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > > That is, for an algebraic integer x, and natural number Ôa where > 1<=a<=n, g_a(x) is an algebraic integer. > > Now consider setting x=0, and notice that gives > > g_1(0) g_2(0)...g_n(0) = P(0) > > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and thats not a trick. > > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > > c_1 c_2...c_n = P(0) > > and the cs are necessarily algebraic integers and factors of the > constant term P(0). > > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > > Then I can substitute and have > > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > > Now let > > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > then > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the as are roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > so let > > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > > Setting x = 0 with the cubic defining the as gives > > a^3 - 3a^2 = 0, so two of the as are 0, and one is 3. > > Since indices are arbitrary let the first two equal 0, and I have > > c_1 = 7, c_2 = 7, and c_3 = 22 > > which is consistent with the constant term of P(x), which is 1078. > > But dividing P(x) by 49, gives > > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > > Algebraic integers are roots of monic polynomials with integer > coefficients. > > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the as is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. > > > Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, its not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. > Thats what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > Its not even really complicated, but there are reasons for people to > get emotional over the issue, as its a mistake at the foundations of > the discipline of mathematics. > Its an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. > > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I dont disagree with it. > > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > > Lets look in detail at that troublesome third term, > Its not troublesome. Its asymmetrical with regard to the rest. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now youre going to cheat, and Im going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. > You say nonunit above as if that applies globally, when youre going > to rely on non-unit status in the ring of algebraic integers. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. > Now you can multiply with that unit, and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But its only a non-unit in the ring of algebraic integers because > its not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > > Notice that the poster didnt say, in the ring of algebraic integers. > Its important that you note that the claim is in the ring of > algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > > The poster is apparently acknowledging that 22 properly is coprime to > 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > > Suppose I divide the latter expression through by w: I get > > 5 a_3(x)/w + 7/w. > > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > > Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > > 5 a_3(x)/w + 7/w > > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > > Can such a w be defined ? > > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > > w_3(x) = GCD(a_3(x), 7). > > Here GCD denotes the Ôgreatest common divisor function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > > Obviously these definitions imply that all of > > (5 a_1(x) + 7)/w_1(x), > > (5 a_2(x) + 7)/w_2(x), and > > (5 a_3(x) + 7)/w_3(x) > > are algebraic integers. > > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. > Essentially the ws the poster is trying to use are unit factors, but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. > Its a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > > w_1(x) * w_2(x) * w_3(x) = 49. > > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > > 2401 x^3 - 147 x^2 + 3x > > is coprime to 7 [except when x itself is divisible by 7], and > > -49*(2401 x^3 - 147 x^2 + 3x) > > is the constant term of the polynomial that the as satisfy, as > you note above. > > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > > What this shows is that it IS possible to divide 49 out of > > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > > in such a way that each factor is an algebraic integer. > > Well, yes, its possible. > But what about the constant terms ? > > It is *not required* that the constant terms be respected. That > Notice the poster doesnt even try at this point, simply resorting to > hand-waving by asserting that the constant terms dont need to be > respected! > The problem is that if you have a constant term that is 7, another > that is 7, and one thats 22, then to get rid of the 7s, you have to > divide out by 7, but thats algebra thats inconvenient to the poster! > So instead the poster just TELLS you that the math doesnt care. > Lets watch to see what else this poster tries on you. > is, even though it is true that > > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isnt the product of the constant terms equal to the > constant term of the product ??? > > Yes. But here is the last key fact. The constant term of > > (r_2(x) + 22)/w_3(x) > > is NOT what you think it is. You think it must be > > 22/w_3(x). > > It isnt. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > > 22/w_3(0). > > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else its actually a function of x. The poster has asserted that the contant term is 22/w_3(0). The poster has asserted explicitely that the constant term is not 22/w_3(x). 22/w_3(0) is not does not vary with x, it has the same value if x=0, x=10, x=pi x = whatever. -William Hughes === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > > g_1(x) g_2(x)...g_n(x) = P(x) > > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. [snip to save space ...] > > Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. No, its not the *appearance* of a contradiction. It is just an outright contradiction - unless someone, somewhere has made a mistake. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, its not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. Overrate ??? The set of algebraic integers has a clear, unambiguous definition. There is a textbook proof that it forms a ring. It is not a field; it is simply a natural extension of the integers. It has some key properties: for example, every algebraic number can be written in the form a/n, where a is an algebraic integer and n is an ordinary integer. And it has the property that any two elements have a greatest common divisor. Unlike the integers, it does NOT have the property of factorization into primes. Which of these characteristics would you describe as overrating the ring of algebraic integers? > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. I disagree. If you throw that out, you lose some of the interesting properties that the ring has. Dedekind devoted intense study to it for good reason. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. A matter of opinion, clearly, rather than a matter of fact. > Thats what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > Its not even really complicated, but there are reasons for people to > get emotional over the issue, as its a mistake at the foundations of > the discipline of mathematics. > Its an error in core. The error is in your non-rigorous application of your own definition of constant term, as shown below. > Or it could be that you have a logical error somewhere in what > you have said above. > > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I dont disagree with it. > > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > > Lets look in detail at that troublesome third term, > Its not troublesome. Its asymmetrical with regard to the rest. Spare me the semantics. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now youre going to cheat, and Im going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. No disagreement there. Thats the *definition*. > You say nonunit above as if that applies globally, when youre going > to rely on non-unit status in the ring of algebraic integers. Sorry that I did not qualify every statement with in the algebraic integers. I thought it was clearly implied. In any case, just to be definite, what I meant was: nonunit *in the ring of algebraic integers*. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. Now its your turn to be imprecise, apparently. What is a unit function ??? > Now you can multiply with that unit, A unit in what ring? Any nonzero complex number is a unit in SOME ring; many are nonunits in others. The word unit in the context you are using it here is not well-defined. You must specify: in what ring? > and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But its only a non-unit in the ring of algebraic integers because > its not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > > Notice that the poster didnt say, in the ring of algebraic integers. OK, Ill say it: Let w be a nonunit factor of 7 in the ring of algebraic integers. I think you know that is what I meant in the first place. > Its important that you note that the claim is in the ring of > algebraic integers. Sure. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > > The poster is apparently acknowledging that 22 properly is coprime to > 7. *In the ring of algebraic integers*. That is clearly implied. > And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > > Suppose I divide the latter expression through by w: I get > > 5 a_3(x)/w + 7/w. > > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > > Remember readers, non-unit in the ring of algebraic integers! No quarrel with that. Do tell us when you find where I start cheating. > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > > 5 a_3(x)/w + 7/w > > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > > Can such a w be defined ? > > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > > w_3(x) = GCD(a_3(x), 7). > > Here GCD denotes the Ôgreatest common divisor function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > > Obviously these definitions imply that all of > > (5 a_1(x) + 7)/w_1(x), > > (5 a_2(x) + 7)/w_2(x), and > > (5 a_3(x) + 7)/w_3(x) > > are algebraic integers. > > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. Wrong! Here is where you make your mistake! You define constant term of a function G(x) to be G(0). So here, the constant term of, for example, (r_3(x) + 22)/w_3(x), is (r_3(0) + 22)/w_3(0) = r_3(0)/w_3(0) + 22/w_3(0). No problem at all with *existence* of a constant term !!! YOUR problem is, the constant term is not what you think it is, or want it to be! > Essentially the ws the poster is trying to use are unit factors, They most certainly are not: not in the ring of algebraic integers. Is THIS where you think I am cheating? Dividing algebraic integers by other (nonunit) algebraic integers? Is that cheating? Remember, I am NOT claiming that 22/w_3(x) is an algebraic integer (unless x = 0). > but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. Its not a technicality, unless you regard the very phrase that defines the ring to be a technicality ! > Its a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! The ring has certain properties that I described above, one of which is the existence of greatest-common-divisors of any two of its elements. It is NOT a neat trick or cheating in any way to make use of known properties! The integers, for example, have the property of unique factorization into primes. Would it be *cheating* to use that property to prove a theorem ??? > Now the crucial thing to show is that > > w_1(x) * w_2(x) * w_3(x) = 49. > > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > > 2401 x^3 - 147 x^2 + 3x > > is coprime to 7 [except when x itself is divisible by 7], and > > -49*(2401 x^3 - 147 x^2 + 3x) > > is the constant term of the polynomial that the as satisfy, as > you note above. > > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > > What this shows is that it IS possible to divide 49 out of > > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > > in such a way that each factor is an algebraic integer. > > Well, yes, its possible. concede everything. > But what about the constant terms ? > > It is *not required* that the constant terms be respected. That > Notice the poster doesnt even try at this point, simply resorting to > hand-waving by asserting that the constant terms dont need to be > respected! Keep reading. I explain very clearly what I meant. > The problem is that if you have a constant term that is 7, another > that is 7, and one thats 22, then to get rid of the 7s, you have to > divide out by 7, but thats algebra thats inconvenient to the poster! > So instead the poster just TELLS you that the math doesnt care. > Lets watch to see what else this poster tries on you. > is, even though it is true that > > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isnt the product of the constant terms equal to the > constant term of the product ??? > > Yes. But here is the last key fact. The constant term of > > (r_2(x) + 22)/w_3(x) > > is NOT what you think it is. You think it must be > > 22/w_3(x). > > It isnt. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > > 22/w_3(0). > > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else its actually a function of x. Read the following very, very carefully, and try to absorb something. I am using YOUR definition of constant term. The constant term of r_3(x)/w_3(x) + 22/w_3(x) is NOT 22/w_3(x). By YOUR OWN DEFINITION, it is 22/w_3(0). Got that? > So the full assertion is that the constant term is NOT CONSTANT, but > is instead a function of x. You are missing the point so incredibly thoroughly. The constant term in question is 22/w_3(0). It IS CONSTANT. Yes, 22/w_3(x) is NOT CONSTANT. But also, by your definition, it is NOT THE CONSTANT TERM. You are deeply confused on this point. As William Hughes has pointed out in another thread, your thinking is based on the idea that you can look at a function, and BY INSPECTION tell what the constant term is. You say, the constant term of r_3(x) + 22 is obviously 22. Thats correct. Then you consider r_3(x)/w_3(x) + 22/w_3(x) and you think: hey, 22/w_3(x) is in the POSITION where the constant term ought to be, so it must BE the constant term. In other words, you just throw YOUR OWN DEFINITION out the window, and declare what you believe to be the constant term BY INSPECTION. And its not. You should have used your own definition consistently and RIGOROUSLY throughout. When you deviated from it, you got into trouble and led yourself down the path to this erroneous conclusion. THIS is what has led you to think there is a contradiction, which in your addled logic leads you to believe there is some kind of problem with the ring of algebraic integers. Again, just to be really, really clear about this. The constant term of (r_3(x) + 22)/w_3(x) is NOT what you think it is, i.e., it is NOT 22/w_3(x). No, by YOUR OWN DEFINITION, the constant term is 22/w_3(0). What you need to do is RIGOROUSLY USE that definition. It does not lead you down the path to the wrong conclusion which you have followed previously. > Remember, when people try to fight mathematics they have to at some > point rely on something that is just wacky. Here you can see that > ultimately the poster is trying to get you to believe that a constant > is in fact a function of x. > But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are > algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = > P(0), and notice, you dont have any functions of x, as x has been set > to 0. > So the poster is using unit factors, because you know that 22 and w_3(x) are coprime in the ring of algebraic integers. That is a CORRECT FACT. However, it is not important. The important thing here is that, when you divide the WHOLE EXPRESSION (r_3(x) + 22) by w_3(x), the result IS an algebraic integer. That is what you require in the first place. There is no reason that the individual terms must be algebraic integers. Heres an analogy in integers. Note that 15 = 13 + 2. Divide both sides by 3. On the left, I get 5, an integer. On the right I get 13/3 + 2/3. Neither of these is an integer. So what? Do I conclude that the right side is not divisible by 3? No!!! The SUM of the two parts still equals 5, an integer. The fact that 2/3 is not an integer is NOT A PROBLEM. You can add two nonintegers and get an integer. Similarly, you can add two non-algebraic-integers and get an algebraic integer. Not a problem at all. > which by a technicality are not > units in the ring of algebraic integers, Saying the DEFINING PROPERTY of the ring is a technicality is utterly absurd. You must see that at least. > unless you do wish to believe > that mathematics is inconsistent, and so what? Your belief would be > wrong. I agree, or at least I agree that what we have discussed here does not imply that mathematics is inconsistent. If you were right, it would. But because you have failed to use YOUR OWN DEFINITION of constant term, you are not right, and mathematics is still safe. > Again: you defined Ôconstant term in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > > Nora B. > Well, I say youre using unit factors, which because they are not > roots of a monic polynomial with integer coefficients are not units in > the ring of algebraic integers. > How do you answer? As above: 22/w_3(x) is not an algebraic integer. I dont claim it is, and there is no reason it SHOULD be. It is NOT the constant term of (r_3(x) + 22)/w_3(x), unless x = 0. Or do you actually still think 22/w_3(x) IS the constant term, in spite of your own definition ?? Try it - use YOUR OWN DEFINITION of constant term - tell me what you get! Nora B. > James Harris === Subject: Re: JSH:Understanding constant terms ... > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > > > The poster is apparently acknowledging that 22 properly is coprime to > 7. > *In the ring of algebraic integers*. That is clearly implied. Actually, with most reasonable definitions of coprime, in any ring that contains both 7 and 22, they are coprime. Standard definition: a and b are coprime if p and q in the ring can be found such that p*a + q*b = 1. Alternative definition: a and b are coprime if the have no common non-unit factor in the ring. With both definitions 7 and 22 are coprime regardless of the ring involved. I have no idea what James means with properly coprime. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH:Understanding constant terms > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. And *that* is precisely what you have done. You have imposed demands on the ring of algebraic integers which are inconsistent with their properties. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. That requirement defines a perfectly valid ring. Whether it is important or not depends on whether it is useful in some mathematical context or not. In your application it is useless. That doesnt speak to the interests of others. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. No, of course not. Rational numbers may not meet that requirement. If your requirements are not satisfied by algebraic integers, dont use them. [snip irrelevant and redundant exposition of fallacious arguments] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH:Understanding constant terms Discussion, linux) > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > Its not even really complicated, but there are reasons for people to > get emotional over the issue, as its a mistake at the foundations of > the discipline of mathematics. > Its an error in core. Er, when did algebraic number theory or Galois theory become part of the foundations of mathematics? I missed that memo. Probably, they dropped me from the mailing list when I moved to a philosophy section thats part of a management department in a technical university. -- Ive been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so Ive gone to marketing books. -- James S. Harris, on when mathematics isnt enough === Subject: Re: final proof Einstein 1905 is crap. > Nice one, Mike. > Androcles I tried hard to incorporate every known formal/informal fallacy exactly like the originator of the thread has done. :) But speaking about fallacies, I think thw Ônon sequitur is the most serious one, that is the conclusion does not follow from the premises, a disconected idea. Mike === Subject: Re: final proof Einstein 1905 is crap. >> Nice one, Mike. >> Androcles > I tried hard to incorporate every known formal/informal fallacy > exactly like the originator of the thread has done. :) > But speaking about fallacies, I think thw Ônon sequitur is the most > serious one, that is the conclusion does not follow from the premises, > a disconected idea. > Mike Actually the commonest fallacy in Special Relativity is to call a conclusion a postulate via the fallacy of circularity, and that is pretty serious. The velocity of light is c in all inertial frames is bandied about as if it were common knowledge, but stems from V = (c+w)/(1+ w/c), and that was a conclusion Einstein drew in section 5 with the help of the equations of transformation developed in section 3 (i.e. the Lorentz transformations) from which he extracts sqrt(1-v^2/c^2). Ref. http://www.fourmilab.ch/etexts/einstein/specrel/www/ The idiots claim to be able to derive the LTs from a conclusion that was itself derived from the LT and circularity is thus established. In fact the original postulate ( which by definition of postulate it cannot be) is light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. The reason it cannot be a postulate is that it is not intuitive, it is testable, and it has no experimental data to support it. It is an hypothesis. That it reaches absurd conclusions indicates it to be a failed hypothesis. Reading Einsteins paper carefully, it becomes obvious he was perpetrating a hoax. He begins with Galilean Relativity, using a magnet and a conductor as an example, carefully avoiding any explicit description but vaguely referring to laws that hold good, claims that although it is irreconcilable with his second hypothesis this is only apparent , further claims that the velocity of light in our theory plays the part, physically, of an infinitely great velocity and produces a mathematical fiction, xi = x/sqrt(1-v^2/c^2) ..... (recall that x = x-vt) The reason Einstein so carefully avoided stating the PoR is that he makes use of it when he says But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v... so he must be careful not to be too explicit in the introduction and preceding sections. To conclude It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain V = (c+w)/(1+w/c) = c. denies him the use of the ray moves [at] c-v... without which he cannot derive the Lorentz transforms. Shubert makes the same fallacy of circularity in his rather long-winded but actually quite simple minded derivation, although he now admits pulling sqrt(1-v^2/c^2) from thin air. He is nothingimportant, of course, although he likes to think he is everythingimportant. Egomania is treatable, I understand, although it may take the services of a trained psychiatrist. Androcles === Subject: Re: final proof Einstein 1905 is crap. > Actually the commonest fallacy in Special Relativity is to call a > conclusion > a postulate via the fallacy of circularity, and that is pretty serious. Well, you are hetting into really dark domains here. Before getting as dvanced as SR, the use of circularity was enforced by those who missinterpreted Newton, or actually desired to present Newtons metaphysics as a science. To be explicit: Newton used a modified version of Keplers third law in order to derive the law of universal gravitation. Please note that without such law, which was purely empirical, the derivation of LUG is impossible. Then, in mechanics books you see the famous Kepler problem and the derivation of Keplers third law using the LUG as given! But of course, if the conclusion is true, all premised in the deduction process must be true otherwise the deduction was unsound in the first place. The same of course hold with Einstein and some claims the constancy of c can be derived , etc. Let me tell you what it boild down to: Math infidels with no understanding of the principles of logic the Greeks established and Aristotle put in a massive organized works called the Organon, have taken over and their misunderstanding and paranoia has resulted in the chaos we see in science todat with inductions presented as deductions, deductions as induction, abductions as inductions and deductions. Question of the day: how to you distinguish between paranoid and genious? Often, both sound as equally ntelligent. My answer: Paranoids are easy to spot is you understand formal and informal fallacies and the ways they can be concealed in what appears to be a sound inference. Watch for: Non sequitur Circularity affirmation of the consequent denial of the antecedant false analogy red herring faulty causality faulty generalization Straw man etc. I belive Einstein has concealed all of the above fallacies in what is a masterpiece of paranoid thinking to result in a circular theory where all predictions will match observations but no prediction will result that will alter the fundamental basis of the theory which is: paranoia. Mike === Subject: Re: final proof Einstein 1905 is crap. Mike: >Newton used a modified version of Keplers third law in order to >derive the law of universal gravitation. Please note that without such >law, which was purely empirical, the derivation of LUG is impossible. Newton did not derive his law of gravitation, nor can it be derived from keplers laws. Newton deduced a general principle from keplers laws and if that principle was to stand a chance of being correct, the first requirement is that keplers laws must be derivable from that principle. >Then, in mechanics books you see the famous Kepler problem and the >derivation of Keplers third law using the LUG as given! Keplers laws _can_ be derived from newtons law of gravity. Its called consistency, not circularity. Newtons law if gravity _cannot_ be derived from keplers laws. Keplers laws say nothing about forces or potentials or even why the planetary motion should be described by his laws. By contrast, the 1/r potential newton proposed can be used to show that stable orbits only occur for forces of the form r^n, where n = +/- 2. >But of course, if the conclusion is true, all premised in the >deduction process must be true otherwise the deduction was unsound in >the first place. >The same of course hold with Einstein and some claims the constancy of >c can be derived , etc. The assertion that `c is a constant cannot be derived, since obviously its possible to create a self-consistent theory of mechanics in which c -> infty, called galilean relativity. Distinguishing between special relativity and galilean relativity is an experimental issue and since galilean relativity makes predictions that are incompatible with what is observed, experiment has determined that there exists some velocity, `c, which is finite. Electromagnetic theory predicts that `c and the speed of light must be the same. [..] >I belive Einstein has concealed all of the above fallacies in what is >a masterpiece of paranoid thinking to result in a circular theory >where all predictions will match observations but no prediction will >result that will alter the fundamental basis of the theory which is: >paranoia. Special relativity has been used to make quite a number of predictions. For example, the standard model is based on special relativity. If relativity was wrong, the standard model would have had zero chance of correctly predicting the mass ratio between the W and Z or the myriad of other predictions it make correctly. While there is more to the standard model than relativity, relativity is the foundation upon which the standard model is constructed. === Subject: Re: final proof Einstein 1905 is crap. >> Actually the commonest fallacy in Special Relativity is to call a >> conclusion >> a postulate via the fallacy of circularity, and that is pretty >> serious. > Well, you are hetting into really dark domains here. Before getting as > dvanced as SR, the use of circularity was enforced by those who > missinterpreted Newton, or actually desired to present Newtons > metaphysics as a science. To be explicit: > Newton used a modified version of Keplers third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Keplers third law using the LUG as given! > But of course, if the conclusion is true, all premised in the > deduction process must be true otherwise the deduction was unsound in > the first place. Hold on a moment. Keplers third law is empirical, youve said. Thats originating in or based on observation or experience. Keplers third law (however modified) is The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes: (P1/P2)^2 = (R1/R2)^3 Are you saying this is NOT what is observed? All I see here is that a simple equation has been determined to describe observation. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. That is a very different matter. Nobody has ever observed light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. There is no empirical foundation to it. In fact the empirical data contradicts it, aside from any logical deduction. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. Youll get no argument from me on that. > Question of the day: how to you distinguish between paranoid and > genious? Often, both sound as equally ntelligent. > My answer: Paranoids are easy to spot is you understand formal and > informal fallacies and the ways they can be concealed in what appears > to be a sound inference. Watch for: > Non sequitur > Circularity > affirmation of the consequent > denial of the antecedant > false analogy > red herring > faulty causality > faulty generalization > Straw man > etc. > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. To you he was paranoid. You are looking into the cause of his ill-concieved ideas and attributing it to a form of mental illness. To me he was a huckster. Without getting into psychology or psychiatry, fields in which I do not claim expertise, for maybe thieves and murderers suffer from mental abberations and it can all be blamed on an abusive childhood, nevertheless I want them locked away in a correctional institution or be taken out of society if it can be justly determined that they are guilty of crime. It is too late to correct Einstein the man, hes lived and died. We are still obligated to eradicate the consequences of his fraud, whatever the cause. Therefore I have to say your analysis is actually --- Non sequitur. But then, so is mine. Whatever the cause, it is our duty to promote freedom of thought in the next generation of budding young physicists and to dissuade them from the indoctrinators that infest this newsgroup. Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. > Actually the commonest fallacy in Special Relativity is to call a >> conclusion >> a postulate via the fallacy of circularity, and that is pretty >> serious. >> Well, you are hetting into really dark domains here. Before getting as > dvanced as SR, the use of circularity was enforced by those who > missinterpreted Newton, or actually desired to present Newtons > metaphysics as a science. To be explicit: > Newton used a modified version of Keplers third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Keplers third law using the LUG as given! > But of course, if the conclusion is true, all premised in the > deduction process must be true otherwise the deduction was unsound in > the first place. > Hold on a moment. Keplers third law is empirical, youve said. > Thats originating in or based on observation or experience. > Keplers third law (however modified) is > The ratio of the squares of the revolutionary periods for two planets > is equal to the ratio of the cubes of their semimajor axes: > (P1/P2)^2 = (R1/R2)^3 > Are you saying this is NOT what is observed? > All I see here is that a simple equation has been determined to describe > observation. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. > That is a very different matter. Nobody has ever observed light is > always propagated in empty space with a definite velocity c which is > independent of the state of motion of the emitting body. > There is no empirical foundation to it. > In fact the empirical data contradicts it, aside from any logical > deduction. Not that hard to do! Use the standard apparatus for testing light velocity, the two, toothed wheels, form which c can be calculated from the difference in their rotation and separation when a beam is shone through. Now stick it on a rocket, and test the sunlight speed when the rocket is travelling away/towards, the sun. Probably cost a few million on the next interplanetary run- FAR too expensive for the DHR budget! (make that too likely to embarass) > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > Youll get no argument from me on that. > Question of the day: how to you distinguish between paranoid and > genious? Often, both sound as equally ntelligent. > My answer: Paranoids are easy to spot is you understand formal and > informal fallacies and the ways they can be concealed in what appears > to be a sound inference. Watch for: > Non sequitur > Circularity > affirmation of the consequent > denial of the antecedant > false analogy > red herring > faulty causality > faulty generalization > Straw man > etc. > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > To you he was paranoid. > You are looking into the cause of his ill-concieved ideas and > attributing > it to a form of mental illness. > To me he was a huckster. > Without getting into psychology or psychiatry, fields in which I do not > claim expertise, for maybe thieves and murderers suffer from mental > abberations and it can all be blamed on an abusive childhood, > nevertheless > I want them locked away in a correctional institution or be taken > out of society if it can be justly determined that they are guilty of > crime. > It is too late to correct Einstein the man, hes lived and died. > We are still obligated to eradicate the consequences of his fraud, > whatever the cause. > Therefore I have to say your analysis is actually --- Non sequitur. > But then, so is mine. Whatever the cause, it is our duty to promote > freedom of thought in the next generation of budding young physicists > and to dissuade them from the indoctrinators that infest this newsgroup. > Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. >> > Actually the commonest fallacy in Special Relativity is to call a > conclusion > a postulate via the fallacy of circularity, and that is pretty > serious. > Well, you are hetting into really dark domains here. Before getting >> as >> dvanced as SR, the use of circularity was enforced by those who >> missinterpreted Newton, or actually desired to present Newtons >> metaphysics as a science. To be explicit: > Newton used a modified version of Keplers third law in order to >> derive the law of universal gravitation. Please note that without >> such >> law, which was purely empirical, the derivation of LUG is >> impossible. > Then, in mechanics books you see the famous Kepler problem and the >> derivation of Keplers third law using the LUG as given! > But of course, if the conclusion is true, all premised in the >> deduction process must be true otherwise the deduction was unsound >> in >> the first place. >> Hold on a moment. Keplers third law is empirical, youve said. >> Thats originating in or based on observation or experience. >> Keplers third law (however modified) is >> The ratio of the squares of the revolutionary periods for two >> planets >> is equal to the ratio of the cubes of their semimajor axes: >> (P1/P2)^2 = (R1/R2)^3 >> Are you saying this is NOT what is observed? >> All I see here is that a simple equation has been determined to >> describe >> observation. > The same of course hold with Einstein and some claims the constancy >> of >> c can be derived , etc. >> That is a very different matter. Nobody has ever observed light is >> always propagated in empty space with a definite velocity c which is >> independent of the state of motion of the emitting body. >> There is no empirical foundation to it. >> In fact the empirical data contradicts it, aside from any logical >> deduction. > Not that hard to do! > Use the standard apparatus for testing light velocity, the two, > toothed wheels, form which c can be calculated from the difference in > their rotation and separation when a beam is shone through. Now stick > it on a rocket, and test the sunlight speed when the rocket is > travelling away/towards, the sun. > Probably cost a few million on the next interplanetary run- FAR too > expensive for the DHR budget! (make that too likely to embarass) The easiest way is Roemers method and Cassini. All it takes is data gathering. The DHRs would say time was increasing and decreasing during the orbit, but it wouldnt be all that convincing. Androcles > Let me tell you what it boild down to: Math infidels with no >> understanding of the principles of logic the Greeks established and >> Aristotle put in a massive organized works called the Organon, have >> taken over and their misunderstanding and paranoia has resulted in >> the >> chaos we see in science todat with inductions presented as >> deductions, >> deductions as induction, abductions as inductions and deductions. >> Youll get no argument from me on that. > Question of the day: how to you distinguish between paranoid and >> genious? Often, both sound as equally ntelligent. > My answer: Paranoids are easy to spot is you understand formal and >> informal fallacies and the ways they can be concealed in what >> appears >> to be a sound inference. Watch for: > Non sequitur >> Circularity >> affirmation of the consequent >> denial of the antecedant >> false analogy >> red herring >> faulty causality >> faulty generalization >> Straw man >> etc. >> I belive Einstein has concealed all of the above fallacies in what >> is >> a masterpiece of paranoid thinking to result in a circular theory >> where all predictions will match observations but no prediction >> will >> result that will alter the fundamental basis of the theory which >> is: >> paranoia. > To you he was paranoid. >> You are looking into the cause of his ill-concieved ideas and >> attributing >> it to a form of mental illness. >> To me he was a huckster. >> Without getting into psychology or psychiatry, fields in which I do >> not >> claim expertise, for maybe thieves and murderers suffer from mental >> abberations and it can all be blamed on an abusive childhood, >> nevertheless >> I want them locked away in a correctional institution or be taken >> out of society if it can be justly determined that they are guilty of >> crime. >> It is too late to correct Einstein the man, hes lived and died. >> We are still obligated to eradicate the consequences of his fraud, >> whatever the cause. >> Therefore I have to say your analysis is actually --- Non sequitur. >> But then, so is mine. Whatever the cause, it is our duty to promote >> freedom of thought in the next generation of budding young physicists >> and to dissuade them from the indoctrinators that infest this >> newsgroup. >> Androcles. >> Mike === Subject: Re: final proof Einstein 1905 is crap. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. > > That is a very different matter. Nobody has ever observed light is > always propagated in empty space with a definite velocity c which is > independent of the state of motion of the emitting body. Androcles is being a retard, as per usual. > There is no empirical foundation to it. More of Androcles not reading the literature, as per usual. > In fact the empirical data contradicts it, aside from any logical > deduction. More of Androcles mistakening his suppositions for reality, as per usual. > Not that hard to do! > Use the standard apparatus for testing light velocity, the two, > toothed wheels, form which c can be calculated from the difference in > their rotation and separation when a beam is shone through. Now stick > it on a rocket, and test the sunlight speed when the rocket is > travelling away/towards, the sun. > Probably cost a few million on the next interplanetary run- FAR too > expensive for the DHR budget! (make that too likely to embarass) Since you have no idea what goes into designing a mission package, such ignorant statements come as no surprise. Why dont you read the literature? Have you ever considered the possibility that you are wrong? === Subject: Re: final proof Einstein 1905 is crap. Jim Greenfield: >Not that hard to do! >Use the standard apparatus for testing light velocity, the two, >toothed wheels, form which c can be calculated from the difference in >their rotation and separation when a beam is shone through. Now stick >it on a rocket, and test the sunlight speed when the rocket is >travelling away/towards, the sun. >Probably cost a few million on the next interplanetary run- FAR too >expensive for the DHR budget! (make that too likely to embarass) Ill be happy to perform the experiment if you pay for it. My guess is that you arent willing to back up your beliefs with cash. === Subject: Re: final proof Einstein 1905 is crap. [snip > Hold on a moment. Keplers third law is empirical, youve said. > Thats originating in or based on observation or experience. > Keplers third law (however modified) is > The ratio of the squares of the revolutionary periods for two planets > is equal to the ratio of the cubes of their semimajor axes: > (P1/P2)^2 = (R1/R2)^3 > Are you saying this is NOT what is observed? > All I see here is that a simple equation has been determined to describe > observation. Actually thats the observed law. Newton did not use that because it will gets you nowhere. Newton used kmT^2 = R^3 where k is a constant. This is not mentioned in most books. The constant k ends up part of G in the LUG. This is what Newton did, he assumed that in the two-body problem: k(m1+m2)T^2 = (R1+R2)^3=R^3 He then assumed that m1>m2 to get: k1m1T^2 ~ R^3 and he assumed again that this hold for all planets so that: km1T1^2 = R1^3 km2T2^2 = R2^3 divide and you get Keplers observed law! This series of assumptions by Newton hides the hypothesis that one of the bodies is fixed in absolute space and is immovable, and that is the SUN. He actually states that hypothesis I: HYPOTHESIS I. That the center of the system of the world is immovable. This is acknowledged by all, while some contend that the earth, others that the sun is fixed in that center. Let us see what may from hence follow. PROPOSITION XI. THEOREM XI. That the common center of gravity of the earth, the sun, and all the planets, is immovable. LOL Without such hypothesis, derivation of the LUG is impossible. Now, math infidels such as Newton and Einstein did not realize that unless bodies trajectories are fixed in space by a magical Ôspirit (actually Newton believed in such spirit) their gravitational dynamics lead to a collapse of the universe in a single point over time. This is why Einstein wanted to include the cosmological constant to prevent that. Yet, everybody is deceived to believe that the derived theories explain gravitation but is deprived of the underline assumptions that basically invoke magic. Most books attempt to circumvent the above using hand waiving arguments and Kolker eats them for breakfast. Yet, what is, is. Mike === Subject: Re: final proof Einstein 1905 is crap. > [snip >> Hold on a moment. Keplers third law is empirical, youve said. >> Thats originating in or based on observation or experience. >> Keplers third law (however modified) is >> The ratio of the squares of the revolutionary periods for two >> planets >> is equal to the ratio of the cubes of their semimajor axes: >> (P1/P2)^2 = (R1/R2)^3 >> Are you saying this is NOT what is observed? >> All I see here is that a simple equation has been determined to >> describe >> observation. >> Actually thats the observed law. Newton did not use that because it > will gets you nowhere. Newton used kmT^2 = R^3 where k is a constant. > This is not mentioned in most books. The constant k ends up part of G > in the LUG. This is what Newton did, he assumed that in the two-body > problem: > k(m1+m2)T^2 = (R1+R2)^3=R^3 > He then assumed that m1>m2 to get: > k1m1T^2 ~ R^3 > and he assumed again that this hold for all planets so that: > km1T1^2 = R1^3 > km2T2^2 = R2^3 > divide and you get Keplers observed law! > This series of assumptions by Newton hides the hypothesis that one of > the bodies is fixed in absolute space and is immovable, and that is > the SUN. He actually states that hypothesis I: > HYPOTHESIS I. > That the center of the system of the world is immovable. > This is acknowledged by all, while some contend that the earth, others > that the sun is fixed in that center. Let us see what may from hence > follow. Einstein does the same thing, calling it the stationary system. However, Newton is careful to call an hypothesis by its name, but Einstein calls his own light speed hypothesis a postulate. > PROPOSITION XI. THEOREM XI. > That the common center of gravity of the earth, the sun, and all the > planets, is immovable. > LOL > Without such hypothesis, derivation of the LUG is impossible. Granted that Newton envisaged a heliocentric universe, nevertheless the solar system can be treated as having a common centre of gravity. He didnt know of the existence of Pluto, the orbit of which is Sol dependent. Moreover, any perturbation of the orbits of the planets by the nearest star are neglible. There is a considerable difference between Pluto being circa 4 light hours from Sol and circa 4 light years from Proxima Centauri, particularly when R^3 is taken into account. (24 * 365.25 )^3 ~= 5,000,000. > Now, math infidels such as Newton and Einstein did not realize that > unless bodies trajectories are fixed in space by a magical Ôspirit > (actually Newton believed in such spirit) their gravitational dynamics > lead to a collapse of the universe in a single point over time. A trajectory cannot be fixed in space by definition of trajectory, which implies motion. I Ôve missed the point you are attempting to make. Until such times as our solar system enters the vicinity of a massive body such as another star, Ill go along with Newtons description, it is an adequate approximation. > This > is why Einstein wanted to include the cosmological constant to prevent > that. > Yet, everybody is deceived to believe that the derived theories > explain gravitation but is deprived of the underline assumptions that > basically invoke magic. Action at a distance IS magic, in that we do not understand it. However, we do observe it and Newton gives a good approximation that will adequately predict the future positions of the planets, even though it will ultimately be chaotic. > Most books attempt to circumvent the above using hand waiving > arguments and Kolker eats them for breakfast. Yet, what is, is. Why bother with Kolker? Hes just another parrot, repeating what hes read. Hes not capable of actually thinking, few are. There are three distinct categories of people that write to this newsgroup. 1) Parrots that can only repeat what theyve read. 2) Egocentrics that want to promote their own pet theory. 3) Rational thinkers. There are three mutually exclusive concepts that compete. a) the velocity of light is medium dependent b) the velocity of light is source dependent c) the velocity of light is observer dependent Kolker is a 1c) Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. > Newton used a modified version of Keplers third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. Not true. A central force with a 1/r potential will give you all of the Keplarian laws. It is quite true that Keplers third law -suggest- to Newton that the inverse square law is correct, but Kepler independent ways of deriving the famous inversesquare law plus all of Keplers empirical laws exist as I have indicated. Refere to Goldsteins book on Mechanics for the mathematical details. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Keplers third law using the LUG as given! Not true, as I have shown. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. Aristotles logic is a restricted subset of first order logic. Furthermore the syllogism is not the ideal form of mathematical proof structure. The Stoics* developed a form of logic** that is much closer to first order logic than Aristotle. Unfortunately, most of their works in logic have not survived. For details refer to -The Development of Logic- by Kneale and Kneale (hom et ux). > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. Utter nonsense. SR is potentially falsifiable. Later ways of expressing SR by means of Minkowski geometry show how stragight forward SR is. The group of transformation (Lorentz Group) which is the bases of SR were established first and independently by Lorentz and Poincare so there are no mathematical mysteries here. Bob Kolker * especially Chrysippus ** modus ponens > Mike === Subject: Re: final proof Einstein 1905 is crap. > > Newton used a modified version of Keplers third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Not true. A central force with a 1/r potential will give you all of the > Keplarian laws. It is quite true that Keplers third law -suggest- to > Newton that the inverse square law is correct, but Kepler independent > ways of deriving the famous inversesquare law plus all of Keplers > empirical laws exist as I have indicated. Refere to Goldsteins book on > Mechanics for the mathematical details. > > Then, in mechanics books you see the famous Kepler problem and the > derivation of Keplers third law using the LUG as given! > Not true, as I have shown. > > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > Aristotles logic is a restricted subset of first order logic. > Furthermore the syllogism is not the ideal form of mathematical proof > structure. The Stoics* developed a form of logic** that is much closer > to first order logic than Aristotle. Unfortunately, most of their works > in logic have not survived. For details refer to -The Development of > Logic- by Kneale and Kneale (hom et ux). > > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > Utter nonsense. SR is potentially falsifiable. Later ways of expressing > SR by means of Minkowski geometry show how stragight forward SR is. The > group of transformation (Lorentz Group) which is the bases of SR were > established first and independently by Lorentz and Poincare so there are > no mathematical mysteries here. Just the biggie! HOW does photon A, emitted from a source stationary ref us, KNOW how fast B, emitted from a moving source at that location, is travelling????????????? Jim G c=c+v > Bob Kolker > * especially Chrysippus > ** modus ponens > > > Mike === Subject: Re: final proof Einstein 1905 is crap. Jim Greenfield: >Just the biggie! >HOW does photon A, emitted from a source stationary ref us, KNOW how >fast B, emitted from a moving source at that location, is >travelling????????????? Why would a photon have to know that? === Subject: Re: final proof Einstein 1905 is crap. > Newton used a modified version of Keplers third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Not true. A central force with a 1/r potential will give you all of the > Keplarian laws. It is quite true that Keplers third law -suggest- to > Newton that the inverse square law is correct, but Kepler independent > ways of deriving the famous inversesquare law plus all of Keplers > empirical laws exist as I have indicated. Refere to Goldsteins book on > Mechanics for the mathematical details. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Keplers third law using the LUG as given! > Not true, as I have shown. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > Aristotles logic is a restricted subset of first order logic. > Furthermore the syllogism is not the ideal form of mathematical proof > structure. The Stoics* developed a form of logic** that is much closer > to first order logic than Aristotle. Unfortunately, most of their works > in logic have not survived. For details refer to -The Development of > Logic- by Kneale and Kneale (hom et ux). > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > Utter nonsense. SR is potentially falsifiable. Not for religious fanatics like Mike (aka Bill Smith, Undeniable, Eleatis etc...) > Later ways of expressing > SR by means of Minkowski geometry show how stragight forward SR is. The > group of transformation (Lorentz Group) which is the bases of SR were > established first and independently by Lorentz and Poincare so there are > no mathematical mysteries here. You are wasting your time with this idiot. Dirk Vdm === Subject: Re: final proof Einstein 1905 is crap. > Shubert makes the same fallacy of circularity in his rather long-winded > but actually quite simple minded derivation, although he now admits > pulling sqrt(1-v^2/c^2) from thin air. I missed his derivation but have noted here that Tom Roberts derivation involves hand-waving when he reaches his penultimate step: the evaluation of a criterion, value one for Newton, over or under for SR and ... polar coordinates? (Dont remember because I didnt actually read that part.) The Newton part was automatic, the SR parts logic trail disappeared and he just said, well make it this particular value and we have SR. SR-cult cretins here have agreed Einsteins 1916/1962 (lets hear it from the idiots) did really idiotic things. AE 1905 was just as bad but less obvious. Ive seen a number of so-called derivations that started with the conclusion. eleaticus He is nothingimportant, of > course, although he likes to think he is everythingimportant. Egomania > is treatable, I understand, although it may take the services of a > trained psychiatrist. > Androcles === Subject: Re: final proof Einstein 1905 is crap. >> Shubert makes the same fallacy of circularity in his rather >> long-winded >> but actually quite simple minded derivation, although he now admits >> pulling sqrt(1-v^2/c^2) from thin air. > I missed his derivation but have noted here that Tom Roberts > derivation > involves hand-waving when he reaches his penultimate step: the > evaluation of > a criterion, value one for Newton, over or under for SR and ... polar > coordinates? (Dont remember because I didnt actually read that > part.) The > Newton part was automatic, the SR parts logic trail disappeared and > he just > said, well make it this particular value and we have SR. Ive carved Roberts up big time in the thread Roberts considers Einstein paper to be irrelevant verbiage. He really is a sad case of fanaticism. Androcles > SR-cult cretins here have agreed Einsteins 1916/1962 (lets hear it > from > the idiots) did really idiotic things. AE 1905 was just as bad but > less > obvious. > Ive seen a number of so-called derivations that started with the > conclusion. > eleaticus > He is nothingimportant, of >> course, although he likes to think he is everythingimportant. >> Egomania >> is treatable, I understand, although it may take the services of a >> trained psychiatrist. >> Androcles === Subject: Re: final proof Einstein 1905 is crap. eleaticus: >Ive seen a number of so-called derivations that started with the >conclusion. on a regular basis. hello.....doctor~ The set of all complex numbers with the usual topology is both separable and 2nd countable. ------------------------------------------------ this is true. because, f : C -> R X R, f(x+yi) = (x,y) so, C and R^2 is homeomorphic. and countable basis of R^2 is {U X V | U=(a,b),V=(c,d), a,b,c,d in Q} so, 2nd countable. so, separable.... um.......right ?? thank you very much for your advice. > ------------------------------------------------ > this is true. > because, f : C -> R X R, f(x+yi) = (x,y) > so, C and R^2 is homeomorphic. > and countable basis of R^2 is > {U X V | U=(a,b),V=(c,d), a,b,c,d in Q} This set is not a basis. but this set is dense in R^2 and countable. Therefore, this set shows that R^2 is separable. Furthermore, Since R^2 is metrizable, R^2 is 2nd countable. (for its known that a metrizable and separable space is second countable.) > so, 2nd countable. so, separable.... > um.......right ?? > thank you very much for your advice. pointed out. I misreaded the set as something else. mina_worlds argument is right. >>------------------------------------------------ >>this is true. >>because, f : C -> R X R, f(x+yi) = (x,y) >>so, C and R^2 is homeomorphic. >>and countable basis of R^2 is >>{U X V | U=(a,b),V=(c,d), a,b,c,d in Q} > This set is not a basis. but this set is dense in R^2 and countable. > Therefore, this set shows that R^2 is separable. > Furthermore, Since R^2 is metrizable, R^2 is 2nd countable. (for its known > that a metrizable and separable space is second countable.) >>so, 2nd countable. so, separable.... >>um.......right ?? >>thank you very much for your advice. How is this set not a basis? this is the set of all products of open intervals with rational endpoints. if B1 is a basis for X and B2 is a basis for Y then the set {AxB|A is in B1,B is in B2} is a basis for XxY. >>------------------------------------------------ >>this is true. >>because, f : C -> R X R, f(x+yi) = (x,y) >>so, C and R^2 is homeomorphic. >>and countable basis of R^2 is >>{U X V | U=(a,b),V=(c,d), a,b,c,d in Q} > This set is not a basis. but this set is dense in R^2 and countable. > Therefore, this set shows that R^2 is separable. > Furthermore, Since R^2 is metrizable, R^2 is 2nd countable. (for its known > that a metrizable and separable space is second countable.) Perhaps its been too long and my recollection is faulty, but if (x,y) is in R^2, and U is an open set of the plane containing (x,y), then isnt there a product (a,b)x(c,d) of intervals with rational endpoints, within U, that contains (x,y)? Surely, if there is a ball around (x,y) with positive radius within U, there is a box with rational vertices as well. Doesnt that count as forming a basis? >>so, 2nd countable. so, separable.... >>um.......right ?? >>thank you very much for your advice. Dale > hello.....doctor~ > The set of all complex numbers with the usual topology > is both separable and 2nd countable. > ------------------------------------------------ > this is true. > because, f : C -> R X R, f(x+yi) = (x,y) > so, C and R^2 is homeomorphic. > and countable basis of R^2 is > {U X V | U=(a,b),V=(c,d), a,b,c,d in Q} > so, 2nd countable. so, separable.... Your proof of second countability is correct. Separability means the existence of a countable dense subset. For R^2, Q^2 naturally embedded is both dense and countable. Igor === Subject: Function naming Im doing some study on a perticular type of functions (in my spare time just for fun). The problem is however that I dont know how these functions are called in English. Could someone provide me with the correct english term and point me perhaps to some more info on this topic. The functions are of the form; y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 This function has two input variables and has an order of two. The number of variables and order can vary accordingly ofcourse. Is there a general term for this kind of functions? === Subject: Re: Function naming > Im doing some study on a perticular type of functions (in my spare time > just for fun). The problem is however that I dont know how these functions > are called in English. Could someone provide me with the correct english > term and point me perhaps to some more info on this topic. > The functions are of the form; > y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 > This function has two input variables and has an order of two. The number of > variables and order can vary accordingly ofcourse. Is there a general term > for this kind of functions? Darius, This is a quadratic (what you call order 2 but in English is usually called degree two) polynomial in two variables. If you vary the number of variables (inputs) then they are called polynomials in several variables. It is impossible to tell your level of mathematical education, so it is not clear where to direct you for further information. Achava === Subject: Re: Function naming function group as I described polynomials. I have some knowledge of math, but I was always under the impression that a polynome was defined as a one variable function, and that these were merely a special case of the multivariable case. Probably has to do with my understanding ;-) Achava Nakhash, the Loving Snake schreef in bericht > Im doing some study on a perticular type of functions (in my spare time > just for fun). The problem is however that I dont know how these functions > are called in English. Could someone provide me with the correct english > term and point me perhaps to some more info on this topic. > The functions are of the form; > y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 > This function has two input variables and has an order of two. The number of > variables and order can vary accordingly ofcourse. Is there a general term > for this kind of functions? > Darius, > This is a quadratic (what you call order 2 but in English is > usually called degree two) polynomial in two variables. If you vary > the number of variables (inputs) then they are called polynomials in > several variables. It is impossible to tell your level of > mathematical education, so it is not clear where to direct you for > further information. > Achava === Subject: Re: Function naming >function group as I described polynomials. I have some knowledge of math, >but I was always under the impression that a polynome was defined as a one >variable function, and that these were merely a special case of the >multivariable case. Probably has to do with my understanding ;-) (Please dont post upside down.) A good resource for definitions is http://mathworld.wolfram.com -- the definition at http://mathworld.wolfram.com/Polynomial.html says A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. I understand that this wouldnt help you when you didnt know the term, but when you do want a definition of a specific term (or facts about a term or concept) its a great starting point. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Function naming >function group as I described polynomials. I have some knowledge of math, >but I was always under the impression that a polynome was defined as a one >variable function, and that these were merely a special case of the >multivariable case. Probably has to do with my understanding ;-) >Im doing some study on a perticular type of functions (in my spare time >just for fun). The problem is however that I dont know how these > >functions >are called in English. Could someone provide me with the correct english >term and point me perhaps to some more info on this topic. >The functions are of the form; >y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 >This function has two input variables and has an order of two. The > >number of >variables and order can vary accordingly ofcourse. Is there a general > >term >for this kind of functions? > To be more specific, I would call it a bivariate quadratic function. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Function naming > function group as I described polynomials. I have some knowledge of math, > but I was always under the impression that a polynome was defined as a one > variable function, and that these were merely a special case of the > multivariable case. Probably has to do with my understanding ;-) You can call them polynomials indeed, but in most cases this is (again indeed) as involving a single variable. To alleviate you could use polynomials with multiple variables for the group of functions. (I have seen the term multinomial used for it, but this creates a different kind of confusion.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Function naming > Im doing some study on a perticular type of functions (in my spare time > just for fun). The problem is however that I dont know how these functions > are called in English. Could someone provide me with the correct english > term and point me perhaps to some more info on this topic. > The functions are of the form; > y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 > This function has two input variables and has an order of two. The number of > variables and order can vary accordingly ofcourse. Is there a general term > for this kind of functions? Its a quadratic (= your order of two) polynomial in two indeterminates (= your input variables). If you dont want to be tied to order and number of indeterminates just call it a polynomial. Actually _as functions_ these things are better called polynomial functions rather than just polynomials, but that may not matter to you. You may also want to Google binary quadratic form. === Subject: Re: Function naming > I dont know how these functions >are called in English. Could someone provide me with the correct english >term and point me perhaps to some more info on this topic. >The functions are of the form; >y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 This particular one is a quadratic in two variables. (The last _three_ terms are all second degree.) >This function has two input variables and has an order of two. The number of >variables and order can vary accordingly ofcourse. Is there a general term >for this kind of functions? More generally, a polynomial of degree N in M variables. degree of 1 -- linear degree of 2 -- quadratic degree of 3 -- cubic degree of 4 -- quartic degree of 5 -- quintic degree of 6 -- sextic etc. But above quadratic the special terms are progressively less used. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Function naming > Im doing some study on a perticular type of functions (in my spare time > just for fun). The problem is however that I dont know how these > functions are called in English. Could someone provide me with the correct > english term and point me perhaps to some more info on this topic. > The functions are of the form; > y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 > This function has two input variables and has an order of two. The number > of variables and order can vary accordingly ofcourse. Is there a general > term for this kind of functions? Id call it quadratic. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Function naming >> Im doing some study on a particular type of function (in my spare time >> just for fun). The problem is however that I dont know how these >> functions are called in English. Could someone provide me with the >> correct English term and point me perhaps to some more info on >> this topic? >> The functions are of the form; >> y = c0 + c1x1 + c2x2 + c3x1x2 + c4x1^2 + c5x2^2 >> This function has two input variables and has an order of two. >> The number of variables and order can vary accordingly of course. The functions are polynomial functions in several variables. >> Is there a general term for these kind of functions? > Id call it quadratic. The example you gave is of a quadratic polynomial function, whose graph is a quadric surface. === Subject: Re: prime numbers and 111111..... > Hi there. > When doing a math problem involving numbers with only 1s, like 11 and > 1111111 etc (which could be any base really), i noticed that n digit 1s > divide by m digit 1s if and only if m is a factor of n. > But I guess this is reducable to prime numbers and is not sagnificant. > Martin Johansen Similar to the previous respondent, A number of repeatable digits, 1s, e.g. 11111 in any base b is representable as the sum of n terms of the geometric progression (b^n-1)/(b - 1) = b^0 + b^1 + b^2 + .. b^(n-1) The polynomial Ôf(b)n defined as f(b)n = b^0 + b^1 + b^2 + .. b^(n-1) i.e. such that f(b)n * (b-1) = (b^n-1) is a cyclotomic polynomial, see Wolfram. It has some very interesting factor properties. Most importantly, if n is composite, the polynomial is composite. Furthermore, it will only have factors of the form 2ln+1 for odd n. e.g. if n = 5, 11111 = 41 * 271, both are of the form 2*l*5 + 1, for 41 l = 4, for 271 l = 27 You can see that a number x such as x=11111 is always such that, when n=5, base 10, 2n|(x-1) and you will get to your interesting observation for some n. f(b)n is one hell of an interesting polynomial. Richard Miller === Subject: Re: third order stationarity behavior of random process >The first order stationarity behavior of random process is that all first >order pdf/pmfs are time-independent; >The second order stationarity behavior of random process is that all joint >pdf/pmfs of any two samples are only dependent on the difference of the two >sampling times, i.e., (t1-t2). >>I dont think those are correct. At least theyre not consistent with >>the definitions Ive seen. For example, in Cox & Miller, The Theory >>of Stochastic Processes: >> ... a process is stationary of order p if all moments up to order p have >> the stationarity property. >> It depends on how one defines moments. If one only uses >> products of the random variables, it is woefully inadequate. >> Moments might not exist; this does not affect stationarity. >Inadequate for what? Are you denying that this is the definition >of stationary of order p in Cox & Miller? Similarly in Bartlett, >An Introduction to Stochastic Processes, section 6.1: > More generally, a process with finite moments will be called stationary > to the rth order if all the moments of order r or less are invariant > under translation. >Or Grimmett and Stirzaker, Probability and Random Processes, sec. 8.2: > X = {X(t): t >= 0} is weakly (or second-order or covariance) stationary if > E(X(t_1)) = E(X(t_2)) > and > cov(X(t_1),X(t_2)) = cov(X(t_1+h),X(t_2+h)) > for all t_1, t_2 and h > 0. The term weakly is very important here. Just as uncorrelated is much weaker than independent, so is covariance stationarity much weaker than stationary. Also, for Gaussian processes, the two are equivalent in both cases, again because joint normal distributions are determined by their moments of order 1 and 2. It is this form of stationarity which is needed for the Loeve-Karhunen representation. Stationary of order k can be defined as above, or as joint distributions at k or fewer time points are preserved under time translation. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Symbolic solution of quadratic matrix equations I need to find a symbolic solution of a quadratic matrix equation X^2 * A + X * B + A = 0 where A is the transpose of (kxk) nonsingular matrix A and B is a (kxk) symmetric matrix. Ive found some papers about numeric solution of such kind of matrix equation but I couldnt find anything about the symbolic solution. If somebody helps me, I will appreciate a lot. === Subject: Re: Symbolic solution of quadratic matrix equations >I need to find a symbolic solution of a quadratic matrix equation > X^2 * A + X * B + A = 0 >where A is the transpose of (kxk) nonsingular matrix A and B is a >(kxk) symmetric matrix. Ive found some papers about numeric solution >of such kind of matrix equation but I couldnt find anything about the >symbolic solution. If somebody helps me, I will appreciate a lot. This is not going to be easy, except in the case where all the matrices commute. You can treat it as a system of k^2 quadratic polynomials in the entries of X, and use try Groebner basis techniques, but Id guess its likely to be rather complicated unless k is very small. A 2 x 2 example should work pretty well, but even 3 x 3 might be very ugly. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Symbolic solution of quadratic matrix equations >>I need to find a symbolic solution of a quadratic matrix equation >> X^2 * A + X * B + A = 0 >>where A is the transpose of (kxk) nonsingular matrix A and B is a >>(kxk) symmetric matrix. >This is not going to be easy, except in the case >where all the matrices commute. You can treat it >as a system of k^2 quadratic polynomials in the entries >of X, and use try Groebner basis techniques, but Id guess >its likely to be rather complicated unless k is very small. >A 2 x 2 example should work pretty well, but even 3 x 3 >might be very ugly. Of course Robert is correct; this is going to be a mess. But it looks like there IS some structure to the answer; I dont have an explanation for it but I offer my findings so someone else can explain the theory making this work. I tried some 2x2 and 3x3 examples, talking to Maple like this: X:=matrix(3,3,[z,y,x,w,v,u,t,s,r]); A:=matrix(3,3,[seq(rand() mod 100, i=1..9)]); B:=matrix(3,3,[seq(rand() mod 100, i=1..9)]); for i to 3 do for j to i-1 do B[i,j]:=B[j,i]:od:od:print(B); evalm(X&*X&*A + X&*B + transpose(A)); eq:={seq(seq(%[i,j],j=1..3),i=1..3)}; lprint(%); and then to Magma like this: Q:=RationalField(); P:=PolynomialRing(Q,9); I:=ideal; Groebner(I); Write(SeeMe,I); The 2x2 case is finished right away; the 3x3 case takes a minute or two. The 3x3 solution looks like this: each of z, y, x, ... can be expressed as a degree-19 polynomial in r with coefficients which are ratios of roughly 150-digit numbers. So there is one solution matrix X for each value of r. And there are 20 possible values of r, namely the roots of a polynomial of degree 20 with integer coefficients. (In the 2x2 case, replace 20 by 6.) The only surprise (to me) is that the polynomial always seems to factor: in the 2x2 case its the product of a quadratic and a quartic with coefficients in the 7- and 15-digit range, respectively. In the 3x3 case theres a degree-8 factor (60-digit coefficients) and a degree-12 factor (90-digit coefficients). Moreover, these polynomials are special: the quartic has a dihedral Galois group, and the two factors in the 3x3 case have galois groups of order just 48. So it appears there are two types of solutions to the original quadratic equation, and that at least for these low degrees, the different solutions can be obtained by some easy root extractions. I admit I dont understand why there is this much of a pattern to what should be a more random-looking polynomial. dave === Subject: Re: Symbolic solution of quadratic matrix equations In answer to Daves question below, I believe I can shed a little more light on this. Rewrite the matrix equation as A.X^2 + B.X + C = 0 (1) Let k and v be an eigenvalue and eigenvector of X so X.v = k.v Then multiplying (1) by v gives (A.k^2 + B.k + C).v = 0 (2) so that det(A.k^2 + B.k + C) = 0 (3) This is then a polynomial equation of degree 2.n for the n eigenvalues of X. Now we take n solutions k_i of (3) we can then solve (2) for v_i. Then X is given by the eigen decomposition formula X = V.K.V^-1 where K = diagonal matrix k_i V = matrix of vectors v_i Which n solutions of (3) should be chosen I dont know. Maybe they all give valid solutions, maybe only some of them do ... >>I need to find a symbolic solution of a quadratic matrix equation >> X^2 * A + X * B + A = 0 >>where A is the transpose of (kxk) nonsingular matrix A and B is a >>(kxk) symmetric matrix. >This is not going to be easy, except in the case >where all the matrices commute. You can treat it >as a system of k^2 quadratic polynomials in the entries >of X, and use try Groebner basis techniques, but Id guess >its likely to be rather complicated unless k is very small. >A 2 x 2 example should work pretty well, but even 3 x 3 >might be very ugly. > Of course Robert is correct; this is going to be a mess. > But it looks like there IS some structure to the answer; > I dont have an explanation for it but I offer my findings > so someone else can explain the theory making this work. > I tried some 2x2 and 3x3 examples, talking to Maple like this: > X:=matrix(3,3,[z,y,x,w,v,u,t,s,r]); > A:=matrix(3,3,[seq(rand() mod 100, i=1..9)]); > B:=matrix(3,3,[seq(rand() mod 100, i=1..9)]); > for i to 3 do for j to i-1 do B[i,j]:=B[j,i]:od:od:print(B); > evalm(X&*X&*A + X&*B + transpose(A)); > eq:={seq(seq(%[i,j],j=1..3),i=1..3)}; lprint(%); > and then to Magma like this: > Q:=RationalField(); > P:=PolynomialRing(Q,9); > I:=ideal; > Groebner(I); > Write(SeeMe,I); > The 2x2 case is finished right away; the 3x3 case takes a minute or two. > The 3x3 solution looks like this: each of z, y, x, ... can be expressed > as a degree-19 polynomial in r with coefficients which are ratios of > roughly 150-digit numbers. So there is one solution matrix X for > each value of r. And there are 20 possible values of r, namely the > roots of a polynomial of degree 20 with integer coefficients. > (In the 2x2 case, replace 20 by 6.) > The only surprise (to me) is that the polynomial always seems to factor: > in the 2x2 case its the product of a quadratic and a quartic with > coefficients in the 7- and 15-digit range, respectively. > In the 3x3 case theres a degree-8 factor (60-digit coefficients) > and a degree-12 factor (90-digit coefficients). Moreover, these > polynomials are special: the quartic has a dihedral Galois group, > and the two factors in the 3x3 case have galois groups of order just 48. > So it appears there are two types of solutions to the original > quadratic equation, and that at least for these low degrees, > the different solutions can be obtained by some easy root extractions. > I admit I dont understand why there is this much of a pattern to > what should be a more random-looking polynomial. > dave === Subject: Re: basic covering space question at 05:50 PM, nojb@fibertel.com.ar (Nicolas Ojeda Baer) said: >Let X -> B, Y -> B be two covering maps. Then X x_B Y -> B is a >covering map. (X x_B Y is the fiber product of X and Y). I know this >is basic, but I cant seem to get it right. Consider these questions: 1. What is the inverse image of a point in B under X x_B Y -> B? 2. How do you define the topology of X x_B Y? Then apply the definition of a covering map. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: In what sense does this vector converge to zero? === >Subject: In what sense does this vector converge to zero? None; convergence is a property of sequences. >Let M be an nxn symmetric positive semidefinite matrix. Suppose v >is an n-tuplet that satisfy 2 conditions (only) in the limit that n >goes to infinity: > v v -> 1 > v M v -> 0. The above has no meaning. You need to replace it with statements about sequences and their limits. Presumably you mean that {vn} and {Mn} are sequences such that vn and Mn are an n-tuplet and an n*n symmetric positive semidefinite matrix and vn vn -> 1 vn Mn vn -> 0. Similarly, you dont have a definition for w but rather a sequence {wn}, wn=Mn vn. In order to talk about limits youl need to imbed all of your vectors into a common vector space and specify what topology youre using on it, at which point you can ask whether {wn} has a limit. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Cardinality of an Infinite set, which is smaller than Alef 0? Without the Axiom of Choice, can anyone prove that Alef 0, is the smallest cardinality of an infinite set? Please note: Assuming that for any pair of non empty sets, the smaller set can be mapped one-to-one to a subset of the larger set, is equivalent to the Axiom of Choice. http://math.vanderbilt.edu/~schectex/ccc/choice.html === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? No; is is the only smallest cardinality of an infinite set, in every other infinite cardinality must have a smaller cardinality. But there can be incomparable infinite cardinalities. >Please note: Assuming that for any pair of non empty sets, the >smaller set can be mapped one-to-one to a subset of the larger >set, is equivalent to the Axiom of Choice. That is part of the definition. What is equivalent ot the Axiom of Choice is that of two sets, on is larger than the other. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? > Without the Axiom of Choice, can anyone prove that Alef 0, is the > smallest cardinality of an infinite set? It depends what you mean. On the one hand if a set A has cardinality less than or equal to aleph_0, then by definition there is an injection f:A -> omega. This means A is wellorderable. If it is also infinite it must have cardinality aleph_0. On the other hand without AC there may be sets which are not comparable with aleph_0 in cardinality. So aleph_0 is minimal among infinite cardinals but not necessarily the minimum without AC. It is the minimum among the cardinals of infinite wellorderable sets. === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? One does not need AC to prove the following: Any subset of a countably infinite set is countable. However, without the axiom of choice, one cannot prove that an infinite set necessarily has a countably infinite subset. A set is Dedekind infinite if and only if it has a countably infinite subset, and it consistent with ZF+not AC that there exist infinite Dedekind finite sets. Thus, countably infinite is a minimal infinite cardinality, but without AC it is not necessarily the only one. Note that aleph0 has a very precise definition to many authors: It is the set of all finite ordinals; the axiom of infinity guarantees its existence. Thus, one does not need the axiom of choice to show that if an ordinal (or cardinal) a < aleph0, then a is finite. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? > One does not need AC to prove the following: Any subset of a countably > infinite set is countable. > However, without the axiom of choice, one cannot prove that an infinite > set necessarily has a countably infinite subset. A set is Dedekind > infinite if and only if it has a countably infinite subset, and it > consistent with ZF+not AC that there exist infinite Dedekind finite sets. I have always been massively confused by this, and I figure this is as good a time to ask as any, since it has been explicitly stated here. The obvious question is, why cant I just pick a countable series of elements from the infinite set? What is it thats preventing me from doing that? Let A be the infinite set, then we know that there exists a member x of the set. Put that x into a subset S. Then we know that there exists a y in A such that y != x, else it would be finite. Put that into a subset S. Continuing, finding, for instance, a z such that z != x, y, you have a countable set. Is it the statements we know that there exists a member ... of the set A that prevent this construction? > Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? > > One does not need AC to prove the following: Any subset of a countably > infinite set is countable. > However, without the axiom of choice, one cannot prove that an infinite > set necessarily has a countably infinite subset. A set is Dedekind > infinite if and only if it has a countably infinite subset, and it > consistent with ZF+not AC that there exist infinite Dedekind finite > sets. > I have always been massively confused by this, and I figure this is as good > a time to ask as any, since it has been explicitly stated here. The obvious > question is, why cant I just pick a countable series of elements from the > infinite set? What is it thats preventing me from doing that? > Let A be the infinite set, then we know that there exists a member x of the > set. Put that x into a subset S. Then we know that there exists a y in A > such that y != x, else it would be finite. Put that into a subset S. > Continuing, finding, for instance, a z such that z != x, y, you have a > countable set. > Is it the statements we know that there exists a member ... of the set A > that prevent this construction? > Stephen J. Herschkorn sjherschko@netscape.net shows that this construction is impossible without the axiom of choice? === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >>Without the Axiom of Choice, can anyone prove that Alef 0, is the >>smallest cardinality of an infinite set? ..................... >> Let A be the infinite set, then we know that there exists a member x of >the >> set. Put that x into a subset S. Then we know that there exists a y in A >> such that y != x, else it would be finite. Put that into a subset S. >> Continuing, finding, for instance, a z such that z != x, y, you have a >> countable set. >> Is it the statements we know that there exists a member ... of the set A >> that prevent this construction? >shows that this construction is impossible without the axiom of choice? It at most requires the countable axiom of choice, and I believe it is weaker than that. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? |shows that this construction is impossible without the axiom of choice? When Paul Cohen proved that if ZF is consistent, then so is ZF+the axiom of choice is false, along with that he also proved that if ZF is consistent then so is ZF+there exists a set S which is infinite but has no countably infinite subset. The proof is in his famous book if youre curious. Here the definition of infinite is that S is infinite if there is a family F of subsets of S that contains the empty set, and such that if X is in F and s is in S, then the union of X with {s} is in F too. This corresponds to the usual definition of finite which is most closely related to the principle of mathematical induction on the natural numbers. Sometimes authors give a different definition of infinite which is called Dedekind infinite which is equivalent to containing a countably infinite subset (S is called Dedekind infinite if there exists a 1-1 correspondence between S and a proper subset of S.) Assuming the axiom of choice the two definitions are equivalent. While not assuming the axiom of choice I believe the first definition I gave is preferred. Keith Ramsay === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >> >However, without the axiom of choice, one cannot prove that an infinite >set necessarily has a countably infinite subset. A set is Dedekind >infinite if and only if it has a countably infinite subset, and it >consistent with ZF+not AC that there exist infinite Dedekind finite > >>sets. >> >>I have always been massively confused by this, and I figure this is as >> >good >>a time to ask as any, since it has been explicitly stated here. The >> >obvious >>question is, why cant I just pick a countable series of elements from the >>infinite set? What is it thats preventing me from doing that? >> >shows that this construction is impossible without the axiom of choice? start, though an explicit model of ZF where there exists a Dedekind set is not given there. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >>However, without the axiom of choice, one cannot prove that an infinite >>set necessarily has a countably infinite subset. A set is Dedekind >>infinite if and only if it has a countably infinite subset, and it >>consistent with ZF+not AC that there exist infinite Dedekind finite >> >sets. >I have always been massively confused by this, and I figure this is as good >a time to ask as any, since it has been explicitly stated here. The obvious >question is, why cant I just pick a countable series of elements from the >infinite set? What is it thats preventing me from doing that? >Let A be the infinite set, then we know that there exists a member x of the >set. Put that x into a subset S. Then we know that there exists a y in A >such that y != x, else it would be finite. Put that into a subset S. >Continuing, finding, for instance, a z such that z != x, y, you have a >countable set. >Is it the statements we know that there exists a member ... of the set A >that prevent this construction? Logical conjunction is a binary operation, so you can use induction to form a subset of any arbitrary *finite* size. Without the axiom of choice, a problem arises here because you want to conjunct infinitely many times. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? ... > However, without the axiom of choice, one cannot prove that an infinite > set necessarily has a countably infinite subset. A set is Dedekind > infinite if and only if it has a countably infinite subset, and it > consistent with ZF+not AC that there exist infinite Dedekind finite > sets. > I have always been massively confused by this, and I figure this is as good > a time to ask as any, since it has been explicitly stated here. The obvious > question is, why cant I just pick a countable series of elements from the > infinite set? What is it thats preventing me from doing that? My gut feeling is that it is the axiom of choice that allows you to do it. If there is no axiom of choice you can not pick such a countable series of elements. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? > > One does not need AC to prove the following: Any subset of a countably > infinite set is countable. > However, without the axiom of choice, one cannot prove that an infinite > set necessarily has a countably infinite subset. A set is Dedekind > infinite if and only if it has a countably infinite subset, and it > consistent with ZF+not AC that there exist infinite Dedekind finite > sets. > I have always been massively confused by this, and I figure this is as good > a time to ask as any, since it has been explicitly stated here. The obvious > question is, why cant I just pick a countable series of elements from the > infinite set? What is it thats preventing me from doing that? > Let A be the infinite set, then we know that there exists a member x of the > set. Put that x into a subset S. Then we know that there exists a y in A > such that y != x, else it would be finite. Put that into a subset S. > Continuing, finding, for instance, a z such that z != x, y, you have a > countable set. > Is it the statements we know that there exists a member ... of the set A > that prevent this construction? > Stephen J. Herschkorn sjherschko@netscape.net The problem is that to choose these countably many elements, you use at least the countable AC. In various (topos) models of set theory, you can even have infinite (that is non-finite) subsets of finite sets. In fact, there are a number of distinct definitions of finite. A regular morass, that disappears under AC. === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >Without the Axiom of Choice, can anyone prove that Alef 0, is the >smallest cardinality of an infinite set? > > One does not need AC to prove the following: Any subset of a countably > infinite set is countable. > However, without the axiom of choice, one cannot prove that an infinite > set necessarily has a countably infinite subset. A set is Dedekind > infinite if and only if it has a countably infinite subset, and it > consistent with ZF+not AC that there exist infinite Dedekind finite > sets. > I have always been massively confused by this, and I figure this is as good > a time to ask as any, since it has been explicitly stated here. The obvious > question is, why cant I just pick a countable series of elements from the > infinite set? What is it thats preventing me from doing that? > Let A be the infinite set, then we know that there exists a member x of the > set. Put that x into a subset S. Then we know that there exists a y in A > such that y != x, else it would be finite. Put that into a subset S. > Continuing, finding, for instance, a z such that z != x, y, you have a > countable set. > Is it the statements we know that there exists a member ... of the set A > that prevent this construction? > Stephen J. Herschkorn sjherschko@netscape.net Youre making infinitely many choices. Basically, you are assuming a choice function for the nonempty subsets of A. === Subject: Re: Cardinality of an Infinite set, which is smaller than Alef 0? >>Without the Axiom of Choice, can anyone prove that Alef 0, is the >>smallest cardinality of an infinite set? >> One does not need AC to prove the following: Any subset of a countably >> infinite set is countable. >> However, without the axiom of choice, one cannot prove that an infinite >> set necessarily has a countably infinite subset. A set is Dedekind >> infinite if and only if it has a countably infinite subset, and it >> consistent with ZF+not AC that there exist infinite Dedekind finite >sets. >I have always been massively confused by this, and I figure this is as good >a time to ask as any, since it has been explicitly stated here. The obvious >question is, why cant I just pick a countable series of elements from the >infinite set? What is it thats preventing me from doing that? >Let A be the infinite set, then we know that there exists a member x of the >set. Put that x into a subset S. Then we know that there exists a y in A >such that y != x, else it would be finite. Put that into a subset S. >Continuing, finding, for instance, a z such that z != x, y, you have a >countable set. So how do you keep choosing the elements? You can choose a finite number of them, but ... . >Is it the statements we know that there exists a member ... of the set A >that prevent this construction? >> Stephen J. Herschkorn sjherschko@netscape.net -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Determinants and commuting matrices (old Hoffman and Kunze Exercise) The last exercise in section 5.4 of Hoffman and Kunzes Linear Algebra asks the reader to prove that, given 4 commuting n x n matrices A, B, C, and D, the determinant of the 2n x 2n matrix A B C D is given by det(AD - BC). Im feeling dense, but I dont see why this should be. Any insight would be appreciated... === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > The last exercise in section 5.4 of Hoffman and Kunzes Linear Algebra > asks the reader to prove that, given 4 commuting n x n matrices A, B, C, > and D, the determinant of the 2n x 2n matrix > A B > C D > is given by det(AD - BC). > Im feeling dense, but I dont see why this should be. Any insight would > be appreciated... If they all commute (including AD-BC), then they have the same invariant subspaces. Try to find a basis where the matrix [ A, B; C, D ] becomes block-diagonal. The determinant of a block diagonal matrix is the product of determinants of each block. Hope this helps. Igor === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g Perhaps it can be done a bit more directly. det ( A, B ; C, D ) = det(A)*det(C)*det(I,A^-1*B; I, C^-1*D) = det (A)*det(C)*det(C^-1*D - A^-1*B) = det(A) * det( D - A^-1 * B * C ) = det(AD - BC) det-ails of why commutativity justifies these steps left to the reader. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > Perhaps it can be done a bit more directly. > det ( A, B ; C, D ) = det(A)*det(C)*det(I,A^-1*B; I, C^-1*D) I dont see how whats below follows from whats above. How do you reduce the determinant of a 2nx2n matrix to the determinant of an nxn matirx? You also get into trouble if A or C is not invertible. Igor > = det (A)*det(C)*det(C^-1*D - A^-1*B) > = det(A) * det( D - A^-1 * B * C ) > = det(AD - BC) > det-ails of why commutativity justifies these steps left to the reader. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g Apologies if this turns into a double-post; Google Groups 2 beta kicked back the first attempt. det( I, M; I, N) = det( I, M; 0, N-M) Subtract the upper rows from the lower ones. To generalize slightly what you pointed out, determinant of a block triangular matrix is the product of the determinants of the diagonal blocks. Yes, you get in trouble if A or C is not invertible, but this can be handled by a generic perturbation argument. Add a small multiple of the identity matrix, sufficient to avoid the singularity, then take the limit as the multiple tends to zero. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g I M I N Subtract the upper rows from the lower ones. Determinant of a block triangular matrix is .... Yes, you get in trouble if A or C is not invertible, but this can be handled by a generic perturbation argument. Add a small multiple of the identity matrix, sufficient to avoid the singularity, then take the limit as the multiple tends to zero. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > The last exercise in section 5.4 of Hoffman and Kunzes Linear Algebra > asks the reader to prove that, given 4 commuting n x n matrices A, B, > C, and D, the determinant of the 2n x 2n matrix > A B > C D > is given by det(AD - BC). > Im feeling dense, but I dont see why this should be. Any insight would be appreciated... If you had a block matrix of the form | P Q | =M | Z R | where Z is the appropriate sized zero matrix, then any term of the det of M expanded about the 1st column of P must contain a zero term if it contains an element or elements of Q. So DetM comprises priducts of the form product n elements of P times product n elements of R i.e. Det(P)Det(R) = Det(PR). If you multiply |A B| by | I -(A^(-1)*B | |C D| | Z I | The determinant of the 2nx2n matrix is unaltered. The product equals | A Z | | C D - CA^(-1)B| So Det(M) = det(A)Det(D - CA^(-1)B = Det(AD -AC A^(-1)B) = Det(AD -CB) for matrices that commute. This assumes that A is invertible but if both A and D were not invertible Det(M)=0 I dont know if this would be the approach envisaged for the excercise === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 Consider A = D = 0, B = C = I. I prefer a perturbation argument to overcome singularity of (for example) A. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 > Consider A = D = 0, B = C = I. Yes, but you are still assuming that the matrices on a diagonal are invertible. You would then swap columns or choose a different multiplying matrix. In any case, it seems that you would have to assume that at least one matrix was non-singualar. > I prefer a perturbation argument to overcome singularity of (for > example) A. Yes, but I suppose you are going beyond elementary methods. I am not familiar with the textbook from which the excercise was taken. Knowing what topics had been covered might suggest how the problem should be solved using commutivity alone or with a peturbation argument if that had been covvered. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 > Consider A = D = 0, B = C = I. Yes, but you are still assuming that the matrices on a diagonal are invertible. You would then swap columns or choose a different multiplying matrix. In any case, it seems that you would have to assume that at least one matrix was non-singualar. > I prefer a perturbation argument to overcome singularity of (for > example) A. Yes, but I suppose you are going beyond elementary methods. I am not familiar with the textbook from which the excercise was taken. Knowing what topics had been covered might suggest how the problem should be solved using commutivity alone or with a peturbation argument if that had been covvered. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account=JrQSJw0AAAA6UtXIbhl36iP_cLZ-hNeT Just to clarify what tools a reader of the textbook might reasonably bring to bear on the exercise, perturbation arguments and even invariant subspaces have not yet been introduced. Topics covered thus far include elementary row operations, vector spaces, bases, and the idea of dimension, linear transformations and their representation by matrices, dual spaces and dual bases, elementary polynomial algebra (ideals of polynomials and some simple results depending therefrom), and determinants of matrices. No canonical forms of any sort have been discussed, though similarity insofar as it relates to changes of bases has been covered. My suspicion, after reading through this thread, is that the solution theyre looking for involves something along the lines of first assuming that both A and D are invertible, proving the result based on something similar I.M. Davidsons line of reasoning, and then showing that it still holds even if A and/or D are both singular. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) >> This assumes that A is invertible but if >> both A and D were not invertible Det(M)=0 >> Consider A = D = 0, B = C = I. >Yes, but you are still assuming that >the matrices on a diagonal are invertible. >You would then swap columns or choose a >different multiplying matrix. >In any case, it seems that you would have to assume >that at least one matrix was non-singualar. >> I prefer a perturbation argument to overcome >singularity of (for >> example) A. >Yes, but I suppose you are going beyond >elementary methods. Not really. The determinant of a matrix being a polynomial in its elements is continuous. As the determinant of A+tI is a polynomial in t, it must be non-singular for all non-zero t which are sufficiently small, and hence A+tI is non-singular. Now pass to the limit as t -> 0. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? >For n = 1, 2, ... let f_n(x) = x^n - x^{n - 1} - ... - x - 1 >= (x^{n + 1} - 2x^n + 1)/(x - 1). Is f_n always irreducible >over Z? Leafing further through my sheaf of photocopies (Ive a bad habit of collecting stuff and not reading it), I found that this has been posed and solved as an American Mathematical Monthly problem: Elementary Problem E 3008 (Vol. 90, No. 7, Aug. - Sep., 1983) Solution of Elementary Problem E 3008 (Vol. 96, No. 2, Feb., 1989) Editorial comment. M. J. DeLeon noted that the result of the problem is contained in the following theorem of Alfred Brauer: if a_1, a_2, ..., a_n are integers with a_1 >= a_2 >= ... > = a_n > 0, then the polynomial x^n - a_1x^{n - 1} - a_2x^{n - 2} - ... - a_n is irreducible over the irrationals. Cf. Alfred Brauer, On algebraic equations with all but one root in the interior of the unit circle, Math. Nachrichten, 4 (1950/51), 250--257. The positive root of the polynomial considered in this problem (or more generally of the polynomial considered in Brauers Theorem) is an example of what is known as a PV-number, which is a real algebraic integer greater than 1 all of whose conjugates lie inside the unit circle. Cf. J. W. S. Cassells, _An Introduction to Diophantine Approx- imation_, Cambridge University Press, 1957, Chapter VIII. -- Angus Rodgers Contains mild peril === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? > is irreducible over the irrationals. a misprint, right? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? >> is irreducible over the irrationals. >a misprint, right? I have discovered a whole new field of mathematics, disproving the ravings of those upstarts Galois and Cantor. Bow down and worship me! Woe to those who dare to suggest that my Word is not Law! Death to the infidel Edgar! -- Angus Rodgers Contains mild peril === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? >> For n = 1, 2, ... let f_n(x) = x^n - x^{n - 1} - ... - x - 1 >> = (x^{n + 1} - 2x^n + 1)/(x - 1). Is f_n always irreducible over Z? > Before anyone wastes too much time on this: the answer is yes > (although I dont know why). > Ernst S. Selmer, On the irreducibility of certain trinomials, > Math. Scand. 4 (1956), 287--302. > I actually had a photocopy of this paper lying around (because > of a question on irreducibility that came up in sci.math a few > years back), but hadnt got around to reading it. > An important criterion, typical of one approach to the problem, > is given by Perron [8]: The polynomial (with integer coefficients) > x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a_n > is irreducible if > |a_1| > 1 + |a_2| + |a_3| + ... + |a_n|. > Applied to f(x) = x^n + ax +- 1, where we substitute x = 1/z, > this shows that f(x) is irreducible for |a| >= 3. When |a| = > 2 and f(+-1) [is not equal to] 0, we can still conclude > irreducibility according to Perron. When |a| = 2 and f(x) has > a rational factor x + 1 or x - 1, the second factor of f(x) > will be irreducible (this is not contained among Perrons > statements, but follows easily from his method). > I dont have access to Perrons paper, but the reference is: > 8. O. Perron, Neue Kriterien fuer die Irreduzibilitat > algebraischer Gleichungen, J. reine angew. Math. 132 (1907), > 288--307. This was discussed on sci.math.research on 1998/03/28, see [1]. For an extensive treatment of irreducibility criteria see Filasetas online book [2]. --Bill Dubuque [2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/ latexbook/ === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? For an extensive treatment of irreducibility criteria see > Filasetas online book [2]. > [2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/ latexbook/ I wasnt able to access this. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? >> For an extensive treatment of irreducibility criteria see >> Filasetas online book [2]. >> [2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/ latexbook/ > I wasnt able to access this. In the past it was accessible without a password, so perhaps if you email Filaseta hell tell you how to access it. --Bill Dubuque === Subject: Re: Is x^n - x^{n - 1} - ... - x - 1 irreducible? > Ernst S. Selmer, On the irreducibility of certain trinomials, > Math. Scand. 4 (1956), 287--302. > An important criterion, typical of one approach to the problem, > is given by Perron [8]: The polynomial (with integer coefficients) > x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a_n > is irreducible if > |a_1| > 1 + |a_2| + |a_3| + ... + |a_n|. >[...] I dont have access to Perrons paper [...] ... but of course theres Google: (Both references note the necessity of the additional condition that a_n is nonzero.) -- Angus Rodgers Contains mild peril === Subject: JSH: Being me I know, many of you probably think its horrible being me, with all these people calling me name, putting up nasty webpages, and spending so much time talking bad about me. Some of you are in your own little world and maybe still think Im wrong, while some of you realize that Im right and STILL wouldnt want to be me considering how much opposition my results have faced and are likely to face. But hey, its actually fun! Like I havent just done arguing with people on math or even just math research as I have an open source project on SourceForge, and I have made some friends (believe it or not) in a few Internet communities. I *thought* I could use the Internet in a groundbreaking way to introduce major results but here I am WAITING ON A JOURNAL, when I said in the past that I wouldnt even use journals! :-) Live and learn. In any event, I also get to go over my own mathematical work, and enjoy talking about it as you might have noticed with me starting new threads--yet again--going over the arguments in different ways. Different looks. Perspective. But, on to something practical, as Ive been looking at BitTorrent clients while Im doing downloads, and I see the download speeds hopping all over the place as the clients--I guess--use various algorithms to try and figure out whats the best connection to make. Anyone here know anything about those algorithms? Im thinking maybe I might make my own BitTorrent client, or go in and fiddle with the algorithms on the one I have as I have Azureus. But its an idle thought, so why not toss it on sci.math? Any of you know anything about the algorithms being used in BitTorrent clients? James Harris === Subject: Re: JSH: Being me > I know, many of you probably think its horrible being me, with all > these people calling me name, putting up nasty webpages, and spending > so much time talking bad about me. I have to be honest, if being you is horrible, those are not the causes I would have listed for it. You receive very little name-calling, the websites mainly say you are obstinately wrong. Ive seen people treated far worse in the public schools. Youve got nothing on some of the politicians out there. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Being me >I know, many of you probably think its horrible being me, with all >these people calling me name, putting up nasty webpages, and spending >so much time talking bad about me. Why in the world would you think anyone cares? And just out of curiosity, who are you adressing here? This happens every once in a while, you seem to be talking to all your other readers, the ones who are not part of the vast conspiracy. What makes you think that they exist? >Some of you are in your own little world and maybe still think Im >wrong, while some of you realize that Im right and STILL wouldnt >want to be me considering how much opposition my results have faced >and are likely to face. Like, what gives you the idea that there are people out there who think youre right? Nobody ever _says_ they think youre right. >But hey, its actually fun! >Like I havent just done arguing with people on math or even just math >research as I have an open source project on SourceForge, and I have >made some friends (believe it or not) in a few Internet communities. That _is_ hard to believe. Why do you keep coming back here? >I *thought* I could use the Internet in a groundbreaking way to >introduce major results but here I am WAITING ON A JOURNAL, when I >said in the past that I wouldnt even use journals! Youve also said in the past that the internet didnt matter, all that mattered was the fact that you were about to be published in a journal. Given just about anything youve ever said, youve said something directly contraditory in the past. ************************ David C. Ullrich === Subject: Re: JSH: Being me > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. Wasnt there that guy from the genius club? What ever happened to him? === Subject: Re: JSH: Being me posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > Wasnt there that guy from the genius club? What ever happened to him? I dont recall Quinn Tyler-Jackson ever saying Harris was right, he objected to how Harris was treated. I dont recall A. Beckwith ever saying Harris was right, he just tried to steal the credit for his paper. === Subject: Re: JSH: Being me Discussion, linux) > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. Just think. Sci.math has *lots* of readers. What are the odds that theyre *all* wrong about Jamess work? -- So I speak before a crowd of the damned, cursed to be unloved throughout time, with only their hatred and bile to comfort them now, having betrayed what should have been their one true lover: Mathematics. -- James Harris reaches a bit === Subject: Re: JSH: Being me > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > Just think. > Sci.math has *lots* of readers. What are the odds that theyre *all* > wrong about Jamess work? Well, I think sci.math readers are stupid. Besides, I do remind you that Im just here goofing off, while I wait. It makes me no never mind to play with you people and see just what I can see. I trace out your neural pathways this way. Like, tomorrow I will come to see who replies to this post, and read information bounced around in their heads. I do this enough that I can build a map, and a model, then I simply test the model. If it fails to respond as you would, then I test again, until it responds as you do. So I build a world and then I can simply test that world, moving the pieces in it as I see fit. And then I know how you will move, as I see fit. I am mostly done. James Harris === Subject: Re: JSH: Being me > Well, I think sci.math readers are stupid. > Besides, I do remind you that Im just here goofing off, while I wait. > It makes me no never mind to play with you people and see just what I > can see. > I trace out your neural pathways this way. > Like, tomorrow I will come to see who replies to this post, and read > information bounced around in their heads. > I do this enough that I can build a map, and a model, then I simply > test the model. > If it fails to respond as you would, then I test again, until it > responds as you do. > So I build a world and then I can simply test that world, moving the > pieces in it as I see fit. And then I know how you will move, as I > see fit. > I am mostly done. > James Harris It appears you have managed to elevate your delusions of grandeur to delusions of divinity. Keep it up. There are special places for you -- even here on earth! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Being me > > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > > Just think. > > Sci.math has *lots* of readers. What are the odds that theyre *all* > wrong about Jamess work? > Well, I think sci.math readers are stupid. You read sci.math, right? === Subject: Re: JSH: Being me > I know, many of you probably think its horrible being me, with all > these people calling me name, putting up nasty webpages, and spending > so much time talking bad about me. The unswerving self-belief (in the face of a mountain of contradictory evidence) must be some comfort, I imagine. > Some of you are in your own little world Project much? > and maybe still think Im > wrong, while some of you realize that Im right and STILL wouldnt > want to be me considering how much opposition my results have faced > and are likely to face. > But hey, its actually fun! Must have an almost narcotic effect, that level of self-belief. > Like I havent just done arguing with people on math or even just math > research as I have an open source project on SourceForge, and I have > made some friends (believe it or not) in a few Internet communities. You should be careful making friends on the Internet James, theres a lot of weirdos out there. > I *thought* I could use the Internet in a groundbreaking way to > introduce major results but here I am WAITING ON A JOURNAL, when I > said in the past that I wouldnt even use journals! Youll be waiting a while, Id wager. > :-) > Live and learn. If only you would! > In any event, I also get to go over my own mathematical work, and > enjoy talking about it as you might have noticed with me starting new > threads--yet again--going over the arguments in different ways. You? Starting new threds? Cant say Id spotted any, no... > Different looks. Perspective. > But, on to something practical, as Ive been looking at BitTorrent > clients while Im doing downloads, and I see the download speeds > hopping all over the place as the clients--I guess--use various > algorithms to try and figure out whats the best connection to make. > Anyone here know anything about those algorithms? Im thinking maybe > I might make my own BitTorrent client, or go in and fiddle with the > algorithms on the one I have as I have Azureus. But its an idle > thought, so why not toss it on sci.math? > Any of you know anything about the algorithms being used in BitTorrent > clients? You have Internet access, learn how to use it. Oh wait, I forgot, youre pathologically incapacle of learning. -- Larry Lard Replies to group please === Subject: Re: JSH: Being me > I know, many of you probably think its horrible being me, > But hey, its actually fun! Would you describe it as . . . bliss? === Subject: Re: JSH: Being me > But, on to something practical, as Ive been looking at BitTorrent > clients while Im doing downloads, and I see the download speeds > hopping all over the place as the clients--I guess--use various > algorithms to try and figure out whats the best connection to make. > Anyone here know anything about those algorithms? Im thinking maybe > I might make my own BitTorrent client, or go in and fiddle with the > algorithms on the one I have as I have Azureus. But its an idle > thought, so why not toss it on sci.math? > Any of you know anything about the algorithms being used in BitTorrent > clients? Sure. What do you want to know? Its a much more naive algorithm than you seem to think it is. http://www.bittorrent.com/protocol.html Ôcid Ôooh === Subject: Re: JSH: Being me jstevh@msn.com says... > I know, many of you probably think its horrible being me, with all > these people calling me name, putting up nasty webpages, and spending > so much time talking bad about me. > Some of you are in your own little world and maybe still think Im > wrong, while some of you realize that Im right and STILL wouldnt > want to be me considering how much opposition my results have faced > and are likely to face. > But hey, its actually fun! > Like I havent just done arguing with people on math or even just math > research as I have an open source project on SourceForge, and I have > made some friends (believe it or not) in a few Internet communities. > I *thought* I could use the Internet in a groundbreaking way to > introduce major results but here I am WAITING ON A JOURNAL, when I > said in the past that I wouldnt even use journals! > :-) > Live and learn. > In any event, I also get to go over my own mathematical work, and > enjoy talking about it as you might have noticed with me starting new > threads--yet again--going over the arguments in different ways. > Different looks. Perspective. > But, on to something practical, as Ive been looking at BitTorrent > clients while Im doing downloads, and I see the download speeds > hopping all over the place as the clients--I guess--use various > algorithms to try and figure out whats the best connection to make. > Anyone here know anything about those algorithms? Im thinking maybe > I might make my own BitTorrent client, or go in and fiddle with the > algorithms on the one I have as I have Azureus. But its an idle > thought, so why not toss it on sci.math? > Any of you know anything about the algorithms being used in BitTorrent > clients? > James Harris Bi-Polar much? === Subject: Re: JSH: Being me Frightening thought. === Subject: Re: JSH: Being me > Frightening thought. Must be hell in there. Dirk Vdm === Subject: Simple Group Theory Question - AGAIN I cant seem to find any way to solve these problems! Let G be a cyclic group of order p when p is prime. Find all the automorphisims of G. === Subject: Re: Simple Group Theory Question - AGAIN >I cant seem to find any way to solve these problems! >Let G be a cyclic group of order p when p is prime. >Find all the automorphisims of G. First, can you tell us which elements of G _generate_ G? ************************ David C. Ullrich === Subject: Re: Simple Group Theory Question - AGAIN > I cant seem to find any way to solve these problems! > Let G be a cyclic group of order p when p is prime. > Find all the automorphisims of G. An automorphism f:G -> G is a homomorphism which maps generators to generators. So fix a generator g so that G = . Since p is a prime, g |-> g^n is an automorphism for 0=ßoor(e^x) So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect. What is the maximum value of n(x) on the interval (1,2)? At what value(s) of x is n(x) largest? Rich === Subject: Re: A question about ßoor(e^x) >Let n(x) be the least integer such that: >1+x+x^2/2+...+x^n/n!>=ßoor(e^x) >So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect. I think youre off by 1 on all of these, e.g. ßoor(e^3) = 20 and sum_{j=0}^8 3^j/j! > 20. >What is the maximum value of n(x) on the interval (1,2)? At what value(s) of x >is n(x) largest? n(ln(k)) does not exist if k is an integer > 1, and n(x) -> infinity as x -> ln(k)+. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A question about ßoor(e^x) >>Let n(x) be the least integer such that: >>1+x+x^2/2+...+x^n/n!>=ßoor(e^x) >>So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect. >I think youre off by 1 on all of these, e.g. >ßoor(e^3) = 20 and sum_{j=0}^8 3^j/j! > 20. Yes. I was thinking of n(x) as the number of terms of e^x needed to compute ßoor(e^x) and did not make the adjustment when I posted the (slightly) different question. >>What is the maximum value of n(x) on the interval (1,2)? At what value(s) >of x >>is n(x) largest? >n(ln(k)) does not exist if k is an integer > 1, and n(x) -> infinity >as x -> ln(k)+. Rich === Subject: Re: A question about ßoor(e^x) > Let n(x) be the least integer such that: > 1+x+x^2/2+...+x^n/n!>=ßoor(e^x) > So n(1)=2, n(2)=5, n(3)=9, n(4)=10, ect. > What is the maximum value of n(x) on the interval (1,2)? At what value(s) of x > is n(x) largest? How about x = ln(3)? Asger. === Subject: adjoining elements to rings posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb What happens when we adjoin the inverse of 2 to the ring Z_12 i.e. what is the field Z_12[x]/(2x-1) isomorphic to? I have reason to believe that it is simply the field of 3 elements, because of the following: In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore 12x-6=0 ==> 6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe that we are talking about F_3 here. However, I cannot prove this. I tried setting up the following maps: j:Z ---> Z_12 pi:Z_12 ---> Z_12[x]/(2x-1) i:Z ---> Z_3 If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is isomorphic to Z_3 by the first isomorphism theorem. I know that ker j = 12Z ker pi = (2x-1) ker i = 3Z But for some reason I cannot compute the kernel of pi o j. Is this even the right track to go on? Similarly I am trying to describe Z[i]/(2+i). Since 2+i=0 we must have 5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5. I know that 2+i is prime in Z[i] and therefore the quotient should be a field. I tried a similar approach to the above but did not get far. I must be going in the wrong direction for these problems. Can someone tell me what is going on? === Subject: Re: adjoining elements to rings >What happens when we adjoin the inverse of 2 to the ring Z_12 i.e. what >is the field Z_12[x]/(2x-1) isomorphic to? I have reason to believe >that it is simply the field of 3 elements, because of the following: >In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore 12x-6=0 ==>6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe that we >are talking about F_3 here. >However, I cannot prove this. I tried setting up the following maps: >j:Z ---> Z_12 >pi:Z_12 ---> Z_12[x]/(2x-1) >i:Z ---> Z_3 >If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is isomorphic >to Z_3 by the first isomorphism theorem. I know that >ker j = 12Z >ker pi = (2x-1) >ker i = 3Z >But for some reason I cannot compute the kernel of pi o j. Is this even >the right track to go on? Well, you know that the kernel contains (3). Since (3) is a maximal ideal of Z, the only possibilities are for the kernel to be all of Z, or else to be just equal to (3). So the question then becomes: is 1 in the kernel? This will happen if and only if 1 = 0 (mod (2x-1)) in Z_{12}[x]. That is, if and only if there exists a polynomial f(x) in Z_{12}[x] such that f(x)(2x-1) = 1. Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then (2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ... (2a_0-a_1)x - a_0. So you must have a_0 = - 1 (mod 12) 2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n 2a_n = 0 (mod 12). So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 = -2. Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12). 2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12). Continuing in this way, a_{i} = -2^i (mod 12). Since we also need 2a_n = 0 (mod 12), that means that -2^{n+1}=0 (mod 12). But this never happens. So no such polynomial exists. Therefore, 1 is not in the kernel of the map. Since the kernel already contains 3, and is not everything, the kernel of pi o j must be (3). HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is isomorphic to F_3; youve only shown that it CONTAINS F_3; you would also have to prove that the induced map from Z is surjective in order to prove that. (I mean, what if it is some other field of characteristic 3?) Slightly easier: you have already shown that 3 is in (2x-1), so the quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1). And since Z_3[x]/(2x-1) is isomorphic to Z_3, you are done. >Similarly I am trying to describe Z[i]/(2+i). This is a bit simpler since Z[i] is not only a domain, but a UFD. > Since 2+i=0 we must have >5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5. I know >that 2+i is prime in Z[i] and therefore the quotient should be a field. Its maximal, and therefore the quotient should be a field (prime and maximal are equivalent in Z[i], but not in general). >I tried a similar approach to the above but did not get far. Yes, it is isomorphic to Z/5Z. Certainly, it is a field a characteristic 5, since a similar approach to the above will establish that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just (5), by noting that 1 is not in (2+i). But you would still have to show that this map is surjective, which is not too hard: you can easily show that every element of Z[i] can be written as (a+bi)(2+i) + r, with r = 0, 1, 2, 3, or 4. This because 5 = (2+i)(2-i). So, given x+yi, with x and y in Z, then we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with t an integer, r=0,1,2,3, or 4; Then x - 2y = (2t-ti)(2+i) + r So (2t-ti)(2+i) + r = (x+yi)-y(2+i) hence (x+yi) = (2t+y - ti)(2+i) + r. Thus, every element in Z[i]/(2+i) is congruent modulo 2+i to 0, 1, 2, 3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective, kernel (5), and you are done. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: adjoining elements to rings posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb >What happens when we adjoin the inverse of 2 to the ring Z_12 i.e. what >is the field Z_12[x]/(2x-1) isomorphic to? I have reason to believe >that it is simply the field of 3 elements, because of the following: >In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore 12x-6=0 >6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe that we >are talking about F_3 here. >However, I cannot prove this. I tried setting up the following maps: >j:Z ---> Z_12 >pi:Z_12 ---> Z_12[x]/(2x-1) >i:Z ---> Z_3 >If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is isomorphic >to Z_3 by the first isomorphism theorem. I know that >ker j = 12Z >ker pi = (2x-1) >ker i = 3Z >But for some reason I cannot compute the kernel of pi o j. Is this even >the right track to go on? > Well, you know that the kernel contains (3). Since (3) is a maximal > ideal of Z, the only possibilities are for the kernel to be all of Z, > or else to be just equal to (3). So the question then becomes: is 1 in > the kernel? > This will happen if and only if 1 = 0 (mod (2x-1)) in Z_{12}[x]. That > is, if and only if there exists a polynomial f(x) in Z_{12}[x] such > that f(x)(2x-1) = 1. > Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then > (2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ... (2a_0-a_1)x - a_0. > So you must have a_0 = - 1 (mod 12) > 2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n > 2a_n = 0 (mod 12). > So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 = -2. > Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12). > 2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12). > Continuing in this way, a_{i} = -2^i (mod 12). > Since we also need 2a_n = 0 (mod 12), that means that -2^{n+1}=0 (mod > 12). But this never happens. So no such polynomial exists. Therefore, > 1 is not in the kernel of the map. Since the kernel already contains > 3, and is not everything, the kernel of pi o j must be (3). > HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is isomorphic to > F_3; youve only shown that it CONTAINS F_3; you would also have to > prove that the induced map from Z is surjective in order to prove > that. (I mean, what if it is some other field of characteristic 3?) > Slightly easier: you have already shown that 3 is in (2x-1), so the > quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1). And since > Z_3[x]/(2x-1) is isomorphic to Z_3, you are done. >Similarly I am trying to describe Z[i]/(2+i). > This is a bit simpler since Z[i] is not only a domain, but a UFD. > Since 2+i=0 we must have >5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5. I know >that 2+i is prime in Z[i] and therefore the quotient should be a field. > Its maximal, and therefore the quotient should be a field (prime and > maximal are equivalent in Z[i], but not in general). >I tried a similar approach to the above but did not get far. > Yes, it is isomorphic to Z/5Z. Certainly, it is a field a > characteristic 5, since a similar approach to the above will establish > that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just (5), by > noting that 1 is not in (2+i). But you would still have to show that > this map is surjective, which is not too hard: you can easily show > that every element of Z[i] can be written as (a+bi)(2+i) + r, with r = > 0, 1, 2, 3, or 4. > This because 5 = (2+i)(2-i). So, given x+yi, with x and y in Z, then > we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with t an > integer, r=0,1,2,3, or 4; Then > x - 2y = (2t-ti)(2+i) + r > So > (2t-ti)(2+i) + r = (x+yi)-y(2+i) > hence > (x+yi) = (2t+y - ti)(2+i) + r. > Thus, every element in Z[i]/(2+i) is congruent modulo 2+i to 0, 1, 2, > 3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective, kernel (5), > and you are done. > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu first. While messing with the second I came across a question that I could not answer. Do you get a different object if you adjoin i to C? I know that R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get something else? I am guessing you get something else since a ring mod an ideal is a field if an only if the ideal is maximal... however (x^2+1) is not maximal in C[x]. But what do you get? === Subject: Re: adjoining elements to rings posting-account=rtZaKw0AAAD069Dyt1QHY9YKfVai02Jb >What happens when we adjoin the inverse of 2 to the ring Z_12 i.e. what >is the field Z_12[x]/(2x-1) isomorphic to? I have reason to believe >that it is simply the field of 3 elements, because of the following: >In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore 12x-6=0 >6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe that we >are talking about F_3 here. >However, I cannot prove this. I tried setting up the following maps: >j:Z ---> Z_12 >pi:Z_12 ---> Z_12[x]/(2x-1) >i:Z ---> Z_3 >If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is isomorphic >to Z_3 by the first isomorphism theorem. I know that >ker j = 12Z >ker pi = (2x-1) >ker i = 3Z >But for some reason I cannot compute the kernel of pi o j. Is this even >the right track to go on? > Well, you know that the kernel contains (3). Since (3) is a maximal > ideal of Z, the only possibilities are for the kernel to be all of Z, > or else to be just equal to (3). So the question then becomes: is 1 in > the kernel? > This will happen if and only if 1 = 0 (mod (2x-1)) in Z_{12}[x]. That > is, if and only if there exists a polynomial f(x) in Z_{12}[x] such > that f(x)(2x-1) = 1. > Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then > (2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ... (2a_0-a_1)x - a_0. > So you must have a_0 = - 1 (mod 12) > 2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n > 2a_n = 0 (mod 12). > So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 = -2. > Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12). > 2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12). > Continuing in this way, a_{i} = -2^i (mod 12). > Since we also need 2a_n = 0 (mod 12), that means that -2^{n+1}=0 (mod > 12). But this never happens. So no such polynomial exists. Therefore, > 1 is not in the kernel of the map. Since the kernel already contains > 3, and is not everything, the kernel of pi o j must be (3). > HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is isomorphic to > F_3; youve only shown that it CONTAINS F_3; you would also have to > prove that the induced map from Z is surjective in order to prove > that. (I mean, what if it is some other field of characteristic 3?) > Slightly easier: you have already shown that 3 is in (2x-1), so the > quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1). And since > Z_3[x]/(2x-1) is isomorphic to Z_3, you are done. >Similarly I am trying to describe Z[i]/(2+i). > This is a bit simpler since Z[i] is not only a domain, but a UFD. > Since 2+i=0 we must have >5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5. I know >that 2+i is prime in Z[i] and therefore the quotient should be a field. > Its maximal, and therefore the quotient should be a field (prime and > maximal are equivalent in Z[i], but not in general). >I tried a similar approach to the above but did not get far. > Yes, it is isomorphic to Z/5Z. Certainly, it is a field a > characteristic 5, since a similar approach to the above will establish > that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just (5), by > noting that 1 is not in (2+i). But you would still have to show that > this map is surjective, which is not too hard: you can easily show > that every element of Z[i] can be written as (a+bi)(2+i) + r, with r = > 0, 1, 2, 3, or 4. > This because 5 = (2+i)(2-i). So, given x+yi, with x and y in Z, then > we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with t an > integer, r=0,1,2,3, or 4; Then > x - 2y = (2t-ti)(2+i) + r > So > (2t-ti)(2+i) + r = (x+yi)-y(2+i) > hence > (x+yi) = (2t+y - ti)(2+i) + r. > Thus, every element in Z[i]/(2+i) is congruent modulo 2+i to 0, 1, 2, > 3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective, kernel (5), > and you are done. > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu first. While messing with the second I came across a question that I could not answer. Do you get a different object if you adjoin i to C? I know that R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get something else? I am guessing you get something else since a ring mod an ideal is a field if an only if the ideal is maximal... however (x^2+1) is not maximal in C[x]. But what do you get? === Subject: Re: adjoining elements to rings days. My association with the Department is that of an alumnus. >first. >While messing with the second I came across a question that I could not >answer. Do you get a different object if you adjoin i to C? Depends what you mean by adjoin i! If you add the complex number i to a ring that already has it, then you just get the same ring back. If you mean taking quotients of certain polynomial rings, thats something else: >I know that >R[x]/(x^2-1)=C, but does C[x]/(x^2+1)=C or do you get something else? I >am guessing you get something else since a ring mod an ideal is a field >if an only if the ideal is maximal... however (x^2+1) is not maximal in >C[x]. Exactly! So C[x]/(x^2+1) CANNOT be a field. > But what do you get? Notice that (x^2+1) = (x+i)(x-i); that (x+i) + (x-i) = 1. So the that C[x]/(x^2+1) is isomoprhic to ( C[x]/(x+i) ) x ( C[x]/(x-i) ) which in turn is isomorphic to C x C, the product of two copies of the complex numbers. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Prime ideal proof hello all, I have been racking my brain for hours trying to prove the following. P is a prime ideal of a ring R iff R / P is a prime ring I wish I could say that I had a good starting point but I dont. Any help would be greatly appreciated. TJ === Subject: Re: Prime ideal proof days. My association with the Department is that of an alumnus. >hello all, >I have been racking my brain for hours trying to prove the following. >P is a prime ideal of a ring R iff R / P is a prime ring What is the definition of Prime Ring? -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Prime ideal proof A ring R is called a prime ring if (0) is a prime ideal of R. -Tj -- >hello all, >I have been racking my brain for hours trying to prove the following. >P is a prime ideal of a ring R iff R / P is a prime ring > What is the definition of Prime Ring? > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Prime ideal proof days. My association with the Department is that of an alumnus. >A ring R is called a prime ring if (0) is a prime ideal of R. In the commutative case, thats called an integral domain... >>I have been racking my brain for hours trying to prove the following. >P is a prime ideal of a ring R iff R / P is a prime ring P is a prime ideal if and only if for every x,y in R, xy in P -> x in P or y in P. ==> Let x+P, y+P in R/P be such that xy + P in (0) in R/P. That means that xy is in P in R. Which means that... <== Let x, y in R such that xy in P. Then (x+P)(y+P) = 0+P in R/P. Therefore... -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Prime ideal proof > A ring R is called a prime ring if (0) is a prime ideal of R. > -Tj > -- Thats normally called an integral domain. Prove that a ring is a prime ring if and only if it has no zero divisors (i.e. whenever xy=0, either x=0 or y=0). Now prove that P is prime if and only if R/P has no zero divisors. === Subject: Re: Prime ideal proof days. My association with the Department is that of an alumnus. >> A ring R is called a prime ring if (0) is a prime ideal of R. >> -Tj >> -- >Thats normally called an integral domain. In the commutative case. But there was no assumption that R was a commutative ring. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Delphi or VB Column generation ? === >Subject: Re: Delphi or VB Column generation ? >Message-id: The Access product we currently have is for optimizing one dimensional stock >cutting, it stores the orders and runs the optimization routine in VBA. >The problem is as I stated, it appears that the code that runs to generate >the valid cutting patterns is malformed or incomplete and excludes some >valid logic that we think should be included. >I need a clean starting point for a simple one dimensional stock cutting >program. Ok, but that still means nothing to me. You need to explain what this actually is. Where I buy lunch, they have 6, 8 and 10 sandwiches they make from 36 loaves of French bread. Would this be a simple example of one dimensional stock cutting? You would not want to make four 8 sandwiches off one loaf because you would be left with an unuseable 4 remnant. >All of the solvers out there have their own way and code to generate the >valid cutting plans / patterns, I want a generic pattern generation / >cutting plan generator that I can modify with our own logic so we can >analyze the impact on adding certain attributes and variables to the mix. But you also said earlier that you want to feed the results to a solver. Do you just want a program/database that creates patterns that will then be optimized, or are you going to do the the whole ball of wax: generate all possible patterns with custom criteria that you will then process with your own optimization algorithm? >might be we need to chat on aol or msn messenger for me to convey my needs >more completely as I am not a mathematician, I am a programmer. Chat sessions might be a problem. Theres scheduling and Ive found live chats to be less productive for technical stuff. I am not a professional mathemetician nor a professional programmer, so keeping the discussion public might be better. And dont be concerned about how complicated this gets. Sci.math has endless topics concerning pure crap so a long thread on a legitimate topic should not be a problem. >Yes I can convert C code to VB Good, because I cant. Seriously, I know VB and Access and Python so we shouldnt have any problems discussing algorithms. === >>Subject: Re: Delphi or VB Column generation ? >>Message-id: The current output is generated simply by putting together all possible >>permutations, even the ones that for obvious reasons (i.e. material >>utilization, etc..) are not good ones therefore increasing the number of >>combinations exponentially and allowing non operationally optimal >patterns >>to be included and selected for usage. >I want a clean starting point that we can go through and INTENTIONALLY >>modify and enhance with LOGIC, not guesswork as it appears to have been >done >>in the spaghetti code in the current application. >I have been reading everything I could get my hands on about the subject >>linear optimization, linear algebra, combinatorics and 1d stock cutting >>solutions but it seems that nobody has ever published a good, >>straightforward way to generate columns / permutations using a computer >for >>submission to a solver of some kind like lp solve. >Unfortunately I cannot post the code as I do not own the copyright, only >>license to use/modify, sorry guys. >> Ok, I thought you said you wanted to fix an existing Access/VB system, >> but later you mentioned growing your own. >> Are you still wanting this done in Access? Access can easily generate >> permutations - provided you want a reasonable scope. For instance, Ive >> got a database query that generates all possible 4 letter words. The >456,976 >> record query is handled easily on modern PCs. But asking for all possible >> 8 letter words gets you in trouble because the query would return 208 >billion >> records. >> And logic is fairly easy to add. I can take my 4 letter word query and ask >> to return only permutations in which the first and last letters are >consonants >> and the middle two letters are vowels. That reduces the the record count >from >> 456,976 to 11,025. But again, trying this with 8 letter words can still >result >> in 121 million records. >at a >> time >> would only be limited by disk space. VB in Access could handle this. If >your >> goal is to prepare a file for some other program, there are probably >better >> choices than VB. >> Let me ask again, what exactly are you looking for? How to generate >> permutations in Access? How to generate them in some programming >> language? What languages do you have available? How good a programmer >> are you? Could you translate a C algorithm into VB? A sample of what you >> want could help identify the scope of the problem. >> === >Subject: Delphi or VB Column generation ? >Message-id: <_LIpd.48$6K5.1@newsread2.news.atl.earthlink.net> >I am doing some research into fixing a simple optimizer written in VB >Access for a friends company. >The product is a few years old and does not exactly provide the best >results. >After some studying of this code it appears that the way that the >column >generation is being done is probably the start of the difficulties. >I have looked but cannot locate a code snippit or function for column >generation that I could start with and modify for my needs or learn >how >others are generating columns and grow my own version. > What do you mean by column generation? > Can you provide an example of what it is currently output vs. what you > would like it to output? Can you quote the code in question so that >those > with some experience in Access/VB dont have to try to read your mind? > It may be that what youre looking for is not in the VB portion but in > the SQL code of the queries involved. > Or it could be both. Ive seen cases where VB programs build temporary > tables with the column generation taking place dynamically based on > a query, which is itself genereated by a SQL statement created inside > a different VB program. > This could get way too complicated for a newsgroup discussion, and no, > you cannot post a copy of the database. But you _can_ post VB code and > if necessary, you can show the queries by opening them in design mode > and selecting SQL view which can be posted. > >Stephen Hauck > > -- > Mensanator > Ace of Clubs >> -- >> Mensanator >> Ace of Clubs -- Mensanator Ace of Clubs === Subject: Re: Infinite number of people toss a coin infinite times >> I said there used to be a theory that the universe was already infinite >> as in no expansion it was _already_ infinite. >> Expansion undermined this theory not because it provided evidence >> of the finiteness of the universe, but because if the universe is under >> expansion there cant be a working completed infinity of the universe >> theory. > I dont see why not. > But then Im not a physicist, Jim, just a simple Texas > set theorist. If theres some reason according to currently > understood physics why an already-infinite universe cant > be expanding, someone please enlighten me. I described this as used to be a theory because I couldnt remember the name of it. I thought at first is was steady-state. But I couldnt reconclile the presence of expansion in steady-state with what I had learned about this. So I kept digging and found: http://www.physics.gmu.edu/classinfo/astr103/CourseNotes/cml_ newt.htm Newtonian Space and Time Cosmology For Newton, space and time are absolute, i.e., the same throughout the Universe and unchanging or unchangeable Absolute space, in its own nature, without relation to anything external, remains always similar and immovable... Absolute, true and mathematical time, of itself, and from its own nature, ßows equably without relation to anything external... Newtonian space time - space and time are absolutes and exist independent of the existence of matter in the Universe; space and time precede the material Universes existence and will survive its demise Decartes and Gottfried Whilhelm Liebniz (1646-1716) did not support Newtons belief in an absolute space and time Liebniz: I hold space to be something purely relative, as time is. Newtonian Cosmology Copernican Principle (Aristarchus, Copernicus, Bondi) - we are not privileged occupants of the universe, so that there is nothing exceptional about our existence and we are consequently not at the Universes center Newtonian Universe - gravity demands that the universe be centerless, edgeless, and of infinite extent in all directions SH: I kept trying to apply the static mathematical Newton infinity to the sold-state theory and it didnt quite fit. Newtons absolute space was unchanging and therefore not undergoing expansion. It seems to me there are traces of Newtons cosmology in the steady-state theory. Stephen === Subject: Re: Infinite number of people toss a coin infinite times > An infinite number of people each toss a coin infinite times, > can you guarantee a new sequence of heads and tails? > Herc Actually, there may be an infinite number of new sequences since: inf + inf = inf Thus, your premise of infinite people and infinite tosses does not guarantee that you are missing an infinite number of people to exaust all possible sequences. Paradoxes of infinity giving finite minds a hard time:) (especially those that look at infinity as a number. lol) Mike === Subject: Re: Infinite number of people toss a coin infinite times posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L right, its possible you can make a new sequence (everyone tossed heads that day), but can you GUARANTEE it? Herc === Subject: Re: Infinite number of people toss a coin infinite times > right, its possible you can make a new sequence (everyone tossed heads > that day), but can you GUARANTEE it? > Herc There is only one thing that you can guarantee in this world with P=1: death some time after birth. Watch out governments that guarantee public debt issues and guaranteed performance mutual funds.:) Mike === Subject: Re: Infinite number of people toss a coin infinite times > Stephen Harris says... >>I said there used to be a theory that the universe was already infinite >>as in no expansion it was _already_ infinite. > Yes, I know. Thats *still* the dominant theory. In an expanding The steady-state theory is not still dominant and I posted some reference quotes to Mike Oliver. > (open) universe, the universe is spatially infinite from the beginning. I accept that this is the explanation. But, I have trouble accepting the fact that it is also claimed that Big Bang originated from an infinitesimal point and then expanded/exploded without the volume of the universe changing. I seem to recall them saying things like after .00000000000000000001 seconds the diameter of the universe was only a few hundred light years. > It isnt that its size is unbounded, but that it is actually > infinite from the beginning. As I said, there are basically two I suppose it is Hawking who uses the finitely unbounded terminology. > models (if we set the cosmological constant to zero, and ignore > weird topologies such as a toroidal universe): (1) the universe is > spatially infinite at all times, and will expand forever, (2) the > universe is spatially finite, and will eventually stop expanding > and recontract to a big crunch. > The possibility that you are considering, that the universe is > finite in size but will expand forever, might be possible by > adjusting the cosmological constant and the matter density. But > expanding does not imply finiteness. > -- > Daryl McCullough > Ithaca, NY === Subject: Re: Infinite number of people toss a coin infinite times Stephen Harris says... >The steady-state theory is not still dominant and I posted some >reference quotes to Mike Oliver. Im sorry---I meant the idea that the universe is infinite. >> (open) universe, the universe is spatially infinite from the beginning. >I accept that this is the explanation. But, I have trouble accepting >the fact that it is also claimed that Big Bang originated from >an infinitesimal point and then expanded/exploded without the volume >of the universe changing. In the unbounded cosmologies, the Big Bang was not a time when the universe was at a single point---it was a time when the universe had infinite density. You might want to look at the cosmology faq entry here: http://www.astro.ucla.edu/~wright/infpoint.html -- Daryl McCullough === Subject: Re: Infinite number of people toss a coin infinite times > Stephen Harris says... >>The steady-state theory is not still dominant and I posted some >>reference quotes to Mike Oliver. > Im sorry---I meant the idea that the universe is infinite. The infinity of the steady-state theory SST is infinitely old and infinitely large. There is no beginning of time as in the Big Bang model. I said the infinity of the steady-state theory is not the same type of infinity as described by the Big Bang model, the universe begins with time. The word infinite is used in both cases but has two different meanings. The SST meaning has a static, absolute, completed sense, which reminds me of how a mathematical infinite plane is stated into existence, no process of expansion of the entire universe but expansion within volumes of the universe that do no exceed their original boundaries. The Big Bang uses a time dependent function for density and size of the universe. This infinity seems relative to me, there is process. And I think that is why Einsteins theory impacted the plausibility of SST. > (open) universe, the universe is spatially infinite from the beginning. >>I accept that this is the explanation. But, I have trouble accepting >>the fact that it is also claimed that Big Bang originated from >>an infinitesimal point and then expanded/exploded without the volume >>of the universe changing. > In the unbounded cosmologies, the Big Bang was not a time when the > universe was at a single point---it was a time when the universe had > infinite density. Im comparing SSt to the Big Bang. There is no infinite density in SSt. Because the same word infinite in both theories does not mean the concept that the word represents is the same in both theories. This answers part of my question within the Big Bang theory context. Note that in the above paragraphs I have been careful to use the term observable Universe rather than Universe. The Universe itself, or the maximum amount of space that we will eventually be able to see given an infinite amount of time, may well be infinite. In quoting a size of the Universe we infer how far we can see in one direction (15 billion light years), and how far we can see in the other direction (15 billion light years) and add the two to get a size (30 billion light years). An age of 15 billion light years in each direction therefore leads us to infer that we are at the centre of a sphere with radius 15 billion light-years, and hence that the Universe is 30 billion light-years across. The trick, however, is that because the Universe is homogeneous and isotropic, every observer must measure a size of the Universe that is 30 billion light years... even ones that are at the edge of our observable Universe! This means that either the Universe is sufficiently curved that space doubles back on itself (like on the surface of a sphere), or that the actual Universe is much larger than the observable one. We currently think that the latter possibility is the case. This quote says the at the moment of creation, space and time were created so that there was a physical dimension. 13.7 billion years ago, the entirety of our universe was compressed into the confines of an atomic nucleus. Known as a singularity, this is the moment before creation when space and time did not exist. According to the prevailing cosmological models that explain our universe, an ineffable explosion, trillions of degrees in temperature on any measurement scale, that was matter and energy but space and time itself. Cosmology theorists combined with the observations of their astronomy colleagues have been able to reconstruct the primordial chronology of events known as the big bang. > You might want to look at the cosmology faq entry here: > http://www.astro.ucla.edu/~wright/infpoint.html > -- > Daryl McCullough There used to be a theory in astronomy/cosmology that the universe was infinite. Then evidence that the universe was expanding caused this theory to be dropped. *** This means the theory SST was dropped with its intepretation of infinity. It does not mean that infinity was dropped from the Big Bang theory. I think the interpretation of infinity in BB is different from SSt becasue SSt used a universe of infinite age/infinitely large while the BB uses an infinity which has a beginning in time. I dont see the state, whatever it was before the big bang as equivalent to the universe as infinitely old. When you challenged my *** quote I clarified with: I said there used to be a theory that the universe was already infinite as in no expansion it was _already_ infinite. Expansion undermined this theory not because it provided evidence of the finiteness of the universe, but because if the universe is under expansion there cant be a working completed infinity of the universe theory. Im distinguishing between the senses of infinity used. Einstein did his work around 1915. He thought the universe was fixed, no expansion. Im saying that there was a theory like SSt which said this fixed size universe was also infinite. This is before Big Bang theory or evidence from Hubble that the universe was expanding; expansion is part of the Big Bang theory tied to a beginning in time. SSt has no beginning in time. Im saying there are two types of infinity involved in this discussion. Actually, the evidence that the universe is expanding is not evidence that it is finite. I did not say that because the universe is expanding it is finite. I said that evidence for expansion meant that the sense of meaning of infinite in the Big Bang theory (which is tied to the beginning of time and space) was evidence to conclude that the sense of meaning of infinite in a proto-SS theory (which ties time and space in a context of _no_ beginning) was no longer as plausible as the Big Bang theory which replaced it. My comment was about the type of conceptual infinities involved in a comparison between the two theories, not an attribution of future finiteness to one. Their difference is static to dynamic. The BB theory is often described as having a finite beginning and is then infinite. The Newtonian cosmological idea has a universe infinitely old. It has no finite beginning nor finite ending (is infinite). The finiteness assertion Im making has to do with the beginning of the universe which seems defintional in both theories. Expansion favors the BB theory. The BB theory that I read has a physical universe evolution with a finite beginning. Contra, an infintely old universe does not have a finite beginning in time evolving into an infinite expression. So Ive said infinity has different meaning between the two theories. To challenge that, one would need to reconcile the meaning of infinity between the two theories. Ive explained my difference in terms of the beginning of time which implies that time was/is not infiite. Explaining why Big Bang also rightly uses the term infinity does not explain how the meanings of infinity are equivalent between the theories. Infinity to me means no beginning and no end. The BB theory says the universe is 13.7 billion years old, not infinitely old. Stephen === Subject: Re: Infinite number of people toss a coin infinite times Stephen Harris says... >The Big Bang uses a time dependent function for density and size of >the universe. This infinity seems relative to me, there is process. And >I think that is why Einsteins theory impacted the plausibility of SST. My point was that the size of the universe *isnt* a function of time in an open Big Bang cosmology. The universe starts out with infinite volume, and always has infinite volume. In a closed Big Bang universe, then yes, the volume of the universe is a function of time. -- Daryl McCullough Ithaca, NY === Subject: [Another Successful Troll by Herc] (no longer mentions): Infinite number of people toss a coin infinite times http://mygate.mailgate.org/mynews/comp/comp.theory/ 47f9ac33199d60366a9c1d786a f8e916.48257%40mygate.mailgate.org Stephen Harris |- This answers part of my question within the Big |- Bang theory context. Note that in the above |- paragraphs I have been careful to use the term |- observable Universe rather than Universe. The |- Universe itself, or the maximum amount of space |- that we will eventually be able to see given an |- infinite amount of time, may well be infinite. In |- quoting a size of the Universe we infer how far we |- can see in one direction (15 billion light years), |- and how far we can see in the other direction (15 |- billion light years) and add the two to get a size |- (30 billion light years). An age of 15 billion |- light years in each direction therefore leads us |- to infer that we are at the centre of a sphere |- with radius 15 billion light-years, and hence that |- the Universe is 30 billion light-years across. |- The trick, however, is that because the Universe |- is homogeneous and isotropic, every observer must |- measure a size of the Universe that is 30 billion |- light years... even ones that are at the edge of |- our observable Universe! This means that either |- the Universe is sufficiently curved that space |- doubles back on itself (like on the surface of a |- sphere), or that the actual Universe is much |- larger than the observable one. We currently think |- that the latter possibility is the case. Um, no, not at all. Observers on the edge of _our_ observable universe are at, or very near, the epoch of the big bang. _Their_ observable universe is much, much smaller, and so stuff there is much closer together. In the limit of our possible observation, their universe collapses to something The universe may be homogeneous and isotropic, but that is merely a hypothesis, not a theory subject to falsifiability, even _in_ theory, since we cant _see_ *that* universe. The one _we_ see grows denser and younger with distance. [It also isnt the least bit isotropic; it has structure at every scale.] xanthian. Its always valuable to remember this recent interchange before wasting a lot of Net bandwidth on another of Hercs inane trolls: Its like talking to someone with amnesia, forget Kent hes a moron who takes antipsychotics all day. -- Herc (gotch@beauty.com) Well, I just did a bit of a search through some of your recentish posts (just a few months ago) in comp.theory. [...] Seems Kent Paul Dolan was right about you, and that it would just be a waste of time for me to continue Ôdiscussing your Ôsolution(s) with you. -- Simon G Best (s.g.best@btopenworld.com) -- === Subject: Re: Infinite number of people toss a coin infinite times > In the unbounded cosmologies, the Big Bang was not a time when the > universe was at a single point---it was a time when the universe had > infinite density. Except it wasnt really a *time*, was it? I mean, there *are* times that get closer and closer to the BB, but according to current cosmology it doesnt make sense to say there was a time *at* which the BB occurred, right? (Be gentle, Im *way* out of my depth here...) Chris Menzel === Subject: Re: Infinite number of people toss a coin infinite times >> Mike Oliver says... >If theres some reason according to currently >understood physics why an already-infinite universe cant >be expanding, someone please enlighten me. It is not current, but I think part of the inconsistent steady-state theory. >> No, there is no reason. > Thats what I thought. Stephen, the point on > which you may be confused is this: In the infinite > case, an expanding universe doesnt actually mean > the universe is getting bigger, just that everythings > moving farther apart. > If you want to know how that can happen without the > universe getting bigger, someones holding an impromptu > seminar on it down at the Hilbert Hotel. All the > conference rooms are booked, but management says > that wont be a problem.... I dont think I remember this incorrectly, but just in case, I will proceed cautiously. I said there was an older SST theory than the modern (Big Bang) which was discarded because of expansion among other things. I think the SST had a static infinity. http://www.pbs.org/wnet/hawking/universes/html/univ_ steady.html Proposed in 1948 by Hermann Bondi, Thomas Gold, and Fred Hoyle, the steady-state theory was based on an extension of something called the perfect cosmological principle. This holds that the universe looks essentially the same from every spot in it and at every time. (This applies only to the universe at large scales; obviously planets, stars, and galaxies are different from the space between them.) Obviously, for the universe to look the same at all times, there could have been no beginning or no end. This struck a philosophical chord with a number of scientists, and the steady-state theory gained many adherents in the 1950s and 1960s. How could the universe continue to look the same when observations show it to be expanding, which would tend to thin out its contents? Supporters of this cosmology balanced the ever-decreasing density that results from the expansion by hypothesizing that matter was continuously created out of nothing. The amount required was undetectably small-about a few atoms for every cubic mile each year. http://www.counterbalance.net/cq-fab/stead-frame.html Statistically speaking, a steady-state universe presents the same face to observers over all time. There is no Big Bang; the Universe existed infinitely far into the past and will exist infinitely far into the future. Galaxies expand out of a given volume, to be replaced by young galaxies newly formed from matter spontaneously generated within the same volume. ------------------------------------------------------------- Olbers Paradox If the universe is infinitely old, infinitely large, and uniform why isnt the sky bright? answer - the universe is not infinitely old, infinitely large, and uniform because expansion can also solve this problem http://phyld.ucr.edu/astronomy/A-28.htm ...Explains and presents supportive evidence for the two opposing theories of the origin of the universe: the big bang theory, which proposes that it began suddenly from nothing, and the steady state theory, which holds it to be infinitely old and large. The time since the Big Bang is the expansion age of the Universe, which is simply how long the Universe has had to expand in order to get to its present size. This is a BB description. If you plug in infinite time and size as used in SST you get: The universe expanded for an infinitely long time (infinitely old) and reached _infinite_ size (infinitely large). The infinitely old means a completed or static concept of time. There is no beginning. Thus there is no start to a universe that grows larger over a time that grows infinitely larger. You cant add to infinity. If the universe is infinitely large, there is no force (gravity) which can reach the end of infinity and start a contraction of the infinitely large universe. Does the universe infinitely age? Which brings up the question of whether time ends upon heat death of the universe or if time continues because there is no end or beginning to infinity--how can something be infinitely old and yet have a finite ending in time. The SSt does not seem to me to have the notion of expanding into infinity, but that it is already infinite. I looked at this situation as a 3-d version of an infinitely ßat plane doesnt grow to infinity, it begins at infinity. Maybe the mathematical infinity is a bit more objective/absolue than the subjective/relative physical universe infinity. Stephen === Subject: Re: Infinite number of people toss a coin infinite times >>Thats what I thought. Stephen, the point on >>which you may be confused is this: In the infinite >>case, an expanding universe doesnt actually mean >>the universe is getting bigger, just that everythings >>moving farther apart. >>If you want to know how that can happen without the >>universe getting bigger, someones holding an impromptu >>seminar on it down at the Hilbert Hotel. All the >>conference rooms are booked, but management says >>that wont be a problem.... > I dont think I remember this incorrectly, but just in case, I will > proceed cautiously. I said there was an older SST theory than the > modern (Big Bang) which was discarded because of expansion > among other things. I think the SST had a static infinity. I may be out of my depth here, but I dont think thats exactly right. AIUI the steady-state theory *did* account for expansion; thats not the difference between SS and BB. In the steady-state picture, as galaxies moved away from one another (expansion), new hydrogen atoms were born ex nihilo and formed new galaxies to fill the intervening space. According to the big-bang, infinite universe version, the volume of the universe is (already) infinite, but its age is finite, and there are no new baby hydrogen atoms to replace the ones that move away. Rather, the density of the universe simply diminishes as galaxies move apart. An aside--there are two wonderful vignettes in Calvinos Le Cosmicomiche, one on each of these themes. The Big Bang one, Tutto in un punto, is IMO perhaps the best in the whole collection, or perhaps comes in behind I colori. The book has been translated into English under the title Cosmicomics and I highly recommend it. === Subject: Re: Infinite number of people toss a coin infinite times >Thats what I thought. Stephen, the point on >which you may be confused is this: In the infinite >case, an expanding universe doesnt actually mean >the universe is getting bigger, just that everythings >moving farther apart. >If you want to know how that can happen without the >universe getting bigger, someones holding an impromptu >seminar on it down at the Hilbert Hotel. All the >conference rooms are booked, but management says >that wont be a problem.... >> I dont think I remember this incorrectly, but just in case, I will >> proceed cautiously. I said there was an older SST theory than the >> modern (Big Bang) which was discarded because of expansion >> among other things. I think the SST had a static infinity. > I may be out of my depth here, but I dont think thats exactly > right. AIUI the steady-state theory *did* account for expansion; > thats not the difference between SS and BB. In the steady-state > picture, as galaxies moved away from one another (expansion), new > hydrogen atoms were born ex nihilo and formed new galaxies to > fill the intervening space. Right, I was ßoundering before I found the appropriate old theory. I will repost my reply. Maybe I will make a literaty excurison into The book has been translated into English under the title Cosmicomics and I highly recommend it. Repost: >> I said there used to be a theory that the universe was already infinite >> as in no expansion it was _already_ infinite. >> Expansion undermined this theory not because it provided evidence >> of the finiteness of the universe, but because if the universe is under >> expansion there cant be a working completed infinity of the universe >> theory. > I dont see why not. I described this as used to be a theory because I couldnt remember the name of it. I thought at first is was steady-state. But I couldnt reconclile the presence of expansion in steady-state with what I had learned about this. So I kept digging and found: http://www.physics.gmu.edu/classinfo/astr103/CourseNotes/cml_ newt.htm Newtonian Space and Time Cosmology For Newton, space and time are absolute, i.e., the same throughout the Universe and unchanging or unchangeable Absolute space, in its own nature, without relation to anything external, remains always similar and immovable... Absolute, true and mathematical time, of itself, and from its own nature, ßows equably without relation to anything external... Newtonian space time - space and time are absolutes and exist independent of the existence of matter in the Universe; space and time precede the material Universes existence and will survive its demise Decartes and Gottfried Whilhelm Liebniz (1646-1716) did not support Newtons belief in an absolute space and time Liebniz: I hold space to be something purely relative, as time is. Newtonian Cosmology Copernican Principle (Aristarchus, Copernicus, Bondi) - we are not privileged occupants of the universe, so that there is nothing exceptional about our existence and we are consequently not at the Universes center Newtonian Universe - gravity demands that the universe be centerless, edgeless, and of infinite extent in all directions SH: I kept trying to apply the static mathematical Newton infinity to the sold-state theory and it didnt quite fit. Newtons absolute space was unchanging and therefore not undergoing expansion. It seems to me there are traces of Newtons cosmology in the steady-state theory. Stephen === Subject: Re: Infinite number of people toss a coin infinite times >>Thats what I thought. Stephen, the point on >>which you may be confused is this: In the infinite >>case, an expanding universe doesnt actually mean >>the universe is getting bigger, just that everythings >>moving farther apart. >If you want to know how that can happen without the >>universe getting bigger, someones holding an impromptu >>seminar on it down at the Hilbert Hotel. All the >>conference rooms are booked, but management says >>that wont be a problem.... > I dont think I remember this incorrectly, but just in case, I will > proceed cautiously. I said there was an older SST theory than the > modern (Big Bang) which was discarded because of expansion > among other things. I think the SST had a static infinity. >> I may be out of my depth here, but I dont think thats exactly >> right. AIUI the steady-state theory *did* account for expansion; >> thats not the difference between SS and BB. In the steady-state >> picture, as galaxies moved away from one another (expansion), new >> hydrogen atoms were born ex nihilo and formed new galaxies to >> fill the intervening space. > Right, I was ßoundering before I found the appropriate old theory. > I will repost my reply. Maybe I will make a literaty excurison into > The book has been translated into English under the > title Cosmicomics and I highly recommend it. > Repost: > I said there used to be a theory that the universe was already infinite > as in no expansion it was _already_ infinite. > Expansion undermined this theory not because it provided evidence > of the finiteness of the universe, but because if the universe is under > expansion there cant be a working completed infinity of the universe > theory. >> I dont see why not. > I described this as used to be a theory because I couldnt > remember the name of it. I thought at first is was steady-state. > But I couldnt reconclile the presence of expansion in steady-state > with what I had learned about this. So I kept digging and found: > http://www.physics.gmu.edu/classinfo/astr103/CourseNotes/cml_ newt.htm > Newtonian Space and Time Cosmology > For Newton, space and time are absolute, i.e., the same throughout the > Universe and unchanging or unchangeable > Absolute space, in its own nature, without relation to anything external, > remains always similar and immovable... Absolute, true and mathematical > time, of itself, and from its own nature, ßows equably without relation > to > anything external... > Newtonian space time - space and time are absolutes and exist independent > of > the existence of matter in the Universe; space and time precede the > material > Universes existence and will survive its demise > Decartes and Gottfried Whilhelm Liebniz (1646-1716) did not support > Newtons belief in an absolute space and time > Liebniz: I hold space to be something purely relative, as time is. > Newtonian Cosmology > Copernican Principle (Aristarchus, Copernicus, Bondi) - we are not > privileged occupants of the universe, so that there is nothing exceptional > about our existence and we are consequently not at the Universes center > Newtonian Universe - gravity demands that the universe be centerless, > edgeless, and of infinite extent in all directions http://www.physics.gmu.edu/classinfo/astr103/CourseNotes/cml_ newt.htm Paradoxes of Newtonian Universe [SH: continued] Olbers paradox (Heinrich Olbers, 1823) - an infinite universe would produce an infinite amount of light at our position, so why is the night sky dark? Probably known since Keplers time; certainly Cheseaux and Halley Probably Newton aware of paradox Gravity paradox (Newton) - an infinite universe would also produce an infinite gravitational attraction at our position Newton assumed that the resolution of the gravity paradox was that stars their collective effects (very unlikely) Solution to both paradoxes comes in modern time > SH: I kept trying to apply the static mathematical Newton infinity to > the sold-state theory and it didnt quite fit. Newtons absolute space > was unchanging and therefore not undergoing expansion. It seems to > me there are traces of Newtons cosmology in the steady-state theory. > Stephen === Subject: Re: Infinite number of people toss a coin infinite times Originator: tchow@MATHSTATION029.MIT.EDU.mit.edu (Timothy Chow) >It is not current, but I think part of the inconsistent steady-state theory. Steady-state was abandoned because of the cosmic microwave background radiation and the lack of any evidence that matter/energy was being created ex nihilo, not because of any mathematical inconsistency. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Infinite number of people toss a coin infinite times >>It is not current, but I think part of the inconsistent steady-state >>theory. > Steady-state was abandoned because of the cosmic microwave background > radiation and the lack of any evidence that matter/energy was being > created > ex nihilo, not because of any mathematical inconsistency. There is no argument against these being valid reasons. There is also a hydrogen argument which supports abandonment of SSt. > -- Does steady-state use the idea of a universe infinitely old and large? Does Big Bang use the idea of the universe which has a finite beginning in time and probably an infinite continuation? I said those ideas of infinity (meanigs) are not consistent. Einstein fudged his results in order to obtain a static universe. Hubbel found evidence for expansion of the universe. I said this undermined SST. Because: Obler asked why there wasnt more starlight if the universe was infinitely old and large. Expansion provides an explanation which challenges the SSt assumption of an infinitely old and large universe. That does not assert that expansion shifts the static infinity of SSt to the dynamic infinity of Big Band. That is done by the interprettion SSt uses of time as having no beginning (infinitely old) vs. having a beginning, BB. That does not seem like a mathematical inconsistency so much as a by choice theoretical lack of agreement. Are you saying those are equivalent? Stephen === Subject: Re: Infinite number of people toss a coin infinite times > -After n tosses, the chance that any two people will have generated > -the same sequence is 2^-n. Since lim n -> oo. 2^-n = 0 there is > -no chance that any two people in the set will generate the same > -sequence indefinitely. > Incorrect. P=0 does not mean no chance, by that logic any infinite > sequence is impossible. there is 1 chance in oo. 1/oo = 0. Thats what the concept of probability density is for when dealing with continuous probability functions. -- Ralph === Subject: Re: Infinite number of people toss a coin infinite times > But your statement that 2 people cannot form the same sequence is > ßawed. > 1 person forming any sequence is P=0, is there no chance of that? Funnily enough, the answer is yes. To illustrate, let me change the problem statement to something equivalent, but easiser to work with: a player randomly chooses a real number in the range [0, 1); for any number n in this range, what is the probability P(n) that the player picks n? (This is the same as the coin tossing problem if we write n down in binary). If P(n) is non-zero then the sum of the probabilities for the different outcomes must be greater than 1 since there are an infinite number of such n. Therefore P(n) must be zero. What you need is a probability density function P(x, dx) which gives the probability of the player choosing a number in the range [x, x + dx). -- Ralph === Subject: Re: Sum of values of cards. Hard. > Im developing a game with 60 cards. Each card has a small integer value > (values up to 20, for example). There are groups of cards sharing the same > value, obviously. > This is a multiplayer game. It can be played by 2, 3, 4, 5 or 6 players. The > full deck of cards is delaed so that... > 1. Each player has the same number of cards (30 for a 2 player game, 20 for > a 3 player game...). > 2. The sum of the values of the cards of each player musy be the same!!! > (Trivial case where all values are the same is not permited). > Haw can I obtain such value distribution? How about having four cards of each denomination up to 15? The 60 cards then add up to 480, and Im sure that with a bit of trial and error you can split them into two sets of 30, each summing to 240, or three sets of 20, each summing to 160, or ... or 6 sets of 10, each summing to 80. Another idea: have 3 of each denomination up to 20, except dont have any elevens, have 6 ones instead. Now the sum of all the cards is 600, and again some trial and error should get you the deals you want. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Prime ideals in Z[x] How do prime ideals in Z[x] look like? I know Z[x] is noetherian, but I couldnt find any prime ideal that would have more than two generators, so... maybe there isnt any? :-) sirix. === Subject: Re: Prime ideals in Z[x] days. My association with the Department is that of an alumnus. >How do prime ideals in Z[x] look like? I know Z[x] is noetherian, but I >couldnt find any prime ideal that would have more than two generators, >so... maybe there isnt any? :-) Here is a post with an answer to that question, from Bill Dubuque: You can safely ignore my quoted response, which is harder than it should be. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Prime ideals in Z[x] >>How do prime ideals in Z[x] look like? I know Z[x] is noetherian, but I >>couldnt find any prime ideal that would have more than two generators, >>so... maybe there isnt any? :-) > Here is a post with an answer to that question, from Bill Dubuque: > You can safely ignore my quoted response, which is harder than it > should be. a curiosity, just to show what will we be working on after geting through Atiyah-McDonald) something like that: If the ring has a big Krull dimension then it is not a UFD. As a matter of fact, bigger the Krull dim is, worse things may happen with any kind of uniqueness. What precisely did he mean? I thought that there is something like Ring has Krull dim=1 iff Ring is UFD but its not the case, as in Z[x] (x) subset (x, 2) and both (x) and (x,2) are prime... sirix. === Subject: Re: Prime ideals in Z[x] days. My association with the Department is that of an alumnus. >How do prime ideals in Z[x] look like? I know Z[x] is noetherian, but I >couldnt find any prime ideal that would have more than two generators, >so... maybe there isnt any? :-) >> Here is a post with an answer to that question, from Bill Dubuque: >> You can safely ignore my quoted response, which is harder than it >> should be. >a curiosity, just to show what will we be working on after geting >through Atiyah-McDonald) something like that: If the ring has a big >Krull dimension then it is not a UFD. As a matter of fact, bigger the >Krull dim is, worse things may happen with any kind of uniqueness. What >precisely did he mean? I thought that there is something like Ring has >Krull dim=1 iff Ring is UFD but its not the case, as in Z[x] (x) >subset (x, 2) and both (x) and (x,2) are prime... No, the Krull dimension is no guarantee; you have there an example of a UFD which has Krull dimension greater than 1; and Z[sqrt(-5)] is an example where the Krull dimension is 1 but the ring is not a UFD. Im not entirely sure what he meant; you can get arbitrarily high Krull dimension and still have a UFD, simply by taking things like Z[x1,....,xn]. Then you have the chain 0< (x1) < (x1,x2) < (x1,x2,x3) < ... < (x1,...,xn) < (2,x1,...,xn) so the dimension is at least n+1. On the other hand, all but one of those prime ideals come from the transcendence degree. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Prime ideals in Z[x] posting-account=erNIBwwAAAAsTnFUogbcaZ25p1YwPrer > No, the Krull dimension is no guarantee; you have there an example of > a UFD which has Krull dimension greater than 1; and Z[sqrt(-5)] is an > example where the Krull dimension is 1 but the ring is not a UFD. > Im not entirely sure what he meant; you can get arbitrarily high > Krull dimension and still have a UFD, simply by taking things like > Z[x1,....,xn]. Then you have the chain > 0< (x1) < (x1,x2) < (x1,x2,x3) < ... < (x1,...,xn) < (2,x1,...,xn) > so the dimension is at least n+1. On the other hand, all but one of > those prime ideals come from the transcendence degree. More generally, a given integral domain R is a UFD if and only if all prime ideals of height 1 are principal. It is easy to show that any prime ideal of height r>0 must be generated by at least r elements. The above tells us is that the converse holds for r=1 <=> R is a UFD. === Subject: Re: Root Finder 12 > You people are savage, and theres no call for it. I suppose > you are one of the anal perfectionist obsessed with keeping on > course down to the Angstrom to offset the Gudermanian. > Newtons Method is an invention of genius. Why not use it? > Newton just didnt come up with the approximations to plug into > it... > or did he? No, but they have been done, and in mathematics, it doesnt matter who does it, just that it is true and can be used. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newtons method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman P.S. I can use cut and paste, too. > > > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If theres a third possibility, Id > welcome your explanation. > > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com > === Subject: Re: Root Finder 12 > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 So, youre saying that t=sqrt(24/25) is a root of x^2+2t-3=0. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newtons method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldnt believe it. >>But I tried it, and it works. >It is found that the roots to the polynomial, >a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] >>T= -----D >> D*N >D=(1,2,3,4,...,n) >> > Ok then, attempt to construct the quadratic formula for your method: > > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > > Clearly you are dead wrong, since the correct ts are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 12 > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 No, this says that t = sqrt(24/25), which is not a root of t^2+2t-3. My claim that your method cant solve quadratic equations is thus proven by your example. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newtons method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. > ex. > at^2+bt+c=0 > N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5 > Q*D=(-c/[b^2+a^2])(b+2a) > (mD-Q)*D=0 > m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a) > T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2) > t^2 =(2/5)(-c/[b^2+a^2])(b+2a) > b+/-{b^2-4ac}^(1/2) > t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=-------------------- > 2a > solve and find the required correction What is this correction? A fudge factor, maybe? -- Christopher Heckman > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldnt believe it. >>But I tried it, and it works. >>[...] >Solve the 6th degree polynomial, >f(t)=t^6 + t - 10=0 >a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) >a[0]=-10 D*N=7 > -a[0] 10 >>T = -----D = ---(1,2,3,4,5,6) >> D*N 7 >t =10/7 t=1.42835 f(1.42835)=-0.071 > > > 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says > that any rational solutions to this polynomial are +/-1, +/-2, +/5, > or +/10. It can only be an approximation. > > Strike Twelve. > > (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- > can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . > > >Solve the 49th degree polynomial, >f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 >a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) >D*N=1+200+16+49=266 > -a[0] 6000 >>T = -----D = ------(1,2,3,..,49) >> D*N 266 >t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 > > > Once again, the Rational Root Theorem says that the only possible rational > roots are integers. > > >>Applying Newtons Method, > > > Ah, so. You arent finding roots after all, only approximations to them. > Youve been told repeatedly that this isnt the same as finding the roots. > > >>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200( 1.15375)^4] > =1.15375 again, so the answer must depend on distant decimal >>places. > > > The problem here is you dont have enough precision to make Newtons > Method work. > > >>[...] >t^2 + 2t - 3 = 0 >a[0]=-3 N=(2,1) D=(1,2) D*N=4 > -a[0] 3 >>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 >> D*N 4 > > > What? Your method cant even solve a quadratic equation? Thats when > you know its really bad. > -- Christopher Heckman === Subject: Re: Root Finder 12 Dodging the issue, heh? I proved your method failed for the general formula, and all you did was post one of the coincidences I mentioned. Why dont you even try to discuss the fact I proved that the general quadratic can not be factored by your method? > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldnt believe it. >>But I tried it, and it works. >It is found that the roots to the polynomial, >a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] >>T= -----D >> D*N >D=(1,2,3,4,...,n) >> > Ok then, attempt to construct the quadratic formula for your method: > > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > > Clearly you are dead wrong, since the correct ts are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 13 > Root Finder 13 Strike Thirteen. > Jon Giffen > Another approach is considered, along with a possibility for > finding the root to an Infinite Series. > It is discovered that the property of the nth degree polynomial, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where > N=(a[1],a[2],a[3],...,a[n]) > T=(t.t^2.t^3,t^4,..., t^n ) The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman > [non sequitor part of the post has been cut] === Subject: Re: Root Finder 13 > Root Finder 13 > Jon Giffen > Another approach is considered, along with a possibility for > finding the root to an Infinite Series. > It is discovered that the property of the nth degree polynomial, Your discovery is wrong, I will explain a few more lines down. > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where > N=(a[1],a[2],a[3],...,a[n]) > T=(t.t^2.t^3,t^4,..., t^n ) > Is, > |C|^2 > |T|^2=(-a[0])^2{-------------------} > |N|^2|C|^2-(N*C)^2 > where > C=(a[1],2a[2],3a[3],4a[4],...,na[n]) > f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 Try changing the constant term of the polynomial to 1000. You will see that your method fails miserably. > a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) > |C|^2=547 |N|^2=48 N*C=153 > |N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 > |C|^2 D > T =(-a[0]){----------------------}^(1/2) --- > |N|^2 |C|^2 - (N*C)^2 |D| where This equation is your method in a nutshell. If it is wrong, your whole method is wrong. Lets say we have two polynomials who differ only in the constant term. Lets call the polynomials W and V and their respective constant terms w and v. Since the equation above has a scalar term of a[0], that equation then implies that if (x*w) is a root of W then that (x*v) is a root of V. That is clearly false, just compare the two polynomials x^5 - 2x - 28 = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first polynomial, which is 2, was found by your method. By the fact that a[0] is scalar in the equation the root was derived from, your method would say the second polynomial would have a root of 237/14=16.93, which is clearly false, since the root is 3. So if you choose any polynomial where your method created an accurate answer, I can find an infinite number polynomials where your method fails miserably by simply changing the constant term. Because of this, your method is fatally ßawed and totally wrong. > D=(1,2,3,4,5,6) and > 100 547 > T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) > 91^(1/2) 2847 > t^6=27.5696 t=1.738097 > t^5=22.9746 t=1.871758 > ---------- > t=1.856637 is the correct root > Notice that the root to a polynomial that is so long, that > it is virtually an infinite Power Series; is found by using > the solution to the Geometric Series, > |C|^2 > |T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{-------------- ----} > |N|^2|C|^2-(N*C)^2 > adding 1 to both sides, > |C|^2 > 1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------ }+1 > |N|^2|C|^2-(N*C)^2 > then the sum S=1/(1-t^2) but > |C|^2 > S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and > |N|^2|C|^2-(N*C)^2 > t ={1 - 1/S}^(1/2) where > N=(a[1],a[2],a[3],...,a[n]) > T=(t,t^2,t^3,t^4,..., t^n ) > C=(a[1],2a[2],3a[3],4a[4],...,na[n]) > to the nth degree power series, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 > Development > Suppose the polynomial, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 > Is expressed as, > a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 > Where c is almost 1 then dividing the two, > a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] > --------------------------------------------------- > a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] > Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, > T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, > N*(K-N) > T is parallel to N - ------- K and > |K-N|^2 > N*(K-N) > [m(N - -------- K) - Q]*N=0 solve this for m. Then > |K-N|^2 > N*(K-N) > T= m(N - -------- K) > |K-N|^2 > Substitute m in the above and take the square of the > magnitude of both sides. Then > (K-N)*(K-N) 0 > |T|^2=Lim (-a[0])^2 ---------------------------- = --- > c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 > applying LHopital two times with respct to c, > |C|^2 > |T|^2=(-a[0])^2{-------------------} > |N|^2 |C|^2-(N*C)^2 > where > d > C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) > c->1 dc > E.O.P. > Jon Giffen > http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 13 >>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 > Try changing the constant term of the polynomial to 1000. You will > see that your method fails miserably. f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t + 1000 = 0 a[0]=1000 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) N*D=33 t^6 = (-1000/33)(6) = -2000/11 (2k+1)(pi) (2k+1)(pi) t = {2000/11}^(1/6)[cos ---------- + i sin -----------] 6 6 Binomial Equation, where k=0,1,2,3,4,5 > Lets say we have two polynomials who differ only in the constant term. > Lets call the polynomials W and V and their respective constant terms > w and v. Since the equation above has a scalar term of a[0], that > equation then implies that if (x*w) is a root of W then that (x*v) is > a root of V. > That is clearly false, just compare the two polynomials x^5 - 2x - 28 > = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first > polynomial, which is 2, was found by your method. By the fact that > a[0] is scalar in the equation the root was derived from, your method > would say the second polynomial would have a root of 237/14=16.93, > which is clearly false, since the root is 3. > So if you choose any polynomial where your method created an accurate > answer, I can find an infinite number polynomials where your method > fails miserably by simply changing the constant term. Because of > this, your method is fatally ßawed and totally wrong. x^5 - 2x - 28 a[0]=-28 N=(-2,0,0,0,1) D=( 1,2,3,4,5) -a[0]/(N*D)=28/3 (t,t^2,t^3,t^4,t^5)=(28/3)(1,2,3,4,5) t^5 = (28/3)(5) t=(140/3)^(1/5)=2.1567 (should be 2) x^5 - 2x - 237 a[0]=-237 N=(-2,0,0,0,1) D=( 1,2,3,4,5) -a[0]/(N*D)=237/3 (t,t^2,t^3,t^4,t^5)=(237/3)(1,2,3,4,5) t^5 = (237/3)(5) t=(1185/3)^(1/5)=3.3061 (should be 3) In general, the root to the nth degree polynomial a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n where T=(t,t^2,t^3,...,t^n) N=(a[1],a[2],a[3],...,a[n]) D=(1,2,3,...,n) is approximated by the identity, -a[0] T = ----- D decoding, N*D -a[0] t = -------------------------- a[1]+2a[2]+3a[3]+...+na[n] -2a[0] t^2= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] -3a[0] t^3= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] . . . -na[0] t^n= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] >>http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 13 Yes I see that squaring all terms leads to a lack of inheritance of the property of negatives in the series. N*C destroys them as well. Root Finder 13 appears to be inferior. Root Finder 12, though, still seem to hold promise. >>Root Finder 13 >>Jon Giffen >>Another approach is considered, along with a possibility for >>finding the root to an Infinite Series. >>It is discovered that the property of the nth degree polynomial, > Your discovery is wrong, I will explain a few more lines down. >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where >>N=(a[1],a[2],a[3],...,a[n]) >>T=(t.t^2.t^3,t^4,..., t^n ) >>Is, >> |C|^2 >>|T|^2=(-a[0])^2{-------------------} >> |N|^2|C|^2-(N*C)^2 >>where >>C=(a[1],2a[2],3a[3],4a[4],...,na[n]) >>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 > Try changing the constant term of the polynomial to 1000. You will > see that your method fails miserably. >>a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) >>|C|^2=547 |N|^2=48 N*C=153 >>|N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 >> |C|^2 D >>T =(-a[0]){----------------------}^(1/2) --- >> |N|^2 |C|^2 - (N*C)^2 |D| where > This equation is your method in a nutshell. If it is wrong, your > whole method is wrong. > Lets say we have two polynomials who differ only in the constant term. > Lets call the polynomials W and V and their respective constant terms > w and v. Since the equation above has a scalar term of a[0], that > equation then implies that if (x*w) is a root of W then that (x*v) is > a root of V. > That is clearly false, just compare the two polynomials x^5 - 2x - 28 > = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first > polynomial, which is 2, was found by your method. By the fact that > a[0] is scalar in the equation the root was derived from, your method > would say the second polynomial would have a root of 237/14=16.93, > which is clearly false, since the root is 3. > So if you choose any polynomial where your method created an accurate > answer, I can find an infinite number polynomials where your method > fails miserably by simply changing the constant term. Because of > this, your method is fatally ßawed and totally wrong. >>D=(1,2,3,4,5,6) and >> 100 547 >>T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) >> 91^(1/2) 2847 >>t^6=27.5696 t=1.738097 >>t^5=22.9746 t=1.871758 >> ---------- >>t=1.856637 is the correct root >>Notice that the root to a polynomial that is so long, that >>it is virtually an infinite Power Series; is found by using >>the solution to the Geometric Series, >> |C|^2 >>|T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------ ------} >> |N|^2|C|^2-(N*C)^2 >>adding 1 to both sides, >> |C|^2 >>1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{---------------- --}+1 >> |N|^2|C|^2-(N*C)^2 >>then the sum S=1/(1-t^2) but >> |C|^2 >>S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and >> |N|^2|C|^2-(N*C)^2 >>t ={1 - 1/S}^(1/2) where >>N=(a[1],a[2],a[3],...,a[n]) >>T=(t,t^2,t^3,t^4,..., t^n ) >>C=(a[1],2a[2],3a[3],4a[4],...,na[n]) >>to the nth degree power series, >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 >>Development >>Suppose the polynomial, >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 >>Is expressed as, >>a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 >>Where c is almost 1 then dividing the two, >>a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] >>--------------------------------------------------- >> a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] >>Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, >>T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, >> N*(K-N) >>T is parallel to N - ------- K and >> |K-N|^2 >> N*(K-N) >>[m(N - -------- K) - Q]*N=0 solve this for m. Then >> |K-N|^2 >> N*(K-N) >>T= m(N - -------- K) >> |K-N|^2 >>Substitute m in the above and take the square of the >>magnitude of both sides. Then >> (K-N)*(K-N) 0 >>|T|^2=Lim (-a[0])^2 ---------------------------- = --- >> c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 >>applying LHopital two times with respct to c, >> |C|^2 >>|T|^2=(-a[0])^2{-------------------} >> |N|^2 |C|^2-(N*C)^2 >>where >> d >>C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) >> c->1 dc >>E.O.P. >>Jon Giffen >>http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Root Finder: Angle Subtending Arc and Chord ANGLE SUBTENDING ARC AND CHORD On a circle of unknown radius, arc A and chord B subtend angle x, which is obtained by the formula, x=5.062792913(1-B/A)^(1/2) B/A x from FORMULA x SHOULD BE 0.995892735 0.324463996 0.314159265 0.983631643 0.647728053 0.628318531 0.963397762 0.968598927 0.942477796 0.935489284 1.28589674 1.256637061 0.900316316 1.598461583 1.570796327 0.858393691 1.905160043 1.884955592 0.810331958 2.204891605 2.199114858 0.756826729 2.496594918 2.513274123 0.698646585 2.77925389 2.827433388 0.636619772 3.051903588 3.141592654 1 <= B/A <= 2/pi Development Let x=angle (unknown) A=arc B=chord then from trigonometry, B x - - = sin(x/2) A 2 and letting t=(x/2)^2, 1 1 1 1 (1-B/A)- ---t + ---t^2 - ---t^3 + ---t^4 - .... 3! 5! 7! 9! then N=(-1/3! , 1/5! , -1/7! , 1/9! , .... ) and -a[0] T = ----- Root Approximation Formula D*N where a[0]=(1-B/A) D=(1,2,3,4,5..) N=(-1/3! , 1/5! , -1/7! , 1/9! , .... ) then oo n 1 2 3 4 D*N = Sum(-1)^n ------- = - --- + --- - --- + --- - .... n=1 (2n+1)! 3! 5! 7! 9! Carrying out the calculation up to n=7, D*N= -0.1505843395 1/(D*N)=-6.640796802 (t,t^2,t^3,t^4,t^5,t^6,t^7)=6.640796802(1-B/A)(1,2,3,4,5,6,7) Selecting the first component, t=6.640796802(1-B/A)(1) but since t=(x/2)^2 , x=2(t^1/2) and x=5.062792913(1-B/A)^(1/2) This sheds merit on -a[0] T = ----- Root Approximation Formula D*N proving its utility in arriving within enough accuracy to require only a few iterations of Newtons Method to obtain a very precise (and accurate) result. Jon Giffen === Subject: Re: Root Finder: Angle Subtending Arc and Chord > ANGLE SUBTENDING ARC AND CHORD > On a circle of unknown radius, arc A and chord B > subtend angle x, which is obtained by the formula, > x=5.062792913(1-B/A)^(1/2) > B/A x from FORMULA x SHOULD BE > 0.995892735 0.324463996 0.314159265 > 0.983631643 0.647728053 0.628318531 > 0.963397762 0.968598927 0.942477796 > 0.935489284 1.28589674 1.256637061 > 0.900316316 1.598461583 1.570796327 > 0.858393691 1.905160043 1.884955592 > 0.810331958 2.204891605 2.199114858 > 0.756826729 2.496594918 2.513274123 > 0.698646585 2.77925389 2.827433388 > 0.636619772 3.051903588 3.141592654 > 1 <= B/A <= 2/pi What this table tells me is that FORMULA, whatever it is, is wrong. Check out the difference between pi and (2143/22)^(1/4). pi = 3.14159265353... (2143/22)^(1/4) = 3.14159265258... Just because two things are close doesnt mean theyre equal. -- Christopher Heckman === Subject: Re: Root Finder: Angle Subtending Arc and Chord > ANGLE SUBTENDING ARC AND CHORD > On a circle of unknown radius, arc A and chord B > subtend angle x, which is obtained by the formula, > x=5.062792913(1-B/A)^(1/2) > B/A x from FORMULA x SHOULD BE > 0.995892735 0.324463996 0.314159265 > 0.983631643 0.647728053 0.628318531 > 0.963397762 0.968598927 0.942477796 > 0.935489284 1.28589674 1.256637061 > 0.900316316 1.598461583 1.570796327 > 0.858393691 1.905160043 1.884955592 > 0.810331958 2.204891605 2.199114858 > 0.756826729 2.496594918 2.513274123 > 0.698646585 2.77925389 2.827433388 > 0.636619772 3.051903588 3.141592654 Using your approximation formula, the worst |relative error| is roughly 3%. > 1 <= B/A <= 2/pi Thats not possible, and so I suppose that you intended to write 2/pi <= B/A <= 1 instead. > Development > Let > x=angle (unknown) > A=arc > B=chord > then from trigonometry, > B x > - - = sin(x/2) > A 2 > and letting t=(x/2)^2, > t=6.640796802(1-B/A)(1) but since > t=(x/2)^2 , x=2(t^1/2) and > x=5.062792913(1-B/A)^(1/2) I fail to see how the coefficient 5.062792913 was obtained. Indeed, it seems clear from the end of your development that we should instead have had x = 2*t^(1/2) = 2*(6.640796802*(1 - B/A))^(1/2) = 5.1539487*(1 - B/A)^(1/2) Using that approximation, relative error exceeds 5% when B/A is near 1. > This sheds merit on > -a[0] > T = ----- Root Approximation Formula > D*N > proving its utility in arriving within enough accuracy to > require only a few iterations of Newtons Method to obtain > a very precise (and accurate) result. Im not sure that it sheds merit. How did 5.06... arise, rather than 5.15...? It is easy to establish that a simple approximation is x = 2*Sqrt(6*(1 - B/A)). The relative error in that approximation is worst, about -6%, when B/A = 2/pi, and its |relative error| decreases to 0 as B/A approaches 1. Instead of the coefficient 2*Sqrt(6), which is about 4.9, if we use a somewhat larger coefficient, then we can reduce worst |relative error|, assuming of course that B/A is restricted to the interval [2/pi, 1]. Specifically, to minimize worst |relative error|, it can be shown that the coefficient should be 5.0504... In any event, its not clear how you arrived at 5.062792913 for the coefficient. Please explain. David Cantrell === Subject: C code for Whittle Estimator Im wondering, does anybody have a C code for Whittle Estimator ? Dmitry === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > Mathematics doesnt always erase confusion in our intuition by > conforming to it; it also does it by _informing_ it. > > It seems unintuitive (at least to me) at first that, even in > principle, you cant trisect every angle using the usual straight edge > and compass; but it is something you can prove; and eventually it > becomes intuitive. > I disagree with this formalist interpretation of mathematics. That is your perogative. But to return to the topic of the thread, it should have no bearing on the correctness or incorrectness of Zenkins argument - his argument fails because it doesnt follow logically; not because he espouses a formalist or other philosophy. > Your intuition was simply wrong at first. That was my point. Your intuition regarding size of a set appears to be wrong as well - at least when you use the standard definition of size of a set. If you want a definititon of size of a set which conforms to your intuitions, youll need to be more specific about what your intuitions require of the term, size of a set. > Its muddled to think that > you just get used to some formal relation. If you dont understand it, > its just syntax. Until you understand it, its just syntax. When you understand it, it becomes incorporated in your new, _correct_ intuition about the result in question. Or dont you think one can acquire new intuitions which extend and correct ones old ones? > (So, I dont like that statement of Von Neumann at > all...) Thats another argument, though, and I dont think its > necessary for me to make it. > Why should the _name_ of the things > we define have any significance in the logical systems we construct? > I never said the names matter. What matters is that they have > reference. I think that should be obvious to anybody who took a logic > class. You have claimed (without particular proof) that cardinality is antimonius - i.e., leads to a _logical_ contradiction. Then if we define the term chasanality identically to cardinality, and list its properties, then it should create a _logical_ contradiciton iff cardinality creates a _logical_ contradicition. I have seen no evidence of such a contradicition. It seems the antimony you are claiming is not a _logical_ contradiction, but instead offends your preconception of what size of a set _should_ mean. The onus is then on you to resolve this for yourself via a different definititon of size of a set. The standard one seems to work well for most mathematicians; in particular, as others have pointed out, it is consistent with ones intuition regarding finite sets, unlike the alternate definition you mentioned. === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > It seems the antimony you are claiming is not a _logical_ > contradiction, but instead offends your preconception of what size of > a set _should_ mean. > The onus is then on you to resolve this for yourself via a different > definititon of size of a set. The standard one seems to work well > for most mathematicians; in particular, as others have pointed out, it > is consistent with ones intuition regarding finite sets, unlike the > alternate definition you mentioned. -- Eray Ozkural === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) >> > Since there is the observed antinomy of the infinitely big, > unfortunately size of a set is far from being an obvious concept. > > What antinomy is that? > Im sure you will find a few philosophers who have better command of >> English than I have. >> >> Cant find anything. Can you name some names? Give a hint of what >> the actual antinomy is? Give some reference more specific than >> Google? > http://ls.poly.edu/~jbain/philmath/philmathlectures/M01. Intro.pdf > There are probably better expositions than these slides, but thats > the best thing I could find. > Certainly, you should check out the rest of the site. That way, > youll see that his solution to the *apparent* paradoxes of infinite > sets is to follow Cantor. As far as I can tell, he presents the slide > you refer to for historical reasons and to motivate the later > discussion on Cantor. > See > . Ive read all the notes, yes. Cantor proposes a solution. Maybe Galileos paradox still persists, though. > In any case, lets not confuse lecture notes with philosophy of > mathematics research. Why not point me to a recent publication on > paradoxes of infinitely big written by a working philosopher of > mathematics? > (Not that Bain is or isnt a working philosopher of mathematics, but > lecture notes arent a good indication of current issues in > philosophy.) Lecture notes, inspired by textbooks he teaches. Philosophy is not caution: I should be looking in the textbooks myself. If the current poster did not think that there could be more preferable formalizations of infinity, he would not suggest looking in Galileos paradox. He would have deemed it solved. -- Eray Ozkural === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) <87zn187lzj.fsf@phiwumbda.org> <87fz2zfzid.fsf@phiwumbda.org> <87d5y1vm2f.fsf@phiwumbda.org> <87wtw7pa62.fsf@phiwumbda.org> Discussion, linux) >> See >> . > Ive read all the notes, yes. Cantor proposes a solution. > Maybe Galileos paradox still persists, though. Yeah, maybe. Except, of course, it doesnt in any reasonable sense. Cantors set theory disposes of Galileos paradox to the same degree that Freges logical foundations overthrew the ontological proof of the existence of God. >> In any case, lets not confuse lecture notes with philosophy of >> mathematics research. Why not point me to a recent publication on >> paradoxes of infinitely big written by a working philosopher of >> mathematics? >> (Not that Bain is or isnt a working philosopher of mathematics, but >> lecture notes arent a good indication of current issues in >> philosophy.) > Lecture notes, inspired by textbooks he teaches. Philosophy is not > caution: I should be looking in the textbooks myself. Philosophy is indeed not like technology. For the philosophy instructor, it is imperative that he puts the issues in context by showing the questions that needed to be answered. Even the mathematician doesnt have to motivate Cantor in the same historical manner as the philosopher. It is simply stupid to look to lecture notes in a philosophy class to measure what the current philosophical questions are---especially when the questions youre looking at are later resolved *in the same notes*. > If the current poster did not think that there could be more > preferable formalizations of infinity, he would not suggest looking > in Galileos paradox. He would have deemed it solved. Current poster? Which current poster? Why should I care about him? -- ÔEvery man who has ever lived holds tight to the belief that for him Therefore, you will marry Guinevere. You do not want advice --- only agreement. Merlin sighed... -- John Steinbeck === Subject: Category Theory books Hi all, Im intrested in buying a good book about category theory. I would like to buy (only)one of the following: 1)Categories for the Working Mathematician Saunders Mac Lane; 2)Conceptual Mathematics: A First Introduction to Categories F. William Lawvere, Stephen H. Schanuel . I need a good piece of advice... I would really appreciate it. TIA. PS: other books about categories are welcome. === Subject: Re: Category Theory books > Hi all, > Im intrested in buying a good book about category theory. I would like to > buy (only)one of the following: > 1)Categories for the Working Mathematician > Saunders Mac Lane; > 2)Conceptual Mathematics: A First Introduction to Categories > F. William Lawvere, Stephen H. Schanuel . > I need a good piece of advice... I would really appreciate it. > TIA. > PS: other books about categories are welcome. It is difficult to give advice here without knowing your current state of mathematical experience. I am glad to have both of them, but in my opinion the are aimed at different audiences. The book (1) is the standard introduction for those students who have already met enough mathematics to be able to appreciate the examples. In other words, the title should be taken seriously to some degree. It will provide you with most of the technical tools you ever need. In contrast (2) can be digested by first year students or sufficiently bright high-school kids. The examples and illustrations are taken from finite sets and variants thereof. It cannot go so deep into technical details, but explains a lot and in a way teaches to Ôthink deeply about the simple things. Perhaps the most useful advice is: try to obtain these books through a nearby (university) library and take a look at them before making a decision. Marc === Subject: Re: Category Theory books >Hi all, >Im intrested in buying a good book about category theory. I would like to >buy (only)one of the following: >1)Categories for the Working Mathematician > Saunders Mac Lane; >2)Conceptual Mathematics: A First Introduction to Categories > F. William Lawvere, Stephen H. Schanuel . >I need a good piece of advice... I would really appreciate it. >TIA. I dont know about the second, but the first is the standard classic on the subject written by one of the founders, and it has had a recent second edition. OTOH I remember vaguely finding online J. Ad.87mek, H. Herrlich and G.E. Strecker -- Abstract and Concrete Categories But I dont have the url. And I dont like it very much -- it is well written but IMHO the emphasis is misplaced. Check also: http://www.geocities.com/alex_stef/mylist.html there are some interesting links there. G. Rodrigues === Subject: Re: Category Theory books >>Im intrested in buying a good book about category theory. I would like to >>buy (only)one of the following: >>1)Categories for the Working Mathematician >> Saunders Mac Lane; >>2)Conceptual Mathematics: A First Introduction to Categories >> F. William Lawvere, Stephen H. Schanuel . ... >I dont know about the second, but the first is the standard classic >on the subject written by one of the founders, and it has had a recent >second edition. The second seems to be written for beginning mathematics students. It starts with simple examples of what sets and maps between sets are. This leads to the idea of what morphisms are and gradually introduces the ideas underlying categories. At one time I think I remember it was published as a notebook of loose pages, or maybe it was bound with those plastic fingers that you fit into perforations down the side of the page, perhaps to make this as inexpensive for students as possible. Then Cambridge sprang for a new version that is paperbound. My copy is buried somewhere and I cant find it at the moment to give a summary of the sections in the book. But for the audience that I think it was intended for I believe it could be a good book. For something that can be as slippery and intangible to a beginner as category theory easily is I think this could lead them to a place where they could begin seeing everything as categories. === Subject: Re: Category Theory books > Hi all, > Im intrested in buying a good book about category theory. I would like to > buy (only)one of the following: > 1)Categories for the Working Mathematician > Saunders Mac Lane; > 2)Conceptual Mathematics: A First Introduction to Categories > F. William Lawvere, Stephen H. Schanuel . > I need a good piece of advice... I would really appreciate it. > TIA. > PS: other books about categories are welcome. While Im not familiar with the latter book, the first is a classic IMO and is one Ive used. (and one I probably ought to buy) === Subject: Re: help finding a closure of a set under som operations > hi... > need some help here... > my world consists of all finite sets of integers. > an inerval [a,b] is the set of all integers between a and b > (inclusive). > given two sets one can define three operations: interestion, addition, > and multiplication by constant. > intersection is the simple set intersection. > the addition of two sets is the set consisting of all sums of one > integer in the first set and the other in the second. > multiplication by a constant results in a set in which every element > is times an integer from the original set. > the family of all intervals is closed under intersections and > additions, as one can easily verify. > the family of all finite arithmetic sequences is closed under > intersections and multiplication by constants. > can you help me(send a ref, search keyword, share ideas, or give me > the result if known) finding the smallest family of finite integer > subset that is closed under all three operations, and includes all > intervals? > amit. Try this (sketch). You are looking at the set of finite subsets of N generated by the intervals and closed under pointwise sum, intersection and pointwise scalar multiplication. Call those sets constructible. I claim it consists of all finite subsets. Suppose every k element set is constructible. Let {a_1,...,a_{k+1)} be a set with a_1 < ... < a_{k+1}. Let n = a_{k+1} - a_1. Then the inductive hypothesis implies that {a_1,...,a_k} is constructible and certainly {n} is. Their pointwise sum is {a_1,...,a_k,a_{k+1}=a_1+n,a_2+n,...}. Now intersect the latter set with the interval [1,...,a_{k+1}]. === Subject: Re: help finding a closure of a set under som operations > hi... > > need some help here... > > my world consists of all finite sets of integers. > an inerval [a,b] is the set of all integers between a and b > (inclusive). > > given two sets one can define three operations: interestion, addition, > and multiplication by constant. > intersection is the simple set intersection. > the addition of two sets is the set consisting of all sums of one > integer in the first set and the other in the second. > multiplication by a constant results in a set in which every element > is times an integer from the original set. > > > the family of all intervals is closed under intersections and > additions, as one can easily verify. > the family of all finite arithmetic sequences is closed under > intersections and multiplication by constants. > > can you help me(send a ref, search keyword, share ideas, or give me > the result if known) finding the smallest family of finite integer > subset that is closed under all three operations, and includes all > intervals? > > > amit. > Try this (sketch). You are looking at the set of finite subsets of N > generated by the intervals and closed under pointwise sum, > intersection and pointwise scalar multiplication. Call those sets > constructible. I claim it consists of all finite subsets. Suppose > every k element set is constructible. Let {a_1,...,a_{k+1)} be a set > with a_1 < ... < a_{k+1}. Let n = a_{k+1} - a_1. Then the inductive > hypothesis implies that {a_1,...,a_k} is constructible and certainly > {n} is. Their pointwise sum is {a_1,...,a_k,a_{k+1}=a_1+n,a_2+n,...}. > Now intersect the latter set with the interval [1,...,a_{k+1}]. That {n} should have been {0,n}. What if you allow only positive integers? Gets much more complicated. You cannot get the set {1,3} for example. === Subject: simple math question. In my book, They resolve that SQRT(56) is equal to 2*SQRT(14). Correct, but what method would you use to come to that conclusion. crzzy1. === Subject: Re: simple math question. > In my book, > They resolve that > SQRT(56) is equal to 2*SQRT(14). > Correct, but what method would you use to come to that conclusion. > crzzy1. sqrt(56) = sqrt (2*2*2*7) = sqrt (2*2) * sqrt (2*7) = 2 * sqrt (14) General approach: To simplfy sqrt(n), express n as a product of prime factors. Wherever some prime factor appears two or more times (like 2 in this case), move out an even number of them under their own square root and simplify. For example, sqrt (2 * 3^7 * 5^2 * 17^3) - whatever that number is - will be = sqrt(3^6 * 5^2 * 17^2) * sqrt(2 * 3 * 17) = 3^3 * 5 * 17 * sqrt(102) === Subject: Re: simple math question. > In my book, > They resolve that > SQRT(56) is equal to 2*SQRT(14). > Correct, but what method would you use to come to that conclusion. > crzzy1. > sqrt(56) = sqrt (2*2*2*7) > = sqrt (2*2) * sqrt (2*7) > = 2 * sqrt (14) > General approach: > To simplfy sqrt(n), express n as a product of prime factors. Wherever some > prime factor appears two or more times (like 2 in this case), move out an > even number of them under their own square root and simplify. > For example, sqrt (2 * 3^7 * 5^2 * 17^3) - whatever that number is - will be > = sqrt(3^6 * 5^2 * 17^2) * sqrt(2 * 3 * 17) > = 3^3 * 5 * 17 * sqrt(102) Excellent explanation. crzzy1. === Subject: Re: SR consistency is crap. > Shame on you: > > http://arxiv.org/abs/astro-ph/0304006 > > http://arxiv.org/abs/astro-ph/0303346 > > Why dont you take a break from these ngs. You post crap. Its > Kopeikin by the way. ÔKopkein is what you think you have heard like > the other stuff you post here. > > Mike > Could mike be trying to insinuate that the above reference say that the > speed of gravitational wave propagation is *not* the same as the speed of > light? If so shame on Mike! Id like to quibble (its fun). I think you should say c not speed of light, since in a medium light moves less than c. c is the speed of light in a vacuum. === Subject: Re: SR consistency is crap. >> J.E.: >> > Then you do not even know what the PoR is. >Im sure you MEANT to say Principle of Relativity which means that >>the laws of physics have the same form in all inertial frames of >>reference. Its a stupid principle which the modern theory makes >>unnecissary. Its a waste of time and just confuses people like you. >> >> Except for the part about androsleaze being confused, I disagree and >> I would say that your comment below is inconsistent with your comment about >> inertial frames. >What IS there to be inconsistent. I dont USE the PoR. I use >Minkowski geometry, so there are EVEN two things TO contradict each >other, just ONE thing, Minkowksi geometry. >> [...] >>I was saying that Minkowksi geometry is based on set theory. SR is >>based on Minkowksi geometry. >> Sure, but without the principle of relativity to connect that geometry >> to physics, all you have is minkowski geometry with no physical meaning. >> I could equally well say that any geometry is based on set theory, but >> in order to decide which one applies to the universe, I also need a >> connection to physical phenomena. That connection is the principle of >> relativity, since what it requires is that physics be unchanged by an >> infinitesimal displacement of the spacetime variables. >I disagree wholeheartedly and completely. What you need is a mapping >between physical events and minkowski geometry, and then a procedure >for computing future observables in the geometry, which can then be >mapped back to observations in reality. PoR is redundant, unecissary >and confusing. Drop it. The symmetries of Minkowski geometry do all >the work FOR you, you dont NEED to do anything else. Did Terrances >clock tick? Find the event. Did Stella see it? Trace a line to find >out what event corresponds to Stella seeing it. What OTHER events >that Stella saw did see observe it BETWEEN, that is ALL you need, the >rest is ßuff and if ßuff confuses people, bag it. Body bag it. >[snip disparagements against Androcles] But that wont obviously tell you whether the resulting theory of physics is one that our universe feels compelled to obey. In Minkowski geometry, an inertial objects path is a straight line through the metric, and the properties of that line are the same regardless of whether or not such an object actually exists. The validity of the exercise depends on the idea that the existence of a real object, or the motion of a real object, has no effect on the lightbeam geometry. light along in their immediate vicinities (Fizeau), and even a reduction in lightspeeds (refractive index). Special relativity does not claim validity for particulate media ... be a particulate medium of some sort. If you are an astronaut, your moving twin ought to be able to declare that the speed of light is reduced where it begins to encounter the glass of your helmet visor, and that that moving glass also produces an offset in the nearby speed of light in your direction of motion. So, while Minkowski geometry might well describe how the principle of relativity would apply to mathematical non-particulate observers exchanging signals, the real geometry of real particulate objects exchanging signals would seem to be different. Mechanims are in play that alter local lightspeeds around real-life observers, that do not exist in special relativity. Start with an empty perfect Minkowski metric and throw a rock through the region, and the rock ought to distort the metric and leave a sort of streak running through the block of spacetime, like an optical imperfection in a block of glass. relativity, then we need a theory that works consistently in curved-spacetime, and a metric for the description of inertial physics that is /not/ purely Minkowskian. Sure, we probably want to start with a ßat background, but we need the physics played out against that ßat background to incorporate localised velocity-dependent distortions. Q: Does SR cope robustly with this sort of velocity-dependent curvature? A: No, apparently not Q: Does GR currently incorporate these effects? A: Apparently no, not yet. Because people insist that it has to reduce to SR. Q: What research do we currently have on methods of amending SR and the Minkowski metric to incorporate velocity-dependent curvature? A: Apparently it cant be done, we seem to need a new type of metric and a new associated theory of how the deviations from the SR metric can also be relativistic. Q: What progress have we made so far on identifying these new relativistic rules and constructing the next-generation theory? A: Apparently none. Instead, we find it easier to insist that we know that spacetime is ßat, and that SR/Minkowski is therefore known to be correct (even though experiment seems to say otherwise) ,and that these issues therefore do not need to be addressed. So, I think that we can play around with Minkowskian geometry to our hearts content, but what we discover still wont necessarily be real physics, because in real life, thats not how real matter seems to behave. Saying that these deviations form the idealised metric are small and dont need to be considered is not a legitimate response, IMO, that would be like saying that small deviations from energy conservation in a theory would be unimportant. If there are localised velocity-dependent deviations from ßat spacetime with real physical matter, we have to know how to deal with them relativistically. With the Minkowski geometry generated by SR, we are using an idealised description of reality whose form we know breaks down when the idealisation is not perfect (usually a sign of a pathological theory), and the usual sanity test that wed do in this situation -- exploring how the theory would come out if the idealisation was lost -- does not seem to have been done. We can attempt to come up with a more realistic theory of relativity by discarding the Minkowski geometry and relationships and deriving a new theory of relativity from scratch -- but if we followed your suggestion of keeping the Minkowskian geometry and discarding the PoR as superßuous, then wed appear to be cutting ourselves off from all hope of progress in this area. IMO wed then be locking ourselves into a theory that seems to be geometrically incompatible with real-life physics, and throwing away the key. IMO, if theres a possibility that we might have made a mistake in adopting SR so wholeheartedly, then we are obligated to check -- either to put SR onto a new stronger footing, or to isolate the problem areas and the parts of physics that would have to be changed in the successor theory. Until thats done, we dont know for sure whether SR is deserves to be considered as hardcore physics, or whether its just a sidebranch that gives a Euclidean thumbnail sketch of some intrinsically non-Euclidean physics, and which provides a crude approximation of parts of the final theory, without being a full subset of it. Minkowskian geometry may be self-consistent as a piece of abstract geometry, but that does not automatically make it physics. It also does not automatically make it consistent with more general correct physics theory, in which some of the idealisations applied within Minkowski geometry might be incompatible with the more general principles being applied (eg GR says that localised energy concentrations warp spacetime, SR requires an arbitrarily-high metric -- one could argue that this should not be a legal situation, on principle, and that perhaps a full general theory ought not to reduce to SR, on principle). Physics theory based upon Minkowskis metric may still turn out to be consistently right in some respects, but consistently wrong in others. To find out, we probably have to step outside the closed Euclidean mindset and see how the problem might look from outside. =Erk= (Eric Baird) : Just look at him. Square. The shape of evil. : -- Plankton, SpongeBob SquarePants === Subject: Re: SR consistency is crap. >> J.E.: > Then you do not even know what the PoR is. >Im sure you MEANT to say Principle of Relativity which means that >>the laws of physics have the same form in all inertial frames of >>reference. Its a stupid principle which the modern theory makes >>unnecissary. Its a waste of time and just confuses people like you. >> >> Except for the part about androsleaze being confused, I disagree and >> I would say that your comment below is inconsistent with your comment about >> inertial frames. >What IS there to be inconsistent. I dont USE the PoR. I use >Minkowski geometry, so there are EVEN two things TO contradict each >other, just ONE thing, Minkowksi geometry. > >> [...] >> >>I was saying that Minkowksi geometry is based on set theory. SR is >>based on Minkowksi geometry. >> >> Sure, but without the principle of relativity to connect that geometry >> to physics, all you have is minkowski geometry with no physical meaning. >> I could equally well say that any geometry is based on set theory, but >> in order to decide which one applies to the universe, I also need a >> connection to physical phenomena. That connection is the principle of >> relativity, since what it requires is that physics be unchanged by an >> infinitesimal displacement of the spacetime variables. >I disagree wholeheartedly and completely. What you need is a mapping >between physical events and minkowski geometry, and then a procedure >for computing future observables in the geometry, which can then be >mapped back to observations in reality. PoR is redundant, unecissary >and confusing. Drop it. The symmetries of Minkowski geometry do all >the work FOR you, you dont NEED to do anything else. Did Terrances >clock tick? Find the event. Did Stella see it? Trace a line to find >out what event corresponds to Stella seeing it. What OTHER events >that Stella saw did see observe it BETWEEN, that is ALL you need, the >rest is ßuff and if ßuff confuses people, bag it. Body bag it. > >[snip disparagements against Androcles] > But that wont obviously tell you whether the resulting theory of > physics is one that our universe feels compelled to obey. There simply is no way to THAT. The best you can hope is the do experiments the distinguish between model/theory pairs that make different predictions, and then everyone picks their favoriate model/theory pair that DOES match all observations. Anything more is impossible and so shouldnt be a standard. > In Minkowski geometry, an inertial objects path is a straight line > through the metric, and the properties of that line are the same > regardless of whether or not such an object actually exists. > The validity of the exercise depends on the idea that the existence of > a real object, or the motion of a real object, has no effect on the > lightbeam geometry. > light along in their immediate vicinities (Fizeau), and even a > reduction in lightspeeds (refractive index). Do you not understand the concept of a vacuum or are you just joking? > Special relativity does not claim validity for particulate media ... > be a particulate medium of some sort. If you are an astronaut, your > moving twin ought to be able to declare that the speed of light is > reduced where it begins to encounter the glass of your helmet visor, > and that that moving glass also produces an offset in the nearby speed > of light in your direction of motion. You can model light through non-vaccums in terms of straight lines in Minkowski geometry too, you just need more than one line. > So, while Minkowski geometry might well describe how the principle of > relativity would apply to mathematical non-particulate observers > exchanging signals, the real geometry of real particulate objects > exchanging signals would seem to be different. Um, does the phrase to within experimental erros mean anything to you. We are talking about 10W signals verus 6W signals, thats a huge difference, there is room for much errors, like extended bodies, not quite instanteous turnaround, etc. > Mechanims are in play > that alter local lightspeeds around real-life observers, that do not > exist in special relativity. Thats nonsense. SR theory doesnt make ANY predictions without a material model. And if you make the RIGHT material model, then OF COURSE you can models to effects of glass, non instant turn arounds, extended bodies, etc. And obviously if you make a SIMPLER model that ignores those things then it wont be 100% accurate. So what? No big deal. Besides, right now I was discussing INTERNAL inconsistencies. > Start with an empty perfect Minkowski > metric and throw a rock through the region, and the rock ought to > distort the metric and leave a sort of streak running through the > block of spacetime, like an optical imperfection in a block of glass. You are a big GR believer arent you? Yes, things affects things, for instance you cant have a laser be absorbed and emitted and yet be a single wavelength, so all my equal separation stuff is impossible, but GET THIS, for large amounts of time, the BIG effect 10 > 6 CAN be observed EVEN dispite the simplicity of the model because we can RESTRICT ourselves to the cases where the other effects are very small, get it? > relativity, then we need a theory that works consistently in > curved-spacetime, I was write you do believe in GR. Well get this, we can ALSO have a background spacetime ßat. > and a metric for the description of inertial physics > that is /not/ purely Minkowskian. Sure, we probably want to start with > a ßat background, but we need the physics played out against that > ßat background to incorporate localised velocity-dependent > distortions. Of EITHER a field OR the metric, I know which Id prefer, ... whichever has the eeasier mathematics of course. > Q: Does SR cope robustly with this sort of velocity-dependent > curvature? > A: No, apparently not Fallacy: Confusing models and theories. > Q: Does GR currently incorporate these effects? > A: Apparently no, not yet. Because people insist that it has to reduce > to SR. Interesting, but Ive never EVER heard that before, so please elaborate and/or provide references. > Q: What research do we currently have on methods of amending SR and > the Minkowski metric to incorporate velocity-dependent curvature? > A: Apparently it cant be done, we seem to need a new type of metric > and a new associated theory of how the deviations from the SR metric > can also be relativistic. I keep imagining everything you say about curvature being about magnetism, you know that velocity dependant force. Rememeber how they made a FIELD theory out of magnetism. Too bad we cant do the same for gravity, wait! We CAN. Whew, you had me worried for a second. > Q: What progress have we made so far on identifying these new > relativistic rules and constructing the next-generation theory? > A: Apparently none. Instead, we find it easier to insist that we know > that spacetime is ßat, and that SR/Minkowski is therefore known to be > correct (even though experiment seems to say otherwise) ,and that > these issues therefore do not need to be addressed. More references to experimental verification of SR from a person who cant distinguish a model from a theory. Go ahead, try and support your claims. > So, I think that we can play around with Minkowskian geometry to our > hearts content, but what we discover still wont necessarily be real > physics, because in real life, thats not how real matter seems to > behave. Yeah real electrons dont have the hyperfine corrections predicted by AFT, how silly of me. > Saying that these deviations form the idealised metric are small and > dont need to be considered is not a legitimate response, IMO, that > would be like saying that small deviations from energy conservation > in a theory would be unimportant. Saying that they have to be large enough to observe and predict IS a requirement, since we cant MAKE a theory match your PERSONAL FEELINGS about deviations. > If there are localised > velocity-dependent deviations from ßat spacetime with real physical > matter, we have to know how to deal with them relativistically. The ONLY hard core proof of curvature (non-ßatness) as opposed to field interaction would be a non-euclidean topology. Do you have evidence for that? > With the Minkowski geometry generated by SR, we are using an idealised > description of reality whose form we know breaks down when the > idealisation is not perfect (usually a sign of a pathological theory), > and the usual sanity test that wed do in this situation -- exploring > how the theory would come out if the idealisation was lost -- does not > seem to have been done. People have done it, their theories are more complicated, and frankly since Ive never heard of a necissity to USE them, Ive never much studied them. > We can attempt to come up with a more realistic theory of relativity > by discarding the Minkowski geometry and relationships and deriving a > new theory of relativity from scratch -- but if we followed your > suggestion of keeping the Minkowskian geometry and discarding the PoR > as superßuous, then wed appear to be cutting ourselves off from all > hope of progress in this area. Yeah, wed never make a field theory based on observations, that didnt happen over ten years ago where the ßat Minkoski metric is UNobservable and ONLY exists to MAKE THE MATH EASY. I must have imagined that. I better alert Cambridge that their paper dont exist I do wonder what those speakers were REALLY saying at those conßicts. wouldnt have known. Either that, or you are WRONG and just attacking a strawman because you dont understand the difference between a model and a theory. > IMO wed then be locking ourselves into a theory that seems to be > geometrically incompatible with real-life physics, and throwing away > the key. Or making math easy. Sheese, since I allow CURVES and ARBITRARY lines, you could KEEP my model and use a different theory, and you could keep the theory and use another model. Sheesh, the advantage of the Minkowski geometry is to CLASSIFY and QUANTIFY certain lines in certain ways. So its just to make the math easy. > IMO, if theres a possibility that we might have made a mistake in > adopting SR so wholeheartedly, then we are obligated to check -- > either to put SR onto a new stronger footing, or to isolate the > problem areas and the parts of physics that would have to be changed > in the successor theory. I wont stop you, if you want to do things that make the math harder that dont affect any observations or experiments we have done or plan to do, go for it. If it turns out we need it, Ill be grateful. But please stop harassing me and making unfounded allegations about my inability to add fields to a model to have more detail, because youre wrong, and I can do it. > Until thats done, we dont know for sure whether SR is deserves to be > considered as hardcore physics, or whether its just a sidebranch that > gives a Euclidean thumbnail sketch of some intrinsically non-Euclidean > physics, and which provides a crude approximation of parts of the > final theory, without being a full subset of it. I have no idea why you mention Euclidean anything with reference to Minkowksi based models. Do you not know what the words mean? Is this a typo? Or did I just miss something? > Minkowskian geometry may be self-consistent as a piece of abstract > geometry, but that does not automatically make it physics. Didnt say it did. But others were saying that certain SR predictions that were predictions of an ELEMENTARY Minkowki model WERE inconsistent, hence the motivation for this VERY SIMPLE model. > It also > does not automatically make it consistent with more general correct > physics theory, in which some of the idealisations applied within > Minkowski geometry might be incompatible with the more general > principles being applied (eg GR says that localised energy > concentrations warp spacetime, SR requires an arbitrarily-high > metric -- one could argue that this should not be a legal situation, > on principle, and that perhaps a full general theory ought not to > reduce to SR, on principle). Metric-fixation. You dont OBSERVE the metric, sorry to disappoint you. > Physics theory based upon Minkowskis metric may still turn out to be > consistently right in some respects, but consistently wrong in > others. To find out, we probably have to step outside the closed > Euclidean mindset and see how the problem might look from outside. No, to find out each side has to make models and theories and then we have to compare to observations. I choose to make models and theoires where the math is EASY for ME to do, if its so hard that I cant make either then I dont accomplish much of ANYTHING. So GO AHEAD and do the math that is HARD for ME, and lets COMPARE to observation. But saying A PRIORI how metric must be related to observation in MY models and MY theories is CHEATING. Make your best model, and compare to my best model, anything else is less than honest. === Subject: Re: SR consistency is crap. >>Id rather teach QM from the start, and get special relativity as a >>limiting case of that, and then Newtonian dynamics as a special case >>of that. >> How do you expect to get special relativity as a limiting case of QM? >> David >comes as a limiting case. Many of the so-called quantum corrections >are actually present in classical SR, so in a sense QM has always been >SR theory. When it comes to gravitational horizons, it seems that the NM energy relationships lead to a classical model that agrees (qualitatively, perhaps also quantitatively) with QM, whereas the SR relationships lead to GRs current description of the horizon as being inescapable, which is apparently not compatible with QM. So in this sense, it does rather seem as though perhaps QM has always been /incompatible/ with SR theory. ;) (SRs concept of how a lightspeed barrier works seems to be too clean to be compatible with QM, GR then inherits certain SR conventions and converts that clean lightspeed barrier into a clean event horizon, which makes the conßict more obvious -- according to QM, information should be able to bleed outwards through the horizon, current GR says that it cant). =Erk= (Eric Baird) : You cant see me, I have my eyes shut! === Subject: Re: SR consistency is crap. >> >>Id rather teach QM from the start, and get special relativity as a >>limiting case of that, and then Newtonian dynamics as a special case >>of that. >> >> How do you expect to get special relativity as a limiting case of QM? >> >> David >comes as a limiting case. Many of the so-called quantum corrections >are actually present in classical SR, so in a sense QM has always been >SR theory. > When it comes to gravitational horizons, it seems that the NM energy > relationships lead to a classical model that agrees (qualitatively, > perhaps also quantitatively) with QM, whereas the SR relationships > lead to GRs current description of the horizon as being inescapable, > which is apparently not compatible with QM. I have absolutely no idea what you are talking about. First, because of time dilation Ive never actually seen anything cross an event horizon and I doubt can, so by Hawking radition the black hole evaporates before you reach it. So the only information inside an event horizon is from the initial collapse and that information is sent out during the collapse. Plus you can get gravitational theories in a ßat SR metric, and the fields can move at any speed relative to the background space, so I dont get your concerns about that either. > So in this sense, it does rather seem as though perhaps QM has always > been /incompatible/ with SR theory. Have you studied a serious SR based model of QM with gravity, there are more than a few to pick from you know, and I dont know any published ones that create problems like you mention. > (SRs concept of how a lightspeed barrier works seems to be too > clean to be compatible with QM, GR then inherits certain SR > conventions and converts that clean lightspeed barrier into a > clean event horizon, which makes the conßict more obvious -- > according to QM, information should be able to bleed outwards through > the horizon, current GR says that it cant). Current GR is a HORRIBLE place to do QM, so use a quantum gravitational theory instead. There are quantum gravitational theories that match all present data AND are even more restrictive than GR since they restrict topologies. === Subject: Re: SR consistency is crap. >> >> I disagree. SR is about a proposed symmetry of nature. >> Explain the twin paradox then. By the way, introducing accelerations >> or boosts is a departure from SR and a move into the Dynamics domain. >Id be happy to explain the twin paradox. Start with the financial >twin paradox that I just explained to eleaticus. And I disagree >about accelerations and boosts being outside SR, altough I understand >why people ignore it. SR is a theory about how the symmetries of >spacetime are similar to the symmetries of Minkowksi geometry. Thats >why the financial twin paradox using an indefinite quadratic form to >determine the rate at which money is sucked out our your pocket and >sent to someone else. >> You got a problem right from the start. SR came to >> challenge the Copernican view of the world (see Hans Reichenbach, The >> Philosophy of Space and Time) the cornerstone of Newtons dynamics. >> Ptolemaic and Copernican systems are kinematically equivalent >> according to Relativity. Yet, while admitting SR, the educational >> status quo refuses to drop Newtonian dynamics, claiming there is ample >> emprirical evidence to support it. Actually, sicne Newtons laws are >> mere tautologies, in non-relativistic limits they will always conform >> to experiment. >> >> Newtonian dynamics are not dropped. They are derived locally, you >> already conceded that below that they are local tautologies. A >> dynamic isnt much without a law about elementary forces. >> >> I took Physics 101 and Modern Phycsi 101 together during the same >> semester as soon as I started undergraduate. The mentor protested and >> argued I should take Modern Physics (Relativity) after I take >> Classical Mechanics, because that was a prerequisite. I protested, I >> argued it must be my choice not theirs. I won. I got a A- in Classical >> and an A+ in Modern. >> >> Good for you. I didnt even realize I liked physics until Modern >> Physics (which for us covered QM, solid state, Stat mech, etc. too) I >> just took it as an easy class since I was good at math. >> Yesy, interesting to see that while it is an easy subject for some >> people it is beyond comprehension for others. >I think its bad teaching. If I had a good background to easily >understand a fixed way of teaching the subject, that doesnt make the >subject inherently hard for others or easy for me, in means by >background and the teaching lined up well. Funnily enough, I found that it was actually very easy to explain GR principles to complete physics newbies, with a suitable choice of words. You just say things like, increasing the strength of gravity in a region seems to make makes space seem more dense and time more rarefied, so that clocks there tick more slowly ... so if we want to create a map of the light-distances in a slice through the region, so that the distances in the map correspond to the distances in the region, we have to extrude the map in order ot be able to cram in the extra space (produce diagram of gravity-well, with a ßourish). People seem to get that. You might get an occasional Gravity makes local time appear to go by slower? Really? Yup! Oh, Okay then But try to explain SR time dilation to the same person, and theyll usually get upset and insist that the thing is rubbish. So untrained people seem to be intuitively happier with supposedly advanced ideas like spacetime curvature than simple ones like SRs Minkowski metric (I found). They seem to find most of the GR-ey principles easier to visualise and accept, stars bend light or gravity slows time or rotating stars pull stuff around with them seem to be easier concepts to take onboard than moving astronauts age slower than each other. Maybe schools should teach the principles of general relativity first, and leave SR to the more advanced students. ;) >> I think this is the way to go. Teach students everything together as >> competing theories and not relativity as an advance extension to >> Newtonian Dynamics. >> >> Id rather teach QM from the start, and get special relativity as a >> limiting case of that, and then Newtonian dynamics as a special case >> of that. >> Well, you are talking 1000 years from now when we will understand >> better how quantum uncertainty gives rise to macrocosmic certainty. >> For now, such approach can result only in confusion. >You seem to have strong opinions about things you obviously dont >know. Bohm did a good job of explaining maro-certaintanty as a >was decades ago. Teaching is always behind the cutting edge. I was taught at school that Newtons prism experiment proved that light was composed of seven colours, and we were made to memorise the seven colours and conduct the experiment ourselves and see the seven colours ourselves. Then we were given coloured filters to play with and were taught that white light was actually composed of three distinct colours, and we were shown how to conduct expeirments to prove /that/. No wonder the poor kids were a bit bewildered. I survived those classes by being arrogant enough to assume that if something sounded wrong it was probably bull. Hopefully anyone going through that syllabus with a real talent for physcs woudl have realised that thge problem was not with them but with the teaching, but I do sometimes wonder how many adepts leave physcs because they think,wrongly, that their misgivings about certain subjects are becuase they arenlt good at the subject. If the people taking up the subject are predominantly people who are more prepared to suspend disbelief than the norm, then that might not be good for the subject. I suppose the people who already have a grounding in physcs before they take the class (eg family background), or those who are pig-headed enough to believe that they are right and the system is wrong, may still get through. I do notice that a strangely high percentage of physics people seem to have physicists or teachers as parents or as elder siblings etc, Perhaps a support network makes it easier for one to survive introductory physics classes without having ones brain scrambled. >> But remember, you still have a problem. SR+GR are >> axiomatic systems. No different from Euclidean geometry in that >> respect. You gain understanding of the theories early on but also you >> raise the doubt in their foundations. >> >> Huh? You want to avoid doubt? Science has doubt, predictions are >> made, the predictions can be compared to data, they might pass or >> fail. >> >> Eventually, someone will lose. >> >> Only theories can lose, not people, please explain more what you mean. >> I meant that popularization of SR, GR will turn out more questioning >> and eventually abandoment. This is the fate of every theory that >> becomes pupolarized. >I seriously doubt that. I think that when the correct SR theory is >finally popularized the incorrect alleged SR theories will FINALLY >be adandoned, and not a moment too soon, yeak! I think that perhaps part of why SR is such a bad subject is that there are probably a lot of pro-SR people strenuously insisting that what they were taught is right, even when it isnt. I think the Penrose/Terrell case illustrates this nicely, we had a situation where professional physcists had supposedly been pushing a wrong result for decades, even though it disagreed with the math, because they had been /taught/ that moving objects are seen to be contracted under SR. Social conditionaing overrode mathematics and geometry. Penrose was a mathematician who snuck his result out as a non-peer-revirewed letter in an obscure local journal, Terrell was an undergraduate who struggled for years to get his paper through peer review. So the matter was eventually tackled by a newbie and a mathematican, not by mature physics people. For some reason, these things always seem to end up being corrected by outsiders, the highly-trained mainstream dont seem to be willing or able to do it themselves. >> Those that are afraid of ending up with a loser keep SR+GR as advance >> subjects. >> >> Id love to have GR and SR as grade-school subjects, does that mean >> Im not afraid of losing? I dont know what you mean by afraid of >> losing or lose, I do hope my predictions match the data, thats >> because if its easy to be wrong, so why bother learning a theory if >> it makes wrong predictions? >> >> Remember what happened to Euclidean geometry? >> >> What happened? Its still around! >> >> As soos as >> they started teaching it, thousands attempted to disprove the Playfair >> axiom (parallel lines never meet). The result was a >> relative-consistency with spherical and hypoerbolic geometries. >> >> Elemetary euclidean geometry was proved consistent, you can drop >> relative-consistency and just say consistent if what you mean is >> that elementary hyperbolic geometry is as consistent as elementary >> Euclidean geometry. And if you mean something other than elementary, >> then that useally means bringing in sets and once youve done that >> then all become equiconsistent with set theory. >> I agree. But I hope you recall that the consistency of Euclidean >> geometry was under question for some time. >Well, the same grounding to elementary euclidean geometry applies to >elementary hyperbolic geometry, and I can do SR predictions with >ELEMENTARY hyperbolic geometry. If you want not instantaneous >accelerations, then I need FULL hyperbolic geometry, which is as >consistent as FULL euclidean geometry. Mmmm, but the GR experience has hopefully taught everyone that mathematical consistency is not necessarily enough -- in physics, a structure also has to be appropriate. If basic Euclidean geometry is consistent, it doesnt neccessarily mean that it is appropriate or sufficient for describing lightbeam geometry in the presence of gravitational fields or accelerating masses (until you start adding dimensions), or even relativiely moving masses. So a theory or model can be completely consistent in its own (artificial) context, but physcally wrong when it comes to attempts to use it to model the behaviour of the real world. Mathematical theorems can appear to be rock-solid, but still be hopelessly wrong (or misunderstood) in the context of attempts to construct physical theory. Nuances of language can be incredibly important. >> The >> same will happen to SR+GR as soon as they became a target of the >> masses. >> >> What will happen to SR+GR (which is GR, unless you meant the same >> will happen to SR and the same will happen to GR as soon as each >> becomes a target of the masses)? And what masses are currently kept >> away from it? >> >> The majority of people fails math and you should know that. The >> majority works using a common sense basis. Anything as advanced as GR >People fail at math because of math is taught badly. Did you know >that you can multiply vectors? Life is easier with that, and hard >without it. I was a math whizz as a pre-adolescent. I nearly failed my eleven plus test because it contained a set of letter-number substitution problems, and I couldnt crack the rolling encryption model they were using. Without having any math background I spent that test filling sheets of paper running through the different conceivable encryption systems that they might have been using (fractional number bases, etc) until I finally came up with the idea of using prime number sequences as a key, and I spent a quarter of the test period working on the problem of how to arbitrarily quantise a range of coordinate-free number surfaces representing different forms of number system before realising that it was going to take me at least another half hour and there were only ten minutes left. So I cheated with the first example (CAT=XXX, DOG=XXX, COG=???), by just doing a dumb substitution, and then realised that every other example worked the same way. I distinctly remember feeling panicky and inadequate, looking around at a hall filled with other elevenyearolds scratching away, and thinking I was the only one there who didnt know how to crack the output of a a prime number encoding system. Anyhow, I only mention this because after starting secondary school and spending years laboriously practicing cross-multiplication day after day (and being accused of cheating when I didnt include working out, which was an alien concept to me at the time), the math part of my brain shrivelled up until I eventually had such a strong mental block to do with anything mathy that I had to drop out of school because I could no longer even add and subtract reliably. My brain was just stalling on me. I had used to go through science subjects just writing down the first number that came into my head and getting the answers right. Then the magic stopped working. The power of aversion therapy I guess. :( (I still cant do math any more, but I have some dim memories of what it was lke when I could, and I remember being distinctly unimpressed with what was inthe textbooks). > will create a reaction and eventually will be dropped, creating more >> problems than the intended solution. Another solution recently >> proposed is to make physics a subject for selected few. >I will concede that fixing math eduation is paramount to fixing >physics education, there isnt a strong reason not to do both by >fixing the education of geometry to be incorporated the cutting edge >from the 19th century, which frankly we have yet to do and thats just >sad sad sad. >> The current approach is to keep the theory for the few and >> feed the supposed prediction to the masses via the media. Silly but >> works due to hype. > Mike >> >> I actively try to teach physics at as early a stage as possible, and >> books on relatively are freely sold in stores, there is no conspiracy >> going on. Anyone can learn. Heck, Im on Usenet because I want to >> help people learn SR, among other reasons, this is a public forum >> where anyone can read it. >> You are wasting your time. If 95% of your student can remember all >> three Newtons laws 2 years after graduation from any level exept grad >> school then you have accomplished your task. You probably aiming for a >> small % while frustrating the rest of the class and ending up with the >> opposite that what you aim, i.e. making people dislike physics. >Physics education should start out with modeling in general, and >comparing models to data, with computers you can bury most of the math >until a later epistological phase that COULD be reserved for the few >that care, where the class discusses the foundations of the models >used earlier. If someone isnt going to remember the basics of how to >make or test a model, then you shouldnt BOTHER teaching science at >all, just drill job skills instead. IMO most people publishing material on SR testing dont seem to display much ability when it comes to being able to correctly compare models. When you see experiment after experiment looking for transverse redshifts, finding them, and then declaring that this result is a significant proof of SR : ... because classical theory does not predict transverse redshifts , one is driven to despair. In the experiments involved, almost every old theory in the books would have predicted a redshift, they just wouldnt normally have /called/ a transverse shift (I think Oliver Lodge referred to the aether theories transverse effects as something like false Doppler) So ... those experimenters needed to be tapped on the shoulder and quietly told that either the supposedly null transverse predictions for other theories didnt relate to the sorts of experiments that they were actually carrying out, or, if we used the word transverse to mean transverse as measured in the lab frame, the only way that the statement would be correct would be if classical theory was defined as being a very limited range of theories that arguably did not include the major C19th theories, or even Newtonian mechanics. Which would make these experiments slightly limited in what they could tell us about how SR compared to other theories. People who think linguistically are possibly more prone to making mistakes when working across different theories with different understood meaning to words, an equation is not physics without that additional context. The critical legalistic small print is usually not actually written down, its instead often assumed that the physics experts already know enough to interpret the understood meanings and usages appropriately without screwing the thing up, and sometimes they get it wrong. Consider this logic trap: - Theory Proponent: : We know that the standard of living is far higher in the US than : in France, because it is known, as a fact, that car ownership is : high in the US, but that nobody in France owns or drives a car. Sceptic: : But I have seen photographs of French roads filled with cars! Theory Proponent: : But those are not really cars! In France, they are more correctly : referred to as voitures, or autos. So I stand by my statement, : car ownership in France is zero, or at least negligible, and the : original statement holds. , and compare that misunderstanding with: SR Proponent: : We know that SR is the right theory, because SR predicts : transverse redshifts, no other theory predicts transverse : redshifts, and transverse redshifts have been found : experimentally. Sceptic: : But almost every older theory would have predicted some sort : of redshift effect in those SR experiments! SR Proponent: : Ah, but those would not have been properly referred to as : /transverse/ redshifts, because according to test theory, in : order to measure a transverse frequency under those other : theories, one should point the detector in a different direction. : So the original statement stands: We conducted our experiment : according to SR protocols, we found a transverse redshift, and : since no earlier theory predicts transverse redshifts, our result : shows that SR is right and earlier theory was wrong. Its like standing up at a food industry conference and stating that The research proves that Italy is the only place in the world where they make Bolognese sauce for pasta, and then when someone produces a can of the stuff made by Heinz in the US, saying, well thats not really bolognese sauce, because in order to earn that name, it has to be made in Bologna, Italy. It makes the original statement a bit pointless. With SR testing this sort of illegal switching of implied definitions in mid-argument seems to be quite common. --------- BTW, This is one of the reasons why Im still an SR sceptic (even though I count myself as a harcore relativist): apart from the fact that many or most of the experimental SR proofs seem to have been badly compromised, and that there still seems to me to be at least one major theoretical loophole that hasnt yet been dealt with (and which IMO should have been tackled decades ago), its the sheer badness of most of the analysis. I dont honestly believe that physics people are usually this bad, without good reason. If SR was wrong, and the community was heroically struggling along with a reference theory that didnt work properly, then perhaps those analyses might be the best that could be achieved without exposing apparent conßicts with SR, and perhaps then wed have a logical reason why the analyses seem to be so consistently compromised. If SR is /not/ a correct physcal theory, then perhaps these repeated screwups make some sort of sense and tell a story. Otherwise, if SR really is correct, I suppose the explanation would have to be that the whole community is just hopeless at these sorts of calculations, period. :( I prefer the heroic interpretation. :) =Erk= (Eric Baird) : Time is money. Time is not money. : Space and time are interchangeable. Space and time are not interchangeable. : Special relativity is true. Special relativity is not true. === Subject: Re: SR consistency is crap. >> I took Physics 101 and Modern Phycsi 101 together during the same >> semester as soon as I started undergraduate. The mentor protested and >> argued I should take Modern Physics (Relativity) after I take >> Classical Mechanics, because that was a prerequisite. I protested, I >> argued it must be my choice not theirs. I won. I got a A- in Classical >> and an A+ in Modern. >> >> Good for you. I didnt even realize I liked physics until Modern >> Physics (which for us covered QM, solid state, Stat mech, etc. too) I >> just took it as an easy class since I was good at math. >> >> Yesy, interesting to see that while it is an easy subject for some >> people it is beyond comprehension for others. >I think its bad teaching. If I had a good background to easily >understand a fixed way of teaching the subject, that doesnt make the >subject inherently hard for others or easy for me, in means by >background and the teaching lined up well. > Funnily enough, I found that it was actually very easy to explain GR > principles to complete physics newbies, with a suitable choice of > words. > You just say things like, increasing the strength of gravity in a > region seems to make makes space seem more dense and time more > rarefied, so that clocks there tick more slowly ... so if we want to > create a map of the light-distances in a slice through the region, so > that the distances in the map correspond to the distances in the > region, we have to extrude the map in order ot be able to cram in the > extra space (produce diagram of gravity-well, with a ßourish). Id be interested in seeing that work in practise, you can really get newbies to see that a parabola in space is a geodesic in spacetime? > People seem to get that. You might get an occasional Gravity makes > local time appear to go by slower? Really? Yup! Oh, Okay then > But try to explain SR time dilation to the same person, and theyll > usually get upset and insist that the thing is rubbish. But how do you know you arent using the wrong words? By thesis is that SR is taught badly. > So untrained people seem to be intuitively happier with supposedly > advanced ideas like spacetime curvature than simple ones like SRs > Minkowski metric (I found). They seem to find most of the GR-ey > principles easier to visualise and accept, stars bend light or > gravity slows time or rotating stars pull stuff around with them > seem to be easier concepts to take onboard than moving astronauts age > slower than each other. And there is an example of bad teaching. Astronauts dont age more slowly than each other, you should discuss what the astronauts see, and if they see each age slower then faster than themself, then who aged more depends on what percentage of the observations were slower versus faster. > Maybe schools should teach the principles of general relativity first, > and leave SR to the more advanced students. ;) Maybe schools should teach things properly. >> I think this is the way to go. Teach students everything together as >> competing theories and not relativity as an advance extension to >> Newtonian Dynamics. >> >> Id rather teach QM from the start, and get special relativity as a >> limiting case of that, and then Newtonian dynamics as a special case >> of that. >> >> Well, you are talking 1000 years from now when we will understand >> better how quantum uncertainty gives rise to macrocosmic certainty. >> For now, such approach can result only in confusion. > >You seem to have strong opinions about things you obviously dont >know. Bohm did a good job of explaining maro-certaintanty as a >was decades ago. Teaching is always behind the cutting edge. > I was taught at school that Newtons prism experiment proved that > light was composed of seven colours, and we were made to memorise the > seven colours and conduct the experiment ourselves and see the seven > colours ourselves. > Then we were given coloured filters to play with and were taught that > white light was actually composed of three distinct colours, and we > were shown how to conduct expeirments to prove /that/. > No wonder the poor kids were a bit bewildered. > I survived those classes by being arrogant enough to assume that if > something sounded wrong it was probably bull. Good, most things are oversimplifications to the point where they are wrong. > Hopefully anyone going through that syllabus with a real talent for > physcs woudl have realised that thge problem was not with them but > with the teaching, but I do sometimes wonder how many adepts leave > physcs because they think,wrongly, that their misgivings about certain > subjects are becuase they arenlt good at the subject. > If the people taking up the subject are predominantly people who are > more prepared to suspend disbelief than the norm, then that might not > be good for the subject. I suppose the people who already have a > grounding in physcs before they take the class (eg family background), > or those who are pig-headed enough to believe that they are right and > the system is wrong, may still get through. You dont have to be simply pig-headed, you can just call your teachers on anything that doesnt make sense and then disregard their assertion if they dont back them up well enough. > I do notice that a strangely high percentage of physics people seem to > have physicists or teachers as parents or as elder siblings etc, > Perhaps a support network makes it easier for one to survive > introductory physics classes without having ones brain scrambled. I took introductory physics by not caring about the subject. >> But remember, you still have a problem. SR+GR are >> axiomatic systems. No different from Euclidean geometry in that >> respect. You gain understanding of the theories early on but also you >> raise the doubt in their foundations. >> >> Huh? You want to avoid doubt? Science has doubt, predictions are >> made, the predictions can be compared to data, they might pass or >> fail. >> >> Eventually, someone will lose. >> >> Only theories can lose, not people, please explain more what you mean. >> >> I meant that popularization of SR, GR will turn out more questioning >> and eventually abandoment. This is the fate of every theory that >> becomes pupolarized. > >I seriously doubt that. I think that when the correct SR theory is >finally popularized the incorrect alleged SR theories will FINALLY >be adandoned, and not a moment too soon, yeak! > I think that perhaps part of why SR is such a bad subject is that > there are probably a lot of pro-SR people strenuously insisting that > what they were taught is right, even when it isnt. > I think the Penrose/Terrell case illustrates this nicely, we had a > situation where professional physcists had supposedly been pushing a > wrong result for decades, even though it disagreed with the math, > because they had been /taught/ that moving objects are seen to be > contracted under SR. > Social conditionaing overrode mathematics and geometry. > Penrose was a mathematician who snuck his result out as a > non-peer-revirewed letter in an obscure local journal, Terrell was an > undergraduate who struggled for years to get his paper through peer > review. > So the matter was eventually tackled by a newbie and a mathematican, > not by mature physics people. > For some reason, these things always seem to end up being corrected by > outsiders, the highly-trained mainstream dont seem to be willing or > able to do it themselves. Ive considered that there might be large numbers of physicists that think clearly about SR, but just assume that everyone else does too and hence dont get invovled in debates about it. The irony is that all of these physicists could compute what people SEE correctly, if only they CARED enough to do so. >> Those that are afraid of ending up with a loser keep SR+GR as advance >> subjects. >> >> Id love to have GR and SR as grade-school subjects, does that mean >> Im not afraid of losing? I dont know what you mean by afraid of >> losing or lose, I do hope my predictions match the data, thats >> because if its easy to be wrong, so why bother learning a theory if >> it makes wrong predictions? >> >> Remember what happened to Euclidean geometry? >> >> What happened? Its still around! >> >> As soos as >> they started teaching it, thousands attempted to disprove the Playfair >> axiom (parallel lines never meet). The result was a >> relative-consistency with spherical and hypoerbolic geometries. >> >> Elemetary euclidean geometry was proved consistent, you can drop >> relative-consistency and just say consistent if what you mean is >> that elementary hyperbolic geometry is as consistent as elementary >> Euclidean geometry. And if you mean something other than elementary, >> then that useally means bringing in sets and once youve done that >> then all become equiconsistent with set theory. >> >> I agree. But I hope you recall that the consistency of Euclidean >> geometry was under question for some time. >Well, the same grounding to elementary euclidean geometry applies to >elementary hyperbolic geometry, and I can do SR predictions with >ELEMENTARY hyperbolic geometry. If you want not instantaneous >accelerations, then I need FULL hyperbolic geometry, which is as >consistent as FULL euclidean geometry. > Mmmm, but the GR experience has hopefully taught everyone that > mathematical consistency is not necessarily enough -- in physics, a > structure also has to be appropriate. > If basic Euclidean geometry is consistent, it doesnt neccessarily > mean that it is appropriate or sufficient for describing lightbeam > geometry in the presence of gravitational fields or accelerating > masses (until you start adding dimensions), or even relativiely moving > masses. What was originally being debated was the internal consistency of SR predictions. Therefore a constructed a slim version of SR without frames or curves, just points and straight lines, which is KNOWN to be consistent, and hence the internal consistency of the SR predictions in my slim model should safe and garanteed. And accelerating bodies can be handled to arbitrary precision. People forget that the accelerating results of SR are KNOWN, and that THAT is HOW we get the gravitational results in GR. > So a theory or model can be completely consistent in its own > (artificial) context, but physcally wrong when it comes to attempts to > use it to model the behaviour of the real world. > Mathematical theorems can appear to be rock-solid, but still be > hopelessly wrong (or misunderstood) in the context of attempts to > construct physical theory. > Nuances of language can be incredibly important. Thats exactly what Im doing, Im contrasting a 4-d Newtonian model that uses the manhattan metric with a minkowksi metric. Each is internally consistent, but Im asserting that the minkowski one would match experiments, which I have GOTTEN to doing yet, because eleaticus and Androcles are acting afraid of the model and I dont think you guys are helping at all. > BTW, This is one of the reasons why Im still an SR sceptic > (even though I count myself as a harcore relativist): apart from the > fact that many or most of the experimental SR proofs seem to have been > badly compromised, and that there still seems to me to be at least one > major theoretical loophole that hasnt yet been dealt with (and which > IMO should have been tackled decades ago), its the sheer badness of > most of the analysis. > I dont honestly believe that physics people are usually this bad, > without good reason. How many people want to test SR for a living? Most good people do that USES the results of SR as a necissary component rather than testing SR as an end to itself. > If SR was wrong, and the community was heroically struggling along > with a reference theory that didnt work properly, then perhaps those > analyses might be the best that could be achieved without exposing > apparent conßicts with SR, and perhaps then wed have a logical > reason why the analyses seem to be so consistently compromised. Its just not consistent, and there arent problems with SR theory, but there *might* be problems with that *some* people think is SR theory, which as Im saying is a problem for physics education to fix. > If SR is /not/ a correct physcal theory, then perhaps these repeated > screwups make some sort of sense and tell a story. > Otherwise, if SR really is correct, I suppose the explanation would > have to be that the whole community is just hopeless at these sorts of > calculations, period. These are fallacies that everyone screws up. Physicis education literature discusses screw ups and how to avoid them in teachers as well as students. For instance look at me, Im not special in any way, just a normal educator, but I dont make these erros you accuse other people of. And there are tons of people like me. I do tend to find frames bad news, for instance if one person hops frames, then the distant events that were called *simultaneous* change in mid hop, so clearly the notion of simultaneous is rather limited in its use. === Subject: Re: SR consistency is crap. <309c03F2sn568U1@uni-berlin.de>: >> How stupid are you going to get? Now you claim Einsteins theories are >> physical theories? >Any theory that leads to testable quantitative predictions about what >happens in the world, under specified circumstance is a physical theory. >The word physics in a physics theory pertains to the domain of >application, not the methodology. >GTR has not yet been falsified by empirical means. Its difficult to say for sure, I think that depends on how the dark matter thing pans out. Dark matter was only postulated to explain why current GTR predictions dont match the available data, so if suitable dark matter is found to /not/ be out there, we might decide, retrospectively, that GTR was discovered to be empirically incorrect years ago. History might decide that dark matter was a brilliant prediction, or was a desperate attempt to salvage a theory that had been experimentally falsified. In a few decades time we might have a better perspective on this. ----------- I guess its also difficult to say when a theory is really falsified experimentally, except with hindsight -- if an experimental result goes against a theory, does subsequent history then say that the theory was officially falsified on the date when that result got through peer review? What if it wasnt taken seriously at the time, or wasnt considered good enough? Or is the theory only falsified on the date when most physicists /believe/ it to have been falsified? This definition is probably more practical in that it allows people at the time to come up with a solid definition of whether a theory is falsified or not, but then we are talking about a value judgment based on a spread of evidence and the prevailing social belief systems, rather than on just hard experimental results. Perhaps we could refer to the theory being discredited than falsified (a theory can be discredited, due to human error, and still be physically correct). Although its nice to think that current experiments support our current theories well, history tells us that this sort of confidence can sometimes be ill-founded. We can /now/ say with the luxury of hindsight that the Michelson/Morley experiment falsified simple crude fixed-aether theory, but contemporary physicists might have felt entitled to say that nobody really took the M&M experiment seriously, and that if you only considered proper independently-verified results, the M&M result didnt count. Nobody else at the time was able to replicate Michelsons figures - Einstein attributed this to Michelsons exceptionally keen eyesight, which (if true) effectively meant that M&Ms extreme precision was based partly on Michelsons being able to see things that his colleagues couldnt. (!) Couple this with the fact that M&M was carried out partly in response to Lorentz pointing out that Michelson had screwed up the math in his earlier experiment in a way that made the experiment look better than it actually had been, and the perception in Europe that any serious work on the subject would be done by people working at the major European institutions, and its easy to see why Michelsons result failed to set the world alight -- it may have looked like an inexplicable and irreproducible result by an incompetent researcher in some backwater institution, and the M&M experimental falsification of simple absolute aether theory may not have been obvious until much later. Considering the avalanche of papers being published on SR&GR nowadays, if someone finds an experiment crucis that blows one or both theories away, and they are in Michelsons situation, Im not sure that we would notice, it might just stay buried in the pile. Its tricky. :( >What is a physics theory. It is a verbal artifact which is used to grind >out testable predictions of what will happen in the world, under >specified condition. It is a predictor. Nothing more, nothing less. >Bob Kolker Hmm. Not a bad description, IMO But I think physical theories also have a use as things that let us understand and appreciate (to some extent) how our universe operates around us. Its easier to get a grasp of how chemistry operates if one has a theory of the structure of the atom that explains how properties and valancies change from element to element. Even if the theory is wrong or artificial in some respects, it has mnemonic value and makes it easier to collate and assemble facts and patterns of data in our minds, which in turn makes it easier to devise newer or more complete theories. Most people in the general public dont need to know the precise atomic weight of oxygen to six decimal places, and dont need to know how many shells etc it has, but knowing the theory that the reason why things burn is because they react with oxygen in the air ... thats useful knowledge in day-to-day life. If you have a house fire, its easier to remember the instructions someone gave you years ago on how to put it out if you understood those instructions in the context of a theory, rather than as a completely random-sounding set of instructions with no obvious context. Theories reduce the number of isolated special-case facts that we need to carry around in our heads, as well as being able to let us predict new things, they can also be useful for the data-compression (and error-correction) of things that we already know. =Erk= (Eric Baird) : Who is Number One? : You are [,] Number Six. : -- The Prisoner === Subject: Re: SR consistency is crap. > <309c03F2sn568U1@uni-berlin.de>: >> How stupid are you going to get? Now you claim Einsteins theories are >> physical theories? >Any theory that leads to testable quantitative predictions about what >happens in the world, under specified circumstance is a physical theory. >The word physics in a physics theory pertains to the domain of >application, not the methodology. >GTR has not yet been falsified by empirical means. > Its difficult to say for sure, I think that depends on how the dark > matter thing pans out. > Dark matter was only postulated to explain why current GTR predictions > dont match the available data, so if suitable dark matter is found to > /not/ be out there, we might decide, retrospectively, that GTR was > discovered to be empirically incorrect years ago. > History might decide that dark matter was a brilliant prediction, or > was a desperate attempt to salvage a theory that had been > experimentally falsified. > In a few decades time we might have a better perspective on this. In a course on extra-galactic astronomy given by Paul Hodge at the University of Washington, I had the obvious pointed out to me. The universe isnt an experiment! You can measure things pertaining to it, but you cant control some parameters to isolate the effects of others. In other words, you cant really falsify a theory using astronomical observations unless the theory doesnt give itself any leeway to deal with how it differs from observation. The predictions of GR, or any other theory of cosmology, are obviously dependent on the observational capabilities of current astronomy. Indeed, astronomy is full of discoveries that involved finding things that explain anomalous observations such as the planet Neptune. John Anderson === Subject: Re: SR consistency is crap. Steve Harris sbharris@ROMAN9.netcom.com: >This is easily proved. The Shroedinger equation is the >non-relativistic limit of the Klein-Gordon *scalar* wave equation, >which explicitly does not include room for spin. However, the pauli equation (which is the schroedinger equation with the electron spin coupled to a magnetic field) _is_ the non-relativistic reduction of the dirac equation. The dirac hamiltonian is, HPsi = (alpha.p + beta m)Psi Using the canonical momentum, p -> p - (e/c)A, one obtains, H = alpha.(p - (e/c)A) + beta m (Where Ive left out the scalar potential, which is irreevant here). The solutions can be decomposed into a two-component spinor, [Phi] Psi = [ ] [Chi] If you plug this in to the hamiltonian and uncouple the two spinors, youll get one equation of the form: HPhi = (1/2m)(delta.p delta.p)Phi where delta are the pauli matrices. You then have the identity, delta.a delta.b = a.b + idelta . (a x b) and since p == p - (e/c)A, you get, HPsi = (1/2m) [ (p - (e/c)A)^2 - (ehbar/c)delta . B ] As you can see, the term containing delta.B contains the spin. This gives the correct magnetic moment (not including the anomalous part). === Subject: Re: SR consistency is crap. > Steve Harris sbharris@ROMAN9.netcom.com: >This is easily proved. The Shroedinger equation is the >non-relativistic limit of the Klein-Gordon *scalar* wave equation, >which explicitly does not include room for spin. > However, the pauli equation (which is the schroedinger equation with > the electron spin coupled to a magnetic field) _is_ the non-relativistic > reduction of the dirac equation. COMMENT: Yes, I never said it wasnt. I missed your point, I suppose. Theres this discussion which goes on periodically (ahem) in physics about the electrons spin. Somehow, since Dirac came out with an equation in which the electrons spin is taken into account and the g-factor doesnt have to be added in by hand, there was for quite some time in physics the myth that the electrons spin is somehow a quantum property, or at least a relativistic one. Neither is true. The electrons spin (spin angular momentum) is quantized in large multiples in a given direction, but only because its so small. Theres nothing there new, or that isnt true of anything elses angular momentum, or any other kind of angular momentum (for example orbital angular momentum). And just because Diracs equation happens to be relativistic, doesnt mean the relativistic treatment does anything special about spin. The fact that the non-relativistic limit of Dirac is Pauli, proves that quite well. The spin energy states of the electron (as non-degenerately appear as separate wave functions in external fields) are described if the equation leaves room for them. Which Paulis does. Which even the Shroedinger equation can do if written as a pair of coupled equations in two unknowns (one for each spin wave function, up or down), with external field term included. This thread contains the idea that *spin* is somehow tied up in, or linked with, the i which appears in the Schroedinger equation. Of course thats not true. Theres an i in the usual way of writing the because its a wave equation and its first-order in time, and any equation first order in time which is going to spit out a wave solution (something like e^i(kx-wt)), is going to need an explicit i in there somewhere, because the i comes out in front when you differentiate with regard to time, once. The classic wave equation (for violin strings or EM waves or whatever) is second-order in time, so it has an i^2, or negative sign. Its the square root of that -1 which ends up on the time-dependent exponents of the solution, and thus gives wave solutions. Take it away and you dont have a diff equation which has waves as solutions. So theres your wave behavior, if its anywhere. The Klein-Gordon equation is a classic wave equation in this sense of having a -1 and a second order time derivative, even though its of course a relativistic equation. Its solutions are matter waves, but they still show up as waves and not other exponential functions, basically due to that -1. The quantum mechanics shows up in the factors you stick in to require that the matter waves wave with the proper frequency w. There has been some discussion of whether or not i is necessary to quantum mechanics, and that Schroedingers genius lay in putting the i in to stand for the time-phase of matter waves. However, I think this is basically a wrong idea, also. Wave functions as written in QM are usually complex, but they dont always need to be in every single instance. The reason psi functions are *usually* complex, as I understand it imperfectly, is because they *usually* describe charged which can be time-reversed with a complex conjugate, because otherwise bad things happen in describing standing-waves (they show up with permanent currents or dipole moments or something). So Shroedinger was stuck needing to use a complex psi for electrons, because otherwise his standing wave solutions came out badly. And working backwards, his equation, as noted, had to have i in it because of his complex solutions and his use of a first order time derivative. But also, in relativistic QM you need a mechanism for relativistic charged matter waves to gain in energy and momentum without gaining in charge, and the complex notation allows a second time-reversed a high-energy matter wave, which otherwise it would gain along with energy and momentum. That keeps an electron which goes by with (say) the energy of 3 electrons, to keep from looking like it has a charge of -3 as though it really was 3 electrons. Basically, if you require that an electron which has the energy of 3 electrons look like it has the charge of just one electron, you need a positron wave solution which adds mass (energy) but subtracts charge. You can see all this in the solutions to the Klein-Gordon equation for pions of +1, -1, or 0 charge. The solutions for charged pions need to be complex, because otherwise theres no way to describe the appear under relativistic energies. However, for uncharged pions, theres no need for such solutions (or rather, conjugate solutinos look the same and are simply additive). So the wave equation for the relativistic neutral pion can be written as a simple real-valued function, with cosines and such, and thats it. So neither relativity, nor QM, nor relativistic QM, really requires complex psi functions in all circumstances. Just most of them. So, what does the i do in quantum mechanics? So far as I can tell, it sort of stands in as an extra dimention (something like polar coordinates) that allows one to specify and add up as a vector, the time directionality of charge transport in a wave, if you need to. But if you have an uncharged wave, like our uncharged pion wave, its not strictly necessary. For an uncharged pion, with no need to correct use a real function that simply describes a matter wave that gains in matter. Thoughts from you physicists on this point are solicited. SBH === Subject: analysis question Let 0 be a point of Lebesgue density of E subset mathbb{R} (i.e. lim_{m(B) -> 0} m(B cap E)/m(B) = 1, where the limit is taken over all balls about 0). Prove that there exists an infinite sequence of points x_n in E, with x_n != 0, and also x_n -> 0 as n -> infinity, such that the sequence also satisfies -x_n in E and 2x_n in E, for all n. any hints appreciated === Subject: Re: analysis question > Let 0 be a point of Lebesgue density of E subset mathbb{R} (i.e. > lim_{m(B) -> 0} m(B cap E)/m(B) = 1, where the limit is taken over all > balls about 0). > Prove that there exists an infinite sequence of points x_n in E, with x_n > != 0, and also x_n -> 0 as n -> infinity, > such that the sequence also satisfies -x_n in E and 2x_n in E, for all n. Please, no TEX when you post here; this is a plain text newsgroup. === Subject: Re: analysis question > > such that the sequence also satisfies -x_n in E and 2x_n in E, for all n. > Please, no TEX when you post here; this is a plain text newsgroup. On the contrary, this is a mathematics newsgroup, where TeX is quite common. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: The joy of plain text > Please, no TEX when you post here; this is a plain text newsgroup. > On the contrary, this is a mathematics newsgroup, where TeX is quite > common. Im not sure what common is supposed to imply: Poorly stated questions are much more common on sci.math than is TeX; that is hardly an argument for them. I was under the impression that sci.math is a plain text ng. Isnt this forum for the worldwide mathematics community at large? This would include many who are unfamiliar with TeX: kids, high-school teachers, all sorts of amateurs and hobbyists, engineers, math Ph.D.s who got their degrees decades ago and are in other fields now, etc. I once knew TeX well enough to write a few papers in it, but I left academia years ago and today I much prefer plain old text to TeX on sci.math. === Subject: Re: The joy of plain text >> Please, no TEX when you post here; this is a plain text newsgroup. >> On the contrary, this is a mathematics newsgroup, where TeX is quite >> common. >Im not sure what common is supposed to imply: Poorly stated questions >are much more common on sci.math than is TeX; that is hardly an argument >for them. >I was under the impression that sci.math is a plain text ng. Isnt this >forum for the worldwide mathematics community at large? This would include >many who are unfamiliar with TeX: kids, high-school teachers, all sorts of >amateurs and hobbyists, engineers, math Ph.D.s who got their degrees >decades ago and are in other fields now, etc. I once knew TeX well enough >to write a few papers in it, but I left academia years ago and today I much >prefer plain old text to TeX on sci.math. That is if the plain old text can intelligently state the problem. Too often, it cannot. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: The joy of plain text > Please, no TEX when you post here; this is a plain text newsgroup. > > On the contrary, this is a mathematics newsgroup, where TeX is quite > common. > Im not sure what common is supposed to imply: Poorly stated questions > are much more common on sci.math than is TeX; that is hardly an argument > for them. > I was under the impression that sci.math is a plain text ng. Isnt this > forum for the worldwide mathematics community at large? This would include > many who are unfamiliar with TeX: kids, high-school teachers, all sorts of > amateurs and hobbyists, engineers, math Ph.D.s who got their degrees > decades ago and are in other fields now, etc. I once knew TeX well enough > to write a few papers in it, but I left academia years ago and today I much > prefer plain old text to TeX on sci.math. Plain text to me means no binaries; no .jpg files, or Word files, etc. TeX is plain text, at least the way I understand the term. Occasionally I post something here that Ive cut Ôn pasted from another source, where it was in TeX. Going through the TeX, especially if its at all long, and manually editing it into a form youd approve is not an option; your options are, I post the TeX, or I dont post at all. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The joy of plain text > Occasionally I post something here that Ive cut Ôn pasted from > another source, where it was in TeX. Going through the TeX, especially > if its at all long, and manually editing it into a form youd approve > is not an option; your options are, I post the TeX, or I dont post > at all. Then dont post it. Unless its simple Tex, likely I may never read it. Of the many many problems in the 7 groups I read, Ive little time to translate hard to read stuff. So if youve not time to make your post readable, why should I take the time to make it readable and then even more time to solve the problem and yet more time to present an answer? You are asking too much of free tutoring. Those who post readable problems are those I read and (when able) answer first. === Subject: Re: The joy of plain text > Occasionally I post something here that Ive cut Ôn pasted from > another source, where it was in TeX. Going through the TeX, especially > if its at all long, and manually editing it into a form youd approve > is not an option; your options are, I post the TeX, or I dont post > at all. > Then dont post it. Unless its simple Tex, likely I may never read it. > Of the many many problems in the 7 groups I read, Ive little time to > translate hard to read stuff. So if youve not time to make your post > readable, why should I take the time to make it readable and then even > more time to solve the problem and yet more time to present an answer? > You are asking too much of free tutoring. Those who post readable > problems are those I read and (when able) answer first. I see that I didnt make myself clear. Most of what I post here is not problems but answers to other peoples problems. Ive been providing, not asking for, the free tutoring for over a decade now (for the most part), though I guess its too much to expect that even regulars contributors such as yourself would have noticed. The typical situation in which Ill post in TeX is when someone asks a research-level question, and I happen to know that theres a paper on the topic already in the literature, and Ill go cut Ôn paste the review from Math Reviews, which review will be written in TeX. If the person who asked the question cant decode the TeX, too bad - how much can she ask of free tutoring? You are hereby excused from reading anything I post in TeX. I think Ill survive. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The joy of plain text >The typical situation in which Ill post in TeX is when someone >asks a research-level question, and I happen to know that theres >a paper on the topic already in the literature, and Ill go cut >n paste the review from Math Reviews, which review will be written >in TeX. For Williams sake, perhaps you could use Zentralblatt instead. Of course, then youd have to schneidnklebb. Lee Rudolph === Subject: Re: The joy of plain text >> Occasionally I post something here that Ive cut Ôn pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if its at all long, and manually editing it into a form youd approve >> is not an option; your options are, I post the TeX, or I dont post >> at all. >Then dont post it. [...] No, please do. Derek Holts post says it all, so I wont labour the point. I just wanted to add another vote for sanity (not that Im claiming to be sane). -- Angus Rodgers Contains mild peril === Subject: Re: The joy of plain text >> Occasionally I post something here that Ive cut Ôn pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if its at all long, and manually editing it into a form youd approve >> is not an option; your options are, I post the TeX, or I dont post >> at all. >Then dont post it. Unless its simple Tex, likely I may never read it. The fact that you may never read it is a poor argument for not posting it. There may be others who do want to read it. My impression is that the most generally preferred method of writing mathematics on this newsgroup is to use a simplified version of TeX, removing all symbols (such as dollars) that are unnecessary for comprehension - for example: alpha^2 + beta^{5/2} = sum_{i=0} ^infty gamma_i ^{-3}. is not difficult to read. How would you prefer that to be written? With a longer chunk of material, cutting and pasting from a TeX document is not ideal, but as long as a handful of people are interested in reading it, it is worth posting it. There is so much total junk on this newsgroup already, that it seems strange to attmept to outlaw potentially interesting material on account of minor shortcomings. Derek Holt. >Of the many many problems in the 7 groups I read, Ive little time to >translate hard to read stuff. So if youve not time to make your post >readable, why should I take the time to make it readable and then even >more time to solve the problem and yet more time to present an answer? >You are asking too much of free tutoring. Those who post readable >problems are those I read and (when able) answer first.