mm-3119 === Subject: JSH: Still retired My last big idea was to use S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) on the factring problem, relying on the reality--the absolute mathematical fact--that the expressions CANNOT have a single factorization. There is no debate on that point. It just cannot only factor S. That's it. I haven't even bothered to step out a solution of the damn thing, relying on the posted work of Tim Peters. I'm done. The big ideas are for young people. That was my last big idea. But I can follow through on it and on my other ideas--or help others follow through. Maybe it's crap. Maybe algebra is some odd little thing that just wishes to protect humanity from the factoring problem being this dinky nothing, so it will fight ot make sure this approach doesn't work. I don't think so, but I say it like that because some of you may have no clue about what mathematics is, so you probably suppose that because a LOT of people are invested in this factoring thing, no way the math can show them to be wrong. But you see, the math doesn't care. People are wrong about all kinds of things. It's just kind of a human thing--being wrong. If you trust, then you can just jump to the conclusion of posters who leap at replying to me, to see what their opinion is. I suggest instead you look to see what their math is. Right now I am sitting in the United States contemplating a war started by the son of a president, a president who was ridiculed for not finishing with Iraq. That president was also ridiculed for saying he wouldn't cut taxes and then doing it later. Seems to me his son just went to the big areaa his father was criticized on, and decided he'd fix it, in his own mindless way. That's the real world. People do weird things, and then a LOT of other people support them on it, as that's what people do. But mathematics IS pure in that no matter what, no matter how many people get together to decide something that is false is true, you can trace out the logical argument, and find a flaw. And it doesn't matter if it's the son of some dude who wants to make good on what he thinks his father failed on--no matter who dies. Mathematics IS pure. But you have to know where the purity actually is. It's not in the value or usefulness that some people put on it, but in the reality of absolute truth. James Harris === Subject: Re: Still retired > My last big idea was to use James, to save your creditability, please factor this number with any thing \ \\ you have that works 269,653,873 === Subject: Re: JSH: Still retired > ... > Nonsense snipped <<< ... > People are wrong about all kinds of things. It's just kind of a human > thing--being wrong. You mean like most of the errors, incorrect statements and useless factoring methods you have been pounding us with for the past decade? Don't you have life where you can do something worthwhile? For an entire decade - you have ZERO results that are worth anything! Funny, since you claim to be the greatest number theorist of all time and thaty the great would tremble at the great thoughts and discoveries. You mentioned that the muse is gone - I contend that the muse was robbing you of something much grander - your life. You have nothing Harris - and the reality is that you know it! If this latest (sqrt incarnation) was worth anything - you'd be spitting the factors of the RSA numbers in our faces! But we all know the TRUTH - don't we! Heck, if you claim you get two factorzations per sqrt, why not use n-th roots and obfuscate matters even more? Please oh please - modify the method to use the 37th root - so we can bask in its beauty! You'll be back - because your delusions of granduer and narcissistic tendaencies already control you, your thought processes and your inability to accept constructive criticism. Go back to the hole from you crawed out of! Have a nice day! -A > ... > More nonsense snipped <<< ... > James Harris === Subject: Re: Still retarted > My last big idea was to use > I'm done. The big ideas are for young people. That was my last big > idea. > But I can follow through on it and on my other ideas--or help others > follow through. > Maybe it's crap. yes, it all was................................. > It's not in the value or usefulness that some people put on it, but in > the reality of absolute truth. and that is in the eye of the beholder, troll. > James Harris === Subject: Re: Still retired > My last big idea was to use > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > on the factring problem, relying on the reality--the absolute > mathematical fact--that the expressions CANNOT have a single > factorization. > There is no debate on that point. > It just cannot only factor S. > That's it. I haven't even bothered to step out a solution of the damn > thing, relying on the posted work of Tim Peters. > I'm done. The big ideas are for young people. That was my last big > idea. > But I can follow through on it and on my other ideas--or help others > follow through. > Maybe it's crap. Maybe algebra is some odd little thing that just > wishes to protect humanity from the factoring problem being this dinky > nothing, so it will fight ot make sure this approach doesn't work. > I don't think so, but I say it like that because some of you may have > no clue about what mathematics is, so you probably suppose that because > a LOT of people are invested in this factoring thing, no way the math > can show them to be wrong. > But you see, the math doesn't care. > People are wrong about all kinds of things. It's just kind of a human > thing--being wrong. > If you trust, then you can just jump to the conclusion of posters who > leap at replying to me, to see what their opinion is. > I suggest instead you look to see what their math is. > Right now I am sitting in the United States contemplating a war started > by the son of a president, a president who was ridiculed for not > finishing with Iraq. That president was also ridiculed for saying he > wouldn't cut taxes and then doing it later. > Seems to me his son just went to the big areaa his father was > criticized on, and decided he'd fix it, in his own mindless way. > That's the real world. People do weird things, and then a LOT of other > people support them on it, as that's what people do. > But mathematics IS pure in that no matter what, no matter how many > people get together to decide something that is false is true, you can > trace out the logical argument, and find a flaw. > And it doesn't matter if it's the son of some dude who wants to make > good on what he thinks his father failed on--no matter who dies. > Mathematics IS pure. But you have to know where the purity actually > is. > It's not in the value or usefulness that some people put on it, but in > the reality of absolute truth. > James Harris What is the point in talking about stuff other than math? I think if you actually knew something about math, you wouldn't have to rely on social crap. Are you sure you're not a politician? I mean, your math is pretty bad \ \\ so you might want to look into something else. Dave === Subject: Re: JSH: Still retired > It's not in the value or usefulness that some people put on it, but in > the reality of absolute truth. > James Harris It just wouldn't be another working day morning without more JSH demagoguery! === Subject: Re: JSH: Still retired [spared sci.crypt and alt.math] [jstevh@msn.com] > My last big idea was to use > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > on the factring problem, relying on the reality--the absolute > mathematical fact--that the expressions CANNOT have a single > factorization. > There is no debate on that point. > It just cannot only factor S. This one is a bit odd: it's literally true that there's no debate on this point, but that's not because there's agreement. It's more that it's so jaw-droppingly inept that it's hard to know what could possibly be said in response. Like, \\umm, well, obviously not\\. You could just as well argue \ \\ that S = (i+j)*(m+n) can't only factor S, because aliens from the planet Contrary may swap our meanings for \\+\\ and \\-\\. Both arguments are silly fun in small doses, \ \\ but that's all. > That's it. I haven't even bothered to step out a solution of the damn > thing, relying on the posted work of Tim Peters. Ewww. I think I finally get something here that's escaped me all along: that business with completing the squares twice and getting an \\only k_i \ \\ and S\\ expression leads to a difference-of-squares equation, and I now finally \ \\ understand that you believe _that_ has something to do with this _other_ business of flipping the signs in your equation for S. Oops! Sorry, they're not at all the same thing -- they're not even related. \ \\ Indeed, that's exactly why it's possible to prove that the form for the difference-of-squares equation you picked could _never_ find a non-trivial factor, but also to prove that the flip-the-signs business _can_ always find \ \\ a non-trivial factor (when you're fortunate enough to pick winning values for the k_i, x and y, and winning values always exist). There's no contradiction here: they're entirely different methods. LOL -- you could just as well have picked (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T out of thin air for the flip-the-signs method. > I'm done. The big ideas are for young people. That was my last big > idea. > But I can follow through on it and on my other ideas--or help others > follow through. > Maybe it's crap. Maybe algebra is some odd little thing that just > wishes to protect humanity from the factoring problem being this dinky > nothing, so it will fight ot make sure this approach doesn't work. > I don't think so, but I say it like that because some of you may have > no clue about what mathematics is, so you probably suppose that because > a LOT of people are invested in this factoring thing, no way the math > can show them to be wrong. > But you see, the math doesn't care. > People are wrong about all kinds of things. It's just kind of a human > thing--being wrong. > If you trust, then you can just jump to the conclusion of posters who > leap at replying to me, to see what their opinion is. > I suggest instead you look to see what their math is. > Right now I am sitting in the United States contemplating a war started > by the son of a president, a president who was ridiculed for not > finishing with Iraq. That president was also ridiculed for saying he > wouldn't cut taxes and then doing it later. > Seems to me his son just went to the big areaa his father was > criticized on, and decided he'd fix it, in his own mindless way. > That's the real world. People do weird things, and then a LOT of other > people support them on it, as that's what people do. > But mathematics IS pure in that no matter what, no matter how many > people get together to decide something that is false is true, you can > trace out the logical argument, and find a flaw. > And it doesn't matter if it's the son of some dude who wants to make > good on what he thinks his father failed on--no matter who dies. > Mathematics IS pure. But you have to know where the purity actually > is. > It's not in the value or usefulness that some people put on it, but in > the reality of absolute truth. > James Harris Huh. Less than two weeks ago you said you were only doing math for the money. Which is right? === Subject: Re: JSH: Still retired http://www3.telus.net/ldh/pix/again.jpg === Subject: Re: JSH: Still retired | http://www3.telus.net/ldh/pix/again.jpg Big time... LOL === Subject: Re: Versions and indistinguishability >You seem to [be] confusing probability with measure on the real line. > >I'm starting to suspect that (at least) one of us is confused >about something here. I wouldn't be so brave as to claim >it's you, of course. Let's see if we agree on the following. >X(t) = 0, for all t \\\\in [0,1] >Y(t) = 0, for all t \\\\in ([0,1]\\\\p) >Y(t) = 1, for t=p >Would you say that those two are versions of eachother or >not, given the definition below? >Two stochastic processes X={X_t}, Y={Y_s} are >versions of each other if >\\\\forall t: P{X_t = Y_t} = 1 No! P{X(p) = Y(p)} = 0, hence it is not true that for all t, P{X(t) = Y(t)} = 1. >And just to be absolutely sure - are they versions of >eachother if we use the definition below? >Two stochastic processes X={X_t}, Y={Y_s} are >versions of each other if >\\\\forall t: P{X_t != Y_t} = 0 This statement , equivalent to the first definition, is also false, since P{X(p) != Y(p)} = 1. Perhaps your confusion is with predicate logic, rather than (or in addition to?) probability vs. measure. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Is Analytic Geometry really Geometry? On Wed, 21 Jun 2006 23:01:39 -0400, Hatto von Aquitanien >How can one discuss points and lines without an a priori sense of a >geometric continuum? >> Very simply. Gene may yet show you how. >I don't believe it is possible to ever show such a thing. The continuum \ \\ is >a priori, and can, therefore, not be excluded from reasoning. http://planetmath.org/encyclopedia/FinitePlane.html === Subject: Re: Is Analytic Geometry really Geometry? > On Wed, 21 Jun 2006 23:01:39 -0400, Hatto von Aquitanien > >>How can one discuss points and lines without an a priori sense of a >>geometric continuum? > > Very simply. Gene may yet show you how. >>I don't believe it is possible to ever show such a thing. The continuum >>is a priori, and can, therefore, not be excluded from reasoning. > http://planetmath.org/encyclopedia/FinitePlane.html Explain that to me without relying on the ordering of symbols on the page/screen, and without any reference to concepts of time such as the word \\repeatedly\\. -- Nil conscire sibi === Subject: Re: Is Analytic Geometry really Geometry? > > On Wed, 21 Jun 2006 23:01:39 -0400, Hatto von Aquitanien > > > > > > >>How can one discuss points and lines without an a priori sense of a > >>geometric continuum? > > > > Very simply. Gene may yet show you how. > > >>I don't believe it is possible to ever show such a thing. The \\ continuum > >>is a priori, and can, therefore, not be excluded from reasoning. > > > > http://planetmath.org/encyclopedia/FinitePlane.html > Explain that to me without relying on the ordering of symbols on the > page/screen, and without any reference to concepts of time such as the > word \\repeatedly\\. It is quite easy in terms of the space being a set of objects, which are called the points of the space( or singleton sets can be the points), and certain subsets of the space as being its lines. For example a set of four objects can be a 3-dimensional object with each one element subset being a point, each two element subset being a one dimensional line and each 3 element subset being a two dimensional face for a \\tetrahedral\\ space. A simplex can be thought of as the the n-dimensional analog of this . See http://mathworld.wolfram.com/Simplex.html === Subject: Re: Is Analytic Geometry really Geometry? On Thu, 22 Jun 2006 05:43:00 -0400, Hatto von Aquitanien >> On Wed, 21 Jun 2006 23:01:39 -0400, Hatto von Aquitanien >> >How can one discuss points and lines without an a priori sense of a >geometric continuum? >> >> Very simply. Gene may yet show you how. >I don't believe it is possible to ever show such a thing. The continuum >is a priori, and can, therefore, not be excluded from reasoning. >> http://planetmath.org/encyclopedia/FinitePlane.html >Explain that to me without relying on the ordering of symbols on the >page/screen, and without any reference to concepts of time such as the >word \\repeatedly\\. Why? === Subject: Re: Is Analytic Geometry really Geometry? > On Thu, 22 Jun 2006 05:43:00 -0400, Hatto von Aquitanien > On Wed, 21 Jun 2006 23:01:39 -0400, Hatto von Aquitanien > > >>How can one discuss points and lines without an a priori sense of a >>geometric continuum? > > Very simply. Gene may yet show you how. >>I don't believe it is possible to ever show such a thing. The \\ continuum >>is a priori, and can, therefore, not be excluded from reasoning. > > http://planetmath.org/encyclopedia/FinitePlane.html >>Explain that to me without relying on the ordering of symbols on the >>page/screen, and without any reference to concepts of time such as the >>word \\repeatedly\\. > Why? To demonstrate that you can \\discuss points and lines without an a priori sense of a geometric continuum?\\. I'm quite capable of applying concepts from one area of thought to another. My contention is that our concept of continuity is a priori. I believe intuitively that the idea of real number continuity is neither in need of proof, nor does it admit of proof. Furthermore, I believe we innately understand certain other geometric principles without any need of instruction. My reason for discussing Analytic Geometry in the first place was to \\ explore the distinction between analytic mathematics and geometric reasoning. For thousands of years, people have been solving problems which we now address with formal mathematics. In some cases the insights expressed in practical and/or ornamental designs are quite profound. They appear to be even more so when we consider that in some cases these works were the creation of illiterate cultures. -- Nil conscire sibi === Subject: Re: sin cos tan help <6736399.1150950897566.JavaMail.jakarta@nitrogen.mathforum.org Here is an \ easy way to remember sin,cos and tan of key > angles: 0, 30, 45, 60, 90 > sin 0 = sqrt(0)/2 > sin 30 = sqrt(1)/2 > sin 45 = sqrt(2)/2 > sin 60 = sqrt(3)/2 > sin 90 = sqrt(4)/2 > cos 0 = sqrt(4)/2 > cos 30 = sqrt(3)/2 > cos 45 = sqrt(2)/2 > cos 60 = sqrt(1)/2 > cos 90 = sqrt(0)/2 Nice Paul!! I am a 44 year old math professor and I would have thought that by now I know all such simple tricks to help students. But I have never seen that simple pattern mentioned, nor noticed it myself. Why doesn't every precalc text point it out to make things easy for the students? I certainly intend to point this out to students in the future. Mike === Subject: Re: sin cos tan help <6736399.1150950897566.JavaMail.jakarta@nitrogen.mathforum.org> Here is \ an easy way to remember sin,cos and tan of key >> angles: 0, 30, 45, 60, 90 >> sin 0 = sqrt(0)/2 >> sin 30 = sqrt(1)/2 >> sin 45 = sqrt(2)/2 >> sin 60 = sqrt(3)/2 >> sin 90 = sqrt(4)/2 >> cos 0 = sqrt(4)/2 >> cos 30 = sqrt(3)/2 >> cos 45 = sqrt(2)/2 >> cos 60 = sqrt(1)/2 >> cos 90 = sqrt(0)/2 > Nice Paul!! I am a 44 year old math professor and I > would have thought that by now I know all such simple > tricks to help students. But I have never seen that > simple pattern mentioned, nor noticed it myself. Why > doesn't every precalc text point it out to make things > easy for the students? I certainly intend to point > this out to students in the future. I've seen this in many texts, and it seems to be fairly well known in my experience, although, having taught at 8 colleges/universities and 2 high schools, maybe I've been around more than you have. I agree that it's very nice and it's something trig. and precalculus texts should mention. In my opinion, it's much better than mnemonics such as \\SOHCAHTOA\\ (which I can never to get it correct), because at least it's based on _some_ mathematical reasoning and/or mathematical pattern recognition. By the way, I used to be very concerned in the early days of calculators, because students could obtain all the exact values of the 30, 45, and 60 degree angles by simply squaring the calculator output, recognizing the equivalent fraction of the resulting decimal [note that the squares of the sine, cosine, tangent of these angles all come out to fractions that any student would recognize (well, they used to be able to recognize) from its decimal expansion], and then taking the square root of the fraction. As the years rolled by, I got to where I didn't worry about this at all, because even when I _taught_ this in class (to help students on standardized tests and to help them for other teachers who were even more insistent about using exact values than I was), the majority of my students had more trouble doing this than just memorizing the values! Dave L. Renfro === Subject: Re: sin cos tan help > > Here is an easy way to remember sin,cos and tan of > key > > angles: 0, 30, 45, 60, 90 > > sin 0 = sqrt(0)/2 > > sin 30 = sqrt(1)/2 > > sin 45 = sqrt(2)/2 > > sin 60 = sqrt(3)/2 > > sin 90 = sqrt(4)/2 > > cos 0 = sqrt(4)/2 > > cos 30 = sqrt(3)/2 > > cos 45 = sqrt(2)/2 > > cos 60 = sqrt(1)/2 > > cos 90 = sqrt(0)/2 > Nice Paul!! I am a 44 year old math professor and I > would have thought > that by now I know all such simple tricks to help > students. But I have > never seen that simple pattern mentioned, nor noticed > it myself. Why > doesn't every precalc text point it out to make > things easy for the > students? I certainly intend to point this out to > students in the > future. > Mike Your welcome :) Glad to share it. === Subject: Re: sin cos tan help <6736399.1150950897566.JavaMail.jakarta@nitrogen.mathforum.org > Here is \ an easy way to remember sin,cos and tan of key > > angles: 0, 30, 45, 60, 90 > > sin 0 = sqrt(0)/2 > > sin 30 = sqrt(1)/2 > > sin 45 = sqrt(2)/2 > > sin 60 = sqrt(3)/2 > > sin 90 = sqrt(4)/2 > > cos 0 = sqrt(4)/2 > > cos 30 = sqrt(3)/2 > > cos 45 = sqrt(2)/2 > > cos 60 = sqrt(1)/2 > > cos 90 = sqrt(0)/2 > Nice Paul!! I am a 44 year old math professor and I would have thought > that by now I know all such simple tricks to help students. But I have > never seen that simple pattern mentioned, nor noticed it myself. Why > doesn't every precalc text point it out to make things easy for the > students? I certainly intend to point this out to students in the > future. I noticed it a few years ago but haven't taught a pre-calculus course since then, so I couldn't show it. BTW, you know about LIATE (for integrating by parts), right? --- Christopher Heckman === Subject: Re: sin cos tan help <6736399.1150950897566.JavaMail.jakarta@nitrogen.mathforum.org > > > \ Here is an easy way to remember sin,cos and tan of key > > > angles: 0, 30, 45, 60, 90 > > > > sin 0 = sqrt(0)/2 > > > sin 30 = sqrt(1)/2 > > > sin 45 = sqrt(2)/2 > > > sin 60 = sqrt(3)/2 > > > sin 90 = sqrt(4)/2 > > > > cos 0 = sqrt(4)/2 > > > cos 30 = sqrt(3)/2 > > > cos 45 = sqrt(2)/2 > > > cos 60 = sqrt(1)/2 > > > cos 90 = sqrt(0)/2 > > Nice Paul!! I am a 44 year old math professor and I would have thought > > that by now I know all such simple tricks to help students. But I have > > never seen that simple pattern mentioned, nor noticed it myself. Why > > doesn't every precalc text point it out to make things easy for the > > students? I certainly intend to point this out to students in the > > future. > I noticed it a few years ago but haven't taught a pre-calculus course > since then, so I couldn't show it. > BTW, you know about LIATE (for integrating by parts), right? I'll post it anyway, for the benefit of other people who haven't seen it yet. If you want to integrate f(x)*g(x), where f and g are \\simple\\ functions, LIATE tells you which one of f and g to integrate, and which to differentiate. L = logarithm = ln(x) I = inverse (trig) = arcsin(x) or arctan(x) (most common) A = algebraic = x^k for some constant k T = (easy) trig = sin(x) or cos(x) E = exponetial = e^x (or a^x in general) As you go from L to E, the functions get easier to integrate. So if you wanted to integrate x^5 * arcsin(x), you should differentiate arcsin(x) and integrate x^4 when you do integration by parts. (I found out about this mneumonic when I was grading Calc II tests, and it took me a while to figure out what it meant.) --- Christopher Heckman === Subject: Re: sin cos tan help >> > > > > Here is an easy way to remember sin,cos and tan of key >> > > angles: 0, 30, 45, 60, 90 >> > > > > sin 0 = sqrt(0)/2 >> > > sin 30 = sqrt(1)/2 >> > > sin 45 = sqrt(2)/2 >> > > sin 60 = sqrt(3)/2 >> > > sin 90 = sqrt(4)/2 >> > > > > cos 0 = sqrt(4)/2 >> > > cos 30 = sqrt(3)/2 >> > > cos 45 = sqrt(2)/2 >> > > cos 60 = sqrt(1)/2 >> > > cos 90 = sqrt(0)/2 >> > > Nice Paul!! I am a 44 year old math professor and I would have \\ thought >> > that by now I know all such simple tricks to help students. But I \\ have >> > never seen that simple pattern mentioned, nor noticed it myself. Why >> > doesn't every precalc text point it out to make things easy for the >> > students? I certainly intend to point this out to students in the >> > future. >> I noticed it a few years ago but haven't taught a pre-calculus course >> since then, so I couldn't show it. >> BTW, you know about LIATE (for integrating by parts), right? > I'll post it anyway, for the benefit of other people who haven't seen > it yet. > If you want to integrate f(x)*g(x), where f and g are \\simple\\ > functions, LIATE tells you which one of f and g to integrate, and which > to differentiate. > L = logarithm = ln(x) > I = inverse (trig) = arcsin(x) or arctan(x) (most common) > A = algebraic = x^k for some constant k > T = (easy) trig = sin(x) or cos(x) > E = exponetial = e^x (or a^x in general) > As you go from L to E, the functions get easier to integrate. So if you > wanted to integrate > x^5 * arcsin(x), > you should differentiate arcsin(x) and integrate x^4 when you do > integration by parts. > (I found out about this mneumonic when I was grading Calc II tests, and > it took me a while to figure out what it meant.) > --- Christopher Heckman Dave === Subject: Re: 19721972..... [hayk] > Proove, that there is a number like > 19721972...1972, > thet it is dividing by 1971. Let N_i be the integer obtained by pasting i copies of 1972 together. So N_1 = 1972 N_2 = 19721972 ... N_(i+1) = N_i * 10000 + 1972 Let M_i = N_i mod 1971. So M_1 = 1 M_2 = 146 ... M_(i+1) = (M_i * 10000 + 1972) mod 1971 = (M_i * 145 + 1) mod 1971 Now I'm just going to give hints: 1. Prove that it's also possible to express M_(i-1) as a function of M_i. (hint: does 145 have a multiplicative inverse modulo 1971?) 2. Prove that for some j > 1, M_j = 1. (hint: the sequence M_i has to fall into a cycle eventually -- prove that M_1 is the first repeated value; hint: if M_i were the first repeated value, and i>1, then it would have two distinct predecessors, contradicting #1) 3. Using #1, prove that for any j such that M_j = 1, it must be that M_(j-1) = 0. A somewhat easier way to do #2 and #3 is to notice that _defining_ M_0 = 0 fits the recurrence, and use the kind of argument in #2 to prove that in the \ \\ sequence M_0, M_1, ... it must be that M_0 is the first repeated value. Or, screw all that, and just look at N_54 ;-) === Subject: Re: 19721972..... > Proove, that there is a number like > 19721972...1972, > thet it is dividing by 1971. thats easy if you add a \\ . \\ in front 4/9 = 0.444444........ 43/99 = 0.4343434343......... etc. === Subject: Re: Four Color Theorem > > > Hypothesis: A simple loopless maximal planar graph with a 5-degree > > > vertex cannot be a > > > minimal counter example to the four color theorem!. > > Actually, that should be a lemma, not a hypothesis. > \\The problem is that a vertex of degree 5 isn't a reducible > configuration. > Technically, no one has been able to figure out how to make it one.\\ > Are you now stating that a vertex of degree 5 is a reducible > configuration? Sheesh. For crying out loud, either continue this privately OR in the Usenet group, but not both. Here's the reply I sent him: No. What I meant was that you need to _prove_ your hypothesis. You can't assume it (which is what \\hypothesis\\ usually means: something you assume is true, and work from there). --- Christopher Heckman === Subject: Re: f(f(x)) = x + a , f - derivable <12457253.1150917932588.JavaMail.jakarta@nitrogen.mathforum.org > How can \ i solve this functional equation: > > f(f(x)) = f(x) + a ? > > And are there any methods of solved like this > > iterated functional equations? > Ok, here is what i've got on the problem: > at frist notice that f us injective, then since it is derivable, thus \\ continious, there exist g(g) = f^{-1} (x). > Now, taking f of both sides of the equation we get that > f(x + a) = f(x) + a (1). That doesn't match either version of the equation. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable <12457253.1150917932588.JavaMail.jakarta@nitrogen.mathforum.org > > How \ can i solve this functional equation: > > > f(f(x)) = f(x) + a ? > > > > And are there any methods of solved like this > > > iterated functional equations? > > Ok, here is what i've got on the problem: > > at frist notice that f us injective, then since it is derivable, thus \ \\ continious, there exist g(g) = f^{-1} (x). > > Now, taking f of both sides of the equation we get that > > f(x + a) = f(x) + a (1). > That doesn't match either version of the equation. Oops, sorry, f(x + a) = f(x) + a is correct. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable > > > How can i solve this functional equation: > > > f(f(x)) = f(x) + a ? > > > > And are there any methods of solved like this > > > iterated functional equations? > > Ok, here is what i've got on the problem: > > at frist notice that f us injective, then since it > is derivable, thus continious, there exist g(g) = > f^{-1} (x). > > Now, taking f of both sides of the equation we get > that > > f(x + a) = f(x) + a (1). > That doesn't match either version of the equation. Why? sorry again, as i replied in the post a bit earlier i actually menat to \ \\ solve the equation f(f(x)) =x + a which is written in the Subject line, so \ \\ taking f of both sides of it f(f(f(x))) = f(x + a),but again f(f(f(x))) = f(x) + a, so f(x+a) = f(x) + a. What's wrong here? > Robert Israel > israel@math.ubc.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable > >> How can i solve this functional equation: > >> f(f(x)) = f(x) + a ? > This doesn't match the Subject line. Which one do > you want? Oh, sorry being not attentive. I wanted to solve this one which was written \ \\ in the subject line: f(f(x)) = x + a. > >> And are there any methods of solved like this > iterated functional equations? > >Let y=f(x), then your equation become f(y) = y+a > >So, f(f(x)) = f(x)+a iff f(x)=x+a > > > >Is it that simple? Out of school for a long time, > and I might have > >forgotten everthing about math. > Not quite that simple. You have f(y) = y+a for all y > in the range of f. > If you didn't assume f is differentiable, you could > have e.g. (if a > 0) > f(x) = |x| + a. > Robert Israel > israel@math.MyUniversity'sInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable <13562916.1150956946802.JavaMail.jakarta@nitrogen.mathforum.org > >> How \ can i solve this functional equation: > > >> f(f(x)) = f(x) + a ? > > This doesn't match the Subject line. Which one do > > you want? > Oh, sorry being not attentive. I wanted to solve this one which was \\ written in the subject line: > f(f(x)) = x + a. I assume a > 0. Note that f can't have any fixed points, and (using continuity) x + a > f(x) > x for all x. Moreover, f is one-to-one, so it is increasing, and since it satisfies f(x+a) = f(f(f(x))) = f(x) + a we only need to define it on [0,a] in such a way that f(a) = f(0) + a, then take f(x + a n) = f(x) + a n for integers n. Let b = f(0), so 0 < b < a and f(b) = a. Let f be any increasing continuous function from [0,b] to [b,a]. Then on [b,a] we want f(x) = f^(-1)(x) + a. This is an increasing function from [b,a] to [a, a+b], so we now have f taking [0,a] to [b,a+b]. Note that f will be differentiable if it is differentiable on (0,b) with strictly positive derivative, differentiable from the right at 0 and from the left at b, and the product of the right derivative at 0 and the left derivative at b is 1. For a specific nontrivial example, take a = 1, b = 4/9, f(x) = n + 4/9 + 2(x-n) - 27/16 (x-n)^2 for n <= x <= n + 4/9 = n + 43/27 - 4/27 sqrt(28 - 27 (x-n)) for n+4/9 <= x <= n+1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable > N/A > > > >> How can i solve this functional equation: > > > >> f(f(x)) = f(x) + a ? > > > > This doesn't match the Subject line. Which one > do > > > you want? > > Oh, sorry being not attentive. I wanted to solve > this one which was written in the subject line: > > f(f(x)) = x + a. > I assume a > 0. > Note that f can't have any fixed points, and (using > continuity) x + a f(x) > x > for all x. Moreover, f is one-to-one, so it is > increasing, and since > it satisfies > f(x+a) = f(f(f(x))) = f(x) + a we only need to define > it on [0,a] in > such a way that > f(a) = f(0) + a, then take f(x + a n) = f(x) + a n > for integers n. > Let b = f(0), so 0 < b < a and f(b) = a. > Let f be any increasing continuous function from > [0,b] to [b,a]. > Then on [b,a] we want f(x) = f^(-1)(x) + a. This is > an increasing > function from [b,a] to [a, a+b], so we now have f > taking [0,a] to > [b,a+b]. Well, Robert, thank you, it seems i understand where you got the lines of \ \\ you solution, but i dont' quite understand whether you are describing the \ \\ whole set of solutions or you are just giving a specific example of such \\ function ? As far as i undertsand you you mean the following way of finding f: 1 step: we bother only with values of f on [0,a] other values we can get by \ \\ the relation f(x + an) = f(x) + an. Well, it ok. 2 step: Take b = f(a), in order to get the values of f on [0,a] we notice \ \\ that f can be any function with given property of diffentiation on [0,b] \\ which takes it values in [b,a] and then to get the values on [b,a] we get \ \\ apply the formula f(x) = f^{-1}(x) + a. (Note that it is neccessary for to be in [b,a] on [0,b] because f(0) = b, \ \\ f(b) = a and f is increasing). In such a way, when we define f: [0,b] -> [b,a] we without problems get the \ \\ values of f on [b,a]. So, we've define f on thw whole [0,a]. 3 step. We shoukd also bother about the differentiation condition, which \\ Robert described here > Note that f will be differentiable if it is > differentiable on (0,b) > with strictly positive > derivative, differentiable from the right at 0 and > from the left at b, > and the product > of the right derivative at 0 and the left derivative > at b is 1. Have i understood you right, Robert ? And could you please look here http://mathforum.org/kb/message.jspa?messageID=4831072&tstart=0 where i tried to reduce the problem to the differential equation (f'(x))^2 = 1 and could you help me to solve it ? Was i right there ? > For a specific nontrivial example, take a = 1, b = > 4/9, > f(x) = n + 4/9 + 2(x-n) - 27/16 (x-n)^2 for n <= x <= > n + 4/9 > = n + 43/27 - 4/27 sqrt(28 - 27 (x-n)) for > )) for n+4/9 <= x <= n+1. > Robert Israel > israel@math.ubc.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada === Subject: Re: f(f(x)) = x + a , f - derivable format=flowed; reply-type=original eugene nous a r\\.8ecemment amicalement signifi\\.8e : > And could you please look here > http://mathforum.org/kb/message.jspa?messageID=4831072&tstart=0 > where i tried to reduce the problem to the differential equation > (f'(x))^2 = 1 and could you help me to solve it ? > Was i right there ? Your mistake is when you write : \\so we get that g'(x) = f'(x) for all x in R. But g'(x) = 1/f'(x), so have the following differntial equation on f:\\ The first part is OK : \\so we get that g'(x) = f'(x) for all x in R\\ The second is wrong : \\But g'(x) = 1/f'(x)\\ In fact f(g(x)) = x implies f'(g(x)) g'(x) = 1 and g'(x) = 1/f'(g(x)) and not g'(x) = 1/f'(x) ... -- Patrick === Subject: Re: f(f(x)) = x + a , f - derivable format=flowed; reply-type=original Robert Israel nous a r\\.8ecemment amicalement signifi\\.8e : > How can i solve this functional equation: > f(f(x)) = f(x) + a ? > This doesn't match the Subject line. Which one do > you want? >> Oh, sorry being not attentive. I wanted to solve this one which was >> written in the subject line: >> f(f(x)) = x + a. > I assume a > 0. > Note that f can't have any fixed points, and (using continuity) x + a > > f(x) > x > for all x. Moreover, f is one-to-one, so it is increasing, and since > it satisfies > f(x+a) = f(f(f(x))) = f(x) + a we only need to define it on [0,a] in > such a way that > f(a) = f(0) + a, then take f(x + a n) = f(x) + a n for integers n. > Let b = f(0), so 0 < b < a and f(b) = a. > Let f be any increasing continuous function from [0,b] to [b,a]. > Then on [b,a] we want f(x) = f^(-1)(x) + a. This is an increasing > function from [b,a] to [a, a+b], so we now have f taking [0,a] to > [b,a+b]. > Note that f will be differentiable if it is differentiable on (0,b) > with strictly positive > derivative, differentiable from the right at 0 and from the left at b, > and the product > of the right derivative at 0 and the left derivative at b is 1. > For a specific nontrivial example, take a = 1, b = 4/9, > f(x) = n + 4/9 + 2(x-n) - 27/16 (x-n)^2 for n <= x <= n + 4/9 > = n + 43/27 - 4/27 sqrt(28 - 27 (x-n)) for n+4/9 <= x <= n+1. I do agree and I'm sorry to have said nearly the same thing a little bit later ;-) -- Patrick > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re[topic modified] : f(f(x)) = f(x) + a , f - derivable format=flowed; reply-type=original eugene nous a r\\.8ecemment amicalement signifi\\.8e : > How can i solve this functional equation: > f(f(x)) = f(x) + a ? I suppose the problem is \\f(f(x)) = f(x) + a\\ and not \\f(f(x)) = x + a\\ I suppose a > 0 1) As a lot of posts said : f(x) = x + a for every x in IM(f) 2) if x is in IM(f), x+a is also in IM(f) 3) if x and x+a are in IM(f) and if f is continuous, [x, x+a] is in IM(f) 4) there are only two cases : 4.1) IM(f) = R and f(x) = x +a for every x in R 4.2) IM(f) = [b, +inf[ or ]b, +inf[ if a >0, In this case, the solution is : Take any fonction g(x) : - defined on ]-inf, b] - whose image is in [b, +inf[ - such that g(b) = b+a and g'(b)=1 The solution is : f(x) = g(x) if x < b f(x) = x+a if x >= b Example : b = 2, a > 0 g(x) = 2 + (x+2a-2)^2/(4a) f(x) = x+a for x >=2 f(x) = 2 + (x+2a-2)^2/(4a) for x < 2 If a < 0 : same approach If a = 0 it is easy to show that f(x) = x is the only differentiable solution (but a lot of other solutions exist if f is just continuous, and not differentiable). -- Patrick === Subject: Re: matrix functions Am 22.06.2006 04:42 schrieb Abstract Dissonance: > Are there any other functions f besides the obvious such that if A is a > matrix then f(A)_(i,j) = f(A_(i,j))? > Is there any simple relationship between exp(A_(i,j)) and exp(A)_(i,j)? > Jon As Robert Israel pointed out, there is none such relation in terms of determination of exp(A[i,j]) -> exp(A)[i,j] However, if A is nilpotent, then examples with nice and \\simple\\ relations can be found (well, whatever someone wants to call \\ \\simple\\...). My favourite example is 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 exp( 0 2 0 0 0 ) = 1 2 1 0 0 0 0 3 0 0 1 3 3 1 0 0 0 0 4 0 1 4 6 4 1 where the first subdiagonals of A and exp(A) are even identical ;-) Gottfried Helms === Subject: Re: Can I get an upper-bound for this series? Oh I am sorry for the misleading wording. Yeah, I actually want \\a simple expression for the sum in terms of M, K, a, \ \\ analogous to the formula for the geometric series\\ if these exists one. With the help provided by many kind minds so far, I am already happy with the geometric series like upper bound. However, if there can be an exact, yet simple, expression for the sum, it will be great. I have already tried many hours to no avail ... :) Patrick >> I came across the following series, but am stuck of finding the >> upper-bound >> of it (if it has one). Can somebody help? >> 1 + a + a^M+a^(M^2)+a^(M^3)+a^(M^4)+......+a^(M^K) >> i.e., >> 1 + summation i=0 to i=K (a^(M^i)) >> where a < 1, M and K are integers and M and K > 1. > I don't think there is an upper bound. The sum can be made greater than > any number you care to specify by choosing K large enough and \\a\\ close > enough to one (whatever the value of M). Is that what you meant? >> If an upper bound does not exist, an expresseion of it in terms of M and \ \\ >> K >> will help as well. > If an upper bound does not exist then how can there be an \\expression > of it\\? > (I'm wondering, actually, if you *really* just want to know if there's > a simple expression for the sum in terms of M, K, a, analogous to the > formula for the geometric series...) === Subject: Re: Does the Calculus rest on Euclid? theorem presented in such works as Gray's _Differential Geometry of \\ Curves > and Surfaces with Mathematica_. I understood Gray's presentation well > enough to correct the serious typos (in the second edition). My question > is: what are the axioms? If you want a rigorous development of real analysis, why not read one? === Subject: Re: Does the Calculus rest on Euclid? >> My concern is not whether I understand the typical proof of the \\ arclength >> theorem presented in such works as Gray's _Differential Geometry of >> Curves >> and Surfaces with Mathematica_. I understood Gray's presentation well >> enough to correct the serious typos (in the second edition). My \\ question >> is: what are the axioms? > If you want a rigorous development of real analysis, why not read one? Such as Scherbert and Bartle? I don't recall it being very axiomatic. \\ But, I'm not sure a book on real analysis (f:R^1->R^1) will address the matter that originally got me thinking about this. I know there was another book suggested in this discussion. Perhaps I will follow up on that suggestion when I find time. I have often found that other people make assumptions about mathematics which I do not make, and they don't make assumptions which I do make. I guess that is a hazard to self directed education. -- Nil conscire sibi === Subject: Re:[topic not modified] : f(f(x)) = x + a , f - derivable format=flowed; reply-type=original eugene nous a r\\.8ecemment amicalement signifi\\.8e : > How can i solve this functional equation: > f(f(x)) = x + a A lot of solutions exist for this functional equation : I suppose a > 0 1) notice that f(x) = x has no solution 2) notice that f(x) = x+a has no solution 3) notice that if f(x0) < x0, then f(y0)> y0+a with y0 = f(x0). So, because f is continuous, it exists x1 (between x0 and y0) such that f(x1) = x1, which is impossible ==> for every x in R, x < f(x) < x+a 4) notice that f'(x)=0 has no solution ==> f is monotonous increasing Then take b in ]0,a[ Take any function g(x) defined on [0,b], monotonous increasing, such that g(b) = a. In order to have f differentiable everywhere in R, we need some conditions more on g : g(x) differentiable g'(b)g'(0) = 1 Then f is easy to define : for x in [0, b[, f(x) = g(x) for x in [b, a[, f(x) = g^[-1](x) + a everywhere else : f(x) = f(x-a[x/a]) + a[x/a] Example : b = a/2 g(x) = (a^2/2)/(a - x) And f : for x in [0, a/2[, f(x) = (a^2/2)/(a - x) for x in [a/2, a[, f(x) = 2a - a^2/(2x) everywhere else : f(x) = f(x-a[x/a]) + a[x/a] -- Patrick === Subject: Re: How many finite geometries are there on n points? Or is there \ \\ at least <21372940.1150882654355.JavaMail.jakarta@nitrogen.mathforum.org \ http://www.math.ohio-state.edu/~fiedorow/math655/HyperEuler.html Pretty, and interesting, but I don't see the connection... David === Subject: Complex Structure of R^2 in Differential Geometry? Alfred Gray introduces what he calls \\the complex structure of R^2\\, J:R^2->R^2, and proceeds to use it in his development of curves in R^2. He goes so far as to claim that it is \\an essential tool\\. I'm wondering \\ how common the use of this idiom of the complex structure is. Here's an implementation of the basic idea in Mathematica: mj = {{0, 1}, {-1, 0}}; (*J represented as a matrix*) mc = {{1, 0}, {0, -1}}; (*complex conjugation matrix*) cp[p_, q_] := {p.(q.mc), p.((q.mc).mj)}(*complex product*) p = {pr, pi}; (*example vector*) q = {qr, qi}; (*example vector*) cp[p, q.mc] (*example application*) {pi qi + pr qr, -pr qi + pi qr} (*result*) The complex product can also be represented as cp[p, q] == {p.(q.mc), Det[{q.mc, p}]}; Clearly, if we normalize these expressions we have cos(theta) and \\ sin(theta) as the real and imaginary parts respectively. The complex product, as well as the e^(i theta) form for vectors appears to be the primary motive for using J. It seems to be used a few times in the subsequent discussion, but not extensively. It's interesting to see that things can be expressed in this way, but I wonder if the trouble of learning to use it is worth the advantage it provides. Is this approach of using J common? -- Nil conscire sibi === Subject: Re: Is there an easy way to show a transposition & n-cycle generate \ \\ S_n? > > > This is a problem I've run into several times. Proving that a > > > transposition and an n-cycle generate S_n > > > (snip) > > > Is there a > > > \\nicer\\ way to prove this fact? > > Below is about the shortest way I know to prove it without induction. > > (snip) > > Given two permutations (1 2 3 4 ... n) and (1 2). > > (snip) > NO! If I meant (1 2 3 ... n) and (1 2) I would have said so, that is > an easier problem than the problem I stated. I said *an n-cycle* and > *a transposition*. We can without loss of generality assume the > n-cycle is (123...n) *or* we can without loss of generality assume the > transposition is (12) but we can't do *both* without loss of > generality, at least not without explaining why, which is the whole > hard part of the proof. The problem is not \\show that an n-cycle and a > transposition of two elements, one of which conveniently gets sent to > the other by the n-cycle, generate S_n\\, which is what you guys seem to > have read it as. Proving (12...n) and (12) generate S_n is trivial, as > you guys pointed out. OK, two comments, one linguistic and one mathematical. For the linguistic point, as you admit yourself, you asked for a proof that S_n is generated by *an n-cycle* and *a transposition*. Well (1,2,...,n) is an n-cycle, and (1,2) is a transposition, so jaapsch has correctly answered the question that you asked. It seems as though what you intended to ask was: let g be any n-cycle and let h be any transposition in S_n; then prove that S_n is generated by g and h. As has already been pointed out, this is false in general - for example, the subgroup of S_4 generated by (1,2,3,4) and (1,3) has order 8, and is not equal to S_4. It turns out that the answer is yes if and only if n is prime (and when n = 1). Derek Holt. === Subject: Re: Is there an easy way to show a transposition & n-cycle generate \ \\ S_n? > > > > This is a problem I've run into several times. Proving that a > > > > transposition and an n-cycle generate S_n > > > > (snip) > > > > Is there a > > > > \\nicer\\ way to prove this fact? > > > > Below is about the shortest way I know to prove it without \ induction. > > > (snip) > > > > Given two permutations (1 2 3 4 ... n) and (1 2). > > > (snip) > > NO! If I meant (1 2 3 ... n) and (1 2) I would have said so, that is > > an easier problem than the problem I stated. I said *an n-cycle* and > > *a transposition*. We can without loss of generality assume the > > n-cycle is (123...n) *or* we can without loss of generality assume the > > transposition is (12) but we can't do *both* without loss of > > generality, at least not without explaining why, which is the whole > > hard part of the proof. The problem is not \\show that an n-cycle and \ \\ a > > transposition of two elements, one of which conveniently gets sent to > > the other by the n-cycle, generate S_n\\, which is what you guys seem \ \\ to > > have read it as. Proving (12...n) and (12) generate S_n is trivial, as > > you guys pointed out. > OK, two comments, one linguistic and one mathematical. > For the linguistic point, as you admit yourself, you asked for a proof > that S_n is generated by *an n-cycle* and *a transposition*. Well > (1,2,...,n) is an n-cycle, and (1,2) is a transposition, so jaapsch has > correctly answered the question that you asked. > It seems as though what you intended to ask was: let g be any n-cycle > and let h be any transposition in S_n; then prove that S_n is generated > by g and h. As has already been pointed out, this is false in general - > for example, the subgroup of S_4 generated by (1,2,3,4) and (1,3) has > order 8, and is not equal to S_4. > It turns out that the answer is yes if and only if n is prime (and > when n = 1). > Derek Holt. Just to add, if one prefers not to get their \\hands dirty\\ constructing the multiplication table as I did, realize the (rigid) symmetries of the square are generated by (1 2 3 4) and (1 3), assuming a consecutive numbering of vertices around the perimeter. This nonabelian group of order 8 is often called the dihedral group D_4 to distinguish it from the quaternion group (the other nonabelian group of order 8). === Subject: Re: Is there an easy way to show a transposition & n-cycle generate \ \\ S_n? > > > > > This is a problem I've run into several times. Proving that \ \\ a > > > > > transposition and an n-cycle generate S_n > > > > > (snip) > > > > > Is there a > > > > > \\nicer\\ way to prove this fact? > > > > (big snippage) > > It turns out that the answer is yes if and only if n is prime (and > > when n = 1). > > Derek Holt. Oh man, I screwed up there, please forgive me I caused anyone to pull a lot of hair out over this. I remember now that when I did this problem on homework some years back, it was assumed n was prime. Here is the context in which this arose on a recent practice qual. A problem in 2 parts: a. Show that (12345) and (12) generate S_5 b. Suppose a quintic polynomial has exactly 2 nonreal roots, show its Galois group is S_5. Part a is trivial of course. For part b, of course the nonreal roots are conjugate, hence the Galois group has a transposition, and by Cauchy's theorem it has an order 5 element, ie a 5-cycle. Given that a transposition and a 5-cycle generate S_5, we're done. I didn't remember the primality condition. The part a seems to be jeering and taunting at me. As if there is some way to show that not only does the Galois group contain an order 5 element, but that it contains an order 5 element that sends one of the nonreal roots to the other. This *seems* like it should be easy, but I wasn't able to show it. So I tried to use the more general result instead... Please forgive a newbie and have patience with me, I apologize sincerely for all the confusion my post caused S.P. === Subject: Re: Closed and bounded - clarification AHA!! Now I see why one must secify both closed and bounded - from your explan: \\As you defined above, the complement of [1,oo) is open, so this interval is closed (but it is not bounded).\\ Guy Corrigall >> Have got myself confused. I've used \\oo\\ to signify infinity. >> An interval I of the reals [a,b] where is a set. I is also closed set, >> because its complement (-oo,a)U(b,+oo) is open. >> OK so far? >> I is also bounded, given a> LUB in I, and is a maximum, because it is a point in I. >> My question is: why use the phrase \\closed and bounded interval\\, when \ \\ >> *any* >> closed interval of the reals is automatically bounded? Why not just >> \\closed >> interval\\? > The entire set of reals is an interval also, and both closed and open. >> My second question is: why is the interval A of the reals [1, oo) >> described >> as \\the *closed* infinite interval\\? Surely it is half open, and not >> closed? Is it because its complement (-oo,1) is open? > As you defined above, the complement of [1,oo) is open, > so this interval is closed (but it is not bounded). >> Am I getting confused between \\closed interval\\ and \\closed set\\? \\ Does >> the >> word \\closed\\ have a different meaning in these two cases? > In these two cases \\closed\\ means topologically closed. > Of course there are other mathematical contexts in which > \\closed\\ means something entirely different, e.g a group > is closed under multiplication, or the complex numbers > form an algebraically closed field. === Subject: Re: Closed and bounded - clarification explan: > \\As you defined above, the complement of [1,oo) is open, > so this interval is closed (but it is not bounded).\\ > Guy Corrigall If it were a matter of saving a word or two, we could describe the closed and bounded intervals of the real line as the \\compact\\ intervals. Compactness is a very useful concept in topology: A set K in topological space X is said to be compact iff every open cover of K has a finite subcover. That is, if we have a family of open sets in X such that K is contained in the union of those open sets, then compactness of K means we can always get a \\cover\\ of K in this sense by using only finitely many of the open sets. At first glance it may be hard to see why compact implies both closedness and boundedness of a subset in the real line, but it does. [Another note: The intervals in the real line are precisely the connected subsets of R.] === Subject: Bezier track path with inversions (Note: This was also posted to comp.graphics.algorithms, but I felt I should go ahead and try posting this here as well) There's a roller coaster simulator out for Windows/Mac called No Limits (http://www.nolimitscoaster.de/ (The site is in English and German) that uses bezier curves to define the track. For banking, you set the angle of banking in degrees for each control point. I'm writing a program to convert tracks made with this program into POV-Ray (http://www.povray.org) scenes. But I'm running into a huge problem with inversions. The problem stems from how banking is calculated in No Limits. You can create a vertical loop with only two control points - one pointing up, the other pointing down. Of course, at both of those points, the track is right-side up, so the banking is 0. Of course, a smooth track will typically have more control points than that. Anything before the vertical will have very little banking, but once you go past the vertical, banking has to be around 180 to get the track to be upside down. So now it comes down to how to implement this. The simplest implementation would just take the cross product of the vector <0,1,0> and the current direction of travel to find an up vector, and then rotate that around the current direction of travel. I then interpolate between the resulting up vectors on each control point to find intermediate up vectors. This works in the second method of constructing a loop as described above, but runs into the problem of vertical track not working. To accomodate for vertical track, No Limits allows you to specify that a given banking angle is \\relative\\ to the previous. Using that information (Plus a hint from someone on the POV-Ray newsgroups) I tried another method. First, I convert all banking angles into relative banking, then I start with an initial up vector, and I step along the bezier. At each step, I cross the current up vector with the current direction of travel to get the \\right\\ vector, then cross that again with the current direction of travel. This finds a new up vector that compensates for a change in vertical slope, which I then rotate around the current direction of travel to find the appropriate banking. I then use that up vector during my next step. This SORT of works for the first method of making a loop described above. The problem is, if the track is changing heading while on a vertical slope or banked, the resulting up vector ends up being a bit banked. Eventually, all track ends up being banked when it shouldn't be. Does anyone have any idea how I could do this? The No Limits website has a demo version that lets you use the editor so you can see exactly how tracks are designed. If anyone can help at all, it would be greatly appreciated, and I may even put your name (if you'd like) in the program credits when I am finished. -DJ (Remove the TRIMBRAKES to reply) === Subject: infinite product Is there a closed form expression for the infinite product: Prod (1+a^n), for n=0..infty, where 0 Is there a closed form expression for the infinite product: > Prod (1+a^n), for n=0..infty, where 0 I'm particularly interested in the value for a=1/16. This is related to the > limiting value of the CORDIC gain. > -Michael. I've found a general result at , formula 59, involing theta functions. Still, does anyone know the exact value for a=1/16? -Michael. === Subject: Re: infinite product Michael JÀrgensen a \\.8ecrit : >> Is there a closed form expression for the infinite product: >> Prod (1+a^n), for n=0..infty, where 0> I'm particularly interested in the value for a=1/16. This is related to > the >> limiting value of the CORDIC gain. >> -Michael. > I've found a general result at > , formula 59, involing > theta functions. > Still, does anyone know the exact value for a=1/16? Maple gives 2.133856336838677365056872920875...., which is not recognized by Plouffe's inverter :-( > -Michael. === Subject: Re: infinite product > Michael JÀrgensen a \\.8ecrit : > > >> Is there a closed form expression for the infinite product: > > >> Prod (1+a^n), for n=0..infty, where 0 > > Still, does anyone know the exact value for a=1/16? > Maple gives 2.133856336838677365056872920875...., which is not > recognized by Plouffe's inverter :-( I just realized, I meant a=1/4, not 1/16. Using my hand calculator, I get 2.7118, but that is not quite enough decimals for Plouffe's inverter. Would you mind trying again using Maple, please :-) -Michael. === Subject: Re: infinite product Michael JÀrgensen a \\.8ecrit : >> Michael JÀrgensen a \\.8ecrit : >> Is there a closed form expression for the infinite product: >> Prod (1+a^n), for n=0..infty, where 0 Still, does anyone know the exact value for a=1/16? >> Maple gives 2.133856336838677365056872920875...., which is not >> recognized by Plouffe's inverter :-( > I just realized, I meant a=1/4, not 1/16. Please make up your mind :-) Using my hand calculator, I get > 2.7118, Correct: it is 2.7118193477269587606910884697079186024433990... (still no match from Plouffe) but that is not quite enough decimals for Plouffe's inverter. Would > you mind trying again using Maple, please :-) > -Michael. === Subject: Re: Even more tensor trouble > So it is somewhat of an art to go from index notation to array > manipulations, but probably worth learning because array calculations are > usually more efficient. Work in index notation until you run into \\ efficiency > problems and then convert to array calculations. Since I start small with i,j,k,l=1,2 I don't think I run into effiency problems :-) But what is, then, the \\(x)\\ guy, also named tensor product, that pops up (e.g.) in the Yang-Baxter equation? It doesn't even commute with one factor being the identity. How does THAT properly look in index notation? (Usually it's just written R#I*I#R*R#I=I#R*R#I*I#R where I use # for the encircled X. Without mentioning ANY indices.) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: An easy matrix quals problem Hi everybody, I'm sure this is supposed to be a really simple problem (for a linear algebra qualifying exam), but I haven't figured it out yet. Can anybody help me? Let A and B be real symmetric matrices, with B positive definite. Show that AB has real eigenvalues. KH === Subject: analysis ! helllo...sir a_n >0 and lim a_n = 0 as n->00 b_n = (a_1) + (a_2)+ .... +(a_n) - {n*(a_n)} (b_n) is bounded above. then, sum a_n [n=1~00] is converge. ----------------------------------------- maybe, i must show that sum a_n is bounded above. so, i need your hint. thank you very much. === Subject: Re: An uncountable countable set and do not exist at all if f is defined to be a surjection. That is right. And why do you believe that a self contradictory approach should prove anything (for instance about the surjectivity of certain mappings)? === Subject: Re: An uncountable countable set mueckenh@rz.fh-augsburg.de says... >> So the resolution is that either f, K_f and M_f are self-contradictory >> and do not exist at all >if f is defined to be a surjection. That is right. And why do you >believe that a self contradictory approach should prove anything (for >instance about the surjectivity of certain mappings)? If an assumption leads to a contradiction, then that assumption must be false. The assumption that there is a surjection from N to P(N) leads to a contradiction. Therefore, it is false that there is a surjection from N to P(N). Therefore, P(N) is uncountable. -- Daryl McCullough Ithaca, NY === Subject: Re: An uncountable countable set > > > A set is required which is the image of \ k if it is not the image \\ of k. > > > > A barber is required who shaves himself if he does not. > > (In case f should be surjective.) > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed \ according \\ to > > > Hessenberg *does* exist. but it is not possible to determine that K(f). > > > If f were a bijection it is required (by > > > the *definition* of bijection) that K(f) (because it is an \\ element > > > of P(N)) is the image of some element of N, but it is not, \\ showing > > > that f is not a bijection. P(N) does not depend on f, But P(N) \\\\ K(f) does. Map f : N --> P(N) \\\\ K(f), and your proof vanishes. > > > so there is > > > no mapping between N and P(N) that is a bijection. Or there is no complete set |N and hence no complete set P(|N). > > Who told you? > The proof above. It does not disprove a surjection but the completeness of |N and P(|N). Why do you think that non-dependence on f is a special feature? The set K does exist for every mapping f. But for every mapping f there is no k to be mapped on K. The set K does not an cannot exist if f is required to be a surjection. Why do you think, this can and must happen for a countable image set M = S(|N) U {K(f)} ? Why do you think the reason is different from that of P(|N) ? > > Map f : N --> P(N) \\\\ K(f). That has not been proven > > impossible. > > Then map g(n+1) = f(n) and g(1) = K(f). There is no proof that this > > cannot be a bijection. > Yes. See above. For *any* mapping it is shown that it is not a \\ bijection. The same is true for any M = S(|N) U {K(f)} defined by a *surjective* mapping f. The set K then does not an cannot exist. Hence, there is no helping g. === Subject: Re: An uncountable countable set > > > > > > > > > > > > A set is required which is the image of k if it is not the \\ image of k. > > > > > A barber is required who shaves himself if he does not. > > > > > > (In case f should be surjective.) > > > > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed \\ according to > > > > Hessenberg *does* exist. > but it is not possible to determine that K(f). Why not? > > > > If f were a bijection it is required (by > > > > the *definition* of bijection) that K(f) (because it is an \\ element > > > > of P(N)) is the image of some element of N, but it is not, \\ showing > > > > that f is not a bijection. P(N) does not depend on f, > But P(N) \\\\ K(f) does. Map f : N --> P(N) \\\\ K(f), and your proof > vanishes. As I have shown already in a much earlier thread, the answer is *no*. > > > > so there is > > > > no mapping between N and P(N) that is a bijection. > Or there is no complete set |N and hence no complete set P(|N). That is altogether another theory. When you talk about current theory there is a complete set N and a complete set P(N). You may of course start your own version of set theory where that is not the case. > > > Who told you? > > > > The proof above. > It does not disprove a surjection but the completeness of |N and P(|N). You are completely wrong. How does it prove that? > Why do you think that non-dependence on f is a special feature? The set > K does exist for every mapping f. But for every mapping f there is no k > to be mapped on K. > The set K does not an cannot exist if f is required to be a surjection. > Why do you think, this can and must happen for a countable image set M > = S(|N) U {K(f)} ? Why do you think the reason is different from that > of P(|N) ? Let's see. Let S be the set of finite subsets of N. And let us have a bijection f: N -> S (they do exist). Now obviously f is not a bijection from N to M(f), that is clear by the construction. On the other hand, we can construct a bijection from N to M(f), as before: g(0) = K(f) g(i) = f(i - 1) when i > 0, given that f is a bijection between N and S, it is easy to see that g is a bijection between N and M(f). Ah, I hear you mutter. But g is not a bijection between N and M(g). No, of course not. g is not even a mapping from N to M(g). > > > Map f : N --> P(N) \\\\ K(f). That has not been \ proven > > > impossible. > > > Then map g(n+1) = f(n) and g(1) = K(f). There is no proof that this > > > cannot be a bijection. > > > > Yes. See above. For *any* mapping it is shown that it is not a \\ bijection. > The same is true for any M = S(|N) U {K(f)} defined by a *surjective* > mapping f. The set K then does not an cannot exist. Hence, there is no > helping g. Note again: M depends on f, a fact that you leave out every time. M in itself is not a set, but for every given f, M(f) is a set. -- 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: An uncountable countable set > > > 1. Given a mapping f: N -> P(N), the set \ K(f) constructed \\ according to > > > > > Hessenberg *does* exist. > > but it is not possible to determine that K(f). > Why not? Because, in case f should be surjective, a number k need be contained in a set in which it must not be contained. > > > > The proof above. > > It does not disprove a surjection but the completeness of |N and \\ P(|N). > You are completely wrong. How does it prove that? Because it is the only explanation of this and other paradoxes. At least you cannot deny that it would resolve he problem. Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of all rational numbers. If you say that it exists, then I can prove that it can be ordered by magnitude without destroying the well-order. Obviously that is impossible. Hence {q_1, q_2, q_3, ...} does not exist. > Let's see. Let S be the set of finite subsets of N. And let us have > a bijection f: N -> S (they do exist). Now obviously f is not a > bijection from N to M(f), that is clear by the construction. Hence by constuction, the set M(f) does not exist, if f is required to be surjective. > Note again: M depends on f, a fact that you leave out every time. M in > itself is not a set, but for every given f, M(f) is a set. And for every given f, this set is not in bijection with |N. === Subject: Re: An uncountable countable set mueckenh@rz.fh-augsburg.de says... >> > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed according to Hessenberg *does* exist. >> > > but it is not possible to determine that K(f). >> Why not? >Because, in case f should be surjective, a number k need be contained >in a set in which it must not be contained. That case is impossible. To prove a case impossible, you only need to show that it leads to a logical contradiction. >Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of >all rational numbers. >If you say that it exists, then I can prove that it can be ordered by >magnitude without destroying the well-order. On the contrary, you cannot prove that. Here's a specific well-ordering: Let r1 and r2 be two rational numbers. Then to figure out whether r1 comes before r2 in the well-ordering: Write r1 in the form p1/q1 where p1 and q1 are integers and q1 > 0 and there are no common factors between p1 and q1. Similarly, write r2 in the form p2/q2. 1. If |p1| + q1 < |p2| + q2, then r1 comes before r2. 2. If |p1| + q1 > |p2| + q2, then r1 comes after r2. 3. If |p1| + q1 = |p2| + q2, and p1 < p2, then r1 comes before r2. 4. If |p1| + q1 = |p2| + q2, and p1 > p2, then r1 comes after r2. This well-ordering gives the following first few terms of the sequence: 0/1, -1/1, +1/1, -2/1, -1/2, +1/2, +2/1 -3/1, -1/3, +1/3, +3/1 -4/1, -3/2, -2/3, -1/4, +2/3, +3/2, +4/1 -5/1, -1/5, +1/5, +5/1 etc. (The first row has |p| + q = 1, the next row has |p| + q = 2, etc.) But notice that this is not ordered by magnitude, since 1/2 < 1/1, but 1/1 comes before 1/2 in the ordering. The ordering by magnitude in not a well-ordering. Obviously that is >impossible. Hence {q_1, q_2, q_3, ...} does not exist. >> Let's see. Let S be the set of finite subsets of N. And let us have >> a bijection f: N -> S (they do exist). Now obviously f is not a >> bijection from N to M(f), that is clear by the construction. >Hence by constuction, the set M(f) does not exist No, that doesn't follow. What follows is that M(f) is *not* a subset of S. Since S contains all finite subsets of N, it follows that M(f) contains some element that is *not* a finite subset of N. -- Daryl McCullough Ithaca, NY === Subject: Re: An uncountable countable set > > > > > > > > A set is required which is the image of k if it is not \ the image \\ of > > > > > k. > > > > > A barber is required who shaves himself if he does not. > > > > (In case f should be surjective.) > > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed \\ according > > > > to > > > > Hessenberg *does* exist. > but it is not possible to determine that K(f). > > > > If f were a bijection it is required (by > > > > the *definition* of bijection) that K(f) (because it is an \\ element > > > > of P(N)) is the image of some element of N, but it is not, \\ showing > > > > that f is not a bijection. P(N) does not depend on f, > But P(N) \\\\ K(f) does. Map f : N --> P(N) \\\\ K(f), and your proof > vanishes. > > > > so there is > > > > no mapping between N and P(N) that is a bijection. > Or there is no complete set |N and hence no complete set P(|N). > > > > Who told you? > > The proof above. > It does not disprove a surjection but the completeness of |N and P(|N). > Why do you think that non-dependence on f is a special feature? The set > K does exist for every mapping f. The mapping f cannot exist if one requires, as Muecken does, that there be some n for which f(n) = {x: x not in f(x)}. That condition makes f an impossibility. > > > Map f : N --> P(N) \\\\ K(f). That has not been proven > > > impossible. Yes it has. Since there are bijections h: P(N) \\\\ K(f) --> P(n), any bijection f:N --> P(N) \\\\ K(f) implies a bijection hof: N --> P(N) Which cannot occur since function hof cannot map onto K(hof). === Subject: Re: An uncountable countable set A *surjective* mapping does not exist. The same \ does Hessenberg. > A *surjective* mapping does exist, but it is not f. There exist a > surjective mapping g -> M(f). Like this one?: 2) 0.12324389 3) 0.23123123 4) 0.85348714 5) 0.11133333 6) 0.31415161 .. 1) 0.24446... This is the proof, that the proof that a surjective mapping f: |N --> [0,1] does not exist, does not exist, isn't it? === Subject: Re: An uncountable countable set > > > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > > > A *surjective* mapping does exist, but it is not f. There exist a > > surjective mapping g -> M(f). > Like this one?: > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > 5) 0.11133333 > 6) 0.31415161 > .. > 1) 0.24446... > This is the proof, that the proof that a surjective mapping f: |N --> > [0,1] does not exist, does not exist, isn't it? This is plain nonsense. -- 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: An uncountable countable set > > > > A *surjective* mapping does not \ exist. The same does Hessenberg. > > > > A *surjective* mapping does exist, but it is not f. There exist \ a > > > surjective mapping g -> M(f). > > Like this one?: > > 2) 0.12324389 > > 3) 0.23123123 > > 4) 0.85348714 > > 5) 0.11133333 > > 6) 0.31415161 > > .. > > 1) 0.24446... > > This is the proof, that the proof that a surjective mapping f: |N -- \ > [0,1] does not exist, does not exist, isn't it? > This is plain nonsense. Maybe. It is, however, the same approach as yours. If it is impossible to find a direct surjective mapping, then first define the set and then try to find a surjection. I can proudly declare in your words: A *surjective* mapping does exist, but it is not f. There exists a surjective mapping g -> M(f). === Subject: Re: An uncountable countable set mueckenh@rz.fh-augsburg.de says... >Maybe. It is, however, the same approach as yours. If it is impossible >to find a direct surjective mapping, then first define the set and then >try to find a surjection. I can proudly declare in your words: > A *surjective* mapping does exist, but it is not f. There exists a >surjective mapping g -> M(f). And those words are perfectly correct: f is not a surjective mapping from N to M(f), but g *is* a surjective mapping from N to M(f). So M(f) is countable. In contrast, there is no surjection from N to P(N). So P(N) is *not* countable. -- Daryl McCullough Ithaca, NY === Subject: Re: An uncountable countable set > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > A *surjective* mapping does exist, but it is not f. There exist a > > surjective mapping g -> M(f). > Like this one?: > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > 5) 0.11133333 > 6) 0.31415161 > .. > 1) 0.24446... > This is the proof, that the proof that a surjective mapping f: |N -- [0,1] \ does not exist, does not exist, isn't it? No. === Subject: Re: An uncountable countable set There is no bijective mapping f from N to M. Correct. What do you think, why this is so? My point is only to show that the impossibility of a mapping involving Hessenberg's trick does not prove anything about cardinalities of sets involved. It shows that the sets |N and P(|N) do not exist. > Note that M is given first, and *then* it is said that no bijective f \\ exists > for that M, with no other condition on f than it has to be bijective. Oh, I see. Let's apply this new insight: 2) 0.12324389 3) 0.23123123 4) 0.85348714 5) 0.11133333 6) 0.31415161 .. 1) 0.24446... So this is the proof, that the proof that a surjective mapping f: |N --> [0,1] does not exist, does not exist, isn't it? > In the well-known proof that P(N) is uncountable, P(N) is a fixed given \ \\ set > before any function is even mentioned. That is the question. > The proof then shows that no matter > which function f is chosen, f will not be a bijection from N to P(N). And my proof shows that no matter what function f is chosen, there will not be a bijection from N to M. You will have obtained from my sketch of Cantor's diagonal proof above that it is *not* allowed to switch the mapping after the set has been fixed. === Subject: Re: An uncountable countable set > > By definition, an infinite set M is uncountable if and only if: > > There is no bijective mapping f from N to M. > > Note that M is given first, and *then* it is said that no bijective f > > exists for that M, with no other condition on f than it has to be > > bijective. > Oh, I see. Let's apply this new insight: > 2) 0.12324389 > 3) 0.23123123 > 4) 0.85348714 > 5) 0.11133333 > 6) 0.31415161 > .. > 1) 0.24446... > So this is the proof, that the proof that a surjective mapping f: |N > --> [0,1] does not exist, does not exist, isn't it? As others have said: Nonsense. A small list of numbers with no definitions, assumptioms, explanations or anything else is, well, a small list of \\ numbers. Your final claim is hard to parse but if you think your small list of \\ numbers with no comments proves that Cantor's diagonal proof is wrong, then: \\ Nonsense. After seeing your replies here and elsewhere, I have decided to not spend \ \\ more time on this discussion. Others have already said several times what I would say but you appear to ignore it. I suggest rereading the replies carefully. > Please note: My name is Mueckenheim or short WM. If you want to be referred to by some name then use it as poster name \\ instead of your e-mail address. Many people reply with a program which automatically inserts your poster \\ name. -- Jens Kruse Andersen === Subject: Re: An uncountable countable set > > By definition, an infinite set M is uncountable if and only if: > > There is no bijective mapping f from N to M. > Correct. What do you think, why this is so? My point is only to show > that the impossibility of a mapping involving Hessenberg's trick does > not prove anything about cardinalities of sets involved. It shows that > the sets |N and P(|N) do not exist. No more than it proves Muecken does not exist. > > Note that M is given first, and *then* it is said that no bijective f \ \\ exists > > for that M, with no other condition on f than it has to be bijective. > > In the well-known proof that P(N) is uncountable, P(N) is a fixed given \ \\ set > > before any function is even mentioned. > That is the question. > > The proof then shows that no matter > > which function f is chosen, f will not be a bijection from N to P(N). > And my proof shows that no matter what function f is chosen, there will > not be a bijection from N to M. You will have obtained from my sketch > of Cantor's diagonal proof above that it is *not* allowed to switch the > mapping after the set has been fixed. Actually in Cantor's proof, one is allowed to construct as many functions as one likes between the fixed set N and the fixed set P(N), as the general anti-diagonal rule works equally well on any of them. Muecken insists on a function and codomain which can only exist if the function is not a bijection, but which, in that case allow a diferent fuction which is a bijection. So that either neither of Muecken's f and M t exist at all and the issue of bijection of N with a non-existent set is irrelevant. Or if f and M do exist, it is because f is not required to be a surjection, in particular K = {x in N: x not in f(x)} is not a value of f, and then f, K and M all exist, and there are plenty of bijections from N to M. === Subject: Re: An uncountable countable set I'm trying to work problems from a book but 2 have me stumped because >the book does not define what the \\nth roots of unity\\ are Exactly what the name suggests; z is an nth root of unity if z^n=1. >besides that it has something to do with the complex numbers While that is generally assumed, you can discuss the nth roots of unity in other fields. -- 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 reply to spamtrap@library.lspace.org === Subject: Re: One is a Unit? In <1j7h92dt16kgj99r5f7ma4jtdgqhq1trjg@4ax.com>, on 06/20/2006 at 06:28 PM, dtfoster@dcn.davis.ca.us said: >I started from basic to relearn some math and settled on integers, >primes and such. I read where one is called a unit. This has >something to do with numbers would not have one and only one integer >solution if it didn't have this special name. No; integers wouldn't have unique prime factorizations if we defined 1 to be a prime, but that has nothing to do with calling it a unit. We call 1 a unit because 1*x=x for any x. -- 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 reply to spamtrap@library.lspace.org === Subject: Re: One is a Unit? > In <1j7h92dt16kgj99r5f7ma4jtdgqhq1trjg@4ax.com>, on 06/20/2006 > at 06:28 PM, dtfoster@dcn.davis.ca.us said: >>I started from basic to relearn some math and settled on integers, >>primes and such. I read where one is called a unit. This has >>something to do with numbers would not have one and only one integer >>solution if it didn't have this special name. > No; integers wouldn't have unique prime factorizations if we defined 1 > to be a prime, but that has nothing to do with calling it a unit. We > call 1 a unit because 1*x=x for any x. No, that's the reason we call 1 \\the unit element\\. We call 1 \\a unit\\ because it has an inverse in Z. We also call -1 a unit for the same reason. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: One is a Unit? In , on 06/21/2006 >you could say that in the world of addition, zero is a \\unit\\, just >like one is a unit in the world of multiplication. whereas in the >world of addition, one not only isn't a unit, it's in fact the _only >\\prime\\_. Why would you not consider 1 and -1 to be equally fundamental in Z? They're both generators. -- 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 reply to spamtrap@library.lspace.org === Subject: Re: One is a Unit? Seems like the field of mathematics is about \ 10 times bigger now > |>than it was in the 60's. > |No, it isn't. > what _is_ the correct ratio, then, and how are you measuring? In terms of the number of papers or number of theorems, I don't know the ratio. I don't think that was what the OP was thinking of, either. I was thinking of (1) the broad outline of the various fields of mathematics, the sort of thing you find at the 2-digit level of the Mathematics Subject Classification. I don't think this has really changed much since the 60's, with a few exceptions, e.g. dynamical systems and computer science. Of course several fields have expanded greatly. (2) the mathematics an undergraduate would encounter, or a graduate student would need to know for qualifying exams. Again, I don't think these have changed much, certainly not in terms of quantity. 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 trigonometric equation Given tan D irrarional tan kD is also irrational. My assertion is that given (tan kD)^1/k irrational (tan kD)^2/k is also irrational. If you donot agree with this assertion kindly give your reason. Yourbinitial reply was very useful. Your additional comments will be highly appreciated. > >Professor Israel, > >slightly modified version of the posting. > >Let A = (tan kD)^(1/k) and B = (tan kD)^(2/k); odd k > 2. > >Situation-1: tan D = u/v, (u, v) = 1, uv > 0, Since tan D is rational > >then tan kD is also rational. > >If A is irrational then tan kD is not a k-th power. Then (tan kD)^2 is > >also rational but not a k-th power. Thererfore, if A is irrational then > >B is also irrational. > Let r = tan kD. This has nothing to do with tan, just with r. If r is > rational but not a k'th power (with k odd), then r^2 is not a k'th power > either. One way to see this: there is some prime p such that the p-adic > order of r is not divisible by k, and the p-adic index of r^2 is twice > that, so still not divisible by k. > >Situation-2: tanD = (u/v)^(1/2), (u,v) = 1 and both are not perfect > >squares.0 >Therefore, tan D is not rational. > OK. But I think you want to say something about tan kD. > And why are these the only two situations you're considering? > >Assertion: Given A irrational B will also be irrational. > Under what assumptions? > Robert Israel israel@math.MyUniversity'sInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Equalizer in the category of rings It is easy to compute the equalizer in the category of rings, its the same way one does it in the category of sets. One takes the kernel of the difference two morphism and one has a ring and then one considers the injection of the kernel to be the equalizer and for the universal property one just defines the unique mapping to be one by which the codomain is restricted.. ok to be more exact.. If p,q : A -> B be ring homomorphism, then the equalizer is E={a in A : p(a)=q(a)}, with e: E -> A, bringing this is a ring as one can easily check and for any ring homomorphism e': E' -> A with pe'=qe' we have the unique ring morphism e'' : E' -> E with e(x) = e'(x), which is in E (as one can check), such that ee''=e' I dont have any question here. I just want to know if there is a reference I can cite if I want to use the equalizer in the category of rings the way I just did here.. ... I dont want to write the whole thing Im writing here in my PhD thesis , so... or are these all quite clear that I dont even need to cite anything for my thesis. (I can of course cite books in category theory, but the ones I know dont show this particularly for rings.. which is probably a trivial matter). === Subject: Are you intelligent? Friends, visit the following link and test your IQ. http://in.geocities.com/prabhuraaman/knowledge/iqtest.htm Bookmark this page as the content keeps changing every time === Subject: Re: Are you intelligent? > Friends, visit the following link and test your IQ. > http://in.geocities.com/prabhuraaman/knowledge/iqtest.htm > Bookmark this page as the content keeps changing every time The link doesn't work. Dave === Subject: Re: Are you intelligent? >>Friends, visit the following link and test your IQ. >>http://in.geocities.com/prabhuraaman/knowledge/iqtest.htm >>Bookmark this page as the content keeps changing every time > The link doesn't work. > Dave You passed. === Subject: Re: Are you intelligent? martin cohen said: > > > >>Friends, visit the following link and test your IQ. > > >>http://in.geocities.com/prabhuraaman/knowledge/iqtest.htm > > >>Bookmark this page as the content keeps changing every time > > > > > > > The link doesn't work. > > > > Dave > > > > > You passed. That's funny. I scored a 43. ;) -- Smiles, Tony === Subject: Re: Functional calculus question On Wed, 21 Jun 2006 17:14:51 +0100, Gon\\.8dalo Rodrigues >On Wed, 21 Jun 2006 08:54:00 -0500, David C. Ullrich > fed this fish to the penguins: >>On Wed, 21 Jun 2006 14:27:45 +0100, Gon\\.8dalo Rodrigues >On Wed, 21 Jun 2006 07:04:59 -0500, David C. Ullrich > fed this fish to the penguins: >>On Tue, 20 Jun 2006 22:51:40 +0100, Gon\\.8dalo Rodrigues >Hi all, I'm reading up on C* and Von-Neumann algebras. Jumping ahead, I have a >question. To fix ideas, fix a probability space X, a Hilbert space H >and an X-spectral measure on H. This triple gives an isometric >*-representation p of L^{\\\\infty}(X) in the C*-algebra B(H) of bounded >linear operators H->H; this is the Borel functional calculus. >>I don't know anything about your actual question, but: >My question is the following: suppose we take the induced map >p:L^{1}(X)->B(H) >>are you certain that there _is_ such an induced map? With >>range contained in B(H)? I would have thought that at best >>an L^1 function corresponded to some unbounded operator. >Since X is a probability space f is integrable iff it is a.e. bounded >>Huh? What does it mean to say that a function is \\a.e. bounded\\? >>(I can only think of two things this might mean: >>(i) for almost every x, {f(x)} is bounded.) >>That seems like a literal translation of \\a.e. bounded\\, >>but that can't be what you mean, because any function >>satisfies that condition.) >>Or maybe you mean >>(ii) There exists c such that |f| <= c a.e. >>That makes more sense, but if you mean (ii) then it's >>simply not true that every L^1(X) function is a.e. bounded. >>I really don't believe that f in L^1(X) induces an element >>of B(H) in a natural way. And I can't make much sense of >>your explanation, since I can't figure out what you mean >>by \\a.e. bounded\\. >>But consider the following example: X is a probability >>space, and H = L^2(X); take the spectral measure so that >>the projection associated with E subset X is just the >>orthogonal projection onto functions supported on E. >>The for f in L^infinity(X), p(f) is just the operator >>of multiplication by f. And this operator is certainly >>_not_ bounded on H for f in L^1(X). >Sorry, too many confusions on my part, so let me try to clear them up. >As you point out, of course if f in L^1 doesn't imply it is in >L^\\\\infty (that's what I mean by a.e. bounded - your ii). So we really >should be thinking of the induced map in the *reverse* direction, e.g. >{Trace-class operators} -> L^1 I should start by saying that I know nothing about any of this - the extent of my knowledge of anything to do with the spectral theorem is that it shows normal operators are equivalent to multiplication operators on an L^2 space. But, having clarified that I don't know what I'm talking about, it seems to me you may be giving up too soon. My objection was to the idea that p maps L^1(X) into B(H); it simply doesn't. But it seems to me that p, or some extension of p which we may as well denote again by p, does map f in L^1(X) to an unbounded densely-defined operator. This is clear in the example where we take H to be L^2(X), for example. What if anything that has to do with trace-class operators I wouldn't know. >whose adjoint is then our original *-representation. I suppose this >can be made to work, but I would be much more interested if there was >L^1 -> {some class of bounded operators} that was a sort of integrable >functional calculus. But as you point out, in the naive way I was >thinking this cannot be made to work - the range falls in the dual of >B(H) or some such. >G. Rodrigues ************************ David C. Ullrich === Subject: Re: Shannon's \\well known\\ result >injected one comment in response to your question about zero length >messages... >> On Wed, 21 Jun 2006 10:34:49 -0500, David C. Ullrich >>In his famous 1948 paper entitled \\A Mathematical Theory of >>Communication\\, >>at the bottom of page 3, Shannon asserts... >>Suppose all sequences of the symbols S1,...,Sn are allowed and these >>symbols >>have durations t1,...,tn. What is the channel capacity? If N(t) >>represents >>the number of sequences of duration t we have >(*) >>N(t) = N(t - t1) + N(t-t2) + ... + N(t-tn) >>The total number is equal to the sum of the numbers of sequences ending \ \\ >>in >>S1,S2,...,Sn and these are N(t-t1), N(t-t2),...,N(t-tn), respectively. >>According to a well-known result in finite differences, N(t) is then >>asymptotic for large t to (X0)^t where X0 is the largest real solution \\ of >>the >>characteristic equation: >(**) >>X^-t1 + X^-t2 + ... + X^-tn = 1 >>My question is... can anyone point me to an explanation of the \\ supposedly >>\\well-known\\ result in finite differences that Shannon is making use \\ of >>here???? >Don't know. I figured out a little bit about this. >First, note that (**) holds for exactly one X > 0 (because >the sum of those powers of X tends to infinity as X -> 0 >and to 0 as X -> infinity). So the largest real root is >simply the unique positive root. >Also note that any function of the form cX^t satisfies (*), >if X is the positive root of (**). So (*) certainly does >not determine N(t), which is to say the answer can't be >_solely_ because of some theorem in finite differences, >there's some property of the initial conditions that comes >in. >When I start thinking about initial conditions I get confused >over whether we're counting the empty string as a message, etc. >I don't think (*) can really hold for all t > 0, but it's >clearly true for t > A, for some A. >I think that we *do* count the empty string as a message, i.e. N(0) = 1. \ \\ My >quasi-information theoretic argument would be that there is a zero length >message, you can \\generate\\ it, but from an informational point of view, \ \\ >there is only one, unlike an N(t), t>0 sequence where there can be a >multiplicity of symbol permutations for a given t. >On your second point about (*) holding for all t > 0, Shannon gives an >example using an alphabet of symbols of length 2, 4, 5, 7, 8 and 10. In >this case, clearly N(1) = 0 and N(3) = 0 since there is no combination of >symbols of those durations, so I agree there can be a few small values of t \ \\ >for which (*) is not true. One way of categorizing these failures is where \ \\ >one or more of the (t-tn) calculations return a \\negative time\\, in other \ \\ >words N(t) t < 0 is undefined (but N(t) t = 0 *is* defined and equals 1 as \ \\ >argued above). I considered more or less all of that. Note that you've said N(1) = 0 and N(0) = 1 here - so N is not non-decreasing? That's where I decided I was confused. >If all symbols were of the same duration, say \\5\\, then you could argue \ \\ that >any t = 5k+m, m = 1,2,3,4 would fail, but I believe this is just a matter \ \\ >of scaling, i.e. re-define \\5\\ as \\1\\. Are we assuming that all the t_j are integers? This would be more efficient if you didn't ignore people's questions, by the way. If we're assuming that all the t's are integers then the problem takes an entirely different flavor (this is related to the question you didn't answer about whether the t_j were rationally independent.) >Say L is the maximum t_j and l is the maximum t_j. (And let's >assume that l = 1, just in case \\l\\ and \\1\\ look the same.) >Now, just because both functions are bounded and bounded away >from 0 on [A, A+L], there exist c > 0 and C < infinity such that >(***) cX^t <= N(t) <= CX^t >for all t in [A, A+L]. This implies that (***) holds for all >t > A, because if (***) holds on the interval [A, B] and >B >= A + L then (*) implies that (***) also holds on the >interval [A, A+B+l/2]. >And that's as far as I got. Seems like there may be a >reason that N(t)/X^t must have a limit at infinity, >and/or a reason the limit must be 1, based on some >inequalities that come from reasoning about counting >messages by considering them as the concatentation of >messages of shorter lengths, but I don't see it. >(***) is closer to an answer than I thought I was >going to get... >> Ah. If the limit _is_ 1 it definitely has to do with >> the combinatorics, and not just (*) and some unspecified >> finite-differences theorem. Consider this example: >> Say t_1 = 2pi, t_2 = 4pi, X is the unique positive >> solution to (**), and let >> N(t) = (2 + sin(t)) X^t. >> Then N is positive, tends to infinity at infinity, >> satisfies (*), but is not asymptotic to c X^t. >> (He didn't say anything about the t_j being rationally >> independent, I suppose? Possibly then something >> does follow...) >> Regarding what this has to do with \\finite differences\\ >> anyway, if we let n(t) = N(t)/X^t then n satisfies the >> equation sum X^(-t_j) n(t-t_j) = n, and since X satisfies >> (**) this says sum X^(-t_j)(n(t) - n(t-t_j)) = 0, which >> is in fact an equation regarding the finite differences >> of n. >> (We also know that n is bounded, so it has a distribution >> Fourier transform. If the t_j are rationally independent >> and we assume that (*) holds for _all_ t it follows >> that the Fourier transform of n is supported at the >> origin, so n is constant. Which means that as I >> suspected at least in this situation (*) cannot >> hold for _all_ t, since N is not continuous.) >>David. >************************ >David C. Ullrich >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Shannon's \\well known\\ result >> In his famous 1948 paper entitled \\A Mathematical Theory of \\ Communication\\, >> at the bottom of page 3, Shannon asserts... >> Suppose all sequences of the symbols S1,...,Sn are allowed and these \\ symbols >> have durations t1,...,tn. What is the channel capacity? If N(t) \\ represents >> the number of sequences of duration t we have >> N(t) = N(t - t1) + N(t-t2) + ... + N(t-tn) >> The total number is equal to the sum of the numbers of sequences ending \ \\ in >> S1,S2,...,Sn and these are N(t-t1), N(t-t2),...,N(t-tn), respectively. >> According to a well-known result in finite differences, N(t) is then >> asymptotic for large t to (X0)^t where X0 is the largest real solution of \ \\ >> the >> characteristic equation: >> X^-t1 + X^-t2 + ... + X^-tn = 1 >> My question is... can anyone point me to an explanation of the \\ supposedly >> \\well-known\\ result in finite differences that Shannon is making use \ \\ of >> here???? >Suppose that Q is a solution of the characteristic equation. >Can you see that N(t) = Q^t is a solution to your first equation? >That in fact if A is any constant then N(t) = A Q^t is a solution? >Can you see that if N_1(t) and N_2(t) are solutions, then so is >N_1(t) + N_2(t)? >It follows that if Q_1, Q_2, ..., Q_n are solutions of >the characteristic equation, then >N(t) = A_1 (Q_1)^t + A_2 (Q_2)^t + ... + A_n (Q_n)^t >is a solution of the 1st equation, where A_1, ..., A_n are arbitrary >constants. But those are certainly not the only solutions... >The well-known result from finite differences (also known as >\\recurrence relations,\\ and under that name you'll find it in books >on discrete mathematics) is that this is in fact the general solution. >It follows that N(t) grows as A_1 |Q_1|^t, where Q_1 is the root >of greatest modulus. How does that follow? And exactly what do you mean by \\grows as\\? If you mean it follows that it's asymptotic to that I don't see why it follows. For example a solution to the recurrence need not be a linear combination of the solutions you mention above; in general the space of solutions to the recurrence is infinite-dimensional. I posted a _proof_ that in your notation we have c Q_1^t <= N(t) <= C Q_1(t) for all large t. This followed just from assuming that N(t) was bounded and bounded away from 0 on a certain bounded interval, which is clear from the fact that it's the number of messages such that whatever. > Why that root is also the greatest real root, >and why A_1 = 1, I don't immediately see, but surely the first fact >has to do with the form of the characteristic equation, and the 2nd >has to do with the initial conditions, which determine the values >of A_1, ..., A_n. When you say \\the values of A_1, .. A_n\\ it sounds like you're saying any solution is a linear combination of those n solutions. This is not so. (For example, it's clear that the actual N(t) in the problem is not continuous). In fact that characteristic equation has exactly one positive root, and any other complex root, however we take those complex powers, has strictly smaller modulus (except in certain degenerate cases where there could be complex roots with the same modulus). This is easy to see. Consider instead the equation for the reciprocal: phi(x) = x^t_1 + ... + x^t_n = 1. Since phi is continuous for x > 0, phi(0+) = 0, phi(+infinity) = infinity, and phi is strictly increasing, there is exactly one x_0 > 0 with phi(x) = 1. Now say z is complex, not positive (possibly negative real), and phi(z) = 1, (with some choice of the powers of z). Since |z^t| = |z|^t it follows that phi(|z|) >= 1, and in fact we have phi(|z|) > 1 unless a miracle happens, said miracle being that z^t_j > 0 for all j (which _can_ happen). Since phi is strictly increasing, phi(|z|) > 1 implies |z| > x_0 (hence 1/|z| < 1\\\\x_0.) ************************ David C. Ullrich === Subject: What is next in the series? 2 more Q 11, 100, 9, 64, 7, _ (1) 8 (b) 36 (c) 48 (d) 81 ------------------------------------------------------------------------- If Jones is a dentist, then his grandfather is from Iowa. If his grandfather is from Iowa, then his father is from Iowa. His father is not from Iowa. (a) Jones is a dentist (b) Jones is not a dentist (c) \\forgot\\ (d) \\forgot\\ ----------------------------------------------------------------------------\ \\ ------------- If John was at the horseraces, he was not at the recital. If John went to horserace, he bought a new watch. John was at the recital. (a) John bought a watch (b) John did not buy a watch (c) John was at the horceraces (d) John was at the horceraces and bought a watch. But do these types of \\if\\ statements true for reverse situation? === Subject: Re: What is next in the series? 2 more Q > 11, 100, 9, 64, 7, _ > (1) 8 > (b) 36 > (c) 48 > (d) 81 Obviously, it's (1), since f(1) = 11, f(2) = 100, f(3) = 9, f(4) = 64, f(5) = 7, and f(6) = 8, where f(x) is given by (253/30)*x^5 - (905/6)*x^4 + (2029/2)*x^3 - (18991/6)*x^2 + (67261/15)*x - 2180. Oh, wait. Maybe it's (2), since g(1) = 11, g(2) = 100, g(3) = 9, g(4) = 64, g(5) = 7, and g(6) = 36, where g(x) is given by (26/3)*x^5 - (463/3)*x^4 + (3103/3)*x^3 - (9653/3)*x^2 + 4548*x - 2208. Then again, maybe it's (3), since h(1) = 11, h(2) = 100, h(3) = 9, h(4) = 64, h(5) = 7, and h(6) = 48, where h(x) is given by (263/30)*x^5 - (935/6)*x^4 + (6257/6)*x^3 - (19441/6)*x^2 + (22877/5)*x - 2220. Or perhaps it's (4), since k(1) = 11, k(2) = 100, k(3) = 9, k(4) = 64, k(5) = 7, and k(6) = 81, where k(x) is given by (217/24)*x^5 - (3839/24)*x^4 + (25589/24)*x^3 - (79249/24)*x^2 + (18603/4)*x - 2253. Dave L. Renfro === Subject: Re: What is next in the series? 2 more Q >> 11, 100, 9, 64, 7, _ >> (1) 8 >> (b) 36 >> (c) 48 >> (d) 81 > Obviously, it's (1), since f(1) = 11, f(2) = 100, > f(3) = 9, f(4) = 64, f(5) = 7, and f(6) = 8, where f(x) > is given by > (253/30)*x^5 - (905/6)*x^4 + (2029/2)*x^3 - (18991/6)*x^2 > + (67261/15)*x - 2180. > [etc] This argument seems to show up every time someone posts one of these integer \ \\ sequences. I think it's nonsense. If you really believed it, you'd have to say that doing any kind of science is pointless, because you can always find \ \\ a polynomial fit to the sun's path in the sky that predicts that it won't rise tomorrow. Unless someone can justify answer (a), (c) or (d) as simply as I can justify answer (b), I think I'm absolutely right to invoke Occam's \ \\ razor and claim that (b) is the /correct/ answer. -- Ben === Subject: Re: What is next in the series? 2 more Q > 11, 100, 9, 64, 7, _ > as I can justify answer (b), I think I'm absolutely right to invoke \\ Occam's > razor and claim that (b) is the /correct/ answer. I can invoke Occam's razor to get yet another simple answer. The pattern is odd, even, odd, even, ... where the odd numbers decrease by 2 and the even numbers decrease by 36. This gives 28 as the next number.... If you want to insist that the correct answer has to be among those given, then I can claim 48 is the correct answer, where the rule is now: The odd numbers decrease by 2, but the even numbers just decrease *by any amount*. Or I could say the the correct answer is 8. The pattern (with extension to the left) is .... 48 13 100 11 100 9 64 7 8 5 -52 ...... Where the even numbers are decreased by 4K * previous odd number as k =...-1, 0,1,2,3,4,... 100 = 48 - 13 * -4 100 = 100 - 11* 0 64 = 100 - 9*4 8 = 64 - 7*8 -52 = 8 - 5*12 etc. Your assumption that the even numbers are squares imposes additional requirments on the sequence. My sequence might be taken as \\simpler\\ than yours because I impose no structure on the even numbers. Until one can give a METRIC which defines what it means for one answer to be \\simpler\\ than another, Occam's razor is not applicable. These kinds of questions are ridiculous. === Subject: Re: What is next in the series? 2 more Q one of these integer sequences. I think it's nonsense. > If you really believed it, you'd have to say that doing > any kind of science is pointless, because you can always > find a polynomial fit to the sun's path in the sky that > predicts that it won't rise tomorrow. Unless someone > can justify answer (a), (c) or (d) as simply as I can > justify answer (b), I think I'm absolutely right to > invoke Occam's razor and claim that (b) is > the /correct/ answer. Actually, the problem with using something like this on a high-stakes multiple choice test is how does the person writing the question _know_for_sure_ that someone taking the test won't find a simpler way to justify one of the other options, then successfully challenge the item itself, thereby causing literally tens of thousands of dollars altogether (at least) wasted on its development, its pre-testing, its statistical analysis, and so on, not to mention the indirect costs arising from the loss of confidence by organizations who use the test and the bad publicity. Dave L. Renfro === Subject: Re: What is next in the series? 2 more Q This argument seems to show up every \ time someone posts > > one of these integer sequences. I think it's nonsense. > > If you really believed it, you'd have to say that doing > > any kind of science is pointless, because you can always > > find a polynomial fit to the sun's path in the sky that > > predicts that it won't rise tomorrow. Unless someone > > can justify answer (a), (c) or (d) as simply as I can > > justify answer (b), I think I'm absolutely right to > > invoke Occam's razor and claim that (b) is > > the /correct/ answer. > Actually, the problem with using something like this > on a high-stakes multiple choice test is how does the > person writing the question _know_for_sure_ that > someone taking the test won't find a simpler way > to justify one of the other options, then successfully > challenge the item itself, thereby causing literally > tens of thousands of dollars altogether (at least) > wasted on its development, its pre-testing, its > statistical analysis, and so on, not to mention the > indirect costs arising from the loss of confidence > by organizations who use the test and the bad publicity. Not to mention the demands from those whose answers were judged wrong that their scores be corrected and that the testing agency send those corrections to every place where the test scores might be sent. This problem is not one of mathematics, or logic, but rather one of aesthetics. The test-taker must try to decide what answer appeared to be the 'nicest' to the one posing the question in the first place. === Subject: Re: What is next in the series? 2 more Q > 11, 100, 9, 64, 7, _ > (1) 8 > (b) 36 > (c) 48 > (d) 81 There is no unique answer to questions such as this. Such questions are ridiculous. I suspect the answer being sought is (b) > ------------------------------------------------------------------------- > If Jones is a dentist, then his grandfather is from Iowa. If his > grandfather is from Iowa, then his father is from Iowa. His father is > not from Iowa. > (a) Jones is a dentist > (b) Jones is not a dentist > (c) \\forgot\\ > (d) \\forgot\\ Hint: Apply the contrapositive to the given statements. > \\ ----------------------------------------------------------------------------\ -\\ ------------ > If John was at the horseraces, he was not at the recital. If John went > to horserace, he bought a new watch. John was at the recital. > (a) John bought a watch > (b) John did not buy a watch > (c) John was at the horceraces > (d) John was at the horceraces and bought a watch. Apply the contrapositive. You will find that none of the answers are correct. The contrapositive of the 1st sentence says if John was at the recital, then he was not at the horseraces. The last sentence then confirms that John was not at the horseraces. Now consider the second sentence. All that can be said is that if John did not buy a watch then he wasn't at the horseraces. We can NOT conclude however, that if he wasn't at the horseraces then he didn't buy a watch. It says nothing about the case where John *did* buy a watch. He might have bought a watch at the recital. === Subject: LaTeX on Linux I am new to Linux. Just installed suse 10.0. I was wondering how I can === Subject: Re: LaTeX on Linux > I am new to Linux. Just installed suse 10.0. I was wondering how I can SuSE 10.0 has LaTeX built-in. Use Software Update to install it, if you used one of the oversimplified installs. --Ron Bruck ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: LaTeX on Linux > I am new to Linux. Just installed suse 10.0. I was wondering how I can comp.text.tex === Subject: Name for a line parallel to an axis? I'm going through Schreiber's _Differential Forms_ and trying to capture \\ key ideas in the form of definitions. I hit something for which I don't have a good name. This is the idea: Assume an orthonormal basis in k-dimensional Euclidian space. Given a point x in this space, fix all but one component, called the ith component, of x, and choose the domain of x to be all of R^1. This defines a line parallel to the the ith coordinate axis. If x were fixed at an integer distance from the origin along any axis other than the ith axis, we could call the line traced out, a coordinate line. (If x were on the ith axis, the line would be the ith axis.) I would not seek to give this arbitrarily positioned line a special name if it were not for what he does next. The next concept he introduces is that of a function of x tracing out a curve on a k-dimensional surface in k+1-space. He parameterizes the ???? line with t and says the curve has one degree of freedom, namely the variation in t. Is there a good term for the ???? line? -- Nil conscire sibi === Subject: Re: Using my analysis, here are 4 under-valued stocks... <4ft4m6F1i7vsmU1@individual.net> I actually erred in saying that my model was only backwards looking. It's actually forwards looking as well! My model is constructed this way: 1. I organize all 500 of the S&P 500 into 5 equal-sized categories, based on earnings yield (Earnings/Price). 2. I then fit a multi-variable model to each of these 5 categories. The independent variables are: Earnings (x1), Book value (x2), Sales (x3), Assets (x4), and Debt (x5). 3. I then regress the 5 categories of stocks. Their market capitalization is regressed against x1:x5. I noticed that the fit/relation is very good. So knowing that the markets and the stocks have future expectations factored into today's market capitalization, my model is forwards looking. === Subject: Re: Using my analysis, here are 4 under-valued stocks... \\The prove is in the pudding.\\ Name the 4 under-valued stocks and we will follow their progress for the next year. Stop teasing us! !! > I actually erred in saying that my model was only backwards looking. > It's actually forwards looking as well! My model is constructed this > way: > 1. I organize all 500 of the S&P 500 into 5 equal-sized categories, > based on earnings yield (Earnings/Price). > 2. I then fit a multi-variable model to each of these 5 categories. > The independent variables are: Earnings (x1), Book value (x2), Sales > (x3), Assets (x4), and Debt (x5). > 3. I then regress the 5 categories of stocks. Their market > capitalization is regressed against x1:x5. I noticed that the > fit/relation is very good. > So knowing that the markets and the stocks have future expectations > factored into today's market capitalization, my model is forwards > looking. === Subject: Re: Using my analysis, here are 4 under-valued stocks... <4ft4m6F1i7vsmU1@individual.net> EK Eastman Kodak Company -2.109848076 GM GENERAL MOTORS -0.792329574 XL XL Capital Ltd. 0.063239447 SUNW Sun Microsystems, Inc. 0.26686982 GOOG Google Inc. 0.340536167 are all over-valued. According to my model, EK and GM's executives should *PAY* you money to buy their stocks!!! > \\The prove is in the pudding.\\ Name the 4 under-valued stocks and we > will follow their progress for the next year. Stop teasing us! !! > > I actually erred in saying that my model was only backwards looking. > > It's actually forwards looking as well! My model is constructed this > > way: > > 1. I organize all 500 of the S&P 500 into 5 equal-sized categories, > > based on earnings yield (Earnings/Price). > > 2. I then fit a multi-variable model to each of these 5 categories. > > The independent variables are: Earnings (x1), Book value (x2), Sales > > (x3), Assets (x4), and Debt (x5). > > 3. I then regress the 5 categories of stocks. Their market > > capitalization is regressed against x1:x5. I noticed that the > > fit/relation is very good. > > So knowing that the markets and the stocks have future expectations > > factored into today's market capitalization, my model is forwards > > looking. === Subject: Re: Using my analysis, here are 4 under-valued stocks... These are your over-valued picks; NOT your under-valued! You maybe right in your assessment and yet, I'd give those stocks a 50/50 chance of going either way. However, if you'd really could pick 4 *under-valued stocks* post them here before they will have achieved their fair value, then after a year you could attach these postings to your resume and your future will be made. Anyone with common sense would hire your expertise! P.S. Stop bragging! Go to work on it (you don't even have to prove any thing to me by posting your 4 under-valued stocks!). Man, you've got it made ;-). I always suspected that someone out there has the answers. P.P.S. Now, if you rally want to go to town find 4 under-valued stocks in the bushes where the real gains (10-baggers) are hiding. Stocks within the SP500 are already misoveranalysed, IMHO. > EK Eastman Kodak Company -2.109848076 > GM GENERAL MOTORS -0.792329574 > XL XL Capital Ltd. 0.063239447 > SUNW Sun Microsystems, Inc. 0.26686982 > GOOG Google Inc. 0.340536167 > are all over-valued. According to my model, EK and GM's executives > should *PAY* you money to buy their stocks!!! >> \\The prove is in the pudding.\\ Name the 4 under-valued stocks and we >> will follow their progress for the next year. Stop teasing us! !! > I actually erred in saying that my model was only backwards looking. > It's actually forwards looking as well! My model is constructed this > way: > 1. I organize all 500 of the S&P 500 into 5 equal-sized categories, > based on earnings yield (Earnings/Price). > 2. I then fit a multi-variable model to each of these 5 categories. > The independent variables are: Earnings (x1), Book value (x2), Sales > (x3), Assets (x4), and Debt (x5). > 3. I then regress the 5 categories of stocks. Their market > capitalization is regressed against x1:x5. I noticed that the > fit/relation is very good. > So knowing that the markets and the stocks have future expectations > factored into today's market capitalization, my model is forwards > looking. === Subject: Re: new question <4499ecbd$0$84074$892e7fe2@authen.yellow.readfreenews.net > > > Hey, \ folks, I have another question. I have a set of 30 points. \\ Three > > > > of these are: > > > > > > -50, 1 > > > > -12.5, 2 > > > > 10, 30 > > > > > > The points describe a curve that is basically exponential, but \ with > > > > various critical inflection points. Up to this time, I have been \\ trying > > > > to model these points by using X = exp{F(Y)}. With X and Y known \\ for > > > > the 30 points, I can calculate F(Y) simply by taking the log of X. \ \\ I > > > > have put many hours (months, actually) into trying to find a curve \ \\ that > > > > fits the graph of F(Y) vs. Y. All the results are unusable. > > > > > > Just today it occurred to me that I might be better off using \ \\ something > > > > like > > > > > > X = F(Y)*exp{G(Y)} + H(Y) > > > > > > This would be a lot more flexible. I've been scratching my head \ on \\ how > > > > to set this up with simultaneous equations, but I'm pretty much at \ \\ a > > > > dead end. If these were parameters, there would be no problem, but \ \\ what > > > > I want to use are functions. Is this sort of thing possible? > > > > > > One thing that makes this seem all the more doable is that the \ fit > > > > doesn't have to be exact. > > > > > > you have left out a key factor, how close does it have to fit \ ? what \\ are > > > the allowable errors ? > > > > (that just makes the exact points into small boxes the curve can go \ \\ through) > > > > you can do this all in excel very easily, it will graph the results \ \\ too. > > > It can calculate total error so you can minimize it per solution. > > The allowable error differs from point to point. So Excel can find > > functions in a manner analogous to finding parameters using > > simultaneous equations? > No. YOU must propose the general form of the fit (for example, f(y) = a > + b*exp(c*y) ), then EXCEL can deal with the problem of minimizing the > sum of squared errors by varying the parameters a, b and c. With a bit > more work, you can, instead, minimize the sum of the absolute errors > (or the maximum absolute error), which may be preferable to working > with sums of squares (less sensitive to outliers, more robust, etc.). > > I've been buying all kinds of math software. > What, specifically, do you have? These may be many times better than > EXCEL. > Good luck, > R.G. Vickson > > I'll buy Excel also, if it can do this. I have Mathematica and various curve fitting programs. I have MS Works as a spreadsheet. === Subject: Re: new question <4499ecbd$0$84074$892e7fe2@authen.yellow.readfreenews.net > > > > Hey, \ folks, I have another question. I have a set of 30 points. \\ Three > > > > > of these are: > > > > > > > > -50, 1 > > > > > -12.5, 2 > > > > > 10, 30 > > > > > > > > The points describe a curve that is basically exponential, \ but \\ with > > > > > various critical inflection points. Up to this time, I have been \ \\ trying > > > > > to model these points by using X = exp{F(Y)}. With X and Y known \ \\ for > > > > > the 30 points, I can calculate F(Y) simply by taking the log of X. \ \\ I > > > > > have put many hours (months, actually) into trying to find a curve \ \\ that > > > > > fits the graph of F(Y) vs. Y. All the results are unusable. > > > > > > > > Just today it occurred to me that I might be better off \ using \\ something > > > > > like > > > > > > > > X = F(Y)*exp{G(Y)} + H(Y) > > > > > > > > This would be a lot more flexible. I've been scratching my \ head on \\ how > > > > > to set this up with simultaneous equations, but I'm pretty much at \ \\ a > > > > > dead end. If these were parameters, there would be no problem, but \ \\ what > > > > > I want to use are functions. Is this sort of thing possible? > > > > > > > > One thing that makes this seem all the more doable is that \ the \\ fit > > > > > doesn't have to be exact. > > > > > > > > > you have left out a key factor, how close does it have \ to fit ? \\ what are > > > > the allowable errors ? > > > > > > (that just makes the exact points into small boxes the curve \ can go \\ through) > > > > > > you can do this all in excel very easily, it will graph the \ results \\ too. > > > > It can calculate total error so you can minimize it per solution. > > > > The allowable error differs from point to point. So Excel can find > > > functions in a manner analogous to finding parameters using > > > simultaneous equations? > > No. YOU must propose the general form of the fit (for example, f(y) = a > > + b*exp(c*y) ), then EXCEL can deal with the problem of minimizing the > > sum of squared errors by varying the parameters a, b and c. With a bit > > more work, you can, instead, minimize the sum of the absolute errors > > (or the maximum absolute error), which may be preferable to working > > with sums of squares (less sensitive to outliers, more robust, etc.). > > > I've been buying all kinds of math software. > > What, specifically, do you have? These may be many times better than > > EXCEL. > > Good luck, > > R.G. Vickson > > > I'll buy Excel also, if it can do this. > I have Mathematica and various curve fitting programs. Mathematica will beat EXCEL many times over, *provided* that it comes with an optimization facility. (I use Maple, which does have powerful optimization packages, so I am not very familiar with Mathematica and thus cannot say exactly what is available in its menus. But, I would be surprised if you could not do optimization.) The real problem is, if I understand correctly, that you want something that no existing computer package can do, viz., automatic curve-fitting. The most you can do is try different functional forms with tuning parameters in them, then vary the parameters to achieve results that satisfy your definition of \\best\\. RGV > I have MS Works > as a spreadsheet. === Subject: Re: new question > Hey, folks, I have another question. I have a set of 30 points. Three > of these are: > -50, 1 > -12.5, 2 > 10, 30 > The points describe a curve that is basically exponential, but with > various critical inflection points. Up to this time, I have been trying > to model these points by using X = exp{F(Y)}. With X and Y known for > the 30 points, I can calculate F(Y) simply by taking the log of X. I > have put many hours (months, actually) into trying to find a curve that > fits the graph of F(Y) vs. Y. All the results are unusable. Right here I'm reaching a dead end. Presumably, the points you've listed above are (X,Y). Well, some of these X's are negative. You're going to have a hard time taking the log of them, unless you're dealing with complex arithmetic. And what do you mean by critical inflection points? Do your points describe a convex or concave curve? If possible, give us all the numbers. And *why* do you want to fit the numbers with an approximate curve? What problem are you trying to solve? Do you want to calculate Y for other \\ values of X? Why not use linear interpolation for that? -Michael. === Subject: Re: new question <449a878f$0$38685$edfadb0f@dread12.news.tele.dk Right here I'm reaching a \ dead end. Presumably, the points you've listed > above are (X,Y). Well, some of these X's are negative. You're going to \\ have > a hard time taking the log of them, unless you're dealing with complex > arithmetic. > And what do you mean by critical inflection points? Do your points \\ describe > a convex or concave curve? If possible, give us all the numbers. > And *why* do you want to fit the numbers with an approximate curve? What > problem are you trying to solve? Do you want to calculate Y for other \\ values > of X? Why not use linear interpolation for that? > -Michael. This problem has to do with a proprietary application, one that will involve a patented program. I'm hesitant to give all the details. However, the problem can be clearly described in terms of a point set for events over an interval of time. This is: X Y 10 30 9.9(10)65 29 9.9(9)8 28 9.9(9)1 27 9.9(8)75 26 9.9(8)5 25 9.9(8)1 24 (.9(7)85 23 9.9(7)75 22 9.9(7)5 21 9.9(5)5 20 9.999 19 9.99 18 9.97 17 9.95 16 9.92 15 9.91 14 9.89 13 9.84 12 9.77 11 9.54 10 9.52 9 9.33 8 9.1 7 8.74 6 8.1 5 3.5 4 -4.75 3 -12.5 2 -50 1 9(3)5 means 9.995. This is crudely an exponential curve of the form Y = a^X where a is some small number. Trial and error with a = e has produced many curves that approximate the point set. So, I have been experimenting with Y = exp(X) In order to make this work, I need to replace X with a function This gives Y = exp(F) Solving for F, I get F = lnY I know the values for Y, so I can graph F vs. Y. This produces an arc from (0, 10ln10) to (30, 0). The arc has four inflection points. Allowable error in Y is ideally not more than 0.1, but 0.5 gives me something that is a lot better than what I have now. It would be usable. Ideally, I would have a single equation that describes this curve. My understanding of interpolation is that it involves a separate equation for each interval. Is that right? I am close to getting this single equation. I have results that are pretty good, but not quite good enough. As I was saying, it has occurred to me that Y = exp(F) might not have enough flexibility. That is why I am now trying to explore Y = F(x)*exp(G(x)) + H(x). === Subject: Expansion of Factorial of a Fraction I looked on the internet and in my textbooks but I can't find what the expansion of a factorial of a fraction looks like. I know that 5!=(5)(4)(3)(2)(1) but what is (1/2)!. Is the definiton of n!=n(n-1)(n-2)(n-3)....(2)(1)? I think that in order to get a complete understanding of how to deal with (1/2)!, I'll have to review the gamma function but I would still like to know how (1/2)! can be expanded. Responses are appreciated. === Subject: Re: Expansion of Factorial of a Fraction > I looked on the internet and in my textbooks but I can't find what the > expansion of a factorial of a fraction looks like. That's only natural because the factorial function is not defined at such values. > I know that > 5!=(5)(4)(3)(2)(1) but what is (1/2)!. Is the definiton of > n!=n(n-1)(n-2)(n-3)....(2)(1)? > I think that in order to get a complete understanding of how to deal > with (1/2)!, I'll have to review the gamma function but I would still > like to know how (1/2)! can be expanded. The abused notation \\a!\\, for a in R\\\\[N U {0}] is nonsense, *unless* it \ \\ is understood that if it has to have any meaning, such a meaning must come \\ from the Gamma function which is the analytic continuation of the factorial function. In a similar spirit: The natural power function is defined for a in R and n in N, as: a^n = {a, iff n=1} {a*a^{n-1}, iff n>1} [1] Informally, a^n = a*a*...*a (n-times) [2] Similarly, you cannot \\expand\\ a^{sqrt{2}}, into a form similar to [2], because [1] is not defined for n in R\\\\N. It can make sense only through the real power function, a^x = exp(log(a)*x), by considering appropriate branches of the complex \\log\\ function, when a \ \\ < 0. > Responses are appreciated. -- Ioannis === Subject: Re: Expansion of Factorial of a Fraction Jonas a \\.8ecrit : > I looked on the internet and in my textbooks but I can't find what the > expansion of a factorial of a fraction looks like. I know that > 5!=(5)(4)(3)(2)(1) but what is (1/2)!. Is the definiton of > n!=n(n-1)(n-2)(n-3)....(2)(1)? > I think that in order to get a complete understanding of how to deal > with (1/2)!, I'll have to review the gamma function This must be a new meaning of \\review\\ I was not aware of. but I would still > like to know how (1/2)! can be expanded. The fact that (1/2)!=sqrt(pi)/2 could give you a hint that your defnition doesn't apply well in that case... > Responses are appreciated. === Subject: Re: Expansion of Factorial of a Fraction As far as I know, (1/2)! doesn't have meaning in and of itself. It is only \ \\ when the correlation between factorials and the gamma function is realized that one can begin to understand (1/2)!. Go straight to the gamma function \ \\ if you really want to see what fractional factorials look like because they \ \\ don't expand in any traditional sense. -Jon >I looked on the internet and in my textbooks but I can't find what the > expansion of a factorial of a fraction looks like. I know that > 5!=(5)(4)(3)(2)(1) but what is (1/2)!. Is the definiton of > n!=n(n-1)(n-2)(n-3)....(2)(1)? > I think that in order to get a complete understanding of how to deal > with (1/2)!, I'll have to review the gamma function but I would still > like to know how (1/2)! can be expanded. > Responses are appreciated. === Subject: Re: JSH: Fighting mathematical research > The so-called \\pure\\ math people dominate and have pushed the idea that > applied is somehow lower than what they do, while they also have > tainted people who focus on applied. You should let the folks at the Courant Institute know about this. I'm sure the world-class applied mathematicians there would appreciate having an unknown amateur explain how the rest of the mathematical world looks down on them and considers them \\tainted.\\ Why, one of them is so \\tainted\\ that he recently was elected to the National Academy of Sciences; not to mention the two who were elected to the American Academy of Arts & Sciences, or the one who won a National Science Foundation CAREER Award, or the two who received Sloan Foundation fellowships... Pretty amazing accomplishments, considering the handicap they face in being \\tainted people who focus on applied.\\ -- Wayne Brown (HPCC #1104) | \\When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.\\ e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, \\Silverlock\\ === Subject: Galois group of primitive 5th root of unity Hi everyone, Here's my question: let w be the primitive 5th root of unity and F=Q(w), then I know G=Gal(F,Q)= where s(w)=w^2, s^2(w)=w^4=w^{-1}, s^3(w)=w^3 and s^4 is the identity. Consider H= I want to find the fixed field of H in F, which I will call E. Because F/Q is Galois, F/E is also Galois so [F:E]=|H|=2 so [E:Q]=2 so I know E is a quadratic field, i.e. E=Q(sqrt something), but I'm having trouble finding that something. Any help would be appreciated. === Subject: Re: Galois group of primitive 5th root of unity days. My association with the Department is that of an alumnus. >Hi everyone, >Here's my question: let w be the primitive 5th root of unity and >F=Q(w), then I know G=Gal(F,Q)= where s(w)=w^2, s^2(w)=w^4=w^{-1}, >s^3(w)=w^3 and s^4 is the identity. Consider H= I want to find the >fixed field of H in F, which I will call E. >Because F/Q is Galois, F/E is also Galois so [F:E]=|H|=2 so [E:Q]=2 so >I know E is a quadratic field, i.e. E=Q(sqrt something), but I'm having >trouble finding that something. Note that w + w^{-1} is fixed by s^2 and is not rational, hence E = Q(w+w^{-1}); so a good idea is to try to find the quadratic polynomial satisfied by w+w^{-1}. If f(x) = x^2 + bx + c is that quadratic polynomial with b and c integers (in general you would have rationals, but since w and w^{-1} are algebraic integers, so is w+w^{-1} so the minimal polynomial will have integer coefficients), then you know that the extension is given as E = Q(sqrt(d)), where the b^2-4c = r^2d, with d a squarefree integer and r an integer. Since w^4 + w^3 + w^2 + w + 1 = 0 you have (w+w^{-1})^2 = w^2 + 2 + w^{3} = 1 + (1+w^2+w^3) = 1 - (w+w^{-1}) so that should give you everything you need to know. If you feel up to it, try to prove the geenral result that if p is a prime and z is a primitive p-th root of unity, then Q(z) contains the quadratic field Q(sqrt(+/- p)), where you take the positive value if p=q (mod 4) and the negative value if p=3 (mod 4). -- \\It's not denial. I'm just very selective about what I accept as reality.\\ --- Calvin (\\Calvin and Hobbes\\) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Like baseball >> But I look back, and no matter what anyone else says, no matter who >> calls me \\crazy\\ or a \\loon\\ or a \\crackpot\\ or a \\crank\\, I \\ stepped up >> to the plate, and I hit some homeruns. > It's just a shame you were on a tennis court at the time. No doubt my coworkers are wondering why I'm laughing so hard right now laugh I've had in quite a while. -- Wayne Brown (HPCC #1104) | \\When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.\\ e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, \\Silverlock\\ === Subject: varimax rotation software Does anyone know where I could find code for the varimax rotation, preferably in C++ or in C? Mike. === Subject: Re: SF: Simpler factoring idea, but does it work? > I thought you were going to quit posting. You _did_ post that > \\baseball\\ thread, right? The one where you said: >> I'm at the end of my run. Mathematics is a young man's game, so far, >> and hopefully, it'll be a young person's game soon enough, as it's past >> time for a woman to step up, and I am looking for her. >> But for now, it's a young man's game and as I'm past 35 I am well past >> my prime, and my time is over. >> [...] >> I am retired. The spirit has died. The Muse has left me. I don't >> feel any more discovery is left in these bones. > Were you lying then? Was he breathing then? Was the Earth orbiting the sun then? (I *tried* to come up with a more obvious question than yours, but I just can't do it.) -- Wayne Brown (HPCC #1104) | \\When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.\\ e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, \\Silverlock\\ === Subject: Re: SF: Simpler factoring idea, but does it work? > But does it work? Amuse me, why should I use your theory (which has been shown to have flaws in it) over methods which are 100% accurate, such as the sieve of Eratosthenes, which, even though might not be the most efficient method available, at least gives us the solution??? === Subject: Re: SF: Simpler factoring idea, but does it work? <4fto9jF1k8kdrU1@news.dfncis.de > I tested 500 composite numbers in the \ range [250000 ... 1000000] made > > up by multiplying two different random primes, with each prime between > > 500 and 1000. I compared your method with Fermat's method - the > > difference of two squares. > > On average Fermat's method tried 8 possibles before finding a factor. > > On average your method tried 500 possibles before finding a factor. > > Your method is about 60 times less efficient than Fermat's method. > > The C++ code I used for testing your method is below. Let me know if > > I have made a mistake in it. > You've made a mistake within your argument. Do you remember the law of > large numbers? > Your test failed to prove that James' methods is worse for really big > numbers as well. After all, the Strong Law of Small Numbers (roughly: \\small numbers can't have all the properties you want them to\\) applies; see See R.K. Guy, \\The strong law of small numbers\\, The American Mathematical Monthly, Vol. 95, No. 8 (1988), pages 697-712 and, while you're at it, \\The second strong law of small numbers\\, Mathematics Magazine, Vol. 63, No. 1 (1990), pages 3-20 for more details. --- Christopher Heckman === Subject: Re: SF: Simpler factoring idea, but does it work? 2**4-1, he'll be cracking those RSA composites in no time. Actually, he came up with the equation 15 = 2*4 + 7*1. --- Christopher Heckman === Subject: Re: SF: Simpler factoring idea, but does it work? > > Give the man his dues. He's managed to factor the number 15! He's up to > > 2**4-1, he'll be cracking those RSA composites in no time. > Actually, he came up with the equation > 15 = 2*4 + 7*1. > --- Christopher Heckman You missed the one... I know that we've gone decimal here in UK since I was at school, but when I was there S = (2 - 7)(1 + 4) = 55 would actually been S = (2 - 7)(1 + 4) = -25 mori === Subject: Re: SF: Simpler factoring idea, but does it work? > After yet another failure with what I call surrogate factoring, where > this time I had been doing some basic algebra wrong, I sat down to > think about it all for a while, and considered that I was quite > reasonably just going in circles, using equations to try and factor > that could only give one answer. > So I started thinking about equations that could give two. > It didn't take long till I was concentrating on: > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > Here the square roots mean that expression can't just factor S, which > is what I call the surrogate, as something else is being factored as > well, but how do I get that something else to be a target composite? > After I posted that equation and started talking about it, a Tim Peters > worked out details following my instructions, but unfortunately, he is > a dedicated, um, \\crank\\ buster you might call it, who spends his time > trying to shoot down my ideas, so when he worked out the equations, and > got to something useable, he promptly began throwing up distracting > posts meant to show it was useless. > However, oddly enough, his results can be used quite simply, where the > first thing is to use some of his equations, to introduce a target > composite, which I call T. > You introduce T using > (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T > Multiply both sides by > k_1*k_4 - k_2*k_3 > to get > (k_1*k_4 + k_2*k_3) = T (k_1*k_4 - k_2*k_3) > and just subtract the left from the right to get > 0 = (T-1)* k_1*k_4 - (T+1)*k_2*k_3 > So > (T-1)* k_1*k_4 = (T+1)*k_2*k_3 > And you have > (k_1*k_4)/(k_2*k_3) = (T+1)/(T-1) > so k_1 and k_4 are integer factors of T+1, and k_2 and k_3 are integer > factors of T-1. > Easy. Just like that you're most of the way to using the equations. > That gives a finite set of possibles for the k's. > For instance, with > So, for instance if T=15, you have > (k_1*k_4)/(k_2*k_3) = 16/14 = 8/7 > so k_1*k_4 = 8, and k_2*k_3 = 7 > and one possible setup then is > k_1 = 2, k_4 = 4, k_2 = 7, k_3 = 1 > so plug those into > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and you need S, and can use any square you like, so it's easiest just > to use > x=y=1, and take the positive result of the square roots to get > S = (2 + 7)(1 + 4) = 45 > which promptly factors it, but I'll continue, as that's just one > possibility, so you have to check for it. > If S were coprime to T, then you now use factors of S, where with > S = g_1*g_2 > you have > k_1*sqrt(x) + k_2*sqrt(y) = g_1 > k_3*sqrt(x) + k_4*sqrt(y) = g_2 > and you just find find squares for x and y that will work to give you > g_1 and g_2, and here's where that second solution from the square > roots comes in, as with each set of squares for x and y that will work, > you just change the sign of one of the square roots, to get the shadow > factorization. Here I said find squares that will work for x and y, but usually I suspect you will find squares of fractions. Note that it's easy enough to solve for sqrt(x), and sqrt(y), as you get sqrt(x) =( k_4*g_1 - k_2*g_2)/(k_1*k_4 - k_3*k_2) and sqrt(y) = (k_3*g_1 - k_1*g_2)/(k_2*k_3 - k_1*k_4) and you can see something familiar in the denominator. So why do you HAVE to use a particular S? I don't know. It sounds good to me that you do, but I suspect that if you don't, but just pick square roots, you'll get nothing better than random behavior, which seems consistent with what some mathematically naive posters apparently tried from what I read in this thread. Remember, the people replying to me are invested in me being wrong. They don't give a damn about whether or not these ideas can be made to work. They only care about supporting their old position. Now I don't know. Maybe these are crap ideas, but I, at least, wish to actually find out. Short story of it is that people like Tim Peters or Dik Winter or Arturo Magidin or David Ullrich couldn't care less about the factoring problem--if I present something that actually is of value. They only care about supporting their social positions as, of course, what happens if I come up with a useful idea? Then they are totally invalidated, and all those posts of theirs come back to haunt them. Don't trust those people. Sure, this idea may be crap, but I can assure you that if it's not, they will not tell you, but instead will do their BEST to hide the fact. Use one S, and iterate through its factors. The bonus is that you get to see those people crushed like little bugs, and watch them squirm as they fight you, try to distract from what you present, and do anything they can to see what they can post to hide the truth--if there is any value in this approach. Otherwise, oh well, it's just another idea among so many others that failed. But I at least care about what's true, versus being invested in fighting no matter what against some guy. > That's it. Remarkably simple, as you go for the hidden factorization. > For instance, still using x=y=1, with my simple example, now take the > negative of ONE of the square roots: > S = (2 - 7)(1 + 4) = 55 > I guess T=15 is too dinky of an example as it just keeps factoring it, > no matter what you do, but at least that still shows the basic idea > here. > To recap, I looked at an expression > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > that can't just be the factorization of a single number because of the > use of square roots. I posted about it and a Tim Peters worked through > some analysis following instructions I gave, which gives up the simple > equation > (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T > which can be solved to get > (k_1*k_4)/(k_2*k_3) = (T+1)/(T-1) > so you can find the k's based on the factors of T+1 and T-1, to plug > back into the original equations and get an S, and then you use the > factors of S to find squares that are solutions for x and y, and then > you do the remarkable trick of just switching signs one at a time to > get to the hidden factorization. > But does it work beyond toy examples like factoring 15? You will NOT get a cogent answer from regular posters who have spent so much time fighting me. These people have lied about my other research. They have no reason not to lie now. They're in for a penny, in for a pound. They will lie until they're broken. Use a single S, and iterate through its factors. Note that in the replies in this thread, that was not done. But they congratulated each other anyway. These people are not researchers!!! They are politicians. They don't give a damn about the factoring problem or any other mathematics, if they think it helps me. They are--anti-civilization. James Harris === Subject: Re: SF: Simpler factoring idea, but does it work? >> so you can find the k's based on the factors of T+1 and T-1, I can do that. >> to plug back into the original equations and get an S, I can do that. >> and then you use the factors of S to find squares that are solutions for \ \\ x and y, I can do that, though I am not fully clear if I just take the integer solutions or if fractional solutions are allowed as well. >> and then you do the remarkable trick of just switching signs one at a \\ time I can do that. >> get to the hidden factorization. How do I do this? The only equation I can see involving T itself is (k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) = T. Where do I plug in the different values of S, x, y, g_1 and g_2 I have derived to solve for factors of T? Previously I was using GCD(S, T) but you say that was wrong. What equation gives the factors of T in terms of S, x, y, g_1 and g_2? I am not a mathematician, just a programmer with some mathematical background and time to spare. rossum === Subject: Re: SF: Simpler factoring idea, but does it work? > Short story of it is that people like Tim Peters or Dik Winter or > Arturo Magidin or David Ullrich couldn't care less about the factoring > problem--if I present something that actually is of value. How do you know? You've never done that. Jose Carlos Santos === Subject: Re: JSH: SF: Simpler factoring idea, but does it work? [added \\JSH:\\ to subject] [jstevh@msn.com] >> Short story of it is that people like Tim Peters or Dik Winter or >> Arturo Magidin or David Ullrich couldn't care less about the factoring >> problem--if I present something that actually is of value. [Jos\\.8e Carlos Santos] > How do you know? You've never done that. Now, now, you're just jealous because he let slip that I took over first place on his Secret Enemies List. I'd like to thank all the little people who helped me achieve this -- but, frankly, you're too little for either James or me to notice ;-) Well, OK, if Rick Decker reads this, I apologize on James's behalf for forgetting that you deserve to be killed too! Rest assued that when the secret prosecutors come for me, yours is the first name I'll give up. OTOH, I'm not sure how Professors Magidin and Ullrich even got on the list of Secret Factoring Enemies. Does James hate them so much that he curses them even for areas in which they're conspicuous mostly by absence? Or is it really that he ... loves them that much? Ewww. I sure hope this isn't his Secret Fantasy Lover list. That would creep me out. Yo, James! You're right about everything! Right right right! That last method with the k's & whatever is fantastic -- all it takes is doing exactly \ \\ what you so clearly said to do. Your only mistake was in underselling it: not only does it factor in sub-exponential time, it actually factors faster \ \\ the larger the composite! Looks like it factors in O(1/N) time, or even O(e^-N) time. Hell, I don't know -- pick any time you like, and that's it! \ \\ Just please, please, for the love of God, take me off that list. Brrrrr. I need a shower. === Subject: Re: JSH: SF: Simpler factoring idea, but does it work? <4fv2otF1kvr10U1@individual.net> <7pednVq4bfYU9wfZnZ2dnUVZ_oydnZ2d@comcast.com [added \\JSH:\\ to \ subject] > [jstevh@msn.com] > >> Short story of it is that people like Tim Peters or Dik Winter or > >> Arturo Magidin or David Ullrich couldn't care less about the factoring > >> problem--if I present something that actually is of value. > [Jos\\.8e Carlos Santos] > > How do you know? You've never done that. > Now, now, you're just jealous because he let slip that I took over first > place on his Secret Enemies List. But there's also his Nefandous Enemies List. Those who's names are not to be spoken. Such as yours truly. So being at the top of the Secret Enemies List doesn't necessarily mean you're at the absolute top. > I'd like to thank all the little people > who helped me achieve this -- but, frankly, you're too little for either > James or me to notice ;-) > Well, OK, if Rick Decker reads this, I apologize on James's behalf for > forgetting that you deserve to be killed too! Rest assued that when the > secret prosecutors come for me, yours is the first name I'll give up. > OTOH, I'm not sure how Professors Magidin and Ullrich even got on the \\ list > of Secret Factoring Enemies. Does James hate them so much that he curses > them even for areas in which they're conspicuous mostly by absence? Or \ \\ is > it really that he ... loves them that much? Ewww. I sure hope this \\ isn't > his Secret Fantasy Lover list. That would creep me out. > Yo, James! You're right about everything! Right right right! That last > method with the k's & whatever is fantastic -- all it takes is doing \\ exactly > what you so clearly said to do. Your only mistake was in underselling \\ it: > not only does it factor in sub-exponential time, it actually factors \\ faster > the larger the composite! Looks like it factors in O(1/N) time, or even > O(e^-N) time. Hell, I don't know -- pick any time you like, and that's \ \\ it! > Just please, please, for the love of God, take me off that list. > Brrrrr. I need a shower. === Subject: Re: JSH: SF: Simpler factoring idea, but does it work? On Thu, 22 Jun 2006 06:01:29 -0400, \\Tim Peters\\ \ OTOH, I'm not sure how Professors Magidin and Ullrich even got on the list \ \\ >of Secret Factoring Enemies. Does James hate them so much that he curses >them even for areas in which they're conspicuous mostly by absence? Yes. >Or is >it really that he ... loves them that much? Ewww. I sure hope this isn't \ \\ >his Secret Fantasy Lover list. That would creep me out. I'm really glad you said that. Just when I was getting over the nightmares about being killed by an angry mob who wants to know why I supressed James' work... >Yo, James! You're right about everything! Right right right! That last >method with the k's & whatever is fantastic -- all it takes is doing \\ exactly >what you so clearly said to do. Your only mistake was in underselling it: \ \\ >not only does it factor in sub-exponential time, it actually factors faster \ \\ >the larger the composite! Looks like it factors in O(1/N) time, or even >O(e^-N) time. Hell, I don't know -- pick any time you like, and that's it! \ \\ >Just please, please, for the love of God, take me off that list. >Brrrrr. I need a shower. ************************ David C. Ullrich === Subject: Re: JSH: SF: Simpler factoring idea, but does it work? [added \\JSH:\\ to subject] [jstevh@msn.com] > Here I said find squares that will work for x and y, but usually I > suspect you will find squares of fractions. Someimes you do, yes. > ... > So why do you HAVE to use a particular S? > I don't know. It sounds good to me that you do, but I suspect that if > you don't, but just pick square roots, you'll get nothing better than > random behavior, which seems consistent with what some mathematically > naive posters apparently tried from what I read in this thread. > Remember, the people replying to me are invested in me being wrong. That's not even true of me, and it's plain bizarre to imagine that \\ \\rossum\\ coherent sense of what you were suggesting) has an axe to grind with you. There's no history there (well, none that I'm aware of -- maybe there is?). > ... > Now I don't know. Maybe these are crap ideas, but I, at least, wish to > actually find out. example harder than 15? Sometimes you sound like you just make up ;-) > [... the usual rants ...] > Use one S, and iterate through its factors. Which is exactly what I guessed (although it was indeed only a guess) you'd \ \\ complain about: I have to confess that James's _theory_ here didn't make sense to me from its start, so I can't predict what's supposed to follow from it. Best guess is that he'll be unhappy you didn't pick a specific S, then march over all two-integer factorizations S=g1*g2, and then use only those g1 and g2 for which k1*sqrtx + k2*sqrty = g1 k3*sqrtx + k4*sqrty = g2 has integer solutions for sqrtx and sqrty. But I'd better let him tell you what's wrong with you himself :-) And so you did, except now you want to throw fractional solutions into this \ \\ too. Fine. I don't know whether that poster wants to take the suggestion, \ \\ but given the purely assholish tone of this reply of yours I doubt they'll want to bother. I'm not going to: as I told you before, I'm not bothering to _test_ any method of yours again unless you first say you've verified it works for all \ \\ 2-digit composites (because few of your methods survive that stage, and if you don't care enough to check that little, no, you're not serious). > ... [the usual revenge fantasies] ... > ... [the usual slanders] ... > Use a single S, and iterate through its factors. Note that in the > replies in this thread, that was not done. As above, I already suggested to \\rossum\\ to take that step, if \\ interested. It's no surprise that he didn't take it to begin with, because your explanations before this have never been clear on this point, and the only (correct) example you ever gave didn't take that step either. Your lack of \ \\ clarity isn't his fault. > But they congratulated each other anyway. Better by far if you thanked him for trying. > These people are not researchers!!! They are politicians. They don't > give a damn about the factoring problem or any other mathematics, if > they think it helps me. > They are--anti-civilization. Sorry, but reporting results truthfully helps mathematics and civilization, \ \\ regardless of whether James Harris, or anyone else, like the results. === Subject: Describing an Operation of Two Ordered Sets Suppose I have two ordered sets: X = (a,b,c,c, d) Y = (d,e,f,g,g) Now I want to create a set from X and Y above: Z = (a, b, c,c, d ,e,f,g,g) Note that whenever the last element of X is equal to the first element of Y, the new set only contain 1 element of it, e.g. only one \\d\\ in Z. How can I express the above set Z in terms of X and Y in the simplest way mathematically? Gundala === Subject: Re: Describing an Operation of Two Ordered Sets > Suppose I have two ordered sets: > X = (a,b,c,c, d) > Y = (d,e,f,g,g) > Now I want to create a set from X and Y above: > Z = (a, b, c,c, d ,e,f,g,g) > Note that whenever the last element of X is equal > to the first element of Y, the new set only contain > 1 element of it, e.g. only one \\d\\ in Z. > How can I express the above set Z in terms of > X and Y in the simplest way mathematically? There's something you're not telling us, because as sets, X = {a,b,c,d} and Y = {d,e,f,g}, so I don't see what the issue is about the duplication you're concerned about. Just define the ordering on Z = X union Y to be the X-elements in the X-ordering followed by the Y-elements in the Y-ordering. I think this only works if (X and Y are disjoint) OR if (the last element of X is the same as the first element of Y and the two sets share no other elements). Dave L. Renfro === Subject: Re: Describing an Operation of Two Ordered Sets days. My association with the Department is that of an alumnus. >Suppose I have two ordered sets: >X = (a,b,c,c, d) >Y = (d,e,f,g,g) I think that what you mean is that you have ->tuples<-, not \\ordered sets\\. (Usually, \\ordered set\\ means a set A together with a total order on it, meaning a binary relation <= such that for all a, b, and c in A you have a<=a; if a<=b and b<=a then a=b; if a<=b and b<= c then a<=c; and for all a and b in A either a<=b or b<=a). >Now I want to create a set from X and Y above: >Z = (a, b, c,c, d ,e,f,g,g) I.e., the tuple obtained by identifying the last entry of X with the first entry of Y and then concatenating. >Note that whenever the last element of X is equal to the first >element of Y, the new set only contain 1 element of it, >e.g. only one \\d\\ in Z. >How can I express the above set Z in terms of X and Y >in the simplest way mathematically? Here is one way, though I don't think it would qualify as \\simplest\\. Drop the parenthesis and consider X and Y as \\words\\ on the elements, i.e., finite (possibly empty) sequences of elements. Given a word X, let first(X) be the first term of X (the empty word if X is empty); let last(X) be the last term of X (the empty word if X is empty); and let trunc(X) be the result of removing the first term from X (the empty word if X is either empty or contains only one term). Define a product * on the collection of all words as follows: X*Y = XY (concatenate X with Y) if last(X) <> first(Y); X*Y = Xtrunc(Y) (concatenate X with the truncation of Y) if last(X) = first(Y). -- \\It's not denial. I'm just very selective about what I accept as reality.\\ --- Calvin (\\Calvin and Hobbes\\) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple proof of FLT ? > On Sat, 17 Jun 2006 22:57:14 EDT, \\Roman B. Binder\\ > >Does somebody dare to think, > >that it can be possible? > > See what could happen after > >applying to both sides c^n: > >once c=kxy -(x+y) > > simply ! > > See my topic: Unexpected shot for FLT > > Ro-bin > I think your proof is unreadable. > > look at first to such Lemma: > grammatically correct would have been: Look first at > this Lemma. > or just put: Lemma 1 > >For every set of x;y; natural numbers we'll > >find such natural c ,that x^n +y^n +c^n > >for n >=3 and prime or odd will be divided by xy. > did you forget n after x;y; ? it's now unclear if > you mean that you can find a c for every n > Set is not the right word here, because order doesn't > matter > in a set. Just put for every x,y where x and y are > natural numbers > We'll find a natural c, such that > for n>=3 and prime or odd. It looks like you mean > that both > n>=3 > n is a prime or n is odd > but n is odd is then redundant > replace 'will be divided' by 'is divisible' > read a math book with proofs in it and see how it's > done. > -- > Wim Benthem english commands and any order: You must be aware, that my Lemma is working not only for primes >=3 but also for all odd numbers >1. And once most of odds are composite of at least two primes so such Lemma \ \\ could have more applications... ( is it correct or You like to \\expand\\ something or \\it would be necessary to add\\ one of Your absolute sentences... ) This is because for every n odd : x^n + y^n + c^n will be: (x+c)Ext + y^n = (y+c)Ext + x^n then once we'll take c = kxy -(x+y) x+c = kxy -y : divisible by y y+c = kxy -x : divisible by x QED * * * in more elementary terms: for n=3 x^3 + y^3 = (x+y)(x^2 -xy +y^2) = (x+y)*Ext then for every n>3 and odd x^n +y^n = (x+y)[ x^(n-1) - x^(n-2) *y + ... + + x^(n-1)/2 *y^(n-1)/2 + ... - x*y(n-2) + y^(n-1)]= = (x+y)*Ext where Ext could be written also generally as: Ext = SUM(k=1 till k=n) (-1)^(k+1) x^(n-k) *y^(k-1) * * * P.S. I am sorry once it was not possible for to find some more simple plots for associated with me mathematical problems in the english literature. (For example FLT) Therefore I used to read more in other languages: thank You for Your english lesson. With Compliments Ro-bin === Subject: Re: Word problem - point of no return >A plane is flying the 2553-mi trip from Los Angeles to Honolulu into a >60-mph headwind. If the speed of the plane in still air is 310 mph, >how far from Los Angeles is the plane's point of no return? >-> A flight's \\point of no return\\ is the point at which the flight >time required to return to Los Angeles is the same as the time required >to continue to Honolulu. First, work out an expression for how long it takes to fly the entire 2553 miles. Second, find an expression for how long it takes to fly x miles from L.A. and x miles back. Third, determine when those two expressions are equal (find x). --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Word problem - point of no return >> A plane is flying the 2553-mi trip from Los Angeles to Honolulu into a >> 60-mph headwind. If the speed of the plane in still air is 310 mph, >> how far from Los Angeles is the plane's point of no return? >Plane takes s = 2553/(310 - 60) hr to get to Honolulu. >At time t, plane is 250t miles from LA. >It would take s - t hours to finish flight and >250t/(310 + 60) hours to return. >Point of no return is at 6 hours when ... You are correct --- not considering the time it takes to turn around but just the plane reversing direction instantly the half-way point is very close to 6 hrs. 5 minutes and 39 sec. or 1523.5645 miles out. Dan >> -> A flight's \\point of no return\\ is the point at which the flight >> time required to return to Los Angeles is the same as the time required >> to continue to Honolulu. === Subject: Re: Word problem - point of no return > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into a > 60-mph headwind. If the speed of the plane in still air is 310 mph, how > far from Los Angeles is the plane's point of no return? > -> A flight's \\point of no return\\ is the point at which the flight \\ time > required to return to Los Angeles is the same as the time required to > continue to Honolulu. With your definition of \\point of no return\\ it's obviously going to depend on what direction you start out in - for example if you go via London then it's going to be quite far... -- Jeremy Boden === Subject: Re: Word problem - point of no return > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into a > 60-mph headwind. If the speed of the plane in still air is 310 mph, > how far from Los Angeles is the plane's point of no return? Plane takes s = 2553/(310 - 60) hr to get to Honolulu. At time t, plane is 250t miles from LA. It would take s - t hours to finish flight and 250t/(310 + 60) hours to return. Point of no return is at 6 hours when ... > -> A flight's \\point of no return\\ is the point at which the flight > time required to return to Los Angeles is the same as the time required > to continue to Honolulu. === Subject: Re: Word problem - point of no return I think they're looking for when the time traveled would equal the time it would take to return, which would be kinda tricky cuz you're heading one direction faster than the other. > > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into a > > 60-mph headwind. If the speed of the plane in still air is 310 mph, > > how far from Los Angeles is the plane's point of no return? > The speed going West is 310 - 60 and the speed going East is 310 + 60. > The point of no return just equates time = distance/speed in the two > directions. > R.G. Vickson > > -> A flight's \\point of no return\\ is the point at which the flight > > time required to return to Los Angeles is the same as the time required > > to continue to Honolulu. === Subject: Re: Word problem - point of no return > I think they're looking for when the time traveled would equal the time > it would take to return, which would be kinda tricky cuz you're heading > one direction faster than the other. Ackshully, it would be kinda easy cuz you just need ta introduce an unknown x for the distance from LA to the point of no return, then write down the times each way and solve for that sucker. Of course, ya need ta acksully do the work. RGV > > > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into \ \\ a > > > 60-mph headwind. If the speed of the plane in still air is 310 mph, > > > how far from Los Angeles is the plane's point of no return? > > The speed going West is 310 - 60 and the speed going East is 310 + 60. > > The point of no return just equates time = distance/speed in the two > > directions. > > R.G. Vickson > > > > -> A flight's \\point of no return\\ is the point at which the \\ flight > > > time required to return to Los Angeles is the same as the time \\ required > > > to continue to Honolulu. === Subject: Re: Word problem - point of no return > I think they're looking for when the time traveled would equal the time > it would take to return, which would be kinda tricky cuz you're heading > one direction faster than the other. > > > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into \ \\ a > > > 60-mph headwind. If the speed of the plane in still air is 310 mph, > > > how far from Los Angeles is the plane's point of no return? > > The speed going West is 310 - 60 and the speed going East is 310 + 60. > > The point of no return just equates time = distance/speed in the two > > directions. > > R.G. Vickson > > > > -> A flight's \\point of no return\\ is the point at which the \\ flight > > > time required to return to Los Angeles is the same as the time \\ required > > > to continue to Honolulu. Let D_w be the distance westbound to Hawaii from point of no return. T_w be time westbound to Hawaii from point of no return. 310 - 60 = speed westbound D_e be the distance eastbound to Los Angeles from point of no return. T_e be time eastbound to Los Angeles from point of no return. 310 + 60 = speed eastbound Distance = speed times time, for constant speed Then D_w + D_e = 2553, distance from LA to Hawaii. T_w = T_e by definition of \\point of not return\\ D_e = (310+60) * T_e, distance = speed times time, eastbound D_w = (310-60) * T_w, distance = speed times time, westbound 4 linear equations in 4 unknowns === Subject: Re: Word problem - point of no return I think they're \ looking for when the time traveled would equal the time > > it would take to return, which would be kinda tricky cuz you're heading > > one direction faster than the other. > > > > A plane is flying the 2553-mi trip from Los Angeles to Honolulu into \ \\ a > > > > 60-mph headwind. If the speed of the plane in still air is 310 \\ mph, > > > > how far from Los Angeles is the plane's point of no return? > > > > The speed going West is 310 - 60 and the speed going East is 310 + \ \\ 60. > > > The point of no return just equates time = distance/speed in the two > > > directions. > > > > R.G. Vickson > > > > > > > -> A flight's \\point of no return\\ is the point at which \ the \\ flight > > > > time required to return to Los Angeles is the same as the time \\ required > > > > to continue to Honolulu. > Let > D_w be the distance westbound to Hawaii from point of no return. > T_w be time westbound to Hawaii from point of no return. > 310 - 60 = speed westbound > D_e be the distance eastbound to Los Angeles from point of no return. > T_e be time eastbound to Los Angeles from point of no return. > 310 + 60 = speed eastbound > Distance = speed times time, for constant speed > Then > D_w + D_e = 2553, distance from LA to Hawaii. > T_w = T_e by definition of \\point of not return\\ > D_e = (310+60) * T_e, distance = speed times time, eastbound > D_w = (310-60) * T_w, distance = speed times time, westbound > 4 linear equations in 4 unknowns The 4 & 4 quickly reduces to, D_w + D_e = 2553: D_w / 250 = D_e / 370. Two equations in two unknowns === Subject: How to decompose this formula I have one question with the following formula : X is discrete FQ:= Sum (re(x)*ln(re(x))), I have value Q1:=Sum(r1(x)*ln(r1(x))) and Q2:=sum(r2(x)*ln(r2(x))), also I know the relation of re(x) , r1(x) and r2(x). re(x)= convolution of r1(x) and r2(x); I tried to use definition and decompose the FQ from current space , but it cannot work. The ln prohibit that method. Is there any way I can use Q1and Q2 to express the FQ? Bin === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net > No, what I said \ was that some of the questions he asked > > were equivalent to that. But hey, given your record to date, > > it's hardly surprising to see such a misrepresentation. > Oh, I see; you weren't kicking him around for asking questions, but for > distinction. I didn't kick him around at all. I explained to him why he'd received some rather brusque responses to some of his queries. > At least you've now admitted your involvement, rather than continue to > try to pass yourself off as a disinterested observer (speaking of > misrepresentation). Interesting mode of argument. Since your approach to this seems to consist entirely of making false accusations, I think I won't bother responding to you any more. === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= > To make this work, you actually have to define and choose a > PARTICULAR model in which every formula is the interpretand > of some numeral s(s(s(....))) in the language. > It doesn't really matter whether this model is standard > since you are only going to be talking about the standard > fragment of it anyway (since all the formulas are finite and > will therefore have finite godel numbers). > But if you add some such non-standard axiom (as ~Con(PA)) > then you will be forced to accept a non-standard model, as > well as a perversion of what you thought you meant by > the predicates (like Proof(.,.)) defined in the theory (e.g. > you will lose the compactness theorem and you may > have to tolerate infinitary proofs). I'm afraid you seem so thoroughly entrenched in your bizarre confusion I see very little point in explaining anything to you, so this will most likely be the last reply you get from me on these issues. The sense in which the G\\.9adel sentence of a theory is self-referential is explained clearly in Torkel's latest book. If you worked through a technical exposition you would also find that the features of the sentence relevant to its self-referentiality are not \\model dependent\\ in the sense that seems to trouble you. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net > We definitely need \ a more concise term. This is the one that best > > matches what I meant: > > a verb, to post controversial or provocative messages in a deliberate > > attempt to provoke flames. > > http://teladesign.com/ma-thesis/glossary.html > 'troll' is a lousy word here. (1) It is really clumsy mixed metaphor to > combine 'troll' (as in fishing, which takes place in water) with > 'flame'. That is not I am 'combining.' I am 'combining' \\to post controversial or provocative messages in a deliberate attempt to provoke flames\\ with \\stirring and perpetuating flame wars\\ (roughly; however it was stated in the backthread). So your complaint is irrelevant. (2) 'troll' is abused on the Internet. A poster often say > things like 'he's a troll' for nothing more than the poster being in > disagreement with the person, so I don't like to perpetuate the 'troll' > rubric. The word has been given a precise definition, so your worries about it being misused are also irrelevant. In this thread, 'troll' is merely a shorthand expression for \\post[ing] controversial or provocative messages in a deliberate attempt to provoke flames\\. If someone misuses the word, he can be pointed back to the definition. So your concern is irrelevant. (3) Again, I don't think you want to limit to saying that > Franzen provoked flames but rather that he also sometimes (often?) > posted flames. Sometimes. A flame is one type of controversial or provocative message, with a high probability of provoking further flames. So 'troll' is a general term that includes trolling, not, and evidence that Franzen flamed would be evidence that he trolled, not evidence against it. Since the above definition matches exactly what I claimed Franken to have been doing, then 'troll' (as defined above) is the best word to use for it. > > Torkel trolled. > You claim. > > IME Virtually every post he made to sci.logic (ie, not > > counting the crossposts from sci.math or the comp lists) was a troll > > (or a flame, which is a type of troll). > You've not come within even a hundred miles of supporting that > accusation. > > I think that, even counting > > the crossposts, the majority of his posts were trolls. > Factor out the crossposts or factor them in, I don't care. You won't be > able to show that virtually all his posts were stirring or perpetuating > flames, even under a very weak definition of 'flame' and even in just > the last five years on sci.logic. And now you mention 'majority'. I > don't feel that you can support even that reduced claim. > MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net That is not I am \ 'combining.' I am 'combining' \\to post controversial > or provocative messages in a deliberate attempt to provoke flames\\ with > \\stirring and perpetuating flame wars\\ (roughly; however it was stated > in the backthread). So your complaint is irrelevant. My remark is a statement of my own aesthetic judgment that 'troll' clashes with 'flames'. I don't require that you consider that relevent if you choose not to. > The word has been given a precise definition, so your worries about it > being misused are also irrelevant. But that you give a more specific definition does not change that the term is likely to be read, despite your definitional stipulation, with much looser understanding of its denotation. I could say that I define 'disingenuous asshole' as meaning 'one who posts the style of George Dance'. So when I say 'George Dance posts as a disingenuous asshole' all I'm saying is tautological, that George Dance posts in the style of George Dance, and that I am not necessarily saying that George Dance is disingenuous or an asshole, only that I have defined the term 'disingenuous asshole' to mean 'posting in the style of George Dance'. But doing so I would be overlooking that no matter how explicit I were in my definition, there will still be people who would take me to be claiming that George Dance is disingenuous and an asshole, even given my protestations that I have stipulated that 'disingenuous asshole' just means 'posting in the style of George Dance no matter what that style may be'. > > > IME Virtually every post he made to sci.logic (ie, not > > > counting the crossposts from sci.math or the comp lists) was a troll > > > (or a flame, which is a type of troll). Still a claim not substantiated by you. If you said something like, \\Very often Franzen made personal attacks and/or posted in a way to foster personal attacks\\, then you might have a reasonable point to be discussed. But saying that VIRTUALLY EVERY post he made was a personal attack or intended to foster personal attacks is just ridiculous overstatement, which serves not to put Franzen's participation as a poster in a fair context but rather to egregiously distort the history of his posting, which includes a lot of good, and, according to his detractors a lot of bad (and even his supporters should not deny that some of his posts were personal attacks). MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net> >\\Modus Ponens\\ allows \ one to derive A from A --> B? You too must be > >>using some highly advanced text! > > As Jeffrey Ketland explained to the poster, he had misunderstood the > > rule; no doubt because, as he'd been told, he hadn't seen the 'sense' > > behind the list of rules he'd been given to memorize the previous week. > So would you say that I was making fun of your use of the term \\Modus > Ponens\\ or \\MP\\? It's possible. I notice your repeated use of scare-quotes, but I've seen those misused too often to make a hasty judgement. === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= >>So would you say that I was making fun of your use of the term \\Modus >>Ponens\\ or \\MP\\? > It's possible. That's silly, there's nothing particularly amusing in the names of the rules of inference. I was just pointing out the absurdity of a rule of inference allowing one to infer A from A --> B, as was Torkel. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= >>So would you say that I was making fun of your use of the term \\Modus >>Ponens\\ or \\MP\\? > It's possible. That's silly, there's nothing particularly amusing in the names of the rules of inference. Obviously if anything I was just pointing out the absurdity of a rule of inference allowing one to infer A from A --> B, as was Torkel. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= >So would you say that I \ was making fun of your use of the term \\Modus > >>Ponens\\ or \\MP\\? > > It's possible. > That's silly, there's nothing particularly amusing in the names of the > rules of inference. You don't find the idea of \\Horseshoe Elimination\\ amusing? Then you have no sense of humour; I think it would crack up a 14-year-old. > Obviously if anything I was just pointing out the > absurdity of a rule of inference allowing one to infer A from A --> B, > as was Torkel. It's not obvious why you were using scare quotes around the term. Why were you using them? And why did you try to hide my point that you were using them? === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= have no sense of humour; I think it would crack up a 14-year-old. It never occurred to me until you mentioned it! Indeed, a fourteen year old would be on the lookout for such juvenile puns. That doesn't mean that Franzen was! It's clear that what Franzen was making fun of was the misapplication of the rule and that application being cited under the banner of the rule. That YOU were on the level of a fourteen year old does not entail that Franzen was. Sheesh. You're really making yourself look ridiculous here. > > Obviously if anything I was just pointing out the > > absurdity of a rule of inference allowing one to infer A from A --> B, > > as was Torkel. > It's not obvious why you were using scare quotes around the term. It was to me! Quote marks were used to emphasize that the poster was making a fallacioius inference under the banner of the rule so named. For crying out loud, just because the word 'elimination' suggested defecation to you does not entail that that pun had anything to do with what Franzen and the others had in mind. You just made the mental association yourself, and assumed that that was what the remarks were about. MoeBlee === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= >>Obviously I was just pointing out the >>absurdity of a rule of inference allowing one to infer A from A --> B, >>as was Torkel. > It's not obvious why you were using scare quotes around the term. I put them there because the rule that allows one to derive A from A --> B is not modus ponens. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= >Obviously I was just \ pointing out the > >>absurdity of a rule of inference allowing one to infer A from A --> B, > >>as was Torkel. > > It's not obvious why you were using scare quotes around the term. > I put them there because the rule that allows one to derive A from A -- B \ is not modus ponens. That explains there first use. It doesn't explain your leaving them in, even after I'd explained that I was not saying that it was. That does sound like you're mocking my use of the term, for some reason. Also left to explain is why, after I'd objected to your trying to hide some of the points that I had made, you'd try to hide that objection as well by doing the same thing again. At this point I am starting to hypothesize about your motives, although my conclusions are still tentative (as they're based on only three of your posts). === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= > Also left to explain is why, after I'd objected to your trying to hide > some of the points that I had made, you'd try to hide that objection as > well by doing the same thing again. My sinister intention was to make fun of your name, obviously. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is?= Also left to explain is \ why, after I'd objected to your trying to hide > > some of the points that I had made, you'd try to hide that objection as > > well by doing the same thing again. > My sinister intention was to make fun of your name, obviously. Not obviously. Snipping words and sentences out of quotes, with no indication that you've doing so, is typical troll behavior; something that your mentor Torkel used to do as well. Add that to some of the other Torkel tricks you've been pulling, and the most obvious explanation is that you've been trying to troll me. === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is?= indication that you've doing so, is typical troll behavior; something > that your mentor Torkel used to do as well. Please provide an example of Franzen intentionally or even unintentionally distorting a poster's post by editing. (Merely snipping is not necessarily distortion, or we'd have to include every word of every post in every reply.) MoeBlee > Add that to some of the > other Torkel tricks you've been pulling, and the most obvious > explanation is that you've been trying to troll me. There's 'troll' used just the way I mentioned. 'troll' used as not much more than an unsubstantiated and general charge to throw at someone when he or she posts in a way that one disapproves. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= What I had in mind was Edward \ Nelson's Intenal Set Theory which I had > > read up sometime back. I distinctly remember Nelson asserting that the > > non-standard integers were internally \\non-standard finite\\. > Well, he was wrong. > \\Internally\\, \\finite\\ is a meaningless adjective because > the distinction simply cannot be drawn; there is no predicate > definable in PA that \\means\\ finite, i.e., that is satisfied by > and only by finite elements. > > Even > > though a non-standard integer x satisfies x > 1, x>2, x>3,...., > and therefore IS infinite, > > there > > is no way to internally express the fact that x exceeds \\all\\ the > > standard integers. > There is no way to express it FINITARILY AS A CONJUNCTION > IN THE LANGUAGE of PA. But simply saying, PRECISELY > AS YOU DID, that \\x satisfies x > 1, x>2, x>3,...., \\ > IS saying that x is infinite. > > Externally you can talk of \\all\\ standard integers > > and so you can see that x is infinite. > All the infinity of things you are seeing, to deduce that from, > are INternal. x>1 is internal. x>2 is internal. All infinity of them > are internal. The fact that you can't have infinite conjunctions > in a finitary language does not stop all infinity of them from being > true, or from having the implication that they jointly have, which > is that x is infinite. Internally. x is internal and the infinite > number > of things that are all smaller than it are also internal, and all those > smaller-thans are internally provable. > > At least this is my vague > > understanding of what I read a long time ago. > Well that's basioally correct. > But the fact that something is not finitarily expressible > in a restricted language is not sufficient to cast doubt > on its truth. To do THAT, you would have to come up with > a model where that same thing was FALSE. > And there are no models of PA in which an element that > is greater than all finite elements is itself also finite. I agree with your post. But Edward Nelson's Internal Set Theory even has a theorem that goes something like \\There is a finite set containing all standard sets\\. Nelson himself called this resutl counter-intuitive and I personally could never come to terms with it. === Subject: Help needed with lattices Hi all, I am a philosopher and I need some help finding out whether a particular diagram is a lattice, more particularly an upper semilattice, or not. And if not, whether there is a mathematical name for the diagram. Would any kind mathematician out there be willing to let me email them the diagram as a Word attachment, so they can reply to me by email? Believe it or not, my own university had not a single mathematician who knew about lattice theory!! David dso371@yahoo.co.uk === Subject: Re: Linear Regression with occasional out-of-range data > We want to do linear regression with occasional out-of-range data > say the range is about 5-10% but happens only 5 data points in 100 data > points captured > What numerical methods or other simple trigonometry would help in this > situation? > We want to anticipate some future values which lie in this straight > line. > Any where I can look? or examples provided? > Jack You could try searching for \\Robust regression\\ or \\Resistant regression\\ Duncan === Subject: Re: Linear Regression with occasional out-of-range data what do you mean by \\out-of-range\\ data ? is the data valid but some values are really large or small ? If the data is not valid there is of course an easy solution ... delete those observations. (usually called \\outliers\\ in the statistical world) Otherwise I agree with the previous post and use some sort of robust regression, perhaps even a non-parametric regession method might work as well ... depends on the data hope that helps === Subject: Re: Linear Regression with occasional out-of-range data > what do you mean by \\out-of-range\\ data ? > is the data valid but some values are really large or small ? If the > data is not valid there is of course an easy solution ... delete those > observations. (usually called \\outliers\\ in the statistical world) > Otherwise I agree with the previous post and use some sort of robust > regression, perhaps even a non-parametric regession method might work > as well ... depends on the data > hope that helps Statistisians also know a censored regression ;) === Subject: Re: union convex set is not convex > Hey, > i have some question about convex set. > - union convex set is not convex, why!! Could you just provide the definition of \\convex\\ that you are using? In all variants I know of, it is not difficult to give examples of two convex sets whose union is not convex, even if their intersection is nonempty. Also note, that the actual statement \\the union of convex sets is in \\ general not convex\\ does allow examples of the union of two convex sets being \\ convex again; it only states, that you cannot expect this to hold all the time. Marc === Subject: Re: union convex set is not convex > The x-axis and the y-axis \ are convex and aren't disjoint. > >> Yet the union of the x-axis with the y-axis isn't convex. > >sorry ,to avoide any mistake > >i mean with Sx a convex set > >Sx={X| X inc R^n} e.g X = {x_i = exp(-t) | 0 >} > Whatever that means, it certainly does no appear to be a convex set. > >Sy {Y| Y inc R^n } e.g Y = {z_i = t exp(-t) | 0 >} > >and not x-axis. > >> > - i want to definie a convex set but the elments are exp-function \\ and > >> > not a simpel point.!! > >> > e.g X = {x_i = exp(t) | 0 X = > >> > {x_0,x_1,...x_n} > >> >> > can you help my. > >> >> > mes salutationes > >> >> > Ismail > >> >> ************************ > David C. Ullrich i have proof the set is convex, what is your reason === Subject: Overview of fields of mathematics The major disciplines within mathematics first arose out of the need to do calculations in commerce, to measure land and to predict astronomical events. These three needs can be roughly related to the broad subdivision of \ \\ mathematics into the study of structure, space and change (i.e. algebra, geometry and analysis). In addition to these three main concerns, there are \ \\ also subdivisions dedicated to exploring links from the heart of mathematics \ \\ to other fields: to logic and other simpler systems (foundations) and to the \ \\ empirical systems of the various sciences (applied mathematics). The study of structure starts with numbers, first the familiar natural numbers and integers and their arithmetical operations, which are recorded in elementary algebra. The deeper properties of whole numbers are studied in \ \\ number theory. The investigation of methods to solve equations leads to the \ \\ field of abstract algebra, which, among other things, studies rings and fields, structures that generalize the properties possessed by everyday numbers. Long standing questions about ruler-and-compass constructions were \ \\ finally settled by Galois theory. The physically important concept of vectors, generalized to vector spaces and studied in linear algebra, belongs \ \\ to the two branches of structure and space. The study of space originates with geometry, first the Euclidean geometry and trigonometry of familiar three-dimensional space (also applying to both more and fewer dimensions), later also generalized to non-Euclidean geometries which play a central role \ \\ in general relativity. The modern fields of differential geometry and algebraic geometry generalize \ \\ geometry in different directions: differential geometry emphasizes the concepts of functions, fiber bundles, derivatives, smoothness, and direction, while in algebraic geometry geometrical objects are described as \ \\ solution sets of polynomial equations. Group theory investigates the concept \ \\ of symmetry abstractly; topology, the greatest growth area in the twentieth \ \\ century, has a focus on the concept of continuity. Both the group theory of \ \\ Lie groups and topology reveal the intimate connections of space, structure \ \\ and change. Understanding and describing change in measurable quantities is \ \\ the common theme of the natural sciences, and calculus was developed as a most useful tool for that. The central concept used to describe a changing variable is that of a function. Many problems lead quite naturally to relations between a quantity and its rate of change, and the methods to solve these are studied in the field of differential equations. The numbers used to represent continuous quantities \ \\ are the real numbers, and the detailed study of their properties and the properties of real-valued functions is known as real analysis. For several reasons, it is convenient to generalise to the complex numbers which are studied in complex analysis. Functional analysis focuses attention on (typically infinite-dimensional) spaces of functions, laying the groundwork \ \\ for quantum mechanics among many other things. Many phenomena in nature can \ \\ be described by dynamical systems; chaos theory makes precise the ways in which many of these systems exhibit unpredictable yet still deterministic behavior. In order to clarify the foundations of mathematics, the fields first of set \ \\ theory and then mathematical logic were developed. Mathematical logic, which \ \\ divides into recursion theory, model theory and proof theory, is now closely \ \\ linked to computer science. When electronic computers were first conceived, \ \\ several essential theoretical concepts were shaped by mathematicians, leading to the fields of computability theory, computational complexity theory, and information theory. Many of those topics are now investigated in \ \\ theoretical computer science. Discrete mathematics is the common name for the fields of mathematics most generally useful in computer science. An important field in applied mathematics is statistics, which uses probability \ \\ theory as a tool and allows the description, analysis and prediction of phenomena where chance plays a part. It is used in all sciences. Numerical analysis investigates methods for efficiently solving a broad range of mathematical problems numerically on computers, beyond human capacities, and \ \\ taking rounding errors and other sources of error into account to obtain credible answers. http://mathematics.ask.dyndns.dk/ === Subject: Re: Overview of fields of mathematics > The major disciplines within mathematics first arose out of the need to \ \\ do > calculations in commerce, to measure land and to predict astronomical > events. If you think that's mathematics. If mathematics involves proving theorems, none of the above qualifies. The physically important concept of > vectors, generalized to vector spaces and studied in linear algebra, \\ belongs > to the two branches of structure and space. So where do modules belong? I can buy a finitely-generated module over a PID as space, I guess, but what about in general? === Subject: Measure and Topology is there some \\invariance of domain\\ theorem for measures? This means: does there exists a bimeasurable (bijective) function R^n -> R^m with the borel or lebesque sigma-algebra? If the answer is no: What about subsets? As points have lebesque measure zero, there certainly are bijective bimeasurable functions f: U -> W of measurable sets in different R^m's, in contrary to topology. But I thought it could be possible, that this is not possible for sets of positive measure. Kilian. === Subject: Re: Measure and Topology > is there some \\invariance of domain\\ theorem for measures? No. > This means: > does there exists a bimeasurable (bijective) function R^n -> R^m with > the borel or lebesque sigma-algebra? Yes. Furthermore, there is such a bijection that preserves Lebesgue measure (n-dimensional to m-dimensional). > If the answer is no: What about subsets? As points have lebesque > measure zero, there certainly are bijective bimeasurable functions f: U > -> W of measurable sets in different R^m's, in contrary to topology. > But I thought it could be possible, that this is not possible for sets > of positive measure. > Kilian. -- G. A. Edgar \\ http://www.math.ohio-state.edu/~edgar/ === Subject: Infinite Induction and the Limits of Curves Hi All - It has been my position that the method of inductive proof is valid, not \\ only for all finite natural n, but for the infinite case as well, given certain precautions. An equality proven inductively, such as f(n)=g(n), always holds \ \\ in the infinite case, but with inequalities we must be careful that the \\ difference that causes the inequality does not have a limit of 0 as n->oo, or the proof \ \\ does only hold for the finite case. Otherwise, if we can inductively prove \ \\ that f(n)>g(n) for all n greater than some finite m, and that f(n)-g(n) does not \ \\ have a limit of 0 as n->oo, then we can say that f(n)>g(n) for all infinite \ \\ n as well. This has important implications. While discussing this in Calculus XOR Probability, Chas offered a counterexample to infinite induction which became an interesting discussion. \ \\ It involved a staircase function from (0,0) to (1,1), in the limit as the \\ number of steps approached oo, and its equivalence to the diagonal line with those \ \\ endpoints, the contradiction coming in the form of two different arc length \ \\ measures for the \\same\\ curve. Well, this led me to devise an improved definition of the curve and demonstrate the difference between the two \\ curves using this definition. I was then challenged to find another curve which WAS \ \\ equivalent to the staircase in the limit given my definition, which I \\ devised, with interesting results, especially regarding the use of infinitesimals. I \ \\ http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm Please enjoy. I look forward to all comments, either here or in email (link \ \\ at bottom of page). I'll update the page with any pertinent ideas or comments. Thanx. Have a nice day! -- Smiles, Tony === Subject: Re: Infinite Induction and the Limits of Curves > Hi All - > It has been my position that the method of inductive proof is valid, not \ \\ only > for all finite natural n, but for the infinite case as well, given certain \ \\ > precautions. There is a valid form of transfinite induction on well ordered sets, but TO's is not valid, as it requires things which contradict the standard properties of limits, such as the commutativity of all limit processes. > An equality proven inductively, such as f(n)=g(n), always holds in > the infinite case Only in TOmatics. > While discussing this in Calculus XOR Probability, Chas offered a > counterexample to infinite induction which became an interesting > discussion. It involved a staircase function from (0,0) to (1,1), in > the limit as the number of steps approached oo, and its equivalence > to the diagonal line with those endpoints, the contradiction coming > in the form of two different arc length measures for the \\same\\ > curve. This is one of the cases in which two limit processes do not commute, the limit of lengths of approximating polygons as the number of segments increases and the limiting set of points as the number of steps increases. The result depends on which limit is taken first. > Well, this led me to devise an improved definition of the > curve A different definition, granted, but it is only in TO's eyes that there is any improvement seen. > and demonstrate the difference between the two curves using > this definition. I was then challenged to find another curve which > WAS equivalent to the staircase in the limit given my definition, > which I devised, with interesting results, especially regarding the > can be viewed at: > http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm As TO's theory of infinitesimals does not agree with any of the standard theroies allowing infinitesimals which all work within standard axiomatic system, and TO has no axiomatic system in which his system can be shown to exist, he is merely hawking vaporware again. > Please enjoy. I look forward to all comments, either here or in email > (link at bottom of page). I'll update the page with any pertinent > ideas or comments. > Thanx. Have a nice day! TO says in the first paragraph of http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm \\ Thus, proving inductively that, for all n greater than 2, 2*n I have an interest in typography and geometry and > wondered if there was any link between the development > of the shapes of some letters, thru geometry, triganometry > and mathematics. > I was wondering whether it was possible to calculate > sin, cos, tan without a calculator, if so how, and if the SOH CAH TOA S sine O opposite - side opposite angle of interest H hypoteneuse - the longest side in a right angled triangle C cosine A adjacent - side next to the angle of interest which is not the hypoteneuse T tangent /|c / C| 5/ |4 / | a /A 3 | |b sinA = opp/hyp = 4/5 cosA = adj/hyp = 3/5 tanA = opp/adj = 4/3 sinC = opp/hyp = 3/5 cosA = adj/hyp = 4/5 tanA = opp/adj = 3/4 ##minty... > following values were given to these early Greek letters, > is it still possible to calculate those angles? > A = 1 > B = 2 > \\.83Á = 3 > \\.83¢ = 4 > E = 5 > H = 8 > \\.83\\.a6 = 9 > I = 10 > K = 20 > \\.83© = 30 > M = 40 > N = 50 > \\.83Â = 60 > O = 70 > P = 100 > \\.83¡ = 200 > T = 300 > Y = 400 > \\.83\\.b3 = 500 > X = 600 > \\.83\\[Micro] = 700 > \\.83\.a6 = 800 > and if so, how? > Mixi === Subject: ~~~~** LAWYER has sex with Pig!!! ****~~~~~ http://www.lawyerresource.info/2006/05/is-lawyer-licensing-necessary.html - Lawyer Chases down pig and has sex with it hilarious video!! === Subject: geometry puzzle OK, you have a sphere with a volume of 1400 cubic centimeters.... what is the volume of the largest cube that can fit inside the sphere with each of the 8 points of the cube touching the sphere? Provide a proof and answer.... it's quite easy really if you have some basic trigonometry knowledge... === Subject: Re: geometry puzzle > OK, you have a sphere with a volume of 1400 cubic centimeters.... > what is the volume of the largest cube that can fit inside the > sphere with each of the 8 points of the cube touching the sphere? > Provide a proof and answer.... it's quite easy really if you have some > basic trigonometry knowledge... Who needs trigonometry? Let the cube have side L. By applying Pythagoras twice, its diagonal is L * sqrt(3), so the sphere's diameter is the same, and its volume simplifies to PI * L^3 *sqrt(3) / 2. Therefore, the volume of the cube is 2800 / (PI * sqrt(3)) or approximately 514.5736... -- === Subject: hypercube puzzle Let's assume you have a cube 10cm^3 = 1000 cm^3 Let's make this a hypercube based on this cube thus being 8000cm^3 OK, this is basically 8 of the original cubes. So if we make these 8 into hypercubes themselves... then what is the final volume? and a twist: if you were to measure the two furtherest points in the cube it would be 10*sqrt(3) right? OK with the final resulting hypercube above represented by volume alone as a simple cube itself... what is the factor (expressed as a square-root) that would multiply by the original 10cm side to get the length of the furtherest distance points of the final cube? i.e. 10cm * sqrt(?) what is that (?) number? === Subject: Re: The list of all natural numbers don't exist <7LadnYvn180m2QvZnZ2dnUVZ_rednZ2d@comcast.com> <4vzlg.7925$lf4.3496@newsread1.news.pas.earthlink.net> <9IBlg.8153$lp.2766@newsread3.news.pas.earthlink.net> sets is NOT provable from ZF-I DOES entail that the statement that > there exist infinite sets is CONSISTENT with ZF-I. I think I might have been incorrect in that particular remark, and I am unlcear on this particular point. In what sense is it a \\fact\\ that ZF-I (ZF without the axiom of infinity) does not prove the statement that there are no infinite sets (call this statement ~I)? If we use ZF to prove the \\fact\\ that ZF-I does not prove ~I, then that would entail ZF proving the consistency of ZF (right?), which we pretty much hope cannot happen (given the second incompleteness theorem). So I think I should say that we presume (not that we know as a fact) that ZF-I does not prove ~I. (In any case, this doesn't affect the rest of my remarks in that post, so my point to Easterly still stands: The statement \\ZF-I is consistent with ~I\\ does not in itself entail \\ZF (i.e., ZF with I) is inconsistent\\.) MoeBlee === Subject: Re: The list of all natural numbers don't exist William Hughes said: > > Omega is a phantom. There is no smallest infinity any more than a \\ largest > > finite. When you have an infinite number and sbtract 1, then it is \\ still > > infinite, therefore a smallest infinite is self contradictory. This is \ \\ the deep > > paradox of infinity. That boundary between the finite and infinite \\ cannot occur > > at any point, or within any finite sequence of points. Successor() can \ \\ only > > bridge the gap between finite and infinite through an infinite number \ \\ of > > iterations. > The problem is that we still have the set of all finite numbers. We > want > something to represent the size of this set. This something cannot > be any finite number (any finite number is clearly too small). Nor can > it be any TO-infinite number (any such number is clearly too large). That's quite true. I really don't see how one can claim to put any size on \ \\ this set, since it would be equal to the largest finite, which clearly doesn't exist, at least in the sense that it can never be specified. It's like \\ trying to find the smallest finite real. It can't be done. But, you could name the \ \\ size of this set, and use that name in formulas to express the size oof \\ subsets of the finite naturals. > Let us call this size Q. Now Q cannot be a TO-natural. And we must > have that Q-1 does not make sense or Q-1=Q. Why does Q-1 not make sense? Take the set of naturals, size Q. Take away element 1. Do you not have one element missing, one fewer element? I think \ \\ that makes perfect sense. > This last is not a contradiction. It is if you hold fundamentally that removing an element froma set always makes it smaller. I do. > Q is not a natural nor a TO-natural, so it is not clear what meaning > we will give to Q-1, and it is certainly not clear that Q-1 must be > different > from Q. If adding elements doesn't increase the size of a set, and removing them doesn't decrease it, in your system, then that system violates a \\ fundamenetal tenet of mine, anyway. Note that decrement is very well defined in my \\ T-riffic number system. If you like omega so much, why not have a largest finite as well? You could \ \\ call it alpha, and say alpha+1=alpha. While you're at it, you might as well \ \\ say alpha+1=omega and omega-1=alpha. That would make as much sense as anything \ \\ esle in transfinite set theory, and at least give it some aesthetic symmetry. :) > -William Hughes -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist Ross A. Finlayson said: > There's no universe in ZF. > Ross But there is a universe, and it's consistent. :) -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist Virgil said: > > David R Tribble said: > > > Mike Kelly said: > > > >> 0 and 1 are only the \\ends\\ of the set [0,1] if there is some \\ sequence > > > >> starting with 0, stepping through every real in the interval and \ \\ ending > > > >> with 1. Maybe you could tell us the second real in that sequence? > > > > > > > > >> There's Lil'un, the unit infinitesimal, the inverse of Big'un. It's \ \\ > > > >> Ross' > > > >> nilpotent infinitesimal and sequential real. That sequence from \\ 0*Lil'un > > > >> to > > > >> Big'un*Lil'un is the set of reals in [0,1]. Since Lil'un is less \ \\ than > > > >> any > > > >> finite difference, no standard real is missed. > > > > > > > > Mike Kelly said: > > > > Lil'un isn't a real number. What is the first real number after 0? > > > > > > Tony is incapable of answering that question. > > > > > > Every real in [0,1] can be expressed as a \\digital fraction\\: > > > x = sum{i=1 to oo} d_i b^(-i), 0 <= d_i < b, for base b. > > > > > > Tony's \\LilUn\\ and Ross's \\iota\\ can't be written in this form, \ \\ so > > > obviously they are not real numbers. So they can't possibly form > > > a sequence of points in the real interval. > > > > > > > > > > Yes, they can, as I've shown you before. In fact, if Big'un=oo, then we \ \\ only > > need sum(i=1->log_b(oo): d_i b^(-i)), so it requires fewer bits than you \ \\ > > claimed. > Every number requiring only finitely many integral bits and no > fractional bits is a natural number , and there is such a natural larger > than any given real. Whch one is that? Omega? ;) > TO's alleged extension of this to a system of \\numbers\\ requiring more > than finitely many integral bits is vaporware. Nope, it's perfectly workable. :D > Given log_b digits to the right of the digital point, Lil'un is > > represented by 0.000...001. That '1' is in the -log_b(Big'un) digit \\ position. > > One would think you'd remember the T-riffics, given how close \\ \\Tribble\\ is to > > Virgil's \\T-errible\\ name for them. :) > Perhaps he only remembers how terrible they actually are and chooses to > ignore everything else about them. That is a very sane and sober > attitude to take about something so ephemeral. Yes, well, then maybe he should get away from the conversation and drown the \ \\ memories of it in a bottle of schnapps, if it's so terrible. -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist Tony Orlow said: > Virgil said: > > > > > David R Tribble said: > > > > Mike Kelly said: > > > > >> 0 and 1 are only the \\ends\\ of the set [0,1] if there is some \ \\ sequence > > > > >> starting with 0, stepping through every real in the interval and \ \\ ending > > > > >> with 1. Maybe you could tell us the second real in that \\ sequence? > > > > > > > > > > > >> There's Lil'un, the unit infinitesimal, the inverse of Big'un. \ \\ It's > > > > >> Ross' > > > > >> nilpotent infinitesimal and sequential real. That sequence from \ \\ 0*Lil'un > > > > >> to > > > > >> Big'un*Lil'un is the set of reals in [0,1]. Since Lil'un is less \ \\ than > > > > >> any > > > > >> finite difference, no standard real is missed. > > > > > > > > > > > Mike Kelly said: > > > > > Lil'un isn't a real number. What is the first real number after \ \\ 0? > > > > > > > > Tony is incapable of answering that question. > > > > > > > > Every real in [0,1] can be expressed as a \\digital fraction\\: > > > > x = sum{i=1 to oo} d_i b^(-i), 0 <= d_i < b, for base b. > > > > > > > > Tony's \\LilUn\\ and Ross's \\iota\\ can't be written in this form, \ \\ so > > > > obviously they are not real numbers. So they can't possibly form > > > > a sequence of points in the real interval. > > > > > > > > > > > > > > Yes, they can, as I've shown you before. In fact, if Big'un=oo, then \ \\ we only > > > need sum(i=1->log_b(oo): d_i b^(-i)), so it requires fewer bits than \ \\ you > > > claimed. > > > > Every number requiring only finitely many integral bits and no > > fractional bits is a natural number , and there is such a natural larger \ \\ > > than any given real. > Whch one is that? Omega? ;) Never mind - I misread this. So....what? > > > > TO's alleged extension of this to a system of \\numbers\\ requiring more \ \\ > > than finitely many integral bits is vaporware. > > > Nope, it's perfectly workable. :D > > > > Given log_b digits to the right of the digital point, Lil'un is > > > represented by 0.000...001. That '1' is in the -log_b(Big'un) digit \ \\ position. > > > One would think you'd remember the T-riffics, given how close \\ \\Tribble\\ is to > > > Virgil's \\T-errible\\ name for them. :) > > > > Perhaps he only remembers how terrible they actually are and chooses to \ \\ > > ignore everything else about them. That is a very sane and sober > > attitude to take about something so ephemeral. > > > Yes, well, then maybe he should get away from the conversation and drown \ \\ the > memories of it in a bottle of schnapps, if it's so terrible. -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist Virgil said: > > Albrecht said: > > > > > > > > > Hi Albrecht, Hi All. Glad to see healthy debate on this issue. \ > > > > It's worth discussing. > > > > > > > > > Hi Tony. Nice to hear from you again. > > > > > > > > > > > > Albrecht, I heartily agree with your logic regarding the unary \ > > > > representation of the naturals and the contradiction between > > > > there being an infinite number of rows, while no column is ever > > > > allowed to be infinite, even though the grid is square at every > > > > point through the set. I would perhaps show it like this: > > > > > > 1 x > > > > > > 2 xx > > > > x > > > > > > 3 xxx > > > > xx > > > > x > > > > > > 4 xxxx > > > > xxx > > > > xx > > > > x > > > > > > etc. > > > > > > This growing square of side n, of which (n^2+n)/2 are x's and \ the > > > > rest empty, where each column is a natural, cannot possibly > > > > become infinite in width and contain an infinite number of > > > > naturals, unless it also becomes infinite in height, therefore > > > > containing a natural with an infinite value. The problem with > > > > explaining this to set theorists is, indeed, that they claim the > > > > set exists all at once, and refuse to look at the formulas which > > > > describe the set in detail. In this case, the formula is x=y, the > > > > indentity function between element count and value. So, in this > > > > respect, we are in sync, because we are viewing infinity from a > > > > quantitative and geometrical perspective, and not according to > > > > Dedekind's set theoretic definition of an infinite set. > > > > > > Yes, the definition that a set is infinite if there is a bijektion > > > to real subsets of itself is nice. But if there is no infinite set, > > > this definition leads to nothing. > > > > > > > There is LUB on the finite naturals > Not in standard mathematics, whatever may be the case in TO's vaporware > versions. That was a typo. It was supposed to be \\There is *no* LUB on the finite naturals\\, of course. That should have been obvious from all the context \ \\ you snipped. > > The set of standard naturals is a finite but > > unboundedly large set. > If it is the *standard* set then one must evaluate its finiteness by the > *standard* test(s) for finiteness, and it fails them, so, whatever it > is, it is not finite. No, I can take you standard set and apply whatever tests I like to it. > > I understand. The proof seems to show that incrementing a finite > > value can never reach an infinite value in any number of steps. But, > > Albrecht, it does! It just takes an infinite number of steps. :D > Then it doesn't by TO's own admission (reach an infinite value in a > finite number of steps)! It doesn't reach an infinite value in a finite number of incrmeents, no. It \ \\ requires an infinite number of incrmeents to turn a finite into an \\ infinite. > In response to someone saying on can not get to the North pole by only > going south, TO would equivalently say \\Yes on can, but one has to go > north to do it\\. You just have to start at Polaris. ;) -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist David R Tribble said: > Mike Kelly said: > >> 0 and 1 are only the \\ends\\ of the set [0,1] if there is some \\ sequence > >> starting with 0, stepping through every real in the interval and \\ ending > >> with 1. Maybe you could tell us the second real in that sequence? > >> There's Lil'un, the unit infinitesimal, the inverse of Big'un. It's \\ Ross' > >> nilpotent infinitesimal and sequential real. That sequence from \\ 0*Lil'un to > >> Big'un*Lil'un is the set of reals in [0,1]. Since Lil'un is less than \ \\ any > >> finite difference, no standard real is missed. > Mike Kelly said: > >> Lil'un isn't a real number. What is the first real number after 0? > David R Tribble said: > >> Tony is incapable of answering that question. > > >> Every real in [0,1] can be expressed as a \\digital fraction\\: > >> x = sum{i=1 to oo} d_i b^(-i), 0 <= d_i < b, for base b. > > >> Tony's \\LilUn\\ and Ross's \\iota\\ can't be written in this form, \ so > >> obviously they are not real numbers. So they can't possibly form > >> a sequence of points in the real interval. > David R Tribble said: Uh, Tony Orlow said this: > > Yes, they can, as I've shown you before. In fact, if Big'un=oo, then we \ \\ only > > need sum(i=1->log_b(oo): d_i b^(-i)), so it requires fewer bits than \\ you > > claimed. Given log_b digits to the right of the digital point, Lil'un \ \\ is > > represented by 0.000...001. That '1' is in the -log_b(Big'un) digit \\ position. > You forgot to mention that BigUn is a number that doesn't equal itself. > Written in base 10, it's divisible by 5, but written in base 3, it's > not. The finite remainder, as I explained before, is insignificant compared to \ \\ the infinite quotient, and can therefore be discarded. What you are reacting to \ \\ is an artifact of the digital number systems. It's really not a problem. In standard math, is 2+Lil'un equal to 2? Yes, any infinitesimal difference can \ \\ be discarded as insignficant, and the same goes for any finite difference when \ \\ comparing infinite quantities in terms of ratio. > Your \\arithmetic\\ only proves my point. oo is not a real number. > Log(oo) is not a real number. BigUn and LilUn are not real numbers. If they are on the real line, they are real numbers. Your real line is unboundedly large but finite. Mine is bounded but infinite. That makes all \ \\ the difference. They're real on my line. > Given that 1/3 = 0.333...333 in your notation, that would be an > \\infinite sequence\\ of 3's ending with a last digit of 3. So then > 3 x 1/3 is 0.999...999, an \\infinite sequence\\ of 9's ending with > a last digit of 9. Yes, when dealing with the quantity on the finite level, there is no reason \ \\ to go further with the digits if you are only going to add infinitesimal quantities, since they don't matter on the finite scale. If we wanted \\ precision on the infinitesimal scale, so that our 1/3 * 3 gave us something equal to 1 \ \\ on the infinitesimal scale, then we have to use subinfinitesimals and say 1/3 \ \\ is 0.333...333:333...333. Then we'll have 0.999...999:999...999, which is indistinguishable from 1 on the infinitesimal scale. However, on the finite \ \\ scale, getting a result like 0.999...999 is not a proble, since on the \\ finite scale that infinitesimal difference of 0.000...001 is indistinguishable from \ \\ 0, and 0.999...999 is indistinguishable from 1. > Most of the rest of us, however, know that 3 x 1/3 = 1. We also know > that \\an infinite sequence of digits with a last digit\\ is a > contradiction in terms, and thus utter nonsense. You are simply > incapable of comprehending \\infinity\\, and that's a pity. You are the one incpabale of comprehending relative infinities, and rules as \ \\ simple as IFR. I think you did all your learning, and now you're in spew \\ mode. Of course 3 * 1/3 = 1. I've shown that works in my system, and there is no contradiction with yours. I'll try to get the paper posted I mentioned to Albrecht, and maybe that will be a good example for you to see how relative \ \\ infinitesimals work when applying the segment-wise definition of the curve \ \\ to Chas' staircase counterexample to infinite induction. > Worse, you continually drag \\unit infinities\\, \\hypernaturals\\, > \\infinitesimals\\, and so forth into the discussion, claiming that they > fit \\quite nicely\\ into standard arithmetic, when it's plainly obvious > that they don't. It's only obvious to those who don't know what I'm talking about. You are incapable of absorbing a system incompatible with the one you've always \\ known, so why try? It's really not a good investment of your energy to put down \\ ideas you don't understand, is it? Wouldn't you feel dumb if you came to realize \ \\ one day that it all DOES actually work? That would be a shame. I would feel bad \ \\ for you. :( -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist David R Tribble said: > >> Then there is the mapping via mirror image of the strings representing \ \\ reals in > >> [0,1) to strings representing the naturals. So, the naturals go \\ ...000, > >> ....001, ...010, ...011, etc. Therefore the reals in [0,1) go \\ 0.000..., > >> 0.100..., 0.0100.., 0.11000..., etc. Then you can tack on 1.000... \\ \\after\\ you > >> get to 0.1111... Now you will complain this is not a bijection, because \ \\ no > >> natural has infinite strings. Well, that's the point I was making \\ above. Any > >> finite natural is finitely expressible. Not every real is. > Mike Kelly said: > >> Uh, so you *agree* that it isn't a bijection? So you *agree* that \\ there > >> is no mapping between the naturals and the reals? Ahh, progress.. > > I declared a mapping between the hypernaturals and the reals in [0,1]. \ \\ That's > > valid, even in your primitive theory. Do you agree that there is such a > > mapping? > You keep \\declaring\\ that, but you've never shown that it means > anything in any logical sense. > 1. What is a \\hypernatural\\ number? It's a counting number which is allowed to assume infinite values. It's the \ \\ value of a digital string with all 0's to the right of the digital point, \ \\ which may have nonzero digits infinitely far to the left of the digital point. > 2. If I've got a hypernatural h, how do I use it to get another > hypernatural? That's a pretty general question. The simplest way is through increment or decrement. In the T-riffic binary number system, increment is performed by inverting the rightmost 0 bit and all 1's to the right of it, and decrement \ \\ is performed by inverting the rightmost 1 bit and all 0's to the right of it. That's a simple example demonstrating that each has a successor and a predecessor. > 3. What is the cardinality of the hypernaturals? (A simple bijection > between them and any other well-defined set will suffice.) > You declare it is c (= card([0,1])), but you've never shown that to > actually be the case. As a digital number, each standard natural may have nonzero digits in any position a finite number of positions to the left of the digital point. The \ \\ hypernaturals allow nonzero bits in infinite positions to the left of the digital point. This is the precise mirror image of the relationship between \ \\ a subset of the rationals, the set of all digit strings with all nonzero \\ digits in finite positions to the right of the digital point, and the reals in \\ [0,1), which may have nonzero digits in infinite positions to the right of the \\ digital point. Therefore, there is a bijection between the hypernaturals and the \\ reals in [0,1), through the exact same set of strings, those digital numbers which \ \\ allow nonzero bits in infinite positions. So, in your parlance, the \\ cardinality is c. See? In my system, the exact numbers of hypernaturals and of reals in [0,1) are \ \\ also equal. That is, the number of unit intervals on the first-order real line is \ \\ equal to the number of infinitesimally-spaced reals in the unit interval. In \ \\ my system, the naturals in [0,Big'un) are mapped to the reals in [0,1) by the mapping formula f(x)=x/Big'un, though that mapping would never be allowed in \ \\ your system. Over the first-order real line, given IFR, we get a total count \ \\ of all reals on that line to be Big'un^2. This agrees with the common method of \ \\ Big'un reals/unit interval times Big'un unit intervals equals Big'un^2 \\ reals. -- Smiles, Tony === Subject: Re: The list of all natural numbers don't exist <7LadnYvn180m2QvZnZ2dnUVZ_rednZ2d@comcast.com> <3vKlg.8443$lp.4183@newsread3.news.pas.earthlink.net> ... > >> Every definition of union and intersection > >> I have seen define these operations as > >> binary operators. > >> Given two sets, we get back one set. > > Then you have never seen the Wikipedia page on the ZF axioms, > > http://en.wikipedia.org/wiki/ZFC#The_axioms: > > \\Union: For any set x, there is a set y such that the elements of y \\ are > > precisely the members of the members of x.\\ > > It might be helpful to give a pointer to the definition of the ZF \\ axioms > > you are using. The ones I've found use the general form in their union > > (or sum set) axiom, not just binary union. > > or on arbitrary intersections, > > \\ http://en.wikipedia.org/wiki/Intersection_(set_theory)#Arbitrary_intersectio\ n\\ s: > > \\If M is a nonempty set whose elements are themselves sets, then x is \ \\ an > > element of the intersection of M if and only if for every element A of > > M, x is an element of A.\\ > OK. You define a new operator and call it union. > Obviously, this new operator is not compatible with > binary union because we know binary union can not > be applied to an infinite set. Binary union is a special case of general union. Of course it is compatible. And why do you say binary union cannot be applied to \\an\\ (?) infinite set? You can take the union of two infinite sets. > Have you also defined a new successor operator? > I don't see how you can. Successor is a unary operator. A new successor operator for what purpose? The successor operator you already accept for finite sets works equally well for infinite sets, assuming one accepts that one can take the union of infinite sets. > There is no way we can use it other than sequentially. I'm not sure what this means. Your use of terms like \\use\\ and \\apply\\ is somewhat misleading; the successor \\operator\\ is not a function call or an algorithm - it simply describes a relationship between two sets. > We have already shown any set formed using > the successor operator is finite. > It should be obvious by now we can never form > an infinite set of natural numbers by applying > the successor function even if we do it an > \\infinite\\ number of times. > How are you forming a set of natural numbers > if you are not using the successor function? Why would we use the successor function??? You keep asking this and it doesn;t make any sense to me. There do exist sets not formed by \\applying\\ the successor function to some other set. Very loosely speaking, the set of natural numbers is the set of all sets that can be \\produced\\ by \\applying\\ the succesor function to the empty set zero or more times. We don't \\produce\\ the set of natural numbers by applying the successor function in any way, shape or form. > We can show any set formed from the union > of a finite ordinal is equivalent to taking the successor > of a natural number. > How can union form an infinite set > of natural numbers when successor can't? Because union isn't the same thing as successor???? These questions seems nonsensical to me. Why shouldn't two different things be different? > I have to question this definition of union if you > are claiming you can form infinite sets of natural > numbers with it. So you are assuming your conclusion. Awesome. > Russell > - Everything you know is wrong. -- mike. === Subject: Re: The list of all natural numbers don't exist >> > ... >> >> Every definition of union and intersection >> >> I have seen define these operations as >> >> binary operators. >> >> Given two sets, we get back one set. >> > > Then you have never seen the Wikipedia page on the ZF axioms, >> > http://en.wikipedia.org/wiki/ZFC#The_axioms: >> > > \\Union: For any set x, there is a set y such that the elements of y \ \\ are >> > precisely the members of the members of x.\\ >> > > It might be helpful to give a pointer to the definition of the ZF >> > axioms >> > you are using. The ones I've found use the general form in their union >> > (or sum set) axiom, not just binary union. >> > > or on arbitrary intersections, >> > \\ http://en.wikipedia.org/wiki/Intersection_(set_theory)#Arbitrary_intersectio\ n\\ s: >> > > \\If M is a nonempty set whose elements are themselves sets, then x \ is \\ >> > an >> > element of the intersection of M if and only if for every element A of >> > M, x is an element of A.\\ >> > OK. You define a new operator and call it union. >> Obviously, this new operator is not compatible with >> binary union because we know binary union can not >> be applied to an infinite set. > Binary union is a special case of general union. Of course it is > compatible. And why do you say binary union cannot be applied to \\an\\ > (?) infinite set? You can take the union of two infinite sets. Binary union can't be applied to \\all\\ the members of an \\infinite\\ \\ class. (See, I have been reading this stuff. It's not a set until you prove its a set. It could be a proper class.) >> Have you also defined a new successor operator? >> I don't see how you can. Successor is a unary operator. > A new successor operator for what purpose? The successor operator you > already accept for finite sets works equally well for infinite sets, Only if I agree there is such a thing as an infinite set, and I don't. There may be infinite classes. > assuming one accepts that one can take the union of infinite sets. Why would I accept we can take the union of \\all\\ members of an infinite class when I can prove the union of any ordinal is finite? >> There is no way we can use it other than sequentially. > I'm not sure what this means. Your use of terms like \\use\\ and \\ \\apply\\ > is somewhat misleading; the successor \\operator\\ is not a function call > or an algorithm - it simply describes a relationship between two sets. To me, it's just a function call. It has a single input and a single output. I like simple functions. This union function you refer to can have an \\infinite\\ number of inputs. I think it still has some bugs. >> We have already shown any set formed using >> the successor operator is finite. >> It should be obvious by now we can never form >> an infinite set of natural numbers by applying >> the successor function even if we do it an >> \\infinite\\ number of times. >> How are you forming a set of natural numbers >> if you are not using the successor function? > Why would we use the successor function??? You keep asking this and it > doesn;t make any sense to me. There do exist sets not formed by > \\applying\\ the successor function to some other set. Natural numbers are defined by the successor function. If a set represents a natural number we can form this set using the successor function. > Very loosely speaking, the set of natural numbers is the set of all > sets that can be \\produced\\ by \\applying\\ the succesor function to \\ the > empty set zero or more times. There is no such set. I am not even sure we can say there is such a class. What do you mean by \\all\\? There is no all. > We don't \\produce\\ the set of natural > numbers by applying the successor function in any way, shape or form. Because it is impossible for us to do so. >> We can show any set formed from the union >> of a finite ordinal is equivalent to taking the successor >> of a natural number. >> How can union form an infinite set >> of natural numbers when successor can't? > Because union isn't the same thing as successor???? These questions > seems nonsensical to me. Why shouldn't two different things be > different? My argument is similar the Set of All Ordinals paradox. http://en.wikipedia.org/wiki/Burali-Forti_paradox If we have a set of all ordinals we can take the successor of this set and prove it is not the set of all ordinals. Supposedly, the Axiom of Regularity fixes this. I wouldn't know, myself. My argument is similar to the set of all ordinals paradox, but it is not the same. My argument is not a paradox, it is a proof that the union of any ordinal is finite. We have already proven the successor of any natural number is a natural number. Replace natural number with ordinal and replace successor with union and we a have a proof that all ordinals are finite. I don't think the Axiom of Regularity fixes this. If your definition of union allows the creation of an infinite set of natural numbers then you need to fix your definition of union. What are you quantifying over? It can't be the set of \\all\\ natural numbers because any set of natural numbers is finite. Most of us agree what \\all\\ means when applied to a finite set. I think \\all\\ is a meaningless term when applied to a class we can't prove is finite. I have shown how even saying there is an \\all\\ is equivalent to saying the class is finite. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist <448D7420.5080600@fastmail.fm> <448D77AC.7050305@fastmail.fm> <448DDFE4.9090306@fastmail.fm> It is really \ funny to produce paradox systems like this. Look at the > > following one: > > There is a sequence of elements starting from on element. Every > > element belongs to his group and reflect his group in consisting of a > > number of elements as the group have. It is some kind of fraktal or > > mirroring the whole structure on lower levels. I try to sketch it: > > x > > xx xx > > x x > > xx > > x > > xxx xxx xxx > > xx xx xx > > x x x > > xxx xxx > > xx xx > > x x > > xxx > > xx > > x > > and so on. (Perhaps better to see with nonproportional font) > > Let's call the whole infinite structure \\structure\\ the groups of > > groups of \\x\\ \\substructures\\ and the the groups of \\x\\ are named > > \\elements\\. > > Now the question: If this structure is completely done at once to have > > an infinite structure, is there an element which is as infinite as the > > whole structure? > No. You have flipped your original diagram. Yes. My argument wasn't quite clear. Let's have the following infintie structure: x ) 1 subgroup of \\x\\ / a set of subgroups x x ) 2 subgroups of \\x\\ / a set of subgroups xx xx ) x x x ) xx xx xx ) 3 subgroups of \\x\\ / a set of subgroups xxx xxx xxx ) x x x x xx xx xx xx xxx xxx xxx xxx xxxx xxxx xxxx xxxx ... Now, the inner structure of every subgroup of \\x\\ mirrors the structure of the subgroups from the first one down to the set of subgroups to which it belongs. Although the whole structure is infinite no subgroup is infinite. I claim: you can't say the whole structure is infinite (in the sense of completed actual infinity) if there is no subgroup infinite as well. But you see always the subgroups as static structures and the whole structure as a dynamic one. In set theory there is no dynamic. Actually completed infinity as you need for an infinite set is an antinomy. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist > > > It is really funny to produce paradox systems like this. Look at the > > > following one: > > > > There is a sequence of elements starting from on element. Every > > > element belongs to his group and reflect his group in consisting of a > > > number of elements as the group have. It is some kind of fraktal or > > > mirroring the whole structure on lower levels. I try to sketch it: > > > > > x > > > > > > xx xx > > > x x > > > > xx > > > x > > > > > > xxx xxx xxx > > > xx xx xx > > > x x x > > > > xxx xxx > > > xx xx > > > x x > > > > xxx > > > xx > > > x > > > > > > and so on. (Perhaps better to see with nonproportional font) > > > > Let's call the whole infinite structure \\structure\\ the groups \ of > > > groups of \\x\\ \\substructures\\ and the the groups of \\x\\ are \\ named > > > \\elements\\. > > > > Now the question: If this structure is completely done at once to \ \\ have > > > an infinite structure, is there an element which is as infinite as \\ the > > > whole structure? > > No. You have flipped your original diagram. > Yes. My argument wasn't quite clear. > Let's have the following infintie structure: > x ) 1 subgroup of \\x\\ / a set of subgroups > x x ) 2 subgroups of \\x\\ / a set of subgroups > xx xx ) > x x x ) > xx xx xx ) 3 subgroups of \\x\\ / a set of subgroups > xxx xxx xxx ) > x x x x > xx xx xx xx > xxx xxx xxx xxx > xxxx xxxx xxxx xxxx > ... > Now, the inner structure of every subgroup of \\x\\ mirrors the structure > of the subgroups from the first one down to the set of subgroups to > which it belongs. > Although the whole structure is infinite no subgroup is infinite. I > claim: you can't say the whole structure is infinite > completed actual infinity) if there is no subgroup infinite as well. Then your definition of a set being infinite is different form the ones in ZFC and NBG, because in those systems infinite sets are requires to exist. > But you see always the subgroups as static structures and the whole > structure as a dynamic one. You may see things that way, but your strabismus does not affect others. > In set theory there is no dynamic. The axiom of infinity in both ZFC and NBG is static. > Actually > completed infinity as you need for an infinite set is an antinomy. Not in ZFC of NBG. === Subject: Re: The list of all natural numbers don't exist <448D7420.5080600@fastmail.fm> <448D77AC.7050305@fastmail.fm> <448DDFE4.9090306@fastmail.fm> > Yes, the \ definition that a set is infinite if there is a bijektion to > >> real subsets of itself is nice. But if there is no infinite set, this > >> definition leads to nothing. > > >> But it is a strong fact for me that the successor > >> operation doesn't lead to infinite numbers. And how will you get the > >> natural numbers than with successor operation? > > Does Albrect claim that the successor \\function\\ does NOT biject the > > set of naturals with a proper subset? > Albrecht believes (wrongly) that an infinite set of naturals can only > exist if it contains \\infinite naturals\\. But he also believes > (rightly) that infinite naturals cannot exist from the Peano axioms. > Therefore, he concludes (wrongly) that the set of all naturals is not > an infinite set. There isn't any infinite set since completed infinity as in form of actual infinite structures like infinite lists, infinite sets, infinite numbers, ... aren't logically thinkable. It is very interesting. There are much people answering on my postings, but don't know my claims. :-) How comical. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist > > >> Yes, the definition that a set is infinite if there is a bijektion \ \\ to > > >> real subsets of itself is nice. But if there is no infinite set, \\ this > > >> definition leads to nothing. > > > > >> But it is a strong fact for me that the successor > > >> operation doesn't lead to infinite numbers. And how will you get the > > >> natural numbers than with successor operation? > > > > Does Albrect claim that the successor \\function\\ does NOT biject \ \\ the > > > set of naturals with a proper subset? > > Albrecht believes (wrongly) that an infinite set of naturals can only > > exist if it contains \\infinite naturals\\. But he also believes > > (rightly) that infinite naturals cannot exist from the Peano axioms. > > Therefore, he concludes (wrongly) that the set of all naturals is not > > an infinite set. > There isn't any infinite set since completed infinity as in form of > actual infinite structures like infinite lists, infinite sets, infinite > numbers, ... aren't logically thinkable. That depends on one's thinking machinery. if yours is incapable of some thoughts, it does not follow that everyone else is similarly handicapped. === Subject: Re: The list of all natural numbers don't exist has not been shown to be contradictory is valid mathematics. Maybe we should be more careful and call it not-definitely- invalid-and-therefore-possibly-valid mathematics? To ask critics for a definite (formal) proof of contradiction seems to me like shifting the burden of proof in a way that is ordinarily frowned upon: for usually you are asked to provide support for your bold claims - and telling us that some axiom systems, like ZFC, actually allow us to formally deduce true - and nothing but true - statements (e.g. about number theory) really is quite a *bold* claim, isn't it? Christian === Subject: Re: The list of all natural numbers don't exist > What can be formally proved from a system of axioms which > > has not been shown to be contradictory is valid mathematics. > Maybe we should be more careful and call it not-definitely- > invalid-and-therefore-possibly-valid mathematics? For axiom systems as thoroughly tested a ZFC and NBG, one can go so far as to say generally-regarded-as-valid-and-will-remain-so-as-long-as -no-contradictions-are-found. === Subject: Re: The list of all natural numbers don't exist >>Maybe we should be more careful and call it not-definitely- >>invalid-and-therefore-possibly-valid mathematics? > For axiom systems as thoroughly tested a ZFC and NBG, one can go so far > as to say generally-regarded-as-valid-and-will-remain-so-as-long-as > -no-contradictions-are-found. Much more can be said: for ZFC there is a clear informal conceptual picture in which all axioms of ZFC are obviously true. Moreover, anyone who accepts the usual set theoretical principles should be equally willing to accept the consistency of ZFC, and the consistency of a whole lot of stronger theories too. This is the source of our conviction in the consistency - actually, much stronger soundness conditions such as arithmetical soundness - of ZFC, not the fact that no one has yet managed to find a contradiction. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The list of all natural numbers don't exist >Maybe we should be more careful and call it not-definitely- >invalid-and-therefore-possibly-valid mathematics? >> For axiom systems as thoroughly tested a ZFC and NBG, one can go so far >> as to say generally-regarded-as-valid-and-will-remain-so-as-long-as >> -no-contradictions-are-found. > Much more can be said: for ZFC there is a clear informal conceptual > picture in which all axioms of ZFC are obviously true. Moreover, anyone > who accepts the usual set theoretical principles should be equally \\ willing > to accept the consistency of ZFC, and the consistency of a whole lot of > stronger theories too. This is the source of our conviction in the > consistency - actually, much stronger soundness conditions such as > arithmetical soundness - of ZFC, not the fact that no one has yet managed > to find a contradiction. I would not be so sure about the real underpinnings of that conviction, that faith. But being neither psychologist nor sociologist I do not want to argue about it here. However, as a mathematician, I would not want to claim the truth of any similar 'convictions' in my field for which I have *no* proof to offer. Much less would I expect the whole field to use such 'convicions' as its very foundation... Christian -- éThe notion of a set is too vague for the continuum hypothesis to have a positive or negative answer.\.82 - Paul Cohen So much for that 'clear informal conceptual picture' of yours... === Subject: Re: The list of all natural numbers don't exist > However, as a mathematician, I would not want to > claim the truth of any similar 'convictions' in my > field for which I have *no* proof to offer. Much > less would I expect the whole field to use such > 'convicions' as its very foundation... Too bad, then, that such convictions are all anyone has to offer. > éThe notion of a set is too vague for the continuum > hypothesis to have a positive or negative answer.\.82 > - Paul Cohen > So much for that 'clear informal conceptual picture' > of yours... How so? Our clear informal conceptual picture might or might not be determinate enough to allow for a solution of the continuum problem. Similarly, our clear informal conceptual picture of the structure of natural numbers might or might not be determinate enough to allow for a solution of this or that arithmetical problem. Such considerations do not affect the observations that these pictures are clear enough for certain basic principles to strike us as true. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The list of all natural numbers don't exist >Maybe we should be more careful and call it not-definitely- >invalid-and-therefore-possibly-valid mathematics? >> For axiom systems as thoroughly tested a ZFC and NBG, one can go so far >> as to say generally-regarded-as-valid-and-will-remain-so-as-long-as >> -no-contradictions-are-found. > Much more can be said: for ZFC there is a clear informal conceptual > picture in which all axioms of ZFC are obviously true. Moreover, anyone > who accepts the usual set theoretical principles should be equally \\ willing > to accept the consistency of ZFC, and the consistency of a whole lot of > stronger theories too. This is the source of our conviction in the > consistency - actually, much stronger soundness conditions such as > arithmetical soundness - of ZFC, not the fact that no one has yet managed > to find a contradiction. I would not be so sure about the real underpinnings of that conviction, that faith. But being neither psychologist nor sociologist I do not want to argue about it here. However, as a mathematician, I would not want to claim the truth of any similar 'convictions' in my field for which I have *no* proof to offer. Much less would I expect the whole field to use such 'convicions' as its very foundation... Christian -- éThe notion of a set is too vague for the continuum hypothesis to have a positive or negative answer.\.82 - Paul Cohen So much for that 'clear informal conceptual picture' of yours... === Subject: Re: The list of all natural numbers don't exist <448DDFE4.9090306@fastmail.fm> <233a6$449a31c0$54482b53$19019@news.hispeed.ch> picture in which all axioms of ZFC are obviously true. Surely this is overstating the case, unless \\clear\\ and \\obvious\\ are claims about what *you* find clear and *you* find obvious. === Subject: Re: The list of all natural numbers don't exist >>Much more can be said: for ZFC there is a clear informal conceptual >>picture in which all axioms of ZFC are obviously true. > Surely this is overstating the case, unless \\clear\\ and \\obvious\\ are > claims about what *you* find clear and *you* find obvious. Sure, the clarity or lack thereof is to a large extent a matter of opinion. After all, some people find the usual idea of the structure of natural numbers unclear. As to obviousness, I would guess that even those who think there's something incoherent or unclear about the cumulative hierarchy would grant that if the idea makes sense at all, the usual set theoretical principles follow, with the possible exception of the replacement axiom. I just wanted to point out that unlike often claimed the confidence in the consistency of ZFC - of those who have that confidence - is not really based on \\inductive evidence\\ but on general conceptual considerations. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \\Wovon man nicht sprechen kann, daruber muss man schweigen\\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The list of all natural numbers don't exist > > ... > >> If, at anytime, the program is about to overwrite > >> the \\last\\ 0 (the rest of the list is all 1's), > >> the program will halt and tell us the input tape > >> had a finite number of 0's and cannot contain > >> a list of all natural numbers. > > >> We allow this program to read and > >> update the entire input tape. > >> What is on the output tape? > > These two paragraphs describe two, mutually incompatible, models of > > computation. > > The first relates to a sequential model, in which it is meaningful to > > ask questions about the state of the tape at each of an ordered \\ sequence > > of states of the computer. > > In that model, the computation never finishes. Each time the program \ \\ is > > about to overwrite a zero the set of remaining zeros is infinite. > > The second paragraph relates to an instantaneous mapping type of > > operation, in which one can actually do something for every element \\ of > > an infinite set and finish. In that model, it is not meaningful to \\ talk > > about the \\last\\ anything. > > I think it is quite likely that combining these behaviors in a single > > computation will lead to contradictions. > > To use that to show a contradiction in set theory based e.g. on the \\ ZF > > axioms, you would need to prove that they imply the existence of a > > theoretical computer that can finish processing an infinite input \\ while > > doing things step by step. > > > >> Saying we overwrite all the 0's on the tape \\all at once\\ > >> doesn't mean there wasn't a last 0. > >> It just means we won't know which 0 was the \\last\\ one. > > It does not, in itself, say whether or not there was a last zero. > > If the input is represents a finite set of natural numbers, there is a > > last zero. If it represents an infinite set of natural numbers there > > isn't. > How do you prove this last statement? > Consider the family of sets I described: > A_0 = {0,1,2,...} > A_1 = {1,2,3,...} > A_2 = {2,3,4,...} > ... > Even if we create all members of this set \\all at once\\, > the empty set is not an element in this set. No one has said that it is in that family of sets, but it is the intersection of that family. > How can you say when we update the 0's on the tape > \\all at once\\ the set of 0's on the output tape > is the empty set? How can you say that when we simultaneously change *every* 0 to a 1 we have not changed every 0 into a 1? === Subject: Re: The list of all natural numbers don't exist > > ... > >> Every definition of union and intersection > >> I have seen define these operations as > >> binary operators. > >> Given two sets, we get back one set. > > Then you have never seen the Wikipedia page on the ZF axioms, > > http://en.wikipedia.org/wiki/ZFC#The_axioms: > > \\Union: For any set x, there is a set y such that the elements of y \\ are > > precisely the members of the members of x.\\ > > It might be helpful to give a pointer to the definition of the ZF \\ axioms > > you are using. The ones I've found use the general form in their union > > (or sum set) axiom, not just binary union. > > or on arbitrary intersections, > > \\ http://en.wikipedia.org/wiki/Intersection_(set_theory)#Arbitrary_intersectio\ > > ns: > > \\If M is a nonempty set whose elements are themselves sets, then x is \ \\ an > > element of the intersection of M if and only if for every element A of > > M, x is an element of A.\\ > OK. You define a new operator and call it union. You miss the point. In ZFC and NBG the general union operator is the ONLY union operator, and there is no binary union operator except as a special case of the general union operator. > Obviously, this new operator is not compatible with > binary union because we know binary union can not > be applied to an infinite set. Binary union is only a specialization of general union. If you choose to believe that binary union is not compatible with the only union operator there is, that is your problem. > Have you also defined a new successor operator? Properly, successor(x) = Union({x,{x}}). > I don't see how you can. Successor is a unary operator. > There is no way we can use it other than sequentially. > We have already shown any set formed using > the successor operator is finite. > It should be obvious by now we can never form > an infinite set of natural numbers by applying > the successor function even if we do it an > \\infinite\\ number of times. On the other hand, there is nothing to prohibit existence of sets closed under successorship (in the sense that the successor of every member is also a member), and N, the set of von Neumann naturals in NBG is a valid example of such a set. > How are you forming a set of natural numbers > if you are not using the successor function? By using the NBG axiom of infinity: There is a set which contains the empty set as an element and contains union( {y, {y}} ) for each of its elements y. as explained in http://en.wikipedia.org/wiki/Von_Neumann-Bernays-G\\.9adel_set_theory > How can union form an infinite set > of natural numbers when successor can't? It is not union alone that does it, it also requires the axiom of infinity. > I have to question this definition of union if you > are claiming you can form infinite sets of natural > numbers with it. Question it all you want to, but ZFC and NBG require it. If you think you have an axiom system better than either of these, show it to us. Otherwise shut up. > Russell > - Everything you know is wrong. === Subject: Re: The list of all natural numbers don't exist > ... >> Every definition of union and intersection >> I have seen define these operations as >> binary operators. >> Given two sets, we get back one set. > Then you have never seen the Wikipedia page on the ZF axioms, > http://en.wikipedia.org/wiki/ZFC#The_axioms: > \\Union: For any set x, there is a set y such that the elements of y \\ are > precisely the members of the members of x.\\ > It might be helpful to give a pointer to the definition of the ZF \\ axioms > you are using. The ones I've found use the general form in their union > (or sum set) axiom, not just binary union. > or on arbitrary intersections, > \\ http://en.wikipedia.org/wiki/Intersection_(set_theory)#Arbitrary_intersectio\ n\\ s: > \\If M is a nonempty set whose elements are themselves sets, then x is \\ an > element of the intersection of M if and only if for every element A of > M, x is an element of A.\\ >> OK. You define a new operator and call it union. > I may be older than most graduate students but I'm not THAT old. I > categorically deny having anything to do with defining the general form > of union, especially when it was new. :-) > According to http://en.wikipedia.org/wiki/Zermelo_set_theory, it was > used in Zermelo set theory, published in 1908, so it is at least that > old. > Of course, it seems new to you because you are reading about it for > the first time now. I know how much I relish the \\newness\\ to me of an > idea the first time I learn it, even if it is ancient. Have fun thinking > about it. Cool! I hope I don't have to wait for Zermelo to answer my questions. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > > ... > >> Every definition of union and intersection > >> I have seen define these operations as > >> binary operators. > >> Given two sets, we get back one set. > > Then you have never seen the Wikipedia page on the ZF axioms, > > http://en.wikipedia.org/wiki/ZFC#The_axioms: > > \\Union: For any set x, there is a set y such that the elements of y \ \\ are > > precisely the members of the members of x.\\ > > It might be helpful to give a pointer to the definition of the ZF \\ axioms > > you are using. The ones I've found use the general form in their \\ union > > (or sum set) axiom, not just binary union. > > or on arbitrary intersections, > > \\ http://en.wikipedia.org/wiki/Intersection_(set_theory)#Arbitrary_intersect > > ions: > > \\If M is a nonempty set whose elements are themselves sets, then x is \ \\ an > > element of the intersection of M if and only if for every element A \\ of > > M, x is an element of A.\\ > > >> OK. You define a new operator and call it union. > > I may be older than most graduate students but I'm not THAT old. I > > categorically deny having anything to do with defining the general form > > of union, especially when it was new. :-) > > According to http://en.wikipedia.org/wiki/Zermelo_set_theory, it was > > used in Zermelo set theory, published in 1908, so it is at least that > > old. > > Of course, it seems new to you because you are reading about it for > > the first time now. I know how much I relish the \\newness\\ to me of \ \\ an > > idea the first time I learn it, even if it is ancient. Have fun \\ thinking > > about it. > Cool! > I hope I don't have to wait for Zermelo to answer my questions. Zermelo died in 1953. As Russell seems determined to ignore answers given by the living, perhaps he should join Zermelo to be able to question him directly. === Subject: Re: The list of all natural numbers don't exist > > > On 18 Jun 2006 04:01:52 \ -0700, georgedance04@yahoo.ca > > > > > said: > > > > > > ... > > > > That's what you appear to be arguing in this case: that my \ only \\ option > > > > > > that makes me non-stupid is to believe the word of the \\ mathematical > > > > > > authorities that infinite sets exist objectively, when all they \ \\ can > > > > > > offer me is evidence that the concept is useful to them in their \ \\ work - > > > > > > the same thing the Aztec priest could offer the peasant. > > > > > > > > Your kidding, right? Ok, I'll play along. The Aztec \ priests \\ were > > > > > mistaken about causes. It was a simple blunder, but, lacking a \ \\ decent > > > > > scientific methododology (and lacking any incentive to give up \\ their own > > > > > claims to knowledge and power), it was an understandable error. \ \\ But the > > > > > flaw in their reasoning is easily revealed -- remove the human > > > > > sacrifice, continue good agricultural practices, and lo, the crops \ \\ keep > > > > > a-coming. > > > > Can you substantiate this claim? I have serious doubts it can be > > > > validated by existing data. Consider: What percentage of all the \\ human > > > > beings that provide us with food, that is grow, herd, hunt, fish \\ and > > > > collect our food, TODAY are devout followers of a religion? What > > > > percentage of the food you and your society eat was prayed for by \ \\ the > > > > people who supplied it? With what certainty can you assert that \\ growing > > > > crops has nothing to do with some sort of ritual religious practice? \ \\ I > > > > seriously think evidence is against you. Look at the Soviet Union \ \\ with > > > > their state imposed atheism. They had terrible famines to the \\ degree > > > > that millions died and they always had to resort to importing food \ \\ from > > > > other countries, the USA were one of their main suppliers. > > > > This should be food for thought (pun intended) > > > > Your example implies that the devout followers of a religion, who \ \\ once > > > lived > > > in the USA, were performing human sacrifice. Can you substatiate \\ this > > > claim? I have doubts. > > Excuse me for asking; have you been diagnosed with some adolescence of > > the mind? I ask because I would rather not use stronger language if > > your obvious intellectual impairment is pathological in nature. For now > > I must assume that you are proffering this idiocy intentionally. I > > implied nothing of the sort. I only expressed serious doubts whether it > > can be easily proven that the Aztec Priests were wrong, or more > > generally that worshiping divinities has no influence on the outcome of > > agriculture. > Your counterclaim, that Aztec priests were not mistaken, does not > follow > from your statements. The percentage of farmers who were religious is What counterclaim? I claim nothing regarding wether they were right or wrong. I am only questioning the assertion that it can be easily proven that they were wrong. You should read and make sure you understood before responding. > impossible to determine. When you say the people who supply the food, > are you referring to the workers or the landowners? The Soviet Union > did not impose but tolerated religious activity after WW2. The Soviet Union did not impose religious activity? Well that is really news for most of us here, thank you for telling us. > If the Aztec Priests were right, then a lack of human sacrifice would > cause > problems for agriculture. Which is very easy to verify experimentally, isn't it? Let's just force a million or so people to adopt the Aztec culture, measure the average crops under \\regular\\ conditions and then see what happens when they are prevented from sacrificing humans to their gods. Why hadn't anybody thought of this before? You really are a genious. > If a country was successfully producing food > then it must also be sacrificing. You really lost me here and anyway, what is your point? What are you trying to achieve, how is your prattling related to the discussion. Try to show some coherence. === Subject: Differential form on Riemann Sphere I'm having a lot of trouble with the basics of forms on Riemann sphere. Let X denote the Riemann Sphere, let z be one coordinate in one chart, and w \ \\ = 1/z be the other coordinate in the other chart. Suppose v is a 1-form on \ \\ X. Denote v_z and v_w, the local expression for the 1-form v in the \\ coordinate z and w, respectively. Suppose v_z = f(z)dz in the coordinate z. \ \\ Let v_w = g(w)dw be the 1-form associated to the other chart. I am supposed \ \\ to show that f is a rational function. Since v is a 1-form, the transformation rule says that g(w) = f(T(w))T'(w), \ \\ where T is the change of coordinates between charts. Of course, T(w) = \ 1/w. So, g(w) = - f(1/w)*1/w^2. But I'm not sure where to really go from here. I always come back to the \ \\ fact that any meromorphic function on the Riemann sphere is a rational \\ function. But f is a function on C, the complex plane, not X. Am I missing something? James