mm-3400 === Subject: Re: Number of abelian groups + exact sequence <291120060142470395%plsperry@sc.rr.com > Do you have any ideas for the following problem: >> How many up to isomoporphism abelian groups A are there, which could be >> included in the following exact sequence: 0 -> Z -> A -> Z/6000Z -> 0 ? look at the cohomology.... Why would he do that, when A is supposed to be abelian? A is 2-generated. You can write it as Z^r(+)T, where T is torsion, and > r is nonnegative. You have information about what T and r can be. You > can figure out what A must be. There are a few possibilities, but you > can forget about cohomology and semi-direct products. i am sorry > but i do not understand why cohomology is not applicable > to the abelian case using cohomology > you get pretty immediately the importance of Ext^1(Z, Z/6000Z) and you start forming pullback squares > and computing which factors work the identity element of Ext^1(Z, Z/6000Z) > for instance > gives the extension Z (+) Z/6000Z i thought this was the point > of using baer sums in such calculations > and was the way one atually ends up proving > that one can write it as Z^r (+) T is this all messed up and incoherent? i have what i think is the correct solution to eugene's problem > and i could post it if comparison is needed > (i am only being cryptic to not answer homework) [...] Ext(Z/6000Z, Z) is isomorphic to Z/6000Z but I don't see that that > helps. The problem is that if 0 -> Z -> A -> Z/6000Z -> 0 and > 0 -> Z -> A' -> Z/6000Z -> 0 are representatives of the same element of > Ext(Z/6000Z, Z) then A and A' are isomorphic. But, as far as I know, A > may be the middle group of some inequivalent extension. So, I suspect > there are _not_ 6000 A's. I agree with Arturo and John that the the way to go is to observe that > A is generated by two elements and thus is the sum of two cyclic > groups. i want to explain this a bit more there are not 6000 extensions and you give one of the main methods of calculating equivalency 0 -> Z -> A -> Z/6000Z -> 0 |id |b |id V V V 0 -> Z -> A' -> Z/6000Z -> 0 over distinguished elements of Ext^1(...) basically the cohomologic construction gets you the results that : Z (+) Z/6000Z is one solution : there correspond (possibly equivalent) solutions to each element of Ext^1 ( which i think is pretty similar to what you need to show for the base Z^r (+) T form modulo a regrouping of factors and use of the definition of the abelian (baer) sum over Ext^1) the way you find the inequivalent ones is follow the factor groups and look at the baer sum functor which is what everyone seems to be telling eugene to do just not in my admittedly overly obtuse way i think the calculations end up being the same i just use the language of categories, functors, and cohomology and the factor calculations are done on the morphisms i do understand this doesn't help eugene if he has not already been introduced to this language though i just wanted to point out that the method can be carried through... =) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Z.Irregular Set Theory. A correction. <456d02d0@news2.lightlink.com> <456d2825@news2.lightlink.com so the axiomatic system would be like below: Primitive e Axioms: > 1) Extensionality: AxAy:x=y <-> (Az:zex<->zey) >> That looks fine - two sets are the same if they contain the same members. > 2) Ax (xex) >> This is different from normal set theory, and not necessarily true as I >> see it, but clear. > 3) E!x:x=QAy (xey) >> I don't understand this. What is the implication? 3) simpley states that there exist only one singlton set, and that set > is Q. and that set > is a member of every set. > There is only one set with size=1? And this single member is a member of > all sets, or rather, the set containing this member is a member of all > sets? What does this mean, and what is this member? If there are more > than one basic element, then there are more than one singleton set, > since each element can alone constitute a set. There is only one null set. Q={Q}. so Q is the only singlton set that is in itself. > Being the only singleton already, it's not surprising that it's the only > one that contains itself, No this is wrong, Q is not the only set that contain itself. Avery set in this theory contain itself as a member, not only Q. but Q is the only set which contain itself as the sole member. All other sets also contain themselfs, but not as the sole member(i.e not as the only member), but why does it? What does that mean? While I've taken to trying to express myself as often as possible here > using the logical predicate, sometimes it helps to state things > verbally, on a mind-to-mind level. :) so to say that there exist P={P} were P=/=Q , then this would be a > contradictory to 3). > so there is no other singleton... so x=Q<-> x|Ay:yex -> y=x Uh. I'm trying to respond, but I'm experiencing cognitive dissonance. Q is the single singleton, right? And, x is Q? And that, then, is > equivalent to the statement that, um, x|Ay:yex -> y=x? So, this is the > set such that, if y is an element of x, then y is x. But, there is only > one such set. What is forall y doing in there? What other sets, > besides the single singleton, are you considering? I'm not sure of the > utility of this rule. But, anyway... you mean what does Avery y, doing here. it means every y that is a member of x, then y is x. Avery here mean that there cannot be any y that is a member of x and y not being x itself. > so x={x} <-> x=Q. > okay Also, this axioms tells you that Q is a member of every set y. > yes, it stated that, but I'm not sure of the motivation. so there is no empty set in this theory, and the least membered set, is > that set which > only contain itself as a member and this set is Q. You can say that Q > is the set in this theory which is empty of members other than itself. I was going to ask whether this was an attempt to rid ourselves of the > null set. I could kind of smell it coming. But, the null set, and the > null string, are not objects to be discarded. They complete the power set. there is no null set in this theory since that would contradict the axiom Ax (xex). however the singlton set Q in this theory can be regarded as something comparable to the null set in ZFC, since it is empty of members other than itself. It's true that certain truths are best illustrated by starting at 1, but > 0 is the origin, when it comes down to it. what do you mean 1, I didn't start with one, I started with zero. but zero here is not the null set you know, it is Q. so 3) is simply read as: there exist a unique x such that x=Q, for > every set y -> xey. Yes, this set containing one element is a set of every set, according to > what you've put forth, but what is the element it contains, and how do > you justify this rule intuitively? What does it mean? what is the member of Q? this is a strange question? the member of Q is Q. I think I made this clear. there is some circularity here, or you can call it cyclicality, but that is what Q is all about, it is the circular element, the thing in itself, that defines itself. in reality the axiom of uniquness which states Q as the only singlton set, which is a member of every set. this axiom was called by me as the axiom of circularity ( or cyclicality ). so for intuitive purposes that you asked for, this axiom is actually stating that there is only one thing which defines itself, and that thing is Q. 4) ExAy:y=/=x ( yex<->P(y) ) >> Again. Can you state this in words? OK, there exist x for every set y other than x, such that y in x if and > only if P(y) is true. > I think you used x in at least two ways there. It's a bit hard to > decipher. Perhaps capital letters for sets, like S and T and N and R? > Are there urelements in your theory? Why not, P(y) stands for any formula in one free variable. so P(y)<->a lion. why not? Example let P(y)<->yey accordingly x will be the set of all sets. since yey is true for all > sets in this theory. > Okay, you have a set of all sets. What are its properties? this is for other theorums in this theory to deal with. let P(y)<->y!ey then x={x} -> x=Q. > Then y!=Q, and I suppose, size(y)<>1. Size(y)<>0? Yea of course y!=Q, since y!=x according to this axiom. but you should understand this axiom and understand why y really refer's to, y is not any member of x, y means any member of x other than x, because in this theory you know every set x is in itself, so y means a member of x that is not x.so the axiom of comprehension really states that there exist a set x composed of two types of elements, the first type is x itself. and the other type of elements is y were y is not x, this axioms states that for every y that is in x then it should fullfill a certain predicate P, or in other way y should make the forumla P(y) a true one. and if y do not fulfil P then y is not in x. Now no y can fulfill be not in itself, since every set in this theory is in itself. therefore we are only left with x as the only member of x. sometimes I call x the host member and y the guest member. so as you said y!=Q, since there is not y at all that is not in itself. let P(y)<->y=Q then x={Q,x} where x=/=Q. > Okay, every set is an addition of zero or more elements to this > singleton. What is its single element? what do you mean by this. let us symbolize the above set as z={Q,z} Now let P(y)<-> y=z v y=Q then by 4) x={Q,z,x} were x=/=Q, x=/=z. and so on , we can form many sets in this theory. > Sure, but I'm still wondering what the utility is in eliminating the > null set. :) it is not utility is necessity, since Ax (xex) then there cannot be a null set in this theory.if there is a null set in this theory then there are sets in this theory which will not have themselfs as members, which violates irregularity. this will lead to Russell paradox when I use comprehension 4. and accordingly the main of this theory, that is to tranverse the universal realm would be unachievable. 5) EN: (Ax:xen ->S(x)eN) >> So far, okay, Peano successor... > S(x)={ y|yexVy=x' } >> The successor of x is the set containing x, or the successor to that >> set? I don't get this one. No x' is the complementary set as defined as below. > Oh, okay. Sorry. the successor of x is the set containing all members of x and x and the > complementary set of x. > x'={y|y!exVy=Q} >> What is Q again? x' is the complementary set of x, and it is the set who'es members are > all sets that are not members of x, except Q which a member of x and > x'. so you see S(x)={ y|yexVy=x' } means what I said above and of course > S(x) contains itself. > see axiom 2. > according to this we have the following . Q={Q} , let me symbolize 0 instead of Q. 0={0} > 1={0,0',1} > 2={0,0',1,1',2} > 3={0,0',1,1',2,2',3} > 4={0,0',1,1',2,2',3,3',4} > . > . > . > . > i={0,0',1,1',2,2',3,3',4,4',......,(i-1),(i-1)',i} > . > . > . > . > w={0,0',1,1',2,2',3,3',4,4',......,(i-1),(i-1)',i,i',.............,w} > w+1={0,0',1,1',2,2',3,3',4,4',......,(i-1),(i-1)',i,i',.............,w,w',w+ 1 } > . > . > . > 2w=={0,0',1,1',2,2',3,3',4,4',...,(i-1),(i-1)',i,i',...,w,w',w+1,(w+1)',.... . ..,2w} > . > . > . According to Axiom 5 , N will be the set of all ordinals. and it is a > set in this theory, unlike in ZFC, were the class of all ordinals is > not a set. This theory has a universe as I demonstrated above, while ZFC doesn't > have. > The set of all ordinals here is a set, while it doesn't exist in ZFC. > And accordingly the cardinality of all cardinals exist here, while it > doesn't in ZFC. Zuhair > Oh. I suppose I see where you're going with this. I'm not much into > ordinals and cardinals. I think they're bogus, personally. There are > reals, discrete subdivisions thereof, and extensions into imaginary > levels, and nonstandard realms. But the divorcing of order from > measure, of counting from the uncountable, is artificial. Indeed, there's a universe out there. Each of us contains an infinitude > of points. It's all measurable, given some measure of metric. > Transfinitological hocus pocus shouldn't prove so distracting at this > point. But, I do wish you the best in these axiomatic endeavors. Onward > and upward, Zuhair! Happy spelunking! :) > Zuhair Is their a problem with this symbology . I mean is it not clear, or unreadable as Moe say. > Zuhair Can you state what you want these to say in English? Granted, I have a >> little problem with this form of logical language. I believe that each >> and every axiom should include an explicit implication, -> or <->, >> for clarity. Can you state each one with such an implication? That might >> clarify things. I'm sorry I haven't kept up with your efforts, but it >> looks like you're trying, which is good. Nice day! >> Tony > === Subject: Re: Circumcircle testing in Barycentric Coordinates > assuming that the coordinate system is such that any two of these three > points > (1,-1,0) > (0,1,-1) > (-1,0,1) > have distance 3 between them. Sorry for not replying... took an extended Holiday break. I need to think about what you said [seems I need to use a slight change in my coord system to enforce the sqrt(3), but do-able if I understand right]. barycentric coordinates for the distance calculations. It may be just as effective to jump out to Cartesian for this. Dan === Subject: Re: The class of transitive sets. > Hi all, Is the class of all transitive sets, a set or a proper class? Zuhair A proper class. It contains all the ordinals. which violate separation in ZFC. Please, elaborate. How can existence of a class, therefore, > nonexistence of a set violate separation axiom? JP AyExAz (zex<->zey/P(z). Axiom of separation. Let y= {z|z is a transitive set} Let P(z)<-> z is a transitive set. Accordinglly z will be the set of all ordinals, which is not a set. which violate separation , since there is a P(z) that defines a subclass of y that is not a set. Zuhair === Subject: Re: The class of transitive sets. > Hi all, Is the class of all transitive sets, a set or a proper class? Zuhair A proper class. It contains all the ordinals. which violate separation in ZFC. Please, elaborate. How can existence of a class, therefore, > nonexistence of a set violate separation axiom? JP AyExAz (zex<->zey/P(z). Axiom of separation. Let y= {z|z is a transitive set} Let P(z)<-> z is a transitive set. Accordinglly z will be the set of all ordinals, which is not a set. which violate separation , since there is a P(z) that defines a > subclass of y that is not a set. Zuhair Yes, but only if the class of transitive sets is a set. Were you doing a proof by contradiction? === Subject: Re: The class of transitive sets. > Hi all, Is the class of all transitive sets, a set or a proper class? Zuhair A proper class. It contains all the ordinals. which violate separation in ZFC. Please, elaborate. How can existence of a class, therefore, > nonexistence of a set violate separation axiom? JP AyExAz (zex<->zey/P(z). Axiom of separation. Let y= {z|z is a transitive set} Let P(z)<-> z is a transitive set. Accordinglly z will be the set of all ordinals, which is not a set. which violate separation , since there is a P(z) that defines a > subclass of y that is not a set. Zuhair Yes, but only if the class of transitive sets is a set. Were you doing > a proof by contradiction? Yes. Zuhair === Subject: Re: MI5 Persecution: Gagged by BBC Ariels editor Gagged by BBC Ariel's editor advertising for the BBC's in-house magazine Ariel. > I requested they run an advert BBC Newsreaders Spying on my Home in the > Personal column. They accepted my instructions > and payment, and the following advert did indeed run in Ariel's issue of 8 > July 1997. Landmark on 17 July 1997, requesting they run exactly > the same advert for 10 issues, enclosing payment. Unfortunately, the > advert did not appear in any further issues of Ariel, > because Ariel's editor Robin Reynolds nixed it. Here is an email from > Elaine Smith of Landmark dated 26 August 1997. > === > Subject: Re: advert - ariel With regard to your E-mail of 22 August 1997 concerning your advertisement > in the Ariel magazine. I was instructed by Ariel to remove your advertisement from the classified > pages as they were not comfortable with the contents of the Internet > address you supplied. What was the internet address? Have been on about this since 1997? === Subject: Re: general ratio test , > > [a good amount on a general ratio test] > > I had thought about your question before and came up with the > following. > > Theorem: Suppose a_n is a positive sequence and a_(n+1)/a_n = 1 - > s/n + o(1/n). (a) If s > 1, then for all t in (1,s) there exists > a C_t > 0 such that a_n < C/n^t for all n, hence sum(n=1,oo) a_n > converges. (b) if s = 1, then no conclusion about the convergence > of sum(n=1,oo) a_n can be drawn. However, if a_(n+1)/a_n = 1 - > 1/n + b_n, where sum (b_n)^2 converges, then sum a_n diverges. That last line should have been However, if a_(n+1)/a_n = 1 - 1/n + b_n, where sum |b_n| < oo, then sum a_n diverges. Below I modify the proof of this part of the theorem. > Lemma: 1. For x < 1, log(1-x) <= -x. > 2. 1/n + 1/(n-1) + ... + 1/N > log(n) - Log(N). Here n > N are > positive integers. > > Pf of theorem: (a) Suppose 1 < t < s. The o(1/n) hypothesis > implies there is N such that a_(n+1)/a_n < 1 - t/n, n > N. > Writing a_(n+1)/a_N as a product of ratios, we get > > a_(n+1) < a_N(1-t/n)(1-t/(n-1))...(1-t/N). > > Take logs and use the Lemma to get > > log(a_(n+1)) < log(a_N) + log(1-t/n) + ... + log(1-t/N) > > < log(a_N) - t(1/n + ... + 1/N) > > < log(a_N) - t(log(n) - log(N)). > > Exponentiating, a_(n+1) < a_N*N^t/n^t for n > N, and this proves > (a). > > For (b), we can look at sum 1/(nlog(n)) and sum 1/(n*log(n)^2) (n > 1). The first series diverges, the second converges, > but in each case a_(n+1)/a_n = 1 - 1/n + o(1/n). Here's the modified proof of the last part of (b): Suppose a_(n+1)/a_n = 1 - 1/n + b_n, where sum |b_n| < oo. This time we use log(1-x) = -x + O(x^2) for small x. Much as above (I'll be brief), log(a_(n+1)) = log(a_N)+log(1-(1/n-b_n))+...+log(1-(1/N-b_N)) = log(a_N) - (1/n-b_n) - ... - (1/N-b_N) (1) + O((1/n-b_n)^2 + ... + (1/N-b_N)^2) (2) (1) = log(a_N) - (1/n + ... + 1/N) + (b_n+...+b_N). Because sum b_m converges, (1) > log(1/n) + constant. Sum |b_m| < oo implies sum b_m^2 < oo; therefore sum [1/m - b_m]^2 < oo, hence (2) > constant. Exponentiate to see a_(n+1) > C/n for all n, hence sum a_n diverges. ---------- Note: The proof shows we can relax the assumption sum |b_n| < oo. If we assume the partial sums of sum b_n are bounded below and sum b_n^2 < oo, then everything works. An example of this would be b_n = (-1)^n/[sqrt(n)log(n)]. === Subject: Re: general ratio test [nice proof] However i think i may have found a quicker route: Given that we know if a[n]/a[n+1] = 1 + S/n + O(1/n^p) converges absolutely if S >1, diverges if S< =1 (by bertrands test which compares raabes against n*log(n) terms) , all i need do is follow that argument and then invert the above a[n+1]/a[n] = 1 - S/n + O(1/n^p) (by expanding the 1/(x+1) ) everything before then still applies and we can deduce that if we can write a[n+1]/a[n] = 1 - S/n + O(1/n^p) then the series is absolutely convergent is S < -1 divergent if S >= -1. Of course there are examples where this or any ratio test can fail and your proof puts a higher condition on it. === Subject: Re: general ratio test [nice proof] > However i think i may have found a quicker route: Given that we know if a[n]/a[n+1] = 1 + S/n + O(1/n^p) converges absolutely if S >1, diverges if S< =1 (by bertrands test which > compares raabes against n*log(n) terms) , all i need do is follow that > argument and then invert the above a[n+1]/a[n] = 1 - S/n + O(1/n^p) (by expanding the 1/(x+1) ) > everything before then still applies and we can deduce that if we can > write a[n+1]/a[n] = 1 - S/n + O(1/n^p) then the series is absolutely convergent > is S < -1 divergent if S >= -1. Of course there are examples where this > or any ratio test can fail and your proof puts a higher condition on it. sorry - meant to say makes the condition more explicit. Because essentially the proof via raabes/bertrand states exactly what you have proven explicitly; that for p>1 a[n+1]/a[n] = 1 + S/n + O(1/n^p) diverges if S>= -1 (and of course O(1/n^p) converges!) Your proof is a nice example of why. === Subject: Re: Zen and...Math?? Zen is beyond language and logic actually. >> I agree because the set of Zen-statements are infinite. > I don't think Zen and mathematics compatable at all. > Zen masters would think mathematics a waste of time. I dont agree because the set of math-statements are finite and math and Zen are both truths ... >> What do you mean by a 'math-statement'? Is 1+1 = 2 a math-statement? >> Stephen > 1+1=1+1 is a Zen statement, if that helps ! > BOfL Not really. I want to know if 1+1 = 2 is a 'math-statement'. If it is, then 1+1=1+1 is also a math-statement, and the set of math-statements is infinite. Stephen === Subject: Re: Zen and...Math?? >> Zen is beyond language and logic actually. >> I agree because the set of Zen-statements are infinite. I don't think Zen and mathematics compatable at all. >> Zen masters would think mathematics a waste of time. >> I dont agree because the set of math-statements are finite and > math and Zen are both truths ... > What do you mean by a 'math-statement'? Is 1+1 = 2 a math-statement? > Stephen > 1+1=1+1 is a Zen statement, if that helps ! > BOfL Not really. I want to know if 1+1 = 2 is a 'math-statement'. > If it is, then 1+1=1+1 is also a math-statement, and the set > of math-statements is infinite. The second version uses arithmetic symbols to illustrate there is only one. That anything else is an illusion.The set of math statements is infinite, because it is circular. Read how Pythagoras responded to a student who introduced the suggestion to resolve the square root of minus one. The proceed with caution ;-))) BOfL Stephen === Subject: Introduction to Data Analysis Hi there, I'm a rusty math major who's completed the basic calculus sequence, linear algebra, discrete mathematics, and intro probability. I've also started differential equations three times, but never finished (the demands of being a father and husband, and working for a living). I still have an aptitude, and hopefully it's not waning; but my familiarity with the field is a little soft. I find myself in the position of having finally received a set of data to analyze in the course of my work. The object of the analysis will be to determine which of a few variables for which I have data drives a single dependent variable, and to model that relationship to allow for more accurate predictions. I'm familiar with curve fitting and regression and the like, and for determining correlation strength to identify which of my potential independent variables actually motivates the dependent variable. I'm also certain that I'm working with a dataset which doesn't include some of the information I expect would be useful for developing the best possible model. Alas, isn't that always the way of things? I've been googling and flipping through the texts I have, looking at other texts, and in general trying to inform myself well enough so that I can produce quality work. The optimum solution, of course, would be to get back in the classroom and develop my knowledge; but that's untenable in the near term. So I come to you. Are there any decent primers available to provide direction on how to approach the data? Of the books available, which are recommended for someone with my background? For instance, I've seen Lyman Ott's An Introduction to Statistical Methods and Data Analysis as well as An Introduction to Multivariate Statistical Analysis by T.W. Anderson. And I have Wackerly's Mathematical Statistics with Applications (for the intro probability course noted above). There is also The Data Analysis BriefBook, available in print and online; but it's more encyclopedic, a reference not an instruction. I realize this is a fairly vague, sweeping question. I'm hoping someone can help nonetheless. Daniel === Subject: Re: Introduction to Data Analysis > Hi there, I'm a rusty math major who's completed the basic calculus sequence, > linear algebra, discrete mathematics, and intro probability. I've also > started differential equations three times, but never finished (the > demands of being a father and husband, and working for a living). I > still have an aptitude, and hopefully it's not waning; but my > familiarity with the field is a little soft. I find myself in the position of having finally received a set of data > to analyze in the course of my work. The object of the analysis will > be to determine which of a few variables for which I have data drives a > single dependent variable, and to model that relationship to allow for > more accurate predictions. I'm familiar with curve fitting and > regression and the like, and for determining correlation strength to > identify which of my potential independent variables actually motivates > the dependent variable. I'm also certain that I'm working with a > dataset which doesn't include some of the information I expect would be > useful for developing the best possible model. Alas, isn't that always > the way of things? I've been googling and flipping through the texts I have, looking at > other texts, and in general trying to inform myself well enough so that > I can produce quality work. The optimum solution, of course, would be > to get back in the classroom and develop my knowledge; but that's > untenable in the near term. So I come to you. Are there any decent primers available to provide direction on how to > approach the data? Of the books available, which are recommended for > someone with my background? For instance, I've seen Lyman Ott's An > Introduction to Statistical Methods and Data Analysis as well as An > Introduction to Multivariate Statistical Analysis by T.W. Anderson. > And I have Wackerly's Mathematical Statistics with Applications (for > the intro probability course noted above). There is also The Data > Analysis BriefBook, available in print and online; but it's more > encyclopedic, a reference not an instruction. I realize this is a fairly vague, sweeping question. I'm hoping > someone can help nonetheless. > There is a classic text, Statistical Methods, by G. W. Snedecor and William G. Cochran, which is enormously useful - lots of examples, not a lot of theory, but great insight and sound guidance on methods. Cochran was one of the giants of 20th century stats. And the book is very readable. One thing it lacks is references to statistical computing packages (SAS, SPSS, Stata, etc.). Another gap is info on survival analysis - for that you might want to look at the text by Elisa Lee, which (unlike some others) is at least readable without much math stat background. Others: the SAS manuals are huge (3600+ pp) and comprehensive - also lots of examples, but not very helpful unless you have access to SAS, and not good if you need an overview. Marcus. > Daniel === Subject: Re: Introduction to Data Analysis > There is a classic text, Statistical Methods, by G. W. Snedecor and > William G. Cochran, which is enormously useful - lots of examples, > not a lot of theory, but great insight and sound guidance on methods. > Cochran was one of the giants of 20th century stats. And the book > is very readable. One thing it lacks is references to statistical > computing packages (SAS, SPSS, Stata, etc.). Another gap is info > on survival analysis - for that you might want to look at the text by > Elisa Lee, which (unlike some others) is at least readable without > much math stat background. Others: the SAS manuals are huge (3600+ pp) and comprehensive - > also lots of examples, but not very helpful unless you have access > to SAS, and not good if you need an overview. Marcus. if a local library would have it. As to whether or not it covers computing packages, I'm not too worried about that. I'm primarily concerned with understanding how to approach the problem, and when I figure out what I'm doing, most packages (even Excel) provide enough documentation to deduce their operation. I'd like to use SciLab for the work, just to have a something to use it on, to get familiarity. === Subject: Re: Introduction to Data Analysis > Hi there, I'm a rusty math major who's completed the basic calculus sequence, > linear algebra, discrete mathematics, and intro probability. I've also > started differential equations three times, but never finished (the > demands of being a father and husband, and working for a living). I > still have an aptitude, and hopefully it's not waning; but my > familiarity with the field is a little soft. I find myself in the position of having finally received a set of data > to analyze in the course of my work. The object of the analysis will > be to determine which of a few variables for which I have data drives a > single dependent variable, and to model that relationship to allow for > more accurate predictions. I'm familiar with curve fitting and > regression and the like, and for determining correlation strength to > identify which of my potential independent variables actually motivates > the dependent variable. I'm also certain that I'm working with a > dataset which doesn't include some of the information I expect would be > useful for developing the best possible model. Alas, isn't that always > the way of things? I've been googling and flipping through the texts I have, looking at > other texts, and in general trying to inform myself well enough so that > I can produce quality work. The optimum solution, of course, would be > to get back in the classroom and develop my knowledge; but that's > untenable in the near term. So I come to you. Are there any decent primers available to provide direction on how to > approach the data? Of the books available, which are recommended for > someone with my background? For instance, I've seen Lyman Ott's An > Introduction to Statistical Methods and Data Analysis as well as An > Introduction to Multivariate Statistical Analysis by T.W. Anderson. > And I have Wackerly's Mathematical Statistics with Applications (for > the intro probability course noted above). There is also The Data > Analysis BriefBook, available in print and online; but it's more > encyclopedic, a reference not an instruction. I realize this is a fairly vague, sweeping question. I'm hoping > someone can help nonetheless. Look through your texts for ANOVA (analysis of variance) and see if that fits the bill for your dependency testing. You might also try posting to sci.stat.math. While there are statistical experts who post here and who might see your query, I think there are more over there. - Randy === Subject: Re: Introduction to Data Analysis might be a better fit than sci.math (after the fact, of course). > Look through your texts for ANOVA (analysis of variance) and > see if that fits the bill for your dependency testing. You might also try posting to sci.stat.math. While there are > statistical experts who post here and who might see your > query, I think there are more over there. > > - Randy === Subject: Re: Concavity of the quotient of a monotone and of a convex function? <14001350.1164830099492.JavaMail.jakarta@nitrogen.mathforum.org>, > > can anybody help me with the following problem? > > Let g:R^n -> R be monotone in every argument and let f:R^n -> R be convex. > Further, let the function h:R^n -> R be defined by h(x) = g(x) / f(x). > > Is there a result saying that h is concave / quasi-concave / strictly > quasi-concave or pseudo-concave or the like? On (0,oo), consider x/x^2. If f, g must live on all of R, consider x/(x^2+1) or x/e^x. === Subject: Re: Concavity of the quotient of a monotone and of a convex function? , > <14001350.1164830099492.JavaMail.jakarta@nitrogen.mathforum.org>, > > > can anybody help me with the following problem? > > Let g:R^n -> R be monotone in every argument and let f:R^n -> R be convex. > Further, let the function h:R^n -> R be defined by h(x) = g(x) / f(x). > > Is there a result saying that h is concave / quasi-concave / strictly > quasi-concave or pseudo-concave or the like? > > On (0,oo), consider x/x^2. If f, g must live on all of R, > consider x/(x^2+1) or x/e^x. Or consider exp(2x_1 + ... + 2x_n)/exp(x_1 + ... + x_n) for any n. === Subject: Re: Concavity of the quotient of a monotone and of a convex function? can anybody help me with the following problem? Let g:R^n -> R be monotone in every argument and let f:R^n -> R be >convex. Further, let the function h:R^n -> R be defined by h(x) = g(x) / >f(x). Is there a result saying that h is concave / quasi-concave / strictly >quasi-concave or pseudo-concave or the like? > Hint: try n = 1, g(x) = exp(2 x + sin(x)) ... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Weighted graph partitioning > Hi. I have a large sparse graph. It also has to be directed, and I assume you're not allowing cycles. > It is weighted. I would like to > partition it according to following rules: > - node without parent starts a new partition > - a child node is added to same partition, if weight is > - over some number, depending on child node > - and this partition has max weight on this child node. Any other > partition must have smaller cut over child node. > - example a has b1 and b2, b1 and b2 both have c1 (lets say each 0.3 > cut). x also has c1 (with 0.4). But c1 gets assigned to a and not to x, > since a controls b1 and b2, therefore 0.6. x controls only 0.4 > - This problem has to do with creating groups of related compaines. > Weights represent ownership.And partitions represent controlling of > group of companies. Root of partion is group leaded. Can you point me to some direction. What algorithm do you suggest? Is > this algorithm known for such finantial problems? You could try a greedy algorithm: The first thing I'd do is to sort the vertices, (i.e., find a function f:V(G) -> {1,2,...,n} so that for every directed edge (u,v), f(u) < f(v)). There should be some fairly well-known algorithms for this. Then, going through the vertices in this order (i = 1, 2, ..., n), do the following: (1) If f(i) has no parents, put it in a new partition. (2) Otherwise calculate how much each partition contributes towards f(i), and put f(i) in the partition which contributes the most to it. The only problem is that there usually is more than one possible ordering. Also, because the algorithm is run greedily, you might miss a more optimum solution. (But that's all the time I've got; I need to leave to teach class.) --- Christopher Heckman === Subject: Re: discrete math with difficult party. Which definition of party graph do you have in mind? I think it's > much more interesting to consider the graphs in which each nonadjacent > pair has *exactly* two common neighbors (as the original poster > intended), as such graphs *seem* to be rare and highly symmetric. Are > there any others at all, besides K_1 and C_4 and the 5-regular graph of > order 16 described in my last message? Yes, the graph K_2 also satisfies the conditions. But that was a silly > oversight. A necessary condition for n to be be the order of such a > graph is that n = d(d+1)/2 + 1, where d is not congruent to 3 modulo 4; > the graph is then d-regular. The aforementioned examples are for d = 0, > 1, 2, and 5. Is there a (strong) party graph of order n for n = 11, > or n = 22, or (since the current examples all have order equal to a > power of 2) n = 4096? Actually, the case n = 11 (d = 4) is easily seen to be impossible: > every 4-regular graph of order 11 contains K_3 or K_{2,3}. How did you know that mina meant that there were only 2 strangers in > any group? That is open to interpretation. The agreed-upon interpretation is: For > any three people A, B, and C, we can find two people (among them) who > don't know each other. The condition could also have been interpretted > as: For any three people A, B, C, there is someone who knows the other > two, but the other two are strangers. This other interpretation limits > the party graphs more than the agreed-upon interpretation. --- Christopher Heckman Let me summarize the problem as I see it. There are 631 people at a party; > 1) In any group of three, there are at least 2 people who do not know > each other. It is not necessary for any member of the group to know > any other member of the group. This means that if A, B, C are any people in the party, then at least > one of the following is true: (1) A doesn't know B > (2) A doesn't know C > (3) B doesn't know A 2) If A &B do not know each other, then at exactly two other people know > A & B. Based on these conditions; Arrange the people in a circle so that each person knows only the person > on the right and the person the left. Someone can know more than 2 people. Try again. This automatically satisfies condition 2). Since the graph is triangle free, > condition 1) is also satisfied. Nope. If the graph is a pentagon, then (2) doesn't hold; there is only > one person who knows both A and B. This distinction was not obvious as the condition was worded. Better would be; ....two people know both A & B. Sorry for the misunderstanding! Bill J --- Christopher Heckman This configuration is valid for groups of any size! > > Bill J === Subject: Bilinear form and non-degeneracy question If ( , ) is a non-degnereate bilinear form on a vector space V, and v_i is a basis for V, and I define c_ij := (v_i, v_j), then why is the matrix (c_ij) invertible? What aspect of the non-degeneracy of ( , ) makes it invertible? James === Subject: Re: Bilinear form and non-degeneracy question >If ( , ) is a non-degnereate bilinear form on a vector space V, and v_i is a basis for V, >and I define c_ij := (v_i, v_j), then why is the matrix (c_ij) invertible? What aspect >of the non-degeneracy of ( , ) makes it invertible? I assume it's a finite-dimensional vector space, otherwise you have an infinite matrix, and I don't know how you'd want to define invertibility for those. Hint: if that matrix C is not invertible, there is some nontrivial X in F^n (where F is your field and V is n-dimensional) such that C X = 0. Use that to find v in V such that (w, v) = 0 for all w. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Bilinear form, non-degenerate, question If ( , ) is a non-degnereate bilinear form on a vector space V, and v_i is a basis for V, and I define c_ij := (v_i, v_j), then why is the matrix (c_ij) invertible? What aspect of the non-degeneracy of ( , ) makes it invertible? James === Subject: Lambert W solution? Hi all- Any help to solve for a in the following: (a + 1)^a = y (1) ? I can see how to solve: (1/a + 1)^a = y (2) but not the one above. Googling helps but not for this particular one. Does (1) permit a LambertW solution? Real or complex, any branch doesn't matter. TIA, MAS === Subject: Re: Lambert W solution? > Hi all- Any help to solve for a in the following: (a + 1)^a = y (1) ? I can see how to solve: (1/a + 1)^a = y (2) but not the one above. Googling helps but not for this particular > one. Does (1) permit a LambertW solution? Real or complex, any branch > doesn't matter. I tried a couple things but couldn't get anything to work, then I tried using Maple, and it couldn't solve (1) either. But then again, it can't do (2), either. --- Christopher Heckman === Subject: Re: Lambert W solution? > Hi all- >> Any help to solve for a in the following: >> (a + 1)^a = y (1) >> ? >> I can see how to solve: >> (1/a + 1)^a = y (2) >> but not the one above. Googling helps but not for this particular >> one. >> Does (1) permit a LambertW solution? Real or complex, any branch >> doesn't matter. I tried a couple things but couldn't get anything to work, >then I tried >using Maple, and it couldn't solve (1) either. But then again, >it can't >do (2), either. Maple can do (2): > _EnvAllSolutions:= true: > solve((1/a+1)^a=y,a); / - (-ln(y) I + 2 Pi _Z2~)/| -I (-ln(y) I + 2 Pi _Z2~) -LambertW(_NN1~, -------------------------) I - ln(y) I y + 2 Pi _Z2~| / The point is that with t = 1/a, (2) becomes t + 1 = y^t But (1) does not seem doable with LambertW. Perhaps you could use a series solution for (1): with s = sqrt(ln(y)), a = s + 1/4*s^2 - 1/96*s^3 + 59/92160*s^5 - 1/2880*s^6 + 2783/20643840*s^7 - 1/24192*s^8 + 1060117/118908518400*s^9 - 1/4838400*s^10 + O(s^11) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Lambert W solution? > Hi all- >> Any help to solve for a in the following: >> (a + 1)^a = y (1) >> [...] >> Does (1) permit a LambertW solution? Real or complex, any branch >> doesn't matter. I tried a couple things but couldn't get anything to work, then I tried >using Maple, and it couldn't solve (1) either. But then again, it can't >do (2), either. Maple can do (2): _EnvAllSolutions:= true: Oh. I didn't know that I had to tell Maple to keep all the solutions. (Why would this not be the default?) --- Christopher Heckman > solve((1/a+1)^a=y,a); / > - (-ln(y) I + 2 Pi _Z2~)/| > -I (-ln(y) I + 2 Pi _Z2~) > -LambertW(_NN1~, -------------------------) I - ln(y) I > y > + 2 Pi _Z2~| > / The point is that with t = 1/a, (2) becomes t + 1 = y^t But (1) does not seem doable with LambertW. Perhaps you could use a series solution for (1): with > s = sqrt(ln(y)), a = s + 1/4*s^2 - 1/96*s^3 + 59/92160*s^5 - 1/2880*s^6 > + 2783/20643840*s^7 - 1/24192*s^8 + 1060117/118908518400*s^9 > - 1/4838400*s^10 + O(s^11) Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Lambert W solution? > Hi all- >> Any help to solve for a in the following: >> (a + 1)^a = y (1) >> ? >> I can see how to solve: >> (1/a + 1)^a = y (2) >> but not the one above. Googling helps but not for this particular >> one. >> Does (1) permit a LambertW solution? Real or complex, any branch >> doesn't matter. I tried a couple things but couldn't get anything to work, >then I tried >using Maple, and it couldn't solve (1) either. But then again, >it can't >do (2), either. Maple can do (2): _EnvAllSolutions:= true: > solve((1/a+1)^a=y,a); / > - (-ln(y) I + 2 Pi _Z2~)/| > -I (-ln(y) I + 2 Pi _Z2~) > -LambertW(_NN1~, -------------------------) I - ln(y) I > y > + 2 Pi _Z2~| > / The point is that with t = 1/a, (2) becomes t + 1 = y^t But (1) does not seem doable with LambertW. Perhaps you could use a series solution for (1): with > s = sqrt(ln(y)), a = s + 1/4*s^2 - 1/96*s^3 + 59/92160*s^5 - 1/2880*s^6 > + 2783/20643840*s^7 - 1/24192*s^8 + 1060117/118908518400*s^9 > - 1/4838400*s^10 + O(s^11) Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Yeah- that's what I was thinking. If we write: (a+1)^a = y as z^(z-1) where z=a+1, then we have: (z^z) / z = y or z lnz = lny +lnz or z lnz -lnz = lny or ... It's curious: ze^z + e^z = y is solvable, but: z lnz + lnz = y is not Oh well, worth a shot. Maybe some closed form in terms of HLW solutions or something, but the series soln's are sufficient. -MAS === Subject: Re: Lambert W solution? > Oh well, worth a shot. Maybe some closed form in terms of HLW > solutions or something, but the series soln's are sufficient. If by HLW you mean the generalized Hyper-Lambert functions, then yes, they indeed can solve it, as follows: (a+1)^a = y => a*(a+1)^a/a = y => a*exp(log((a+1)^a/a)) = y => a = HLW(log((x+1)^x/x),y) => a = HLW(x*log(x+1) - log(x),y) The equation can also be solved using a different HLW function: (a+1)*(a+1)^(a-1) = y => (a+1)*exp(log((a+1)^(a-1))) = y => a+1 = HLW(log((x+1)^(x-1)),y) => a = HLW((x-1)*log(x+1),y) - 1 > -MAS -- Ioannis ------- The best way to predict reality, is to know exactly what you DON'T want. === Subject: Re: Lambert W solution? > Hi all- Any help to solve for a in the following: (a + 1)^a = y (1) ? I can see how to solve: (1/a + 1)^a = y (2) but not the one above. Googling helps but not for this particular > one. Does (1) permit a LambertW solution? Real or complex, any branch > doesn't matter. I tried a couple things but couldn't get anything to work, then I tried > using Maple, and it couldn't solve (1) either. But then again, it can't > do (2), either. > > --- Christopher Heckman MAS === Subject: ugly limit help -- trinomial distn Can anyone help me evaluate a limit? let LambdaX, LambdaY in (0, infty) Let Px = LambdaX / n; Py = LambdaY/n Let x,y nonnegative integers I want to evaluate the limit as n goes to infinity of the trinomial density f(x,y, n-(x+y); Px, Py, 1-(Px+Py)) That is, f(x1,x2,x3) = [n!/(x1! x2! x3!)] * P1^x1 * P2^x2 * P3^x3 Where P1 = probability of event 1, etc. Simple algebra gives: f(x,y, n-(x+y)) = [ LambdaX^x / x!][ LambdaY^y / y!][ 1 / n^(x+y)] [ n! / (n-(x+y))!][ 1 - Lambdax/n - lambdaY/n]^(n-x-y) A little more algebra allows cancellation of the n^-(x+y) and leaves: f(x,y, n-(x+y)) = [ LambdaX^x / x!][ LambdaY^y / y!][ n! / (n-(x+y))!][ (n- LambdaX - LambdaY)^n-(x+y) ][ n^-n] But, well, I can't seem to satisfy myself as to what that limit ought to be. Ideally it would have marginal distributions of Poisson type. There are pieces there that go to zero (n! / n^n) and others that look a lot like the limit definition of exponent, but I can't solve it. Any help is greatly appreciated. === Subject: Re: ugly limit help -- trinomial distn > Can anyone help me evaluate a limit? > > let LambdaX, LambdaY in (0, infty) > Let Px = LambdaX / n; Py = LambdaY/n > Let x,y nonnegative integers > > I want to evaluate the limit as n goes to infinity of the trinomial > density f(x,y, n-(x+y); Px, Py, 1-(Px+Py)) > > That is, f(x1,x2,x3) = [n!/(x1! x2! x3!)] * P1^x1 * P2^x2 * P3^x3 > Where P1 = probability of event 1, etc. > > > Simple algebra gives: > f(x,y, n-(x+y)) = [ LambdaX^x / x!][ LambdaY^y / y!][ 1 / n^(x+y)] [ n! > / (n-(x+y))!][ 1 - Lambdax/n - lambdaY/n]^(n-x-y) > > A little more algebra allows cancellation of the n^-(x+y) and leaves: > f(x,y, n-(x+y)) = [ LambdaX^x / x!][ LambdaY^y / y!][ n! / (n-(x+y))!][ > (n- LambdaX - LambdaY)^n-(x+y) ][ n^-n] > > > But, well, I can't seem to satisfy myself as to what that limit ought > to be. Ideally it would have marginal distributions of Poisson type. > There are pieces there that go to zero (n! / n^n) and others that look > a lot like the limit definition of exponent, but I can't solve it. Any > help is greatly appreciated. > Using different notation, you're looking at n!/[x!y!(n-(x+y))!]*(a/n)^x*(b/n)^y*(1-(a+b)/n)^(n-(x+y)). We're fixing x, y and letting n -> oo, right? Using Stirling, I'm getting what I think you want to see: exp(-(a+b)) a^x b^y / (x!y!) === Subject: Re: AC is inconsistent? the second-order language of set theory can be proved in NBG. But to get from there to the fact that it actually can be proved in > NBG, you need the 1-consistency of ZF. Yes. But obviously the information that some statement in the > second-order language of set theory can be proved in NBG is contained > in ZFC in the same sense that, say, the information about > uncountability of reals is. Seriously, though, for all mathematical purposes ZFC and NBG are > interchangeable. Presumably because we're only interested in what statements in the > first-order language of set theory they can prove. But in the context > of category theory we might encounter some statements in the > second-order language of set theory which can be proved in NBG but > can't even be stated in ZF. Right. However, any such result can be recast as a result about > formulas defining the relevant proper classes. > Is that a theorem? What about a pi^1_2 sentence? > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Closed form for a PDE? I am dealing with the following system of equations: x(t)=a(t)*x(0)+b(t)*x(1) dy(t)/dt + df(x(t),y(t))/dy = 0 (for any t from 0 to 1) Under the following constraints: f(x(0),y(0)) = f(x(1),y(1)) = 0 df(x(t),y(t))/dy@{y=y(0),x=x(0)} = 0 df(x(t),y(t))/dy@{y=y(1),x=x(1)} = 0 a(0)=1, a(1)=0 b(0)=0, b(1)=1 Everything happens in the time interval 0 to 1, and out of this interval is of no interest. Given functions are: a(.), b(.), e(.,.) Unknowns are: x(1) and consequently x(t) and hence df(x(t),y(t)) are the unknowns. We are only interested in all or at least some solutions of x(1) with respect to the given functions. If it cannot be solved, can we at least talk about some properties that x(1) must have to be a solution for this system? H.M. === Subject: Re: Closed form for a PDE? I am dealing with the following system of equations: x(t)=a(t)*x(0)+b(t)*x(1) >dy(t)/dt + df(x(t),y(t))/dy = 0 (for any t from 0 to 1) Under the following constraints: f(x(0),y(0)) = f(x(1),y(1)) = 0 >df(x(t),y(t))/dy@{y=y(0),x=x(0)} = 0 >df(x(t),y(t))/dy@{y=y(1),x=x(1)} = 0 >a(0)=1, a(1)=0 >b(0)=0, b(1)=1 Everything happens in the time interval 0 to 1, and out of this >interval is of no interest. Given functions are: >a(.), b(.), e(.,.) [ later noted that e should be f; also x(0) and w(0) (which I assume means y(0)) are known. ] >Unknowns are: >x(1) and consequently x(t) and hence df(x(t),y(t)) are the unknowns. And y(t), I suppose? >We are only interested in all or at least some solutions of x(1) with >respect to the given functions. If it cannot be solved, can we >at least >talk about some properties that x(1) must have to be a solution for >this system? x(1) is a single parameter. The differential equation part is a first order ODE (not PDE) for y(t). Assuming everything is analytic and y doesn't go off to infinity in the interval [0,1], the initial value problem y'(t) + f_2(a(t) x(0) + b(t) x(1), y(t)) = 0 [ where f_2 is the partial derivative with respect to the second variable ] y(0) given has a solution, and y(1) becomes an analytic function of the parameter x(1). Only in rather special cases will there be a closed form solution. But then you seem to have two requirements on y(1): f(x(1), y(1)) = 0 f_2(x(1), y(1)) = 0 In general, we wouldn't expect two equations with only one unknown to have a solution. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Closed form for a PDE? By e(.,.) I meant f(.,.) I am dealing with the following system of equations: x(t)=a(t)*x(0)+b(t)*x(1) > dy(t)/dt + df(x(t),y(t))/dy = 0 (for any t from 0 to 1) Under the following constraints: f(x(0),y(0)) = f(x(1),y(1)) = 0 > df(x(t),y(t))/dy@{y=y(0),x=x(0)} = 0 > df(x(t),y(t))/dy@{y=y(1),x=x(1)} = 0 > a(0)=1, a(1)=0 > b(0)=0, b(1)=1 Everything happens in the time interval 0 to 1, and out of this > interval is of no interest. Given functions are: > a(.), b(.), e(.,.) Unknowns are: > x(1) and consequently x(t) and hence df(x(t),y(t)) are the unknowns. We are only interested in all or at least some solutions of x(1) with > respect to the given functions. If it cannot be solved, can we at least > talk about some properties that x(1) must have to be a solution for > this system? > > > H.M. === Subject: Re: Closed form for a PDE? Sorry, I forgot one more thing... x(0) and w(0) are given too. > By e(.,.) I meant f(.,.) I am dealing with the following system of equations: x(t)=a(t)*x(0)+b(t)*x(1) > dy(t)/dt + df(x(t),y(t))/dy = 0 (for any t from 0 to 1) Under the following constraints: f(x(0),y(0)) = f(x(1),y(1)) = 0 > df(x(t),y(t))/dy@{y=y(0),x=x(0)} = 0 > df(x(t),y(t))/dy@{y=y(1),x=x(1)} = 0 > a(0)=1, a(1)=0 > b(0)=0, b(1)=1 Everything happens in the time interval 0 to 1, and out of this > interval is of no interest. Given functions are: > a(.), b(.), e(.,.) Unknowns are: > x(1) and consequently x(t) and hence df(x(t),y(t)) are the unknowns. We are only interested in all or at least some solutions of x(1) with > respect to the given functions. If it cannot be solved, can we at least > talk about some properties that x(1) must have to be a solution for > this system? > > > H.M. === Subject: Re: Need German equivalent of crank > Any native German speakers in the house? I'm looking for the German > equivalent, if there is one, of crank, in the usual sci.math sense of > that word. Cuckoo. Since the Swiss invented the whole subject, not Americans or Germans, Since that's where William Tell stories came from. > http://en.wikipedia.org/wiki/Talk:Crank_%28person%29 > Crankosity, despite its great importance and worldwide prevalence, is > treated at Wikipedia only in English, Danish, and Chinese. Now this > room is full of experts on the subject, from all over the globe. Get to > work, you guys. === Subject: Re: Need German equivalent of crank Can you please explain? I can see that cuckoo relates to kooky which is referred to as derived from cuckoo http://en.wikipedia.org/wiki/Crank_(person). But what the Swiss have to do with it - other than cuckoo clocks - and the well known line in The Third Man. Nick > Any native German speakers in the house? I'm looking for the German >> equivalent, if there is one, of crank, in the usual sci.math sense of >> that word. Cuckoo. Since the Swiss invented > the whole subject, not Americans or Germans, > Since that's where William Tell stories came from. >> http://en.wikipedia.org/wiki/Talk:Crank_%28person%29 >> Crankosity, despite its great importance and worldwide prevalence, is >> treated at Wikipedia only in English, Danish, and Chinese. Now this >> room is full of experts on the subject, from all over the globe. Get to >> work, you guys. > === Subject: Re: Need German equivalent of crank referred to as derived from cuckoo > http://en.wikipedia.org/wiki/Crank_(person). But what the Swiss have to do with it - other than cuckoo clocks - and the > well known line in The Third Man. Because the Swiss invented precision timing, Which has nothing inherently to do with either ptrcision or timing, Since it's eniirely included in what is now called Chain reactions and Chaos Theory, Nick >> Any native German speakers in the house? I'm looking for the German >> equivalent, if there is one, of crank, in the usual sci.math sense of >> that word. Cuckoo. Since the Swiss invented > the whole subject, not Americans or Germans, > Since that's where William Tell stories came from. >> http://en.wikipedia.org/wiki/Talk:Crank_%28person%29 >> Crankosity, despite its great importance and worldwide prevalence, is >> treated at Wikipedia only in English, Danish, and Chinese. Now this >> room is full of experts on the subject, from all over the globe. Get to >> work, you guys. > === Subject: Re: Need German equivalent of crank > Any native German speakers in the house? I'm looking for the German > equivalent, if there is one, of crank, in the usual sci.math sense > of > that word. > http://en.wikipedia.org/wiki/Talk:Crank_%28person%29 > Crankosity, despite its great importance and worldwide prevalence, is > treated at Wikipedia only in English, Danish, and Chinese. Now this > room is full of experts on the subject, from all over the globe. Get > to > work, you guys. Observe that the Danish page doesn't actually give a Danish term. I > think the term is only established in English, like many other > internet-related terms. preserved on Usenet to use the word crank with the derogatory meaning > dates from August 1982 (*). However, it even features in my 1964 > English-Dutch dictionary. Not as derogatory as nowadays, but certainly > with derogatory connotations. According to http://en.wikipedia.org/wiki/Crank_(person) Etymology The term crank in general refers to something bent or twisted, as with mechanical devices that are called cranks. In 1906, Nature offered essentially the same definition which is used here: A crank is defined as a man who cannot be turned. ÖNature, 8 Nov 1906, 25/2 Going further back http://www.answers.com/topic/crank-person-old Etymology The word crank can be traced back to 1833, where the Oxford English Dictionary describes it as A person with a mental twist; one who is apt to take up eccentric notions or impracticable projects, probably related to An eccentric notion or action; a mental twist put into practice. An item in Nature magazine in 1906 noted that A crank is defined as a man who cannot be turned. These men are all cranks. The reference for the OED is given as Oxford English Dictionary, 2nd edition (1989) Nick PS A crank might be defined by some people as someone who spends half an hour on Google trying to find information to respond to this and similar questions. === Subject: Re: Need German equivalent of crank > Oh, the things sci.math can do! >> i wonder what the turkish equivalent of troll is... It seems to derive more from the verb troll than the noun troll. > Trolling is a fishing term (pulling a baited line through water, > hoping that a fish will bite the bait). Back to the main topic, though ... Eccentric/whimsical doesn't > quite capture the nature of crankness. Eccentric/whimsical is a nice > kind of insane; crankdom is a psychotic type of insane. Clearly there is a very imperceptible boundary between brilliance and madness. And, I believe that some contributions demonstrate that quite well. Believing that you are right and everyone else is wrong, despite evidence to the contrary. Nick === Subject: Re: Need German equivalent of crank > Any native German speakers in the house? I'm looking for the German >> equivalent, if there is one, of crank, in the usual sci.math sense of >> that word. >> http://en.wikipedia.org/wiki/Talk:Crank_%28person%29 >> Crankosity, despite its great importance and worldwide prevalence, is >> treated at Wikipedia only in English, Danish, and Chinese. Now this >> room is full of experts on the subject, from all over the globe. Get to >> work, you guys. >> Der Sonderling - see http://www.dict.cc/german-english/Sonderling.html >> crank >> weirdo >> odd chap >> original >> character >> odd customer >> queer fellow >> singular man >> nerd [sl.] >> odd character >> an odd person >> queer customer >> strange character I don't think you can use the terms > queer fellow or queer customer as synonyms for crank > these days. The OP had a particular context in mind where someone > was quirky but not necessarily away with the fairies > in other words Ein Monomane mit einer Mathemacke > My preferred online dictionary http://dict.leo.org gives the meanings of der Sonderling as: crank eccentric nerd (slang) queer fellow This doesn't mean that these words are synonymous in English. Indeed the delight of translating from one language to another is choosing the word that appears to suit the situation. And if it was as easy as simple as looking up a dictionary online or otherwise linguists would be out of a job. Linguistics might incorporate some aspects of mathematics - but not for nothing is the study of languages an art and not a science. Nick === Subject: double integral of exp^max(b^2 x^2, a^2 y^2) dy dx Please help me with this question, Question : .bc .bc exp ^max(b^2 x^2, a^2 y^2) dy dx where b^2 x^2, a^2 y^2 represent the rectangle. The limits of outer .bc is 0~a The limits of inner .bc is 0~b ---------------------------------------------------------------------------- - === Subject: Re: double integral of exp^max(b^2 x^2, a^2 y^2) dy dx > Please help me with this question, > Question : ? ? exp ^max(b^2 x^2, a^2 y^2) dy dx where b^2 x^2, a^2 y^2 represent the rectangle. The limits of outer ? is 0~a > The limits of inner ? is 0~b > ---------------------------------------------------------------------------- - I assume the ?'s are supposed to be integrals, so that you have: int(int(e^max(b^2 x^2, a^2 y^2), y=0..b),x=0..a). A natural thing to do is to split the rectangle in two pieces, one where b^2 x^2 > a^2 y^2, and another where the reverse inequality is true. For the part (R1) where b^2 x^2 > a^2 y^2, which implies b x > a y, so R1 consists of the points in the rectangle with the additional constraint that y < b/a x. Also, max(b^2 x^2, a^2 y^2) = b^2 x^2 here, so the integral over R1 becomes int(int(e^(b^2 x^2), y=0..(b/a x)), x=0..a), and you should be able to integrate this, since the first integral you do is with respect to y. For the other part (R2), you will want to change the order of integration to dx dy. --- Christopher Heckman === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com I do not think that James has ever said he is looking for an analytic > function. He is looking for a perfect continuous function. It seems > likely that James has only a vague idea what he means by that phrase. In order for James to achieve the level of fame, wealth and babes that > he is seeking, he will have to come up with a continuous analytic > function that parallels for the prime number function pi() the > relationship that the gamma function has to the factorial function on > integers. So where gamma(n) = (n -1)! wjhen n is a positive integer, James' new > function (I will call it jamma in his honor) should be defined, > continuous, and differentiable over all real numbers (or even better, > all complex numbers) and have a form similar to jamma(p) = f(pi(p)) when evaluated at any prime positive integer p, where f() is a simple > algebraic function on pi(). Perhaps it should be called bernie in honor of Georg Friedrich > Bernhard Riemann. He expressed pi() as an expression involving (among > other things) the zeroes of the Riemann zeta function. And /that/ is > why the zeta function is relevant to the distribution of primes. I wouldn't care except my fear is that Riemann failed, otherwise hey call it bernie. As for the rest of what you said, yup, but that's obvious if you read through a decent translation of Riemann's own notes. Heck, it's obvious if you know a little German, just by looking at his own words. > John Derbyshire, /Prime/ /Obsession/ , explains what Riemann's formula > says. Derbyshire's heroic achievement is to make sense of it to a > reader with no more than an imperfect grasp of high school math. > Harold Edwards's book on the Riemann zeta function takes a more > advanced reader through the proof. > I highly recommended this post as one of the few intelligent responses in this entire thread. I don't personally recommend any of the pop books on the subject though, as I think it makes more sense just to read through a translation of Riemann, unless you know German, and then you should just read what he himself said. Much of it isn't even all that complicated despite the posturing that posters continually do over the Riemann hypothesis, where probably they do so because they think it makes them seem more intelligent to do so. I recommend readers go to the source. === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com I do not think that James has ever said he is looking for an analytic > function. He is looking for a perfect continuous function. It seems > likely that James has only a vague idea what he means by that phrase. In order for James to achieve the level of fame, wealth and babes that > he is seeking, he will have to come up with a continuous analytic > function that parallels for the prime number function pi() the > relationship that the gamma function has to the factorial function on > integers. So where gamma(n) = (n -1)! wjhen n is a positive integer, James' new > function (I will call it jamma in his honor) should be defined, > continuous, and differentiable over all real numbers (or even better, > all complex numbers) and have a form similar to jamma(p) = f(pi(p)) when evaluated at any prime positive integer p, where f() is a simple > algebraic function on pi(). Perhaps it should be called bernie in honor of Georg Friedrich > Bernhard Riemann. He expressed pi() as an expression involving (among > other things) the zeroes of the Riemann zeta function. And /that/ is > why the zeta function is relevant to the distribution of primes. I wouldn't care except my fear is that Riemann failed, otherwise hey > call it bernie. As for the rest of what you said, yup, but that's obvious if you read > through a decent translation of Riemann's own notes. Heck, it's obvious if you know a little German, just by looking at his > own words. John Derbyshire, /Prime/ /Obsession/ , explains what Riemann's formula > says. Derbyshire's heroic achievement is to make sense of it to a > reader with no more than an imperfect grasp of high school math. > Harold Edwards's book on the Riemann zeta function takes a more > advanced reader through the proof. > I highly recommended this post as one of the few intelligent responses > in this entire thread. I don't personally recommend any of the pop books on the subject > though, as I think it makes more sense just to read through a > translation of Riemann, unless you know German, and then you should > just read what he himself said. Much of it isn't even all that complicated despite the posturing that > posters continually do over the Riemann hypothesis, where probably they > do so because they think it makes them seem more intelligent to do so. I recommend readers go to the source. > Can you clarify your position on the use of complex numbers? === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com I do not think that James has ever said he is looking for an analytic > function. He is looking for a perfect continuous function. It seems > likely that James has only a vague idea what he means by that phrase. In order for James to achieve the level of fame, wealth and babes that > he is seeking, he will have to come up with a continuous analytic > function that parallels for the prime number function pi() the > relationship that the gamma function has to the factorial function on > integers. So where gamma(n) = (n -1)! wjhen n is a positive integer, James' new > function (I will call it jamma in his honor) should be defined, > continuous, and differentiable over all real numbers (or even better, > all complex numbers) and have a form similar to jamma(p) = f(pi(p)) when evaluated at any prime positive integer p, where f() is a simple > algebraic function on pi(). Perhaps it should be called bernie in honor of Georg Friedrich > Bernhard Riemann. He expressed pi() as an expression involving (among > other things) the zeroes of the Riemann zeta function. And /that/ is > why the zeta function is relevant to the distribution of primes. I wouldn't care except my fear is that Riemann failed, otherwise hey > call it bernie. > Can you explain just what it is that you fear Riemann failed at? - William Hughes === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com I do not think that James has ever said he is looking for an analytic > function. He is looking for a perfect continuous function. It seems > likely that James has only a vague idea what he means by that phrase. In order for James to achieve the level of fame, wealth and babes that > he is seeking, he will have to come up with a continuous analytic > function that parallels for the prime number function pi() the > relationship that the gamma function has to the factorial function on > integers. So where gamma(n) = (n -1)! wjhen n is a positive integer, James' new > function (I will call it jamma in his honor) should be defined, > continuous, and differentiable over all real numbers (or even better, > all complex numbers) and have a form similar to jamma(p) = f(pi(p)) when evaluated at any prime positive integer p, where f() is a simple > algebraic function on pi(). Perhaps it should be called bernie in honor of Georg Friedrich > Bernhard Riemann. He expressed pi() as an expression involving (among > other things) the zeroes of the Riemann zeta function. And /that/ is > why the zeta function is relevant to the distribution of primes. I wouldn't care except my fear is that Riemann failed, otherwise hey > call it bernie. > Can you explain just what it is that you fear Riemann failed at? - William Hughes You have to understand the mystery and why there is a big deal about the prime distribution and conjectures like Riemann's related to it, as there is this odd thing that you can get functions like x/ln(x) near pi(x)--but not exact. So you have the prime number theorem, but why? Why the gap? Well, I found that you can define the prime counting function multi-variably with a partial difference equation--eureka!!! BUT for it to be exact you have to have a special constraint, where you have P(x,y) and sum your partial difference equation with a special constraint that if y>sqrt(x) you reset back to sqrt(x). So there's this weird thing that even with a difference equation, if you don't do this special thing, what do you get? A gap. More enterprising young mathematicians might explore the size of that gap with the difference equation, while I would guess that it is roughly the size of what is shown in the prime number theorem. continuous function, which shows--almost shockingly--that it wasn't about discrete or continuous, but about a special constraint--that is, special rules. Wim Bethem talked about integrating the partial differential equation with the same constraint, and it seems reasonable at this point to conjecture that doing so will finally give a function that maps perfectly to the prime distribution at discrete points while being continuous. BUT it cannot be the Riemann function or anything like it--unless his research has something that embodies the special rule. Looking over it, last time I did was a few years ago, but I doubt anything has changed, you find his work does not, so his function cannot match the prime distribution perfectly. And then you have a back door to evaluating the Riemann Hypothesis, which is to prove that the continuous function that follows from integrating the partial differential equation that follows from my prime counting function with the special constraint gives you the prime distribution perfectly at discrete points. So then no need for the prime number theorem as you have a perfect function and the reason for why it wasn't perfect before. And you have a way to disprove the Riemann Hypothesis and get a million dollars from those Clay people JUST by integrating the partial differential equation that follows from my research. So for years now I've known that Riemann probably failed. But I could see that easily enough just by glancing over his actual notes. He does some sketchy things but he was brainstorming. Not his fault that some people grabbed a hold of his speculations and went wild with them, so I don't fault Riemann. === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com But I could see that easily enough just by glancing over his actual > notes. He does some sketchy things but he was brainstorming. > No, he was actually thinking... > Not his fault that some people grabbed a hold of his speculations and > went wild with them, so I don't fault Riemann. How very big of you. I'm sure Riemann would be relieved to hear this. J === Subject: Re: JSH: Under review [jstevh@msn.com] > You have to understand the mystery and why there is a big deal about > the prime distribution and conjectures like Riemann's related to it, as > there is this odd thing that you can get functions like x/ln(x) near > pi(x)--but not exact. So you have the prime number theorem, but why? Why the gap? Well, I found that you can define the prime counting function > multi-variably with a partial difference equation--eureka!!! BUT for > it to be exact you have to have a special constraint, where you have > P(x,y) and sum your partial difference equation with a special > constraint that if y>sqrt(x) you reset back to sqrt(x). So there's this weird thing that even with a difference equation, if > you don't do this special thing, what do you get? A gap. Not true. For your spelling of your difference equation, it's good enough to force y = min(x, y) on entry. It still computes pi() correctly then. This is easiest to prove by deriving your formula from Legendre's. It's prettier and easier all around to consider a related formula that leaves its arguments alone and doesn't use square roots anywhere: Q(x, y) = floor(x) - 1 - sum for k=2 to y of [(Q(x/k, k-1) - Q(k-1, k-1)) * (Q(k, k-1) - Q(k-1, k-1))] Then no special constraints of any kind are needed to ensure the following for natural x and y arguments: Q(x, y) = pi(x) provided 1 <= sqrt(x) <= y <= x [1] /and/ the perhaps surprising: Q(x, y) = pi(y) provided 1 <= x <= y [2] The last is especially interesting, since it implies: Q(1, y) = pi(y) for all y >= 1 [3'] Although not implied by [2], it so happens that [3'] can be strengthened to (because Q(1, i) = 0 for all integer i <= 0): Q(1, y) = pi(y) for all integer y [3] For example, Q(1, 1000) = pi(1000) = 168. Also true that: 168 = Q(2, 1000) = Q(3, 1000) = ... = Q(999, 1000) = Q(1000, 1000) and 168 = Q(1000, 31) = Q(1000, 32) = ... = Q(1000, 998) = Q(1000, 999) Q(1000, i) for i < 31 computes the same result as your P(1000, i). In fact, with some possible minor endcase exceptions, Q(x, y) = your P(x, y) whenever pi(y) <= pi(x) (there is no analogue to the pretty [2] for your P(x, y) when pi(y) > pi(x), /because/ you artificially force y <= sqrt(x)). Alas, to understand how to derive Q, you'd first need to understand how to derive your messier formula from Legendre's. You can then get Q by removing logically unnecessary steps from that derivation. Here's a complete Python implementation, using a memo so it doesn't take forever to run: d = {} def Q(x, y): key = x, y if key in d: return d[key] result = int(x)-1 - sum((Q(x/k, k-1) - Q(k-1, k-1)) * (Q(k, k-1) - Q(k-1, k-1)) for k in xrange(2, y+1)) d[key] = result return result You can use that to verify the examples above. Proof is easiest, again, by deriving Q from Legendre's recurrence. There are no special constraints in Legendre's recurrence, so if you don't go out of your way to add constraints, there's no need for constraints in stuff derived from it either. > More enterprising young mathematicians might explore the size of that > gap with the difference equation, while I would guess that it is > roughly the size of what is shown in the prime number theorem. continuous function, which shows--almost shockingly--that it wasn't > about discrete or continuous, but about a special constraint--that is, > special rules. As above, the gap you perceive was an artifact of /introducing/ needless special constraints. I don't know whether you introduced them to begin with because you were looking for speed, or because they happened to worm around bugs in your early implementations, but as the Q function here shows it's actually easier to define a similar formula without constraints and end up with no gaps. Of course Q is also a minor respelling of Legendre's recurrence, and I don't claim it's of any importance whatsoever. Although, as an exercise in extreme obfuscation, it has a lot going for it ;-) > ... === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com> <6eidnc78ffAyLPPYnZ2dnUVZ_oqdnZ2d@comcast.com Q(x, y) = floor(x) - 1 - > sum for k=2 to y of [(Q(x/k, k-1) - Q(k-1, k-1)) * > (Q(k, k-1) - Q(k-1, k-1))] > === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com> <6eidnc78ffAyLPPYnZ2dnUVZ_oqdnZ2d@comcast.com Q(x, y) = floor(x) - 1 - > sum for k=2 to y of [(Q(x/k, k-1) - Q(k-1, k-1)) * > (Q(k, k-1) - Q(k-1, k-1))] > === Subject: Re: JSH: Under review [Tim Peters] >> ... >> Q(x, y) = floor(x) - 1 - >> sum for k=2 to y of [(Q(x/k, k-1) - Q(k-1, k-1)) * >> (Q(k, k-1) - Q(k-1, k-1))] [willo_thewisp@hotmail.com] You're welcome! I had the advantage of doing some grad work in recursive function theory some decades ago, and if you're attracted to dense formulas that accomplish surprising things with trivial arithmetic, that field is a rich source of awesome entertainment. One of my favorites is related to Ackermann's function. I derived it when writing compiler recursion stress tests a long time ago, although I'm sure someone else came up with it earlier. Here i, j and k are naturals (all integers >= 0): T(i, j, k) = j+1 if k=0 i if j=0 and k=1 0 if j=0 and k=2 1 if j=0 and k>2 T(i, T(i, j-1, k), k-1) otherwise (k>0 and j>0) Obviously, T(i, j, 0) = successor(j) = j+1. Less obviously, T(i, j, 1) = i+j, as can be seen by noting that for k=1, the last expression is essentially the recursive definition of addition in terms of the successor function (i+j = successor(i+(j-1))). Hint: that was a big hint :-) For points, what are better-known names for T(i, j, 2) and T(i, j, 3)? For bonus points, T(i, j, 4)? === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com> <6eidnc78ffAyLPPYnZ2dnUVZ_oqdnZ2d@comcast.com> > ... >> Q(x, y) = floor(x) - 1 - >> sum for k=2 to y of [(Q(x/k, k-1) - Q(k-1, k-1)) * >> (Q(k, k-1) - Q(k-1, k-1))] > [willo_thewisp@hotmail.com] You're welcome! I had the advantage of doing some grad work in recursive > function theory some decades ago, and if you're attracted to dense formulas > that accomplish surprising things with trivial arithmetic, that field is a > rich source of awesome entertainment. One of my favorites is related to Ackermann's function. I derived it when > writing compiler recursion stress tests a long time ago, although I'm sure > someone else came up with it earlier. Here i, j and k are naturals (all > integers >= 0): T(i, j, k) = j+1 if k=0 > i if j=0 and k=1 > 0 if j=0 and k=2 > 1 if j=0 and k>2 > T(i, T(i, j-1, k), k-1) otherwise (k>0 and j>0) Obviously, T(i, j, 0) = successor(j) = j+1. Less obviously, T(i, j, 1) = i+j, as can be seen by noting that for k=1, the > last expression is essentially the recursive definition of addition in terms > of the successor function (i+j = successor(i+(j-1))). Hint: that was a big > hint :-) For points, what are better-known names for T(i, j, 2) and T(i, j, 3)? For > bonus points, T(i, j, 4)? T(i,j,2) should be i*j (or something like it), and T(i,j,3) should be i^j or j^i (or something like that). Using Conway's arrow notation, T(i,j,4) should be i ^^ j (or maybe j ^^ i). --- Christopher Heckman === Subject: Re: JSH: Under review [Tim Peters] >> ... >> One of my favorites is related to Ackermann's function. I derived >> it when writing compiler recursion stress tests a long time ago, >> although I'm sure someone else came up with it earlier. Here i, j >> and k are naturals (all integers >= 0): >> T(i, j, k) = j+1 if k=0 >> i if j=0 and k=1 >> 0 if j=0 and k=2 >> 1 if j=0 and k>2 >> T(i, T(i, j-1, k), k-1) otherwise (k>0 and j>0) >> Obviously, T(i, j, 0) = successor(j) = j+1. >> Less obviously, T(i, j, 1) = i+j, as can be seen by noting that for >> k=1, the last expression is essentially the recursive definition of >> addition in terms of the successor function (i+j = >> successor(i+(j-1))). Hint: that was a big hint :-) >> For points, what are better-known names for T(i, j, 2) and T(i, j, 3)? >> For bonus points, T(i, j, 4)? [Proginoskes] > T(i,j,2) should be i*j (or something like it), and T(i,j,3) should be > i^j or j^i (or something like that). Wow -- I'd love to see how you answer multiple choice questions ;-) In a wordier way, letting op_k be the infix binary operator defined by T(i, j, k), the last clause says: i op_k j = i op_(k-1) (i op_k (j-1)) If you know that, as was given, op_1 is +, then: i op_2 j = i + (i op_2 (j-1)) follows. If you're guessing that op_2 is something like *, then does: i*j = i + i*(j-1) ? Yup! In addition, when j=0 and k=2, i*0 = 0 is returned by the special cases at the top. Similarly for i op_3 j = i^j. > Using Conway's arrow notation, T(i,j,4) should be i ^^ j (or maybe > j ^^ i). Right, except that i^^j is Knuth's notation (in Conway's notation, it's i -> j -> 2). Either way, T(i, j, 4) is an exponential tower with j copies of i, aka tetration: i^^0 = T(i, 0, 4) = 1 i^^1 = T(i, 1, 4) = i i^^2 = T(i, 2, 4) = i^i i^^3 = T(i, 3, 4) = i^(i^i) i^^4 = T(i, 4, 4) = i^(i^(i^i) ... T is most naturally related to Knuth's notation, as for k >= 4, T(i, j, k) is the same as Knuth's k-2 arrows operator applied to i and j. For example, T(3, 2, 5) = 3^^^2 = [a tower of 2 3's connected by ^^] 3^^3 = [a tower of 3 3's connected by ^] 3^(3^3) = 3^27 = 7625597484987 As the recursion reduces ^^^ to repeated instance of ^^, and ^^ to repeated instances of ^, it also similarly reduces the exponentiations to repeated multiplications, the multiplications to repeated additions, and the additions to repeated application of the successor function. It gets to 7625597484987 in the example by adding 1 at a time. Cool :-) === Subject: Re: JSH: Under review > You have to understand the mystery and why there is a big deal about > the prime distribution and conjectures like Riemann's related to it, > as > there is this odd thing that you can get functions like x/ln(x) near > pi(x)--but not exact. So you have the prime number theorem, but > why? Why the gap? Well, I found that you can define the prime counting function > multi-variably with a partial difference equation--eureka!!! BUT > for > it to be exact you have to have a special constraint, where you have > P(x,y) and sum your partial difference equation with a special > constraint that if y>sqrt(x) you reset back to sqrt(x). So there's this weird thing that even with a difference equation, if > you don't do this special thing, what do you get? A gap. More enterprising young mathematicians might explore the size of > that > gap with the difference equation, while I would guess that it is > roughly the size of what is shown in the prime number theorem. continuous function, which shows--almost shockingly--that it wasn't > about discrete or continuous, but about a special constraint--that > is, > special rules. Wim Bethem talked about integrating the partial differential > equation > with the same constraint, and it seems reasonable at this point to > conjecture that doing so will finally give a function that maps > perfectly to the prime distribution at discrete points while being > continuous. BUT it cannot be the Riemann function or anything like it--unless > his > research has something that embodies the special rule. Looking over it, last time I did was a few years ago, but I doubt > anything has changed, you find his work does not, so his function > cannot match the prime distribution perfectly. And then you have a back door to evaluating the Riemann Hypothesis, > which is to prove that the continuous function that follows from > integrating the partial differential equation that follows from my > prime counting function with the special constraint gives you the > prime > distribution perfectly at discrete points. So then no need for the prime number theorem as you have a perfect > function and the reason for why it wasn't perfect before. And you have a way to disprove the Riemann Hypothesis and get a > million > dollars from those Clay people JUST by integrating the partial > differential equation that follows from my research. So for years now I've known that Riemann probably failed. But I could see that easily enough just by glancing over his actual > notes. He does some sketchy things but he was brainstorming. Not his fault that some people grabbed a hold of his speculations > and > went wild with them, so I don't fault Riemann. James, here is a quote from Epilimnion posting in one of your threads in August 2000. It is as true today as it was then: James, I have no idea what your day job is. But you should be a politician. You really should. * You are _very_ thick skinned. * You persevere in the face of adversity. * You can spout bull in unending quantities. * Your natural stupidity is boundless. * You have the ability to talk in sound bites. * You can turn _every_ defeat into a glorious victory. I am sure you could become a great politician like... erm... like Richard Nixon, for example. I might add a couple of other useful attributes that you possess: * When asked a straightforward question, you have no qualms about not answering it. * Your utter inability to explain anything clearly makes it very difficult to pin you down. -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com> <4t7imfF12io1sU1@mid.individual.net > James, here is a quote from Epilimnion posting in one of your threads > in August 2000. It is as true today as it was then: James, I have no idea what your day job is. But you should be a > politician. You really should. * You are _very_ thick skinned. > * You persevere in the face of adversity. > * You can spout bull in unending quantities. > * Your natural stupidity is boundless. > * You have the ability to talk in sound bites. > * You can turn _every_ defeat into a glorious victory. I am sure you could become a great politician like... erm... like > Richard Nixon, for example. I might add a couple of other useful attributes that you possess: * When asked a straightforward question, you have no qualms about not > answering it. * Your utter inability to explain anything clearly makes it very > difficult to pin you down. -- > Clive Tooth Given that he has such a well planned agenda, would you say he is Republican? But seriously, one of JSH's best qualification for being a politician is that he is cut off from reality and also that he has not the slightest bit of curiosity for the truth. Also, I should add his great concern for the younger generations, not to mention his penchant for wanting to punish others that don't fit into his world view of what is right. Yeah, he must be Republican ;>) J === Subject: Re: JSH: Under review <3japm21o59em0h9qokj370p7k8us5859ls@4ax.com> <4t50rfF11rkfdU1@mid.individual.net> <291120061247414739%chenrich@monmouth.com I do not think that James has ever said he is looking for an analytic > function. He is looking for a perfect continuous function. It seems > likely that James has only a vague idea what he means by that phrase. In order for James to achieve the level of fame, wealth and babes that > he is seeking, he will have to come up with a continuous analytic > function that parallels for the prime number function pi() the > relationship that the gamma function has to the factorial function on > integers. So where gamma(n) = (n -1)! wjhen n is a positive integer, James' new > function (I will call it jamma in his honor) should be defined, > continuous, and differentiable over all real numbers (or even better, > all complex numbers) and have a form similar to jamma(p) = f(pi(p)) when evaluated at any prime positive integer p, where f() is a simple > algebraic function on pi(). Perhaps it should be called bernie in honor of Georg Friedrich > Bernhard Riemann. He expressed pi() as an expression involving (among > other things) the zeroes of the Riemann zeta function. And /that/ is > why the zeta function is relevant to the distribution of primes. I wouldn't care except my fear is that Riemann failed, otherwise hey > call it bernie. > Can you explain just what it is that you fear Riemann failed at? - William Hughes You have to understand the mystery and why there is a big deal about > the prime distribution and conjectures like Riemann's related to it, as > there is this odd thing that you can get functions like x/ln(x) near > pi(x)--but not exact. So you have the prime number theorem, but why? Why the gap? Well, I found that you can define the prime counting function > multi-variably with a partial difference equation--eureka!!! BUT for > it to be exact you have to have a special constraint, where you have > P(x,y) and sum your partial difference equation with a special > constraint that if y>sqrt(x) you reset back to sqrt(x). So there's this weird thing that even with a difference equation, if > you don't do this special thing, what do you get? A gap. More enterprising young mathematicians might explore the size of that > gap with the difference equation, while I would guess that it is > roughly the size of what is shown in the prime number theorem. continuous function, which shows--almost shockingly--that it wasn't > about discrete or continuous, but about a special constraint--that is, > special rules. Wim Bethem talked about integrating the partial differential equation > with the same constraint, and it seems reasonable at this point to > conjecture that doing so will finally give a function that maps > perfectly to the prime distribution at discrete points while being > continuous. BUT it cannot be the Riemann function or anything like it--unless his > research has something that embodies the special rule. Looking over it, last time I did was a few years ago, but I doubt > anything has changed, you find his work does not, so his function > cannot match the prime distribution perfectly. And then you have a back door to evaluating the Riemann Hypothesis, > which is to prove that the continuous function that follows from > integrating the partial differential equation that follows from my > prime counting function with the special constraint gives you the prime > distribution perfectly at discrete points. So then no need for the prime number theorem as you have a perfect > function and the reason for why it wasn't perfect before. And you have a way to disprove the Riemann Hypothesis and get a million > dollars from those Clay people JUST by integrating the partial > differential equation that follows from my research. So for years now I've known that Riemann probably failed. But I could see that easily enough just by glancing over his actual > notes. He does some sketchy things but he was brainstorming. Not his fault that some people grabbed a hold of his speculations and > went wild with them, so I don't fault Riemann. > This does not answer the question of just what it is that you fear Riemann failed at. It is clear that you think Riemann tried to do something related to the evaluation of pi(n) and you fear that he failed. However it is not clear what you think Riemann tried to do, So. What do you think that Riemann tried to do? What is the Riemann function? - William Hughes === Subject: Re: please recommend good numerical integration package in C or Matlab > computing the integral of x^a in (0 , 1], where a=[-0.9 ,-0.1] > Now let's look at Matlab's quadv results: >> the first column are the a's, the second column are the results from >> Matlab, the third column are the theoretical values: >> -1.000000e-01 9.664831e-01 1.111111e+00 >> -2.000000e-01 9.241422e-01 1.250000e+00 >> -3.000000e-01 8.048907e-01 1.428571e+00 >> -4.000000e-01 7.327810e-01 1.666667e+00 >> -5.000000e-01 6.788767e-01 2.000000e+00 >> -6.000000e-01 5.157902e-01 2.500000e+00 >> -7.000000e-01 4.686996e-01 3.333333e+00 >> -8.000000e-01 4.278468e-01 5.000000e+00 >> -9.000000e-01 3.748324e-01 1.000000e+01 You can also try quadl in MATLAB for a higher-order algorithm: > for a=-.9:.1:-.1 > f=@(x)x.^a; > disp(sprintf('%g %g',a,quadl(f,0,1))); > end > Warning: Minimum step size reached; singularity possible. >> In quadl at 95 > In qtest at 3 > -0.9 9.874 > Warning: Minimum step size reached; singularity possible. >> In quadl at 95 > In qtest at 3 > -0.8 4.99921 > -0.7 3.33333 > -0.6 2.5 > -0.5 2 > -0.4 1.66667 > -0.3 1.42857 > -0.2 1.25 > -0.1 1.11111 FWIW, I couldn't reproduce your posted numbers using quadv or quad. hope that helps, > -Ben Hinkle > 9.874 is still far off... === Subject: Re: please recommend good numerical integration package in C or Matlab http://www.cs.kuleuven.be/~nines/software/CUBPACK/ This may be useful. === Subject: Re: please recommend good numerical integration package in C or Matlab > Hi Mike, don't know, what you did - but I believe very much in matlab and of course > tried it. My results are: > Warning: Minimum step size reached; singularity possible. >> In quadv at 84 > -0.900000 9.895613 1.000000e+01 > -0.800000 4.996894 5.000000e+00 > -0.700000 3.333284 3.333333e+00 > -0.600000 2.500008 2.500000e+00 > -0.500000 2.000011 2.000000e+00 > -0.400000 1.666676 1.666667e+00 > -0.300000 1.428580 1.428571e+00 > -0.200000 1.250014 1.250000e+00 > -0.100000 1.111122 1.111111e+00 Using > for k=-.9:.1:-.1, w=quadv(@(x)myfun(x,k),0,1); fprintf('%f %fn',k,w); end > and > function w=myfun(x,a) > w=x^a; > return There are some errors, but matlab is not too far away! > -0.900000 9.895613 1.000000e+01 9.895613 from 10 that's not far away? === Subject: Axiom of choice and the three spheres. The decomposition of a sphere into pieces and assembly of the pieces into two spheres congruent to the first in [1] is often called the Banach-Tarski paradox. I used that phrase to find the reference to their paper. Yet their construction is not a paradox. It is not a contradiction either logically or intuitively. Reason: we have no right to expect that disassembling a measurable set A into non-measurable sets, then reassembling them into a measurable set B implies that mu(B) = mu(A). Not having read the paper, for all I know their purpose was to give an example of this. To mount my hobby horse: the Axiom of Choice should not be the whipping boy of every disagreeable result in set theory. There is usually a better explanation for problems than blaming the axiom of choice, if more [1] Banach and Tarski, Sur la d.8ecomposition des ensembles de points en parties respectivement congruentes, Fundamenta Mathematicae, 6, (1924), 244-277. -- Michael Press === Subject: Re: Axiom of choice and the three spheres. >The decomposition of a sphere into pieces and assembly >of the pieces into two spheres congruent to the first >in [1] is often called the Banach-Tarski paradox. I >used that phrase to find the reference to their paper. >Yet their construction is not a paradox. It is not a >contradiction either logically or intuitively. It's lucky that the question of whether something is intuitive is not well-defined... >Reason: >we have no right to expect that disassembling a >measurable set A into non-measurable sets, then >reassembling them into a measurable set B implies that >mu(B) = mu(A). Not having read the paper, for all I >know their purpose was to give an example of this. To mount my hobby horse: the Axiom of Choice should not >be the whipping boy of every disagreeable result in set >theory. There is usually a better explanation for >problems than blaming the axiom of choice, if more I'm missing your point here. First, I don't see people blaming AC for Banach-Tarski, nor saying that BT is disagreeable... Anyway, without AC you can't prove the existence of a non-measurable set, hence you certainly can't prove BT. So if we _were_ going to assign blame for BT I don't see how AC gets off the hook so easily. >[1] Banach and Tarski, Sur la d.8ecomposition des >ensembles de points en parties respectivement >congruentes, Fundamenta Mathematicae, 6, (1924), >244-277. ************************ David C. Ullrich === Subject: Re: Axiom of choice and the three spheres. , > >The decomposition of a sphere into pieces and assembly >of the pieces into two spheres congruent to the first >in [1] is often called the Banach-Tarski paradox. I >used that phrase to find the reference to their paper. >Yet their construction is not a paradox. It is not a >contradiction either logically or intuitively. > > It's lucky that the question of whether something > is intuitive is not well-defined... > >Reason: >we have no right to expect that disassembling a >measurable set A into non-measurable sets, then >reassembling them into a measurable set B implies that >mu(B) = mu(A). Not having read the paper, for all I >know their purpose was to give an example of this. To mount my hobby horse: the Axiom of Choice should not >be the whipping boy of every disagreeable result in set >theory. There is usually a better explanation for >problems than blaming the axiom of choice, if more > > I'm missing your point here. First, I don't see people > blaming AC for Banach-Tarski, nor saying that BT > is disagreeable... Folks often delineate when they invoke AC. I keep getting the impression that AC wears the scarlet letter. When analysts started constructing continuous nowhere differentiable functions others found the results disagreeable. I believe `monsters' came into the discussion. > Anyway, without AC you can't prove the existence of > a non-measurable set, hence you certainly can't prove > BT. So if we _were_ going to assign blame for BT > I don't see how AC gets off the hook so easily. Yes, I went off track there. Without AC all subsets of R^n are measurable? That is a convenient realm for proving things; no worries about that hypothesis. :) >[1] Banach and Tarski, Sur la d.8ecomposition des >ensembles de points en parties respectivement >congruentes, Fundamenta Mathematicae, 6, (1924), >244-277. -- Michael Press === Subject: Re: Axiom of choice and the three spheres. > The decomposition of a sphere into pieces and assembly > of the pieces into two spheres congruent to the first > in [1] is often called the Banach-Tarski paradox. I > used that phrase to find the reference to their paper. > Yet their construction is not a paradox. It is not a > contradiction either logically or intuitively. Reason: > we have no right to expect that disassembling a > measurable set A into non-measurable sets, then > reassembling them into a measurable set B implies that > mu(B) = mu(A). Not having read the paper, for all I > know their purpose was to give an example of this. > > To mount my hobby horse: the Axiom of Choice should not > be the whipping boy of every disagreeable result in set > theory. There is usually a better explanation for > problems than blaming the axiom of choice, if more > > [1] Banach and Tarski, Sur la d.8ecomposition des > ensembles de points en parties respectivement > congruentes, Fundamenta Mathematicae, 6, (1924), > 244-277. You would enjoy reading The Banach-Tarski Paradox by Stan Wagon. It's quite accessible and discusses the paradox in detail. It is a paradox in the sense that is is logically counterintuitive. === Subject: Re: Axiom of choice and the three spheres. > > The decomposition of a sphere into pieces and assembly > of the pieces into two spheres congruent to the first > in [1] is often called the Banach-Tarski paradox. I > used that phrase to find the reference to their paper. > Yet their construction is not a paradox. It is not a > contradiction either logically or intuitively. Reason: > we have no right to expect that disassembling a > measurable set A into non-measurable sets, then > reassembling them into a measurable set B implies that > mu(B) = mu(A). Not having read the paper, for all I > know their purpose was to give an example of this. > > To mount my hobby horse: the Axiom of Choice should not > be the whipping boy of every disagreeable result in set > theory. There is usually a better explanation for > problems than blaming the axiom of choice, if more > > [1] Banach and Tarski, Sur la d.8ecomposition des > ensembles de points en parties respectivement > congruentes, Fundamenta Mathematicae, 6, (1924), > 244-277. > > You would enjoy reading The Banach-Tarski Paradox by Stan Wagon. It's > quite accessible and discusses the paradox in detail. > > It is a paradox in the sense that is is logically counterintuitive. Well, OP just got finished saying he didn't find it counterintuitive. Maybe he has a better intuition. Maybe what's paradoxical is that there are non-measurable sets. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Axiom of choice and the three spheres. > > > The decomposition of a sphere into pieces and assembly > of the pieces into two spheres congruent to the first > in [1] is often called the Banach-Tarski paradox. I > used that phrase to find the reference to their paper. > Yet their construction is not a paradox. It is not a > contradiction either logically or intuitively. > > You would enjoy reading The Banach-Tarski Paradox by Stan Wagon. It's > quite accessible and discusses the paradox in detail. > > It is a paradox in the sense that is is logically counterintuitive. > > Well, OP just got finished saying he didn't find it counterintuitive. > Maybe he has a better intuition. Dare I say it that probably everyone has marginally different intuitions. To flatly assert, as the OP has done, that something which when described to 99% of the world's population would cause them to question its truth is intuitive is utterly preposterous. > Maybe what's paradoxical is that there are non-measurable sets. Or even if such intangible monsters can created, that they can be constructed, using such a simple method, from a measurable set. Phil -- Home taping is killing big business profits. We left this side blank so you can help. -- Dead Kennedys, written upon the B-side of tapes of /In God We Trust, Inc./. === Subject: Re: Axiom of choice and the three spheres. > > > The decomposition of a sphere into pieces and assembly > of the pieces into two spheres congruent to the first > in [1] is often called the Banach-Tarski paradox. I > used that phrase to find the reference to their paper. > Yet their construction is not a paradox. It is not a > contradiction either logically or intuitively. > > You would enjoy reading The Banach-Tarski Paradox by Stan Wagon. It's > quite accessible and discusses the paradox in detail. > > It is a paradox in the sense that is is logically counterintuitive. > > Well, OP just got finished saying he didn't find it counterintuitive. > Maybe he has a better intuition. > > Dare I say it that probably everyone has marginally different > intuitions. To flatly assert, as the OP has done, that something > which when described to 99% of the world's population would cause > them to question its truth is intuitive is utterly preposterous. It is intuitive when one is told that the pieces cannot be measured. What is preposterous? You seem to have taken me to have said that the usual presentation is not counter-intuitive. It is purposely described popularly so as to elicit the reaction it gets. People are advised that it does not work on a discrete topology. They are advised that the pieces are `wild', but are not informed that the pieces are not measurable. I restated it as a matter-of-fact state of affairs. Terms of a conditionally convergent series are rearranged so as to get any result you specify. A measureable set is rearranged so as to produce a measureable set with different measure. > Maybe what's paradoxical is that there are non-measurable sets. > > Or even if such intangible monsters can created, that they can be > constructed, using such a simple method, from a measurable set. -- Michael Press === Subject: Re: Axiom of choice and the three spheres. <87d575z89g.fsf@nonospaz.fatphil.org> reaction it gets. People are advised that it does not > work on a discrete topology. They are advised that the > pieces are `wild', but are not informed that the pieces > are not measurable. But surely it's obvious that, if such a decomposition is possible, then the pieces can't all be measurable? Measure is countably additive (indeed, we only need finite additivity here), and a fairly standard result/exercise in introductory measure theory is to show that Lipschitz mappings preserve Lebesgue measure (and rigid motions are certainly Lipschitz mappings). Dave L. Renfro === Subject: Re: Axiom of choice and the three spheres. <87d575z89g.fsf@nonospaz.fatphil.org> (snip) > It [The Banach Tarski Pardox] is intuitive when one is told that the pieces cannot > be measured. Is it intuitive that such decompostions exist in dimensions 1 and 2? After all, R^1 and R^2 have non-measurable sets - but the analogue of the Banach-Tarski Paradox in fact fails there. Something more than simple non-measurability is needed. What seems paradoxical (in the sense of surprising and initially implausible), even to most mathematicians I would guess, is that the Banach-Tarski Paradox works with *finitely many* pieces reassembled with rigid motions. I don't think people would have been that impressed if they pulled it off with denumerably many pieces. Your comparison with conditionally convergent series is apt in some ways but a bit misleading in others since measure is more akin to absolute convergence. If you object to the word paradox here, do you object to Simpon's paradox in probability or Arrow's paradox in social choice theory? Ultimately it is a matter of taste - but I see nothing wrong with refering to certain mathematical theorems or phenomena in a way that emphasizes their surprising nature. -John Coleman p.s. - I second the recommendation of Stan Wagon's book on BT. It is very well written. === Subject: Re: Axiom of choice and the three spheres. , > > (snip) > > It [The Banach Tarski Pardox] is intuitive when one is told that the pieces cannot > be measured. > > Is it intuitive that such decompostions exist in dimensions 1 and 2? > After all, R^1 and R^2 have non-measurable sets - but the analogue of > the Banach-Tarski Paradox in fact fails there. Something more than > simple non-measurability is needed. What seems paradoxical (in the > sense of surprising and initially implausible), even to most > mathematicians I would guess, is that the Banach-Tarski Paradox works > with *finitely many* pieces reassembled with rigid motions. I don't > think people would have been that impressed if they pulled it off with > denumerably many pieces. Your comparison with conditionally convergent > series is apt in some ways but a bit misleading in others since measure > is more akin to absolute convergence. If you object to the word > paradox here, do you object to Simpon's paradox in probability or > Arrow's paradox in social choice theory? Ultimately it is a matter of > taste - but I see nothing wrong with refering to certain mathematical > theorems or phenomena in a way that emphasizes their surprising nature. > > p.s. - I second the recommendation of Stan Wagon's book on BT. It is > very well written. Please, I am not raising objections. I think that their construction is marvelous. It is to this end that I consider to what degree it is paradoxical and what degree it is a marvel. Measure is additive. Who would expect non-measurable sets assembled into a measurable set to always have the same measure? Nothing paradoxical there. -- Michael Press === Subject: Re: Axiom of choice and the three spheres. <87d575z89g.fsf@nonospaz.fatphil.org> (snip) It [The Banach Tarski Pardox] is intuitive when one is told that the pieces cannot > be measured. Is it intuitive that such decompostions exist in dimensions 1 and 2? > After all, R^1 and R^2 have non-measurable sets - but the analogue of > the Banach-Tarski Paradox in fact fails there. Something more than > simple non-measurability is needed. What seems paradoxical (in the > sense of surprising and initially implausible), even to most > mathematicians I would guess, is that the Banach-Tarski Paradox works > with *finitely many* pieces reassembled with rigid motions. I don't > think people would have been that impressed if they pulled it off with > denumerably many pieces. Your comparison with conditionally convergent > series is apt in some ways but a bit misleading in others since measure > is more akin to absolute convergence. If you object to the word > paradox here, do you object to Simpon's paradox in probability or > Arrow's paradox in social choice theory? Ultimately it is a matter of > taste - but I see nothing wrong with refering to certain mathematical > theorems or phenomena in a way that emphasizes their surprising nature. p.s. - I second the recommendation of Stan Wagon's book on BT. It is > very well written. Please, I am not raising objections. I think that their > construction is marvelous. It is to this end that I > consider to what degree it is paradoxical and what > degree it is a marvel. Measure is additive. Who would > expect non-measurable sets assembled into a measurable > set to always have the same measure? Nothing > paradoxical there. I know that you were not objecting to the construction itself, but I took you as objecting to the fairly established use of the word paradox in mathematics to describe things that are (to many) counter-intuitive and surprising but not actually contradictory. By the way - to answer your question Who would expect non-measurable sets assembled into a measurable set to always have the same measure? an answer could be a mathematician whose intuition about how measurable and non-measurable sets works is based upon how things work in R^1, R^2 where such paradoxes don't exist. All you need is the existence of a finitely additive invariant measure (that extends basic definitions of area) to rule out a BT paradox, and Vitali's proof of the existence of non-measurable sets depended so heavily on the use of countable additivity that it came as a surprise that this assumption could be dropped in higher dimensions. Also - it wasn't just intuitions about *measure* that this seemed to go against but also intuitions about *isometries*. How could finitely many isometries allow for something this wierd? -John Coleman > -- > Michael Press === Subject: Re: Axiom of choice and the three spheres. Maybe what's paradoxical is that there are non-measurable sets. Not so much paradoxical as just plain false. ----------------------------------------------------------------------- > Bill Taylor W.Taylor@math.canterbury.ac.nz > ----------------------------------------------------------------------- ***************************************************************** Hi: To call a mathematical result plain false seems a rather risky thing to do if done without giving any sound reasoning for it: it is not that hard to construct a non-measurable set in R...with the axiom of choice, of course. Now if AC is NOT used then things can be different, and this includes measure theory itself, I guess. Tonio === Subject: Re: Axiom of choice and the three spheres. > > Maybe what's paradoxical is that there are non-measurable sets. > > Not so much paradoxical as just plain false. In a measure theory which accepts the axiom of choice, there are unmeasurable sets. This can be seen by constructing, via that axiom, a partition of the unit interval into countably many subsets each of which would have to have the same measure if any of them were to be found measurable. === Subject: Re: Small Proper Classes? > For example, the construction of a non-measurable set in ZFC uses the > Axiom of Choice. There are other theories, such as ZF + All sets are > measurable in which AC is false. In such this theory, that same > collection of reals used to create a non-measurable set in ZFC is > (obviously) not a set at all in ZF + All sets are measurable. Could > such a collection be considered a proper class with respect to this > theory, since it cannot be a set? Another (perhaps more obvious) reason why your non-measurable collection cannot be considered a class in ZF+~AC: Although subclasses of proper classes may be either a poper class or a set, the subclass of a set is always a set. The set of reals is still a set, even in ZF+~AC. Since your non-measurable collection is a subset of the reals, it would have to be a set if it existed at all. Since its existence contradicts ~AC, it can't exist in any form, whether as a set or as a proper class. Hope that helps. Jonathan Hoyle Eastman Kodak === Subject: Breakdown of unitarity of the S-Matrix in complex open systems Including our pocket universe. Hence signal nonlocality. All physical Hilbert spaces accessible in real experiments with real detectors are finite. Consequently the ODLRO emergence of new order demands that conservation of probability and unitarity of the physical S-matrix is an illusion. Suppose we have a quantum computer system described by a database state space of N qubits. This is a space of dim 2^N. The sum of probabilities pi(N) adds to 1 for all times such that the dimension is N. But now a phase transition happens. The system is open. The data base is now N + n qubits from the ODLRO emergence of new order. We now have a larger state space by a factor of 2^n. The original state space is a small sub-volume of the new one. The sum of the initial pi(N) is now MUCH LESS than 1. BTW it can go the other way as well if we destroy possibilities in the phase transition. That is, the actual mathematics of Hilbert space, also differential geometry are idealizations. Our actual measurements and experiences are always of coarse-grained finite resolution. As far as the laws of mathematics refer to reality, they are not certain, as far as they are certain, they do not refer to reality. Einstein === Subject: Four variable Diophine equations Is there a general way to solve four variable diophine equations? I want to solve the specific case (z^2)*x^3-(z^2)*w^2=y^2 === Subject: Re: Four variable Diophine equations > Is there a general way to solve four variable diophine equations? > I want to solve the specific case (z^2)*x^3-(z^2)*w^2=y^2 (z^2)*x^3-(z^2)*w^2= z^2(x^3 -w^2) z^2 is always a square (x^3 -w^2) = t^2 => w^2 +t^2 = x^3 => (a+ib)^3(a-ib)^3 = x^3 So you can express, w,t,x in terms of a and b which ultimately gives a parametric solution in z,a,b === Subject: Re: Four variable Diophine equations Sorry, I meant Diophantine. === Subject: Re: Inductors and lamps puzzle Ok, it's in. Just out of curiosity, Alec, can you offer some thoughts on > how > you came up with the complete sequence, its generating function and the > expression 2*A005840(n) + n - 2? The main credit should go to Gerry Meyerson. been counting circuits with at most n resistors while A005840 is counting circuits with exactly n resistors. I think that makes the next number in our sequence 94. Let b(n)=A005840(n) if n>0 and b(0)=0, and let a(n) be the sequence discussed in this thread. How to calculate a(4), for example? a(4)=b(4) + 4 b(3) + 6 b(2) + 4 b(1) + b(0), because there is one choice of 4 resistors out of 4, 4 choices of 3 resistors out of 4, 6 choices of 2 resistors out of 4, 4 choices of 1 resistor out of 4, and for every such choice the number of circuits constracted from them is b(..). In general, we get formula a(n) = sum C(n,k)*b(n) for k from 0 to n. In other words, sequence a(0), a(1),... is the inverse binomial transform of the sequence b(0), b(1), ... Now, if g_b(x) = sum b(n)/n!*x^n for n from 0 to infinity, it is called an exponential generating function for b, and g_a(x) = sum a(n)/n!*x^n is the exponential generating function for a, and a(0),a(1),... is the inverse binomial transform of b(0),b(1),... then it is well known and it is easy to check that g_a(x)=exp(x)*g_b(x). g_b(x) can be obtain by subtracting 1 from the exponential generating function for A005840. That's how I get g_a(x). It can be rewritten in the form 2*(x-1)*exp(x)/(exp(x)-2) + (x-2)*exp(x) for n>0. The first term gives 2*b(n), and the second term gives (n-2), so a(n)=2*b(n)+n-2 It also true for n=0. Alec === Subject: Re: Inductors and lamps puzzle Ok, it's in. Just out of curiosity, Alec, can you offer some thoughts on > how > you came up with the complete sequence, its generating function and the > expression 2*A005840(n) + n - 2? > > The main credit should go to Gerry Meyerson. If you're referring to me, then 1. the spelling is Myerson, and 2. you're far too kind. While you may have found it routine to get from my remark to your formulas, I'd say you did the lion's share of the work. Well, you and whoever did the original work on A005840. Well done. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Inductors and lamps puzzle >> The main credit should go to Gerry Meyerson. If you're referring to me, then 1. the spelling is Myerson, and Sorry about that. I don't understand how that could happen. Certainly I meant to write Gerry Myerson, Alec === Subject: Re: Hom(V,V), linear algebra, and dual space question ... > Let V be an inner product space over R (the real numbers) with inner > product (*,*). I will let = denote isomorphic to. Then Hom_R(V,V) = V* (x) V = V (x) V. Assuming V is finite dimensional and the inner product is nonsingular, the inner product defines an isomorphism between V* and V, as follows. An element v of V defines this linear form on V (or, in other words, this element of V*): w mapsto w.v The mapping v mapsto w is the isomorphism V to V*. this multilinear algebra is messy. Tedious, yes, but not messy :) === Subject: eigenpairs for matrix powers A basic theorem (which is easy to prove0 in linear algebra states If (lambda,x) is an eigenpair of A, then (lambda^k,x) is an eigenpair of A^k. Is the converse of this theorem also true? That is, is it true that If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an eigenpair of A. ??? -- % Randy Yates % Remember the good old 1980's, when %% Fuquay-Varina, NC % things were so uncomplicated? %%% 919-577-9882 % 'Ticket To The Moon' %%%% % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr === Subject: Re: eigenpairs for matrix powers If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an > eigenpair of A. > Try A=[[0,1],[1,0]]. === Subject: Re: eigenpairs for matrix powers > If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an >> eigenpair of A. >> > Try A=[[0,1],[1,0]]. What is true is that if (gamma, x) is an eigenpair of A^k [which I assume means gamma is the eigenvalue and x the eigenvector], then at least one of the k'th roots [in the complex numbers] of gamma is an eigenvalue of A, and x is a linear combination of eigenvectors of A for such eigenvalues. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: eigenpairs for matrix powers > If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an eigenpair of A. >> Try A=[[0,1],[1,0]]. What is true is that if (gamma, x) is an eigenpair of A^k > [which I assume means gamma is the eigenvalue and x the > eigenvector], then at least one of the k'th roots [in the > complex numbers] of gamma is an eigenvalue of A, and x is a > linear combination of eigenvectors of A for such eigenvalues. Hi Robert, I see that in a cloudy way. Let me present my real, higher-level problem. This is HW, so if you have a problem helping me out, you can refuse. We are to show that if A is symmetric, then ||A||_2 = rho(A), that is, the two-norm of A is equal to the spectral radius of A. Our professor had us prove in the immediately previous problem that if A=A^T and Ax = lambda x, then A^kx = lambda^k x (as I stated initially). However, to prove the result on the norm, it seems that we need the converse. Am I wrong, and can you point me to the way to solve this? -- % Randy Yates % How's life on earth? %% Fuquay-Varina, NC % ... What is it worth? %%% 919-577-9882 % 'Mission (A World Record)', %%%% % *A New World Record*, ELO http://home.earthlink.net/~yatescr === Subject: Re: eigenpairs for matrix powers If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an eigenpair of A. Try A=[[0,1],[1,0]]. What is true is that if (gamma, x) is an eigenpair of A^k > [which I assume means gamma is the eigenvalue and x the > eigenvector], then at least one of the k'th roots [in the > complex numbers] of gamma is an eigenvalue of A, and x is a > linear combination of eigenvectors of A for such eigenvalues. Hi Robert, I see that in a cloudy way. Let me present my real, higher-level > problem. This is HW, so if you have a problem helping me out, > you can refuse. We are to show that if A is symmetric, then ||A||_2 = rho(A), > that is, the two-norm of A is equal to the spectral radius of > A. Our professor had us prove in the immediately previous > problem that if A=A^T and Ax = lambda x, then A^kx = lambda^k x > (as I stated initially). However, to prove the result on the norm, > it seems that we need the converse. Am I wrong, and can you point me to the way to solve this? The converse is not true. Why do you think you need it? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: eigenpairs for matrix powers > If (gamma,x) is an eigenpair of A^k, then ((gamma)^(1/k),x) is an > eigenpair of A. >> Try A=[[0,1],[1,0]]. >> What is true is that if (gamma, x) is an eigenpair of A^k >> [which I assume means gamma is the eigenvalue and x the >> eigenvector], then at least one of the k'th roots [in the >> complex numbers] of gamma is an eigenvalue of A, and x is a >> linear combination of eigenvectors of A for such eigenvalues. >> Hi Robert, >> I see that in a cloudy way. Let me present my real, higher-level >> problem. This is HW, so if you have a problem helping me out, >> you can refuse. >> We are to show that if A is symmetric, then ||A||_2 = rho(A), >> that is, the two-norm of A is equal to the spectral radius of >> A. >> Our professor had us prove in the immediately previous >> problem that if A=A^T and Ax = lambda x, then A^kx = lambda^k x >> (as I stated initially). However, to prove the result on the norm, >> it seems that we need the converse. >> Am I wrong, and can you point me to the way to solve this? The converse is not true. Why do you think you need it? By definition, the 2-norm of a matrix A is norm{A}_2 = sqrt{rho(A^T*A)}. If A is symmetric, then this means that norm{A}_2 = sqrt{rho(A^2)} = sqrt{|alpha|}, where alpha is an eigenvalue of A^2 with the largest magnitude. So we know alpha is the eigenvalue of A^2 with the largest magnitude. It seems like the next step is to be able to show that sqrt{alpha} is then the eigenvalue of A with the largest magnitude, in which case norm{A}_2 = rho(A), which was to be shown. So if we can show that A^2 x = alpha x Longrightarrow Ax = sqrt{alpha}x, then we'd have the missing link to complete the proof. Where am I going wrong? -- % Randy Yates % The dreamer, the unwoken fool - %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow... %%% 919-577-9882 % %%%% % 'Eldorado Overture', *Eldorado*, ELO http://home.earthlink.net/~yatescr === Subject: Re: eigenpairs for matrix powers > If (gamma,x) is an eigenpair of A^k, then >((gamma)^(1/k),x) is an >> eigenpair of A. > Try A=[[0,1],[1,0]]. > What is true is that if (gamma, x) is an eigenpair of A^k [which I assume means gamma is the eigenvalue and x the eigenvector], then at least one of the k'th roots [in the complex numbers] of gamma is an eigenvalue of A, and x is a linear combination of eigenvectors of A for such eigenvalues. > Hi Robert, > I see that in a cloudy way. Let me present my real, higher-level problem. This is HW, so if you have a problem helping me out, you can refuse. > We are to show that if A is symmetric, then ||A||_2 = rho(A), that is, the two-norm of A is equal to the spectral radius of A. > Our professor had us prove in the immediately previous problem that if A=A^T and Ax = lambda x, then A^kx = lambda^k x (as I stated initially). However, to prove the result on the norm, it seems that we need the converse. > Am I wrong, and can you point me to the way to solve this? >> The converse is not true. Why do you think you need it? By definition, the 2-norm of a matrix A is norm{A}_2 = sqrt{rho(A^T*A)}. If A is symmetric, then this means that norm{A}_2 = sqrt{rho(A^2)} > = sqrt{|alpha|}, where alpha is an eigenvalue of A^2 with the largest magnitude. So we know alpha is the eigenvalue of A^2 with the largest magnitude. It seems like the next step is to be able to show that sqrt{alpha} is >then the eigenvalue of A with the largest magnitude, in which case Except that it might be - sqrt(alpha) rather than sqrt(alpha) that is the eigenvalue with largest magnitude. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Generational sizes. Hi All, I want to introduce the idea of generational size of a set. For sets A and B, if set B is defined to be the range of a function f, who's domain is A, then set B is called a generational set from A by function f, and symbolized as B_f(A). Example: A={0,1,2,3,.....} f:A->B, f(x)=2x B_f(A) = {y|y=f(x):xeA} Accordingly B_f(A) is the set of all even numbers,that is generated from A by f. so B_f(A) is a generational set. f is called the generational function of B from A. B is also called the generated set, while A is called the source set. the generational size is symbolized as g.size, and f is called the generational function symbolized as g.f. so g.f of B from A = f in the above example. So in cases when we have any two sets A and B, were the generational function of one of them from the other is known, then we can determine the generational size of them as follows: For sets A and B_f(A) 1) g.size A = g.size B <-> f is bijective from A to B. 2) g.size A > g.size B <-> f is strictlly serjective from A to B. 3) g.size A < g.size B <-> f is a ONE-MANY RELATION from A to B. ( f can be called as the inverse of a strictly serjective g.function defined from B to A ). f is called strictly serjective, if it is serjective and not bijective. For any two arbitrary sets A and B, the generational size of one of them from the other can be determined without knowing the generational function if and only if sets A and B are homomorphic. A is homomorphic to B <-> (Every serjective function from one of them to the other is also injective)XOR (Every serjective function from one of them to the other is also not bijective) IN symboles A is homomorphic to B <-> (Af:f=serj(A,B)->f=inj(A,B) v f=serj(B,A) -> f =inj(B,A) ) Xor (Af:f=serj(A,B)->f=~inj(A,B) v f=serj(B,A) -> f =~inj(B,A) ) Examples: 1)A is finite and B is infinite or B is finite and A is infinite. 2)A and B are finite sets. 3) A is the power set of B or B is the power set of A. So for all homomorphic sets, g.size is the same as cardinality. But for heteromorphic sets ( heteromorphic =~homomorphic ), this is not always the case. All infinite sets of equal cardinality are heteromorphic. In the case of heteromorphic sets, it is impossible to determine their g.sizes without knowing the generational function of one of them from the other. Applications. Suppose that we have the set of all parents P, and the set of all children C. Now if every two parents have one child, i.e the g.f of C from P is a 2-1 serjective function. then we can say that the generational size of P is bigger than that of C. This idea ,in these circumstances, seems more plausable than the general cardinality as a measure of the relative sizes of P and C . By using cardinality, then we have: |P|=|C|. which is counter-intuitive, since it is clear that the number of children should be smaller than the number of parents, because the number of children is generationally determined by their parents desire of family size of having one child. However if we are confronted with two sets, one of parents and one of their children and we do not know how many each pair of parents have of children. then we cannot determine their g.sizes. In these circumstances we resort to the more general cardinality of Cantor, by saying that |P|=|C|. So the idea of generational size is a more sensitive measure of infinite size comparison than that of Cantor's, but require more than the mere knowledge of what set members are to determine it. In addition to knowing the set members we should know the generational function between them. So having sufficient information to determine the generational size of sets, then this method is a more a sensitive measure of set size than cardinality, however not having such sufficient information , then Cantor's cardinality is another resort we can go to. Zuhair === Subject: Re: Generational sizes. > For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). Example: A={0,1,2,3,.....} f:A->B, f(x)=2x B_f(A) = {y|y=f(x):xeA} Accordingly B_f(A) is the set of all even numbers,that is generated > from A by f. so B_f(A) is a generational set. f is called the > generational function of B from A. B is also called the generated set, while A is called the source set. > the generational size is symbolized as g.size, and f is called the > generational function symbolized as g.f. so g.f of B from A = f in the > above example. As Moe says, the idea of range of f is very old. But here you have added your own incoherence. What do you mean f is the generational function of B from A? In your own example, we can see that there are infinitely many functions from A onto B. Let g be defined by: g(0) = 94 g(47) = 0 g(n) = 2n if n is neither 0 nor 47 Why isn't g the generational function of B from A? -- I'm the theory guy. Other people are the experimental people. If you push me on details I get annoyed, as I'm the theory guy. I'm the theoretical amateur mathematician. --James S. Harris, poet === Subject: Re: Generational sizes. <87k61czn0q.fsf@phiwumbda.org > For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). Example: A={0,1,2,3,.....} f:A->B, f(x)=2x B_f(A) = {y|y=f(x):xeA} Accordingly B_f(A) is the set of all even numbers,that is generated > from A by f. so B_f(A) is a generational set. f is called the > generational function of B from A. B is also called the generated set, while A is called the source set. > the generational size is symbolized as g.size, and f is called the > generational function symbolized as g.f. so g.f of B from A = f in the > above example. As Moe says, the idea of range of f is very old. But here you have > added your own incoherence. What do you mean f is the generational > function of B from A? first: I am not saying that I am introducing the idea of range. I know that this is an old idea, I am introducing the idea of generational size of a set, this is what is new. second, there is not need to speak in this offending manner, if you don't know something just ask wihtout violation to ethics,as you did by saying the wordyour own incoherene. If you want to understand what is said, seriouselly then read it carfully in a non critical manner, and then after you understand it, then start critisicim. A function is called generational from A to B, if B is defined as the range of f from A. since if this happens then B is symbolized as B_f(A), it means that B is the range of f who's domain is A. then every f is generational from A to C_f(A). The intuitive base for this is the simple everyone agreeable idea, that you get from me what I gave you so if y gives x , $10, then x received from y,$10, not less not more. Now let me illustrate that idea in a more convincing manner. suppose that there is a countable infinite number of banks belonging to American banks company. so we have set B={b1,b2,b3,.........} Now Let us have another countably infinite set of borrowers P={p1,p2,p3,........} Let every bi lended $2 to every pi, i.e b1 lends $2 to p1, b2 lends $2 to p2,.....etc Now the set of dollers per persons will be D={ D1-P1,D2-P1, D1-P2,D2-P2,......,D1-Pi,D2-Pi,........} Now what is the size of D as compared to B. obviouselly it is double that of B. Now suppose that each person pi should return back the $2 he received from the bank bi, after one month. IF Cantor is right, then each person ill return one dollar to the bank he took money from. why? because according to Cantor D is equal in size to B , then this amount to every person in P having only one dollar from the bank in B. Another example: Suppose you have a countable infinity of Pregnant ladies. and each lady has TWO foetuses in here uterus. so we have two sets, the set of pregnant woman,P, were P={ w1,w2,w3,......} and the set of babies inside the uteruses of these pregnant women,B, were B={b0,b1,b2,........}. Let b0 and b1 be in the uterus of w1, b2 and b3 be in the uterus of w2, and in general let bi be in the uterus of w(floor(i/2)). Now the question is what is the size of B as compared to the size of P. If you go like cantor and say that since there is bijection between P and B, then the size of P is the same as that of B, then this is equivalent to saying that avery lady wi is having one baby bi. which is not the case. It is true that there is bijection between P and B, but that bijection is irrelavent to the relative size comparison between these two sets, since it is not the BIOLOGICAL determining function of the existance of babies from their mothers. The biological function here is let f:B->P, f(x)=floor(x/2). Then the generational RELATION of babies from mothers is f^(-1):P->B. Now the g.size B > g.size P. since f^(-1) which is the generational relation of B from P, is a Many-one relation. This relation is what determine the relative sizes of P and B, since NATURE chose this relation in this case, the bijective function between B and P has no biological significance, it is only an abstraction. So the relative size of any set B as compared to set A is determined by the function Or relation whoe's domain is A and B is its Range, AS SPECIFIED by the question. i.e if B is defined explicitelly as the range of f whoe's domain is A, then B is symbolized as B_f(A). and it is this function and only this function which is explicitelly specified in the question that can determine the relative size comparison between A and B. However if f is not specified, i.e we are encountered with a situation that A and B without any one of them being defined as the range of the other, then we have many and sometimes infinte serjective functions and inverse of serjective functions each one of them can be a generational function, so which one determine the comparison. The answer is that if A and B are homomorphic ( see upthread for definition of homomorphism) then ANY serjective function between A and B can be the generational function determining their relative size comparison. What if A and B are heteromorphic. In this case, without the generational function being specified by the examinar, then it is impossible to know which one of them is the g.function. and accordinglly we cannot determine exactly there generational sizes. However in the last case, I can make a certain approximation as a salvage proceedure, and so, in the case where A and B are heteromorphic and no generational function is specified, then it is the bijective function between them that is the generational function. As an example imagine you have two sets one is countable infinite set of pregnant women and another also a countable infinite set of their babies, but we do not know exactly how many babies each pregnant women has, then in this situation we can say that the bijective function between them is the generational function. This idea is not so easy, and I don't beleive that it was introduced before. According to this idea Cantors cardianlities would only be a gross estimation due to ignorance of the generational relation between two sets. Zuhair In your own example, we can see that there are infinitely many > functions from A onto B. Let g be defined by: g(0) = 94 > g(47) = 0 > g(n) = 2n if n is neither 0 nor 47 Why isn't g the generational function of B from A? -- > I'm the theory guy. > Other people are the experimental people. > If you push me on details I get annoyed, as I'm the theory guy. > I'm the theoretical amateur mathematician. --James S. Harris, poet === Subject: Re: Generational sizes. <87k61czn0q.fsf@phiwumbda.org As an example imagine you have two sets one is countable infinite set > of pregnant women and another also a countable infinite set of their > babies, but we do not know exactly how many babies each pregnant women > has, An interesting situation indeed. In this case, it would seem that at least there's plenty of work for the obstetric industry. So is it your guess that in this case we'll need a countable or an uncountable number of ultrasound machines? MoeBlee === Subject: Re: Generational sizes. > Hi All, I want to introduce the idea of generational size of a set. For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). This is already done in chapter one set theory. B is called the image of A under the function f. In symbols B = f|[A]|, or B = FA. You've introduced an idea that is at least about a hundred and twenty five years old. MoeBlee === Subject: Re: Generational sizes. > Hi All, I want to introduce the idea of generational size of a set. For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). This is already done in chapter one set theory. B is called the image > of A under the function f. In symbols B = f|[A]|, or B = FA. You've introduced an idea that is at least about a hundred and twenty > five years old. MoeBlee huhuha, Moe you are only reading what you want to object to. the idea is not about a range of set, I know this is old, the idea is generational size, is generational size as I presented it an old idea. Zuhair === Subject: Re: Generational sizes. > huhuha, Moe you are only reading what you want to object to. the idea > is not about a range of set, I know this is old, the idea is > generational size, is generational size as I presented it an old idea. Oh, no, I'm well aware that your ideas only get more and more profound and dazzlingly original as your posts unfold. Sometimes I can't finish reading your posts as they leave me reeling from the sheer complexity and ingeniousness that flow from your mathematically virtuosic mind. Please forgive me for not always comprehending all the profundities of all your posts. MoeBlee === Subject: Re: Generational sizes. > Hi All, Hi Zuhair - I saw your thoughts in Galileo's Paradox. Not bad. This idea also is not entirely bad, and is similar to IFR, up to a point. > > I want to introduce the idea of generational size of a set. > > For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). > > Example: A={0,1,2,3,.....} > > f:A->B, f(x)=2x > > B_f(A) = {y|y=f(x):xeA} > > Accordingly B_f(A) is the set of all even numbers,that is generated > from A by f. so B_f(A) is a generational set. f is called the > generational function of B from A. Up to here it's good. It corresponds to the bijective mapping function in the Inverse Function Rule. > > B is also called the generated set, while A is called the source set. > the generational size is symbolized as g.size, and f is called the > generational function symbolized as g.f. so g.f of B from A = f in the > above example. Doesn't f map the elements of A to those of B? Are you saying the size is equal to f? Isn't the relative size of the set mapped using f(x)=2x 1/2 of the original, as a subset with density 1/2? Perhaps I misunderstand this part. > > So in cases when we have any two sets A and B, were the generational > function of one of them from the other is known, then we can determine > the generational size of them as follows: > > For sets A and B_f(A) such that: B_f(A) = {y|y=f(x):xeA} > > 1) g.size A = g.size B <-> f is bijective from A to B. > 2) g.size A > g.size B <-> f is strictlly serjective from A to B. > 3) g.size A < g.size B <-> f is a ONE-MANY RELATION from A to B. ( f > can be called as the inverse of a strictly serjective g.function > defined from B to A ). If B_f contains all y such that y=f(x) and x e A, then isn't the function necessarily bijective? That appears to be what you expressed above, no? In my system, using IFR, we employ functions that are bijective over a given range, be it finite, or over the real line, using a unit infinity as that range. It's called the inverse function rule because the inverse of f, g: f(g(x))=g(f(x))=x, over the value range of the generated set, gives the count of the source set, when the source set is the naturals. > > f is called strictly serjective, if it is serjective and not bijective. > > For any two arbitrary sets A and B, the generational size of one of > them from the other can be determined without knowing the generational > function if and only if sets A and B are homomorphic. > > A is homomorphic to B <-> (Every serjective function from one of them > to the other is also injective)XOR (Every serjective function from one > of them to the other is also not bijective) > > IN symboles > > A is homomorphic to B <-> (Af:f=serj(A,B)->f=inj(A,B) v f=serj(B,A) - f =inj(B,A) ) Xor > > (Af:f=serj(A,B)->f=~inj(A,B) v f=serj(B,A) -> f =~inj(B,A) ) > > Examples: > > 1)A is finite and B is infinite or B is finite and A is infinite. > 2)A and B are finite sets. > 3) A is the power set of B or B is the power set of A. > > So for all homomorphic sets, g.size is the same as cardinality. > > But for heteromorphic sets ( heteromorphic =~homomorphic ), this is not > always the case. > > All infinite sets of equal cardinality are heteromorphic. > > In the case of heteromorphic sets, it is impossible to determine their > g.sizes without knowing the generational function of one of them from > the other. Okay, but do you have an example of this calculation, first showing it works in the finite case, and then generalizing fruitfully to the infinite case? > > Applications. > > Suppose that we have the set of all parents P, and the set of all > children C. > > Now if every two parents have one child, i.e the g.f of C from P is a > 2-1 serjective function. then > we can say that the generational size of P is bigger than that of C. I guess you mean if every parent has one child, not every pair, since one could have children with multiple people, and still have only one child for any given pair. > > This idea ,in these circumstances, seems more plausable than the > general cardinality as a measure of the relative sizes of P and C . How do you calculate, say, the relative number of squares to naturals over the real line? > > By using cardinality, then we have: |P|=|C|. which is > counter-intuitive, since it is clear that the number of children should > be smaller than the number of parents, because the number of children > is generationally determined by their parents desire of family size of > having one child. Sure. > > However if we are confronted with two sets, one of parents and one of > their children and we do not know how many each pair of parents have of > children. then we cannot determine their g.sizes. > In these circumstances we resort to the more general cardinality of > Cantor, by saying that > |P|=|C|. Or, even if you know the mapping function, but choose to ignore the relationship it expresses. > > So the idea of generational size is a more sensitive measure of > infinite size comparison than that of Cantor's, but require more than > the mere knowledge of what set members are to determine it. In addition > to knowing the set members we should know the generational function > between them. Agreed. > > So having sufficient information to determine the generational size of > sets, then this method is a more a sensitive measure of set size than > cardinality, however not having such sufficient information , then > Cantor's cardinality is another resort we can go to. > > Zuhair > Can you give an example of determining, say, the number of decimal vs. binary strings of arbitrary length, or the number of powers of 2 vs. naturals? IFR is good for that. :) === Subject: Re: Generational sizes. <456f32f8@news2.lightlink.com Hi All, Hi Zuhair - I saw your thoughts in Galileo's Paradox. Not bad. This idea also is > not entirely bad, and is similar to IFR, up to a point. > I want to introduce the idea of generational size of a set. For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). Example: A={0,1,2,3,.....} f:A->B, f(x)=2x B_f(A) = {y|y=f(x):xeA} Accordingly B_f(A) is the set of all even numbers,that is generated > from A by f. so B_f(A) is a generational set. f is called the > generational function of B from A. Up to here it's good. It corresponds to the bijective mapping function > in the Inverse Function Rule. > B is also called the generated set, while A is called the source set. > the generational size is symbolized as g.size, and f is called the > generational function symbolized as g.f. so g.f of B from A = f in the > above example. Doesn't f map the elements of A to those of B? Are you saying the size > is equal to f? Isn't the relative size of the set mapped using f(x)=2x > 1/2 of the original, as a subset with density 1/2? Perhaps I > misunderstand this part. > So in cases when we have any two sets A and B, were the generational > function of one of them from the other is known, then we can determine > the generational size of them as follows: For sets A and B_f(A) > such that: > B_f(A) = {y|y=f(x):xeA} 1) g.size A = g.size B <-> f is bijective from A to B. > 2) g.size A > g.size B <-> f is strictlly serjective from A to B. > 3) g.size A < g.size B <-> f is a ONE-MANY RELATION from A to B. ( f > can be called as the inverse of a strictly serjective g.function > defined from B to A ). If B_f contains all y such that y=f(x) and x e A, then isn't the > function necessarily bijective? That appears to be what you expressed > above, no? No f is not necessarily bijective, it might be bijective , it might be strictly serjective i.e serjective but not injective and thus not bijecive. Examples: Let N be the source set, and let N_f(N) be the generational set from N. Now, I will give two examples when f is bijective and f is not bijective. 1) Let f:N->N_f(N), f(x)=x. Now N_f(N)={y|y=f(x):xeN} Now: g.size N = g.size N_f(N). because f is bijective from N to N_f(N). 2) Let h:N->N_g(N), h(x)=floor(x/2). Now N_h(N)={y|y=h(x):xeN} Now g.size N > g.size N_h(N), why because g is serjective from N to N_h(N). In my system, using IFR, we employ functions that are bijective over a > given range, be it finite, or over the real line, using a unit infinity > as that range. It's called the inverse function rule because the inverse > of f, g: f(g(x))=g(f(x))=x, over the value range of the generated set, > gives the count of the source set, when the source set is the naturals. > f is called strictly serjective, if it is serjective and not bijective. For any two arbitrary sets A and B, the generational size of one of > them from the other can be determined without knowing the generational > function if and only if sets A and B are homomorphic. A is homomorphic to B <-> (Every serjective function from one of them > to the other is also injective)XOR (Every serjective function from one > of them to the other is also not bijective) IN symboles A is homomorphic to B <-> (Af:f=serj(A,B)->f=inj(A,B) v f=serj(B,A) - f =inj(B,A) ) Xor (Af:f=serj(A,B)->f=~inj(A,B) v f=serj(B,A) -> f =~inj(B,A) ) Examples: 1)A is finite and B is infinite or B is finite and A is infinite. > 2)A and B are finite sets. > 3) A is the power set of B or B is the power set of A. So for all homomorphic sets, g.size is the same as cardinality. But for heteromorphic sets ( heteromorphic =~homomorphic ), this is not > always the case. All infinite sets of equal cardinality are heteromorphic. In the case of heteromorphic sets, it is impossible to determine their > g.sizes without knowing the generational function of one of them from > the other. Okay, but do you have an example of this calculation, first showing it > works in the finite case, and then generalizing fruitfully to the > infinite case? OK, I will illustrate that later. > Applications. Suppose that we have the set of all parents P, and the set of all > children C. Now if every two parents have one child, i.e the g.f of C from P is a > 2-1 serjective function. then > we can say that the generational size of P is bigger than that of C. I guess you mean if every parent has one child, not every pair, since > one could have children with multiple people, and still have only one > child for any given pair. No, I mean if every pair of parents have only one biological child. > This idea ,in these circumstances, seems more plausable than the > general cardinality as a measure of the relative sizes of P and C . How do you calculate, say, the relative number of squares to naturals > over the real line? > By using cardinality, then we have: |P|=|C|. which is > counter-intuitive, since it is clear that the number of children should > be smaller than the number of parents, because the number of children > is generationally determined by their parents desire of family size of > having one child. Sure. > However if we are confronted with two sets, one of parents and one of > their children and we do not know how many each pair of parents have of > children. then we cannot determine their g.sizes. > In these circumstances we resort to the more general cardinality of > Cantor, by saying that > |P|=|C|. Or, even if you know the mapping function, but choose to ignore the > relationship it expresses. > So the idea of generational size is a more sensitive measure of > infinite size comparison than that of Cantor's, but require more than > the mere knowledge of what set members are to determine it. In addition > to knowing the set members we should know the generational function > between them. Agreed. > So having sufficient information to determine the generational size of > sets, then this method is a more a sensitive measure of set size than > cardinality, however not having such sufficient information , then > Cantor's cardinality is another resort we can go to. Zuhair > Can you give an example of determining, say, the number of decimal vs. > binary strings of arbitrary length, or the number of powers of 2 vs. > naturals? IFR is good for that. :) === Subject: Re: Generational sizes. > Hi All, I want to introduce the idea of generational size of a set. For sets A and B, if set B is defined to be the range of a function f, > who's domain is A, then set B is called a generational set from A by > function f, and symbolized as B_f(A). Example: A={0,1,2,3,.....} f:A->B, f(x)=2x B_f(A) = {y|y=f(x):xeA} Accordingly B_f(A) is the set of all even numbers,that is generated > from A by f. so B_f(A) is a generational set. f is called the > generational function of B from A. B is also called the generated set, while A is called the source set. > the generational size is symbolized as g.size, and f is called the > generational function symbolized as g.f. so g.f of B from A = f in the > above example. So in cases when we have any two sets A and B, were the generational > function of one of them from the other is known, then we can determine > the generational size of them as follows: For sets A and B_f(A) 1) g.size A = g.size B <-> f is bijective from A to B. > 2) g.size A > g.size B <-> f is strictlly serjective from A to B. > 3) g.size A < g.size B <-> f is a ONE-MANY RELATION from A to B. ( f > can be called as the inverse of a strictly serjective g.function > defined from B to A ). f is called strictly serjective, if it is serjective and not bijective. For any two arbitrary sets A and B, the generational size of one of > them from the other can be determined without knowing the generational > function if and only if sets A and B are homomorphic. A is homomorphic to B <-> (Every serjective function from one of them > to the other is also injective)XOR (Every serjective function from one > of them to the other is also not bijective) IN symboles A is homomorphic to B <-> (Af:f=serj(A,B)->f=inj(A,B) v f=serj(B,A) - f =inj(B,A) ) Xor (Af:f=serj(A,B)->f=~inj(A,B) v f=serj(B,A) -> f =~inj(B,A) ) Examples: 1)A is finite and B is infinite or B is finite and A is infinite. > 2)A and B are finite sets. > 3) A is the power set of B or B is the power set of A. So for all homomorphic sets, g.size is the same as cardinality. But for heteromorphic sets ( heteromorphic =~homomorphic ), this is not > always the case. All infinite sets of equal cardinality are heteromorphic. In the case of heteromorphic sets, it is impossible to determine their > g.sizes without knowing the generational function of one of them from > the other. Applications. Suppose that we have the set of all parents P, and the set of all > children C. Now if every two parents have one child, i.e the g.f of C from P is a > 2-1 serjective function. then > we can say that the generational size of P is bigger than that of C. This idea ,in these circumstances, seems more plausable than the > general cardinality as a measure of the relative sizes of P and C . By using cardinality, then we have: |P|=|C|. which is > counter-intuitive, since it is clear that the number of children should > be smaller than the number of parents, because the number of children > is generationally determined by their parents desire of family size of > having one child. However if we are confronted with two sets, one of parents and one of > their children and we do not know how many each pair of parents have of > children. then we cannot determine their g.sizes. > In these circumstances we resort to the more general cardinality of > Cantor, by saying that > |P|=|C|. So the idea of generational size is a more sensitive measure of > infinite size comparison than that of Cantor's, but require more than > the mere knowledge of what set members are to determine it. In addition > to knowing the set members we should know the generational function > between them. So having sufficient information to determine the generational size of > sets, then this method is a more a sensitive measure of set size than > cardinality, however not having such sufficient information , then > Cantor's cardinality is another resort we can go to. Zuhair I forgot to mentioned how do we determine, the relative g.size comparison for any two homomorphic set. The answer is any serjective function between A and B is a generational function. IF two heteromorphic sets with no known g.function between them, then it is the bijective function between them that is the generational function. Zuhair === Subject: Re: puzzle. 1/n contains ooo in period?? Originator: msb@shell.vex.net (Mark Brader) Don McDonald posts near-gibberish containing the following: > Contains triple ooo in period.. 1/n.. > > 1001 ---- 0.0009990010 > 1111 ---- 0.0009000900 > 9009 ---- 0.0001110001 > 9999 ---- 0.0001000100 > 10001 ---- 0.0000999900 > 10002 ---- 0.0000999800 > 10003 ---- 0.0000999700 > 90009 ---- 0.0000111100 > 111109 ---- 0.0000090002 > 111108 ---- 0.0000090003 > 111107 ---- 0.0000090003 > 111106 ---- 0.0000090004 > 111105 ---- 0.0000090005 > 111104 ---- 0.0000090006 > 111103 ---- 0.0000090007 > 111102 ---- 0.0000090007 > 111101 ---- 0.0000090008 > 111100 ---- 0.0000090009 > 11109 ---- 0.0000900171 > 11108 ---- 0.0000900252 > 11107 ---- 0.0000900333 > 11106 ---- 0.0000900414 > 11105 ---- 0.0000900495 > 11104 ---- 0.0000900576 > 11103 ---- 0.0000900657 > 11102 ---- 0.0000900739 > 11101 ---- 0.0000900820 Of the 27 reciprocals indicated, 25 do contain a triple 0 in their period, although obviously it's not always visible from the short expansions shown above. But the other two are clearly wrong: 1/111,104 = 0.000009000(576036866359447004608294930875)... 1/11,104 = .00009(00576368876080691642651296829971181556195965417 867435158501440922190201729106628242074927953890489913544668587 896253602305475504322766570605187319884726224783861671469740634)... -- Mark Brader, Toronto | Mark is probably right about something, msb@vex.net | but I forget what -- Rayan Zachariassen === Subject: Re: puzzle. 1/n contains ooo in period?? <12msnp4hcnql48b@corp.supernews.com> [nz.general trimmed as this mathematical puzzle has nothing to do with NZ] > Don McDonald posts near-gibberish containing the following: > Contains triple ooo in period.. 1/n.. 1001 ---- 0.0009990010 > 1111 ---- 0.0009000900 > 9009 ---- 0.0001110001 > 9999 ---- 0.0001000100 (further examples snipped) If Don is looking for n whose reciprocal, expressed in decimal, has 000 in its period, he's overlooked many examples. Among those that are quite easy to see: 3003 0.000333000333... 3333 0.000300030003... Any integer > 1000 that is coprime to 10 will also serve as n, e.g. 1003 0.0009970089730807577268195413758723828514456630109670987038 8833499501495513459621136590229312063808574277168494516450 6480558325024925224327018943170488534396809571286141575274 1774675972083748753738783649052841475573280159521435692921 2362911266201395812562313060817547357926221335992023928215 5892323030907278165503489531405782652043868394815553339980 0598205383848454636091724825523429710867397806580259222333 000997... 1017 0.00098328416912487708947885939036381514257620452310717797 44346116027531956735496558505408062930186823992133726647 000983... 1023 0.000977517106549364613880742913000977... as will some other integers e.g. 2038 0.0(00490677134445534838076545632973503434739941118743866535819430 81452404317958783120706575073601570166830225711481844946025515210 99116781157998037291462217860647693817468105986261040235525024533 85672227674190382728164867517173699705593719332679097154072620215 89793915603532875368007850834151128557409224730127576054955839057 89990186457311089303238469087340529931305201177625122669283611383 70951913640824337585868498527968596663395485770363101079489695780 17664376840039254170755642787046123650637880274779195289499509322 86555446516192345436702649656526005888125613346418056918547595682 04121687929342492639842983316977428851815505397448478900883218842 00196270853778213935230618253189401373895976447497546614327772325 80961727183513248282630029440628066732090284592737978410206084396 4671246319921491658488714425907752698724239450441609421 --->00098135426889106967615309126594700686947988223748773307163886162 90480863591756624141315014720314033366045142296368989205103042198 23356231599607458292443572129538763493621197252208047105) 2098 0.0004766444232602478551000953... 2114 0.0(00473036896877956480605487228003784295175023651844843897824030 27436140018921475875118259224219489120151371807 --->00094607379375591296121097445600756859035) === Subject: Re: puzzle. 1/n contains ooo in period?? <12msnp4hcnql48b@corp.supernews.com ... above. But the other two are clearly wrong: > ... 291066 ... Uh, oh. multiplier in a reversible hash function. Didn't want to redo it. James === Subject: puzzle. 1/n contains ooo in period?? Reciprocals 1001 1111 9009 9999 10001/2/3 90009 111109/08/7/6/5/4/3/2/1/0. 11109.................?????????? query, a few thoughts. sci calculator. i may do an excel spreadsheet.***************xx Contains triple ooo in period.. 1/n.. 1001 ---- 0.0009990010 1111 ---- 0.0009000900 9009 ---- 0.0001110001 9999 ---- 0.0001000100 10001 ---- 0.0000999900 10002 ---- 0.0000999800 10003 ---- 0.0000999700 90009 ---- 0.0000111100 111109 ---- 0.0000090002 111108 ---- 0.0000090003 111107 ---- 0.0000090003 111106 ---- 0.0000090004 111105 ---- 0.0000090005 111104 ---- 0.0000090006 111103 ---- 0.0000090007 111102 ---- 0.0000090007 111101 ---- 0.0000090008 111100 ---- 0.0000090009 11109 ---- 0.0000900171 11108 ---- 0.0000900252 11107 ---- 0.0000900333 11106 ---- 0.0000900414 11105 ---- 0.0000900495 11104 ---- 0.0000900576 11103 ---- 0.0000900657 11102 ---- 0.0000900739 11101 ---- 0.0000900820 > dear Dr mike th, Editor nzmm, greetings, Reciprocals 1001 1111 10001 111109 etc.: dr mike > thomas nzmm, nz math, nz m. > i am just thinking this morning 28.11.06 > [tickle.maths] what is the first 2 or 3 [reciprocals of integer] > that contain triple o-o-o in the period? > but not terminating decimal e.g. 2^n*5^m. a pretty simple puzzle i think. Reciprocals 1001 1111 9009 9999 10001/2/3 90009 > 111109/08/7/6/5/4/3/2/1/0. query, a few thoughts. > sci calculator. > i may do an excel spreadsheet. > > etc.: dr mike thomas nzmm, nz math Don S. McDonald === Subject: decimal expansion of 1/n contains string 000 in period?? integer sequences. decimal expansion of. 1/n contains string 000 in period?? integer sequences. Reciprocals 1001 1111 9009 9999 10001/2/3 90009 111109/08/7/6/5/4/3/2/1/0. > 11109.................?????????? query, a few thoughts. > sci calculator. > i may do an excel spreadsheet.***************xx > Contains triple ooo in period.. 1/n.. 1001 ---- 0.0009990010 > 1111 ---- 0.0009000900 > 9009 ---- 0.0001110001 > 9999 ---- 0.0001000100 > 10001 ---- 0.0000999900 > 10002 ---- 0.0000999800 > 10003 ---- 0.0000999700 > 90009 ---- 0.0000111100 > 111109 ---- 0.0000090002 > 111108 ---- 0.0000090003 > 111107 ---- 0.0000090003 > 111106 ---- 0.0000090004 > 111105 ---- 0.0000090005 > 111104 ---- 0.0000090006 > 111103 ---- 0.0000090007 > 111102 ---- 0.0000090007 > 111101 ---- 0.0000090008 > 111100 ---- 0.0000090009 > 11109 ---- 0.0000900171 > 11108 ---- 0.0000900252 > 11107 ---- 0.0000900333 > 11106 ---- 0.0000900414 > 11105 ---- 0.0000900495 > 11104 ---- 0.0000900576 > 11103 ---- 0.0000900657 > 11102 ---- 0.0000900739 > 11101 ---- 0.0000900820 > dear Dr mike th, Editor nzmm, greetings, Reciprocals 1001 1111 10001 111109 etc.: dr mike > thomas nzmm, nz math, nz m. i am just thinking this morning 28.11.06 > [tickle.maths] what is the first 2 or 3 [reciprocals of integer] > that contain triple o-o-o in the period? > but not terminating decimal e.g. 2^n*5^m. a pretty simple puzzle i think. Reciprocals 1001 1111 9009 9999 10001/2/3 90009 > 111109/08/7/6/5/4/3/2/1/0. query, a few thoughts. > sci calculator. > i may do an excel spreadsheet. > > etc.: dr mike thomas nzmm, nz math > > Don S. McDonald === Subject: Space is not a vacuum but Dirac's sea of positrons and that is gravity itself <4so8p2F10smqjU2@mid.individual.net> <4spi5bF10pgsfU1@mid.individual.net > You've ignored this does not annihilate stance in my first question. > But since I find it here I now have to ask you is the positron that you > speak of the same as the positron that wiki speaks of? the electron. The positron has an electric charge of +1, a spin of 1/2, > and the same mass as an electron. When a low-energy positron collides > with a low-energy electron, annihilation occurs, resulting in the > production of two gamma ray photons (see electron-positron > annihilation). > - http://en.wikipedia.org/wiki/Positron > I have not ignored your annihilation quandry. There is a huge difference between production of positrons in a accelerator and then The positrons I speak of is from the Dirac Sea. Wikipedia has an excellent page on it where I quote the first paragraph: --- quoting Wikipedia on Dirac Sea --- Dirac sea Jump to: navigation, search The Dirac sea is a theoretical model of the vacuum as an infinite sea physicist Paul Dirac in 1930 to explain the anomalous negative-energy quantum states predicted by the Dirac equation for relativistic electrons. The positron, the antimatter counterpart of the electron, was originally conceived of as a hole in the Dirac sea, well before its experimental discovery in 1932. Dirac, Einstein and others recognised that it is related to the 'metaphysical' aether [1]: ... with the new theory of electrodynamics we are rather forced to have an aether. - P.A.M. Dirac, 'Is There An Aether?,' Nature, v.168, 1951, p.906. The equation relating energy, mass and momentum in special relativity is: E2 = p2c2 + m2c4, equation reduces to E2 = m2c4, which is usually quoted as the familiar E = mc2. However, this is a simplification because, while x*x = x2, we can also see that (-x)*(-x)= x2. Therefore, the correct equation to use to relate energy and mass in the Hamiltonian of the Dirac equation is: E = ? mc2. Here the negative solution is antimatter, discovered by Carl Anderson as the positron. The interpretation of this result requires a Dirac sea, showing that the Dirac equation is not merely a combination of special relativity and quantum field theory, but it also implies that --- end quoting --- The sense in which I speak of positrons is the sense of Dirac's Sea. Where SPACE is the same as a sea or ocean of positrons. And this space does not annihilate with the electrons of ordinary matter. Why? I co-mingled without annihilation. This is how a Monopole would act, according to Dirac on pages 45,46 of this Directions in Physics book. So in answer to your question, Space full of positrons, or Sea of Positrons or Ocean of Positrons is very different from a positron produced in a experimental accelerator lab which annihilates a electron. It has been proven by experimental physicists that the vacuum of Space is loaded with positrons and these are called holes. So Space is not empty and not a vacuum but can be drawn out of as many positrons as one desires to have positrons. Dirac never had the Atom Totality theory to work with. But if you put the Dirac Sea of Positrons with the Atom Totality theory then you solve what gravity is. Have a look at the Wikipedia page on Dirac Sea for it has alot more information about Space being positrons. Archimedes Plutonium www.iw.net/~a plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Space is not a vacuum but Dirac's sea of positrons and that is gravity itself <4so8p2F10smqjU2@mid.individual.net> <4spi5bF10pgsfU1@mid.individual.net You've ignored this does not annihilate stance in my first question. > But since I find it here I now have to ask you is the positron that you > speak of the same as the positron that wiki speaks of? the electron. The positron has an electric charge of +1, a spin of 1/2, > and the same mass as an electron. When a low-energy positron collides > with a low-energy electron, annihilation occurs, resulting in the > production of two gamma ray photons (see electron-positron > annihilation). > - http://en.wikipedia.org/wiki/Positron > I have not ignored your annihilation quandry. There is a huge > difference between production of positrons in a accelerator and then The positrons I speak of is from the Dirac Sea. Wikipedia has an > excellent page on it where I quote the first paragraph: > --- quoting Wikipedia on Dirac Sea --- > Dirac sea > Jump to: navigation, search The Dirac sea is a theoretical model of the vacuum as an infinite sea > physicist Paul Dirac in 1930 to explain the anomalous negative-energy > quantum states predicted by the Dirac equation for relativistic > electrons. The positron, the antimatter counterpart of the electron, > was originally conceived of as a hole in the Dirac sea, well before its > experimental discovery in 1932. Dirac, Einstein and others recognised > that it is related to the 'metaphysical' aether [1]: ... with the new theory of electrodynamics we are rather forced to have > an aether. - P.A.M. Dirac, 'Is There An Aether?,' Nature, v.168, > 1951, p.906. The equation relating energy, mass and momentum in special relativity > is: E2 = p2c2 + m2c4, equation reduces to E2 = m2c4, which is usually quoted as the familiar > E = mc2. However, this is a simplification because, while x*x = x2, we > can also see that (-x)*(-x)= x2. Therefore, the correct equation to use > to relate energy and mass in the Hamiltonian of the Dirac equation is: E = ± mc2. Here the negative solution is antimatter, discovered by Carl Anderson > as the positron. The interpretation of this result requires a Dirac > sea, showing that the Dirac equation is not merely a combination of > special relativity and quantum field theory, but it also implies that > --- end quoting --- The sense in which I speak of positrons is the sense of Dirac's Sea. > Where SPACE is the same as a sea or ocean of positrons. And this space > does not annihilate with the electrons of ordinary matter. Why? I > co-mingled without annihilation. This is how a Monopole would act, > according to Dirac on pages 45,46 of this Directions in Physics book. So in answer to your question, Space full of positrons, or Sea of > Positrons or Ocean of Positrons is very different from a positron > produced in a experimental accelerator lab which annihilates a > electron. It has been proven by experimental physicists that the vacuum of Space > is loaded with positrons and these are called holes. So Space is not > empty and not a vacuum but can be drawn out of as many positrons as one > desires to have positrons. Dirac never had the Atom Totality theory to work with. But if you put > the Dirac Sea of Positrons with the Atom Totality theory then you solve > what gravity is. Have a look at the Wikipedia page on Dirac Sea for it has alot more > information about Space being positrons. Archimedes Plutonium > www.iw.net/~a plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies I've spent a little bit of time reviewing the Dirac Sea. It's a bit confusing how to get from its abstract supposition to the actuality of the positron and still see it the way that it was conceived. Is this a sea of electron-positron pairs? No, I suppose not; that would be more the modern quantum view. A sea of negative energy... It's quite confusing which the electron is and which the positron is and if they existence and it is admitted to anihillate with the electron I am fully confused. from http://www.answers.com/topic/dirac-sea I see: In relativistic quantum mechanics, the completely filled, negative energy electron state that comprises a vacuum. If a negative energy electron is promoted to a positive energy state, the hole is perceived as a positron. This implies an inversion of what you are claiming. I am looking for something similar and here Dirac simply is getting electrons for free. The special places are where the electron isn't. These are positrons. I need to stew on this for a while. You've done a nice job of holding up your argument. Some shy away and fade out but you are owning your argument and you deserve credit for that. I can't say that I adopt your model but am happy to admit that some of it may be helpful for my own. That is what we should all be doing here. Scavenging and composting, exploring and building. I am trying to play with what looks like a gravity/thermodynamic construction of a simplex oscillator on a thread titled The Unity Problem. But getting charge into it is pushing toward something like a plows through space. This sort of impedance could set up some standing be acceptable under the delta emitter since it is already an endlessly active source. Anyhow, your Dirac Sea model is a little bit akin to -Tim === Subject: Re: Space is not a vacuum but Dirac's sea of positrons and that is gravity itself On 29 Nov 2006 20:34:48 -0800, a_plutonium You've ignored this does not annihilate stance in my first question. >> But since I find it here I now have to ask you is the positron that you >> speak of the same as the positron that wiki speaks of? >> the electron. The positron has an electric charge of +1, a spin of 1/2, >> and the same mass as an electron. When a low-energy positron collides >> with a low-energy electron, annihilation occurs, resulting in the >> production of two gamma ray photons (see electron-positron >> annihilation). >> - http://en.wikipedia.org/wiki/Positron I have not ignored your annihilation quandry. There is a huge >difference between production of positrons in a accelerator and then The positrons I speak of is from the Dirac Sea. Wikipedia has an >excellent page on it where I quote the first paragraph: >--- quoting Wikipedia on Dirac Sea --- >Dirac sea >Jump to: navigation, search The Dirac sea is a theoretical model of the vacuum as an infinite sea >physicist Paul Dirac in 1930 to explain the anomalous negative-energy >quantum states predicted by the Dirac equation for relativistic >electrons. The positron, the antimatter counterpart of the electron, >was originally conceived of as a hole in the Dirac sea, well before its >experimental discovery in 1932. Dirac, Einstein and others recognised >that it is related to the 'metaphysical' aether [1]: ... with the new theory of electrodynamics we are rather forced to have >an aether. - P.A.M. Dirac, 'Is There An Aether?,' Nature, v.168, >1951, p.906. The equation relating energy, mass and momentum in special relativity >is: E2 = p2c2 + m2c4, equation reduces to E2 = m2c4, which is usually quoted as the familiar >E = mc2. However, this is a simplification because, while x*x = x2, we >can also see that (-x)*(-x)= x2. Therefore, the correct equation to use >to relate energy and mass in the Hamiltonian of the Dirac equation is: E = ± mc2. Here the negative solution is antimatter, discovered by Carl Anderson >as the positron. The interpretation of this result requires a Dirac >sea, showing that the Dirac equation is not merely a combination of >special relativity and quantum field theory, but it also implies that >--- end quoting --- The sense in which I speak of positrons is the sense of Dirac's Sea. >Where SPACE is the same as a sea or ocean of positrons. And this space >does not annihilate with the electrons of ordinary matter. Why? I >co-mingled without annihilation. This is how a Monopole would act, >according to Dirac on pages 45,46 of this Directions in Physics book. So in answer to your question, Space full of positrons, or Sea of >Positrons or Ocean of Positrons is very different from a positron >produced in a experimental accelerator lab which annihilates a >electron. It has been proven by experimental physicists that the vacuum of Space >is loaded with positrons and these are called holes. So Space is not >empty and not a vacuum but can be drawn out of as many positrons as one >desires to have positrons. Dirac never had the Atom Totality theory to work with. But if you put >the Dirac Sea of Positrons with the Atom Totality theory then you solve >what gravity is. Have a look at the Wikipedia page on Dirac Sea for it has alot more >information about Space being positrons. Archimedes Plutonium >www.iw.net/~a_plutonium >whole entire Universe is just one big atom >where dots of the electron-dot-cloud are galaxies If you want to see the content of quantum vacuum see my paper #1 on http://www.dualspace.net Dirac deduced a sea of electrons simply from the minus sign in the total energy equation. He did very little with it. My analysis demanded to know how empty space could have 8.8uuF/m. The results are amazing, a dozen significant values compared to Dirac's hole or electron going backward. Vastly stiffer than steel, transmission velocity = c, cell size is Compton WL x alpha. I'm sorry it requires a fair amount of study. The stars of the universe came from expelled electrons at speed of light. The expansion by 1/alphacubed = 2.5 million reduces the density to that of iron. The energy density expands to iron's density at the speed of light. Excuse the turgid exposition. (I don't buy the one big atom, though). John Polasek === Subject: Re: Humans Lack Genetics to Understand Physics game on TV. Did Dick Feynman ever do that? No, Since Fetmann was by his own > admission, a mathematician, not a phycist, Not so, he said he worked through intuition. Einstein commented similarly, that his strength was insight into nature, and that math was a means to an end. It would be interesting to survey professional scientists, to find out how many regard themselves primarily as mathematicians. > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than physics. OK -- Rich === Subject: Re: Humans Lack Genetics to Understand Physics game on TV. Did Dick Feynman ever do that? No, Since Fetmann was by his own > admission, a mathematician, not a phycist, Not so, he said he worked through intuition. Einstein > commented similarly, that his strength was insight into > nature, and that math was a means to an end. It would be interesting to survey professional scientists, > to find out how many regard themselves primarily as mathematicians. Not many, after Cauchy kicked most of the cranks out of the math club, > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than physics. > > OK > > > -- > Rich === Subject: generalized product rule Hello. I try to prove nth derivative of the product h of two function f and g: h^(n)(x) = sum( [n,k]*f^(k)(x)*g^(n-k)(x), from k = 0 to n ) where [n,k] = n! / (k! * (n-k)!) I proved the base case n = 1, but get stuck in the induction step. I suppose the above formula holds for n, and I try to obtain the n+1 formula: h^(n+1) = sum( [n,k] * (f^(k)*g^(n-k))', from k = 0 to n ) = sum( [n,k] * [ (f^(k+1)*g^(n-k) + f^(k)*g^(n+1-k) ], from k = 0 to n ) = sum( [n,k] * [ (f^(k+1)*g^(n-k) ], from k = 0 to n ) + sum( [n,k] * [ f^(k)*g^(n+1-k) ], from k = 0 to n ) what is the next step? === Subject: Re: generalized product rule > Hello. I try to prove nth derivative of the product h of two function > f and g: > > h^(n)(x) = sum( [n,k]*f^(k)(x)*g^(n-k)(x), from k = 0 to n ) > > where [n,k] = n! / (k! * (n-k)!) Fix n and call the unknown coefficients in the formula c_k. Now fix j in {0,1,...,n}. Set f(x) = x^j, g(x) = x^(n-j). Then h(x) = x^n. Hence h^n(0) = n! = sum(k=0,n) c_kf^(k)(0)*g^(n-k)(0). But clearly the last sum has only one nonzero term, namely when k = j. That term is c_j*j!*(n-j)! Therefore c_j = n!/(j!*(n-j)!). === Subject: Re: generalized product rule > Hello. I try to prove nth derivative of the product h of two function > f and g: h^(n)(x) = sum( [n,k]*f^(k)(x)*g^(n-k)(x), from k = 0 to n ) where [n,k] = n! / (k! * (n-k)!) I proved the base case n = 1, but get stuck in the induction step. I > suppose the above formula holds for n, and I try to obtain the n+1 > formula: h^(n+1) = sum( [n,k] * (f^(k)*g^(n-k))', from k = 0 to n ) No, it isn't; this is a cross between two true formulas. You know that h^(n) = sum([n,k] f^(k) g^(n-k), for i = 0 to n) and want to show that h^(n+1) = sum([n+1,k] f^(k) g^(n+1-k), for i = 0 to n+1). > what is the next step? Take your assumption and take the derivative of that: h^(n+1) = diff( sum([n,k] f^(k) g^(n-k), for i = 0 to n)) = sum([n,k] diff(f^(k) * g^(n-k)), for i = 0 to n)) = sum([n,k] (f^(k)*g^(n-k+1) + g^(n-k)*f^(k+1)), i = 0 to n)) Now the messy part: You need to combine these terms so that you get the formula you want. This relies heavily on the fact that [n+1,k] = [n,k] + [n,k-1]. After regrouping, you're done. --- Christopher Heckman === Subject: solution to a complex equation.... This is a calculation for the present value of a growing annuity AND a certain amount of money in the future. The use of the equation is to figure out how much of initial cash-distribution you can take from your retirement fund (or an annuity), given a certain growth rate in the cash-distribution, rate of return, time until last distribution, and amount of money you want to have left over... PV = C * [1 + (1+g)^t/(1+r)^t] / (r-g) + FV / (1+r)^t it is easy to solve this equation for FV or for C...however, I've been trying to solve it for r and g, and it's just driving me nuts. Can anyone here solve this equation for r, and also for g? (that is, two separate solutions solving for those different variables). Note, PV and FV are one variable not two. === Subject: Re: solution to a complex equation.... > This is a calculation for the present value of a growing annuity AND a > certain amount of money in the future. The use of the equation is to > figure out how much of initial cash-distribution you can take from your > retirement fund (or an annuity), given a certain growth rate in the > cash-distribution, rate of return, time until last distribution, and > amount of money you want to have left over... > > PV = C * [1 + (1+g)^t/(1+r)^t] / (r-g) + FV / (1+r)^t > > it is easy to solve this equation for FV or for C...however, I've been > trying to solve it for r and g, and it's just driving me nuts. There is no general solutions for either r or g, except for a very few small integer values of t. The usual methods of solution for either g or r are by using numerical approximation techiques, given numerical values for all the other variables except the one to be found. === Subject: hint on derivative problem Suppose f has a finite derivative in (a,b) and f(x)-->oo as x-->b from the left. Prove that the lim (f)'(x) does not exist or is infinite as As an example, I'm thinking of something like f(x) = 1/x, then the derivative tends to -infinity as x goes to zero. Or some situation where the derivative is oscillating real fast at a point (e.g., sin(1/x)) so that the limit of the derivative does not exist. But can someone give me a hint on how to prove this? It is pretty abstract in that I don't have anything concrete to work with (e.g., to find two subsequences converging to different limits to show the limit does not exist). === Subject: Re: hint on derivative problem > Suppose f has a finite derivative in (a,b) and f(x)-->oo as x-->b from > the left. Prove that the lim (f)'(x) does not exist or is infinite as > x-->b from the left. As an example, I'm thinking of something like f(x) = 1/x, then the > derivative tends to -infinity as x goes to zero. This isn't an example of what you're trying to prove, since f(x) -> -infinity as x approaches 0 from the left. A better example is f(x) = -1/x, where b = 0. > Or some situation > where the derivative is oscillating real fast at a point (e.g., > sin(1/x)) so that the limit of the derivative does not exist. But can someone give me a hint on how to prove this? It is pretty > abstract in that I don't have anything concrete to work with (e.g., to > find two subsequences converging to different limits to show the limit > does not exist). I'd try contradiction here; if the conclusion is false, then lim(f'(x)) exists as x approaches b from the left. It looks like something along the lines of the Mean Value Theorem is required here. --- Christopher Heckman === Subject: Re: Unimodality of Dirichlet distribution Herman, Can I combine multiple Dirichlets to construct a distribution with multi finite peak? Or, is there any ready known distribution in good performance? Sum(vi)=1, i=1..N and several peaks (might > N) I want to use it in an inverse problem. Rebecca >Hi gurus, >I'm confused at the following problem: >Is it true that N-dimensional Dirichlet distribution can at most have >one peak? It can have at most one finite peak. It can have up to > N infinite peaks, depending on the parameters. If it has > an infinite peak, it has no finite peak. There is only one max extrema value that Dirichlet distribution is able >to describe? For finite values, yes. >So, a multivariate probability distribution (with all elements add up >to one) can not be described by one dirichlet distribution. Am I right? No. Jamaica -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Adaptive euler heun method for soliving a diff eqn Hi , I have the following code.............. function [t y] = adapt1(FunFcn,tspan,y0,abstol,reltol,hmax,hmin) h = sqrt(hmax*hmin); i=0; yold = y0; told = tspan(1); tfinal = tspan(2); tnew = told + h; while tnew <= tfinal k1 = feval(FunFcn, told, yold); % Note k1 is a column vector yeuler = yold + h*k1; % yeuler is a column vector k2 = feval(FunFcn, tnew, yeuler); yheun = yold + .5*h*k1 + .5*h*k2; % column vector reject = 0; for j=1:length(yheun) est(j) = abs(yheun(j)-yeuler(j))/h; tol(j) = max(reltol*abs(yheun(j)),abstol); gam(j) = tol(j)/est(j); end if max(est-tol) > 0 reject =1; end gamma = min(gam); if reject == 0 i = i+1; tnew = told + h; tr(i) = tnew; y(i,:) = yheun'; told = tnew; yold = yheun; ht = min(h*0.8*gamma,hmax); h = min(ht,5*h); else ht = max(h*0.8*gamma,hmin); h = max(ht,h/10); % Do not allow step size smaller than hmin % and do not decrease step size by more than a factor of 10 end end function F = testfcn(t,y) F = 19/80*y(1,:).*y(1,:)*exp(-t/4); Executed using the foll command [y t] = adapt1('testfcn',[0 15],0,0.00001,0.01,0.00001,1) However I am stuck in some infinite loop..............dont know how to get it fixed......... I even need to decide on the right initial h.............. === Subject: ? nilpotent Hi: Are there ways other than actually multiplying a matrix to check if it is nilpotent? by Cheng Cosine Nov/30/2k6 NC === Subject: Re: ? nilpotent > Hi: Are there ways other than actually multiplying a matrix to check if it is nilpotent? by Cheng Cosine > Nov/30/2k6 NC I think the quickest way in practice would be to check that the characteristic polynomial is x^n, which is what others have been suggesting. To do this, start with any nonzero (column) vector v1, and compute v2 = A v1, v3 = A v2, ... If you get to v_{n+1} ( = A^n v) with all vi nonzero, then A is not nilpotent. Otherwise, you get v_{r+1} = 0 for some r. If r=n then A is nilpotent. If r < n, then let V1 be the subspace spanned by v1,...,vr. Choose v_{r+1} not in V1, and start again computing v_{r+2} = A v_{r+1} etc. This time you stop as soon as you find some v_{s+1} in V1, and if you get to v_{n+1} without this happening, then A is not nilpotent. If s=n then A is nilpotent. If s < n, set V2 to be subspace spanned by v1,...,vs, etc etc. You echelonize the subspace spanned by v1,v2,v3,... as you go in order to facilitate membership testing in the subspace. As I said above, this is checking that the characteristic polynomial of A is x^n. This will not tell you the smallest m with A^m = 0. For that, you need the minimal polynomial, which can be computed in a similar fashion, but is slightly more complicated, because you cannot stop when you find a vi in the subspace found so far, but have to carry on and locate the power of A which kills the vector. Derek Holt. === Subject: Re: ? nilpotent > Hi: Are there ways other than actually multiplying a matrix to check if it is nilpotent? by Cheng Cosine > Nov/30/2k6 NC I think the quickest way in practice would be to check that the characteristic polynomial is x^n, which is what others have been suggesting. To do this, start with any nonzero (column) vector v1, and compute v2 = A v1, v3 = A v2, ... If you get to v_{n+1} ( = A^n v) with all vi nonzero, then A is not nilpotent. Otherwise, you get v_{r+1} = 0 for some r. If r=n then A is nilpotent. If r < n, then let V1 be the subspace spanned by v1,...,vr. Choose v_{r+1} not in V1, and start again computing v_{r+2} = A v_{r+1} etc. This time you stop as soon as you find some v_{s+1} in V1, and if you get to v_{n+1} without this happening, then A is not nilpotent. If s=n then A is nilpotent. If s < n, set V2 to be subspace spanned by v1,...,vs, etc etc. You echelonize the subspace spanned by v1,v2,v3,... as you go in order to facilitate membership testing in the subspace. As I said above, this is checking that the characteristic polynomial of A is x^n. This will not tell you the smallest m with A^m = 0. For that, you need the minimal polynomial, which can be computed in a similar fashion, but is slightly more complicated, because you cannot stop when you find a vi in the subspace found so far, but have to carry on and locate the power of A which kills the vector. Derek Holt. === Subject: Re: ? nilpotent > >>Hi: >> Are there ways other than actually multiplying a matrix >>to check if it is nilpotent? >>by Cheng Cosine >> Nov/30/2k6 NC > > > I think the quickest way in practice would be to check that the > characteristic polynomial is x^n, which is what others have been > suggesting. To do this, start with any nonzero (column) vector v1, and > compute > > v2 = A v1, v3 = A v2, ... > > If you get to v_{n+1} ( = A^n v) with all vi nonzero, then A is not > nilpotent. > > Otherwise, you get v_{r+1} = 0 for some r. > > If r=n then A is nilpotent. If r < n, then let V1 be the subspace > spanned by v1,...,vr. Choose v_{r+1} not in V1, and start again > computing v_{r+2} = A v_{r+1} etc. This time you stop as soon as you > find some v_{s+1} in V1, and if you get to v_{n+1} without this > happening, then A is not nilpotent. > > If s=n then A is nilpotent. If s < n, set V2 to be subspace spanned by > v1,...,vs, etc etc. You echelonize the subspace spanned by v1,v2,v3,... > as you go in order to facilitate membership testing in the subspace. > > As I said above, this is checking that the characteristic polynomial of > A is x^n. This will not tell you the smallest m with A^m = 0. For that, > you need the minimal polynomial, which can be computed in a similar > fashion, but is slightly more complicated, because you cannot stop when > you find a vi in the subspace found so far, but have to carry on and > locate the power of A which kills the vector. > > Derek Holt. > It seems to me that this is effectively the same as computing the powers of A. === Subject: Re: ? nilpotent >Hi: >> Are there ways other than actually multiplying a matrix >>to check if it is nilpotent? >>by Cheng Cosine >> Nov/30/2k6 NC > I think the quickest way in practice would be to check that the > characteristic polynomial is x^n, which is what others have been > suggesting. To do this, start with any nonzero (column) vector v1, and > compute v2 = A v1, v3 = A v2, ... If you get to v_{n+1} ( = A^n v) with all vi nonzero, then A is not > nilpotent. Otherwise, you get v_{r+1} = 0 for some r. If r=n then A is nilpotent. If r < n, then let V1 be the subspace > spanned by v1,...,vr. Choose v_{r+1} not in V1, and start again > computing v_{r+2} = A v_{r+1} etc. This time you stop as soon as you > find some v_{s+1} in V1, and if you get to v_{n+1} without this > happening, then A is not nilpotent. If s=n then A is nilpotent. If s < n, set V2 to be subspace spanned by > v1,...,vs, etc etc. You echelonize the subspace spanned by v1,v2,v3,... > as you go in order to facilitate membership testing in the subspace. As I said above, this is checking that the characteristic polynomial of > A is x^n. This will not tell you the smallest m with A^m = 0. For that, > you need the minimal polynomial, which can be computed in a similar > fashion, but is slightly more complicated, because you cannot stop when > you find a vi in the subspace found so far, but have to carry on and > locate the power of A which kills the vector. Derek Holt. > It seems to me that this is effectively the same as computing the powers > of A. I think what I describe is faster than simply computing powers for large values of n. I am computing A v for at most n column vectors v. In addition I have to do a certain amount of echelonization. To compute A^n explicitly, you have to do about log(n) matrix multiplications, and each individual matrix multiplication involves computing Av for n column vectors v. I realize this is an oversimplification. For example matrix multiplication is highly parallelizable, and there are faster methods (like Strassen) for doing it. Simply computing powers might also be quicker if it turend out that there was a very small m with A^m = 0. Derek Holt. === Subject: Re: ? nilpotent > Hi: Are there ways other than actually multiplying a matrix to check if it is nilpotent? by Cheng Cosine > Nov/30/2k6 NC >>I think the quickest way in practice would be to check that the characteristic polynomial is x^n, which is what others have been suggesting. To do this, start with any nonzero (column) vector v1, and compute >v2 = A v1, v3 = A v2, ... >If you get to v_{n+1} ( = A^n v) with all vi nonzero, then A is not nilpotent. >Otherwise, you get v_{r+1} = 0 for some r. >If r=n then A is nilpotent. If r < n, then let V1 be the subspace spanned by v1,...,vr. Choose v_{r+1} not in V1, and start again computing v_{r+2} = A v_{r+1} etc. This time you stop as soon as you find some v_{s+1} in V1, and if you get to v_{n+1} without this happening, then A is not nilpotent. >If s=n then A is nilpotent. If s < n, set V2 to be subspace spanned by v1,...,vs, etc etc. You echelonize the subspace spanned by v1,v2,v3,... as you go in order to facilitate membership testing in the subspace. >As I said above, this is checking that the characteristic polynomial of A is x^n. This will not tell you the smallest m with A^m = 0. For that, you need the minimal polynomial, which can be computed in a similar fashion, but is slightly more complicated, because you cannot stop when you find a vi in the subspace found so far, but have to carry on and locate the power of A which kills the vector. >Derek Holt. It seems to me that this is effectively the same as computing the powers >>of A. > > > I think what I describe is faster than simply computing powers for > large values of n. I am computing A v for at most n column vectors v. > In addition I have to do a certain amount of echelonization. > > To compute A^n explicitly, you have to do about log(n) matrix > multiplications, and each individual matrix multiplication involves > computing Av for n column vectors v. Yes, you are correct. Sorry for the misunderstanding. Stephen === Subject: Re: ? nilpotent > Hi: > > Are there ways other than actually multiplying a matrix > > to check if it is nilpotent? > > by Cheng Cosine > Nov/30/2k6 NC Being singular is a necessary, but not sufficient, condition. === Subject: Re: ? nilpotent > Hi: Are there ways other than actually multiplying a matrix to check if it is nilpotent? by Cheng Cosine > Nov/30/2k6 NC ****************************************** Hi: If zero is its only eigenvalue, say. Tonio === Subject: Re: ? nilpotent >> Are there ways other than actually multiplying a matrix >> to check if it is nilpotent? > If zero is its only eigenvalue, say. ... in an algebraically complete extension of the field you're working with. Jose Carlos Santos === Subject: Re: ? nilpotent <4t80fqF12vhf0U1@mid.individual.net >> Are there ways other than actually multiplying a matrix >> to check if it is nilpotent? > If zero is its only eigenvalue, say. ... in an algebraically complete extension of the field you're working > with. > Jose Carlos Santos ******************************************** Uh? What does algebraically complete extension mean here, please? Anyway, I think there's no need to go out of the field we're working on: zero is the unique eigenvector of a given operator (matrix) iff x^m is the operator's char. polynomial, and this is always defined over any field. Perhaps I'm missing something here... Tonio === Subject: Re: ? nilpotent days. My association with the Department is that of an alumnus. >> Are there ways other than actually multiplying a matrix > to check if it is nilpotent? If zero is its only eigenvalue, say. >> ... in an algebraically complete extension of the field you're working >> with. >> Jose Carlos Santos >******************************************** >Uh? What does algebraically complete extension mean here, please? An extension which is algebraically closed, presumably. >Anyway, I think there's no need to go out of the field we're working >on: zero is the unique eigenvector of a given operator (matrix) iff x^m >is the operator's char. Consider the matrix over the real numbers ( 0 -1 0) ( 1 0 0) ( 0 0 0) It has characteristic polynomial x(x^2+1), and therefore, 0 is the only eigenvalue (over the field in question). But the matrix is not nilpotent. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: ? nilpotent <4t80fqF12vhf0U1@mid.individual.net> ........................................................... > An extension which is algebraically closed, presumably. Anyway, I think there's no need to go out of the field we're working >on: zero is the unique eigenvector of a given operator (matrix) iff x^m >is the operator's char. Consider the matrix over the real numbers ( 0 -1 0) > ( 1 0 0) > ( 0 0 0) It has characteristic polynomial x(x^2+1), and therefore, 0 is the > only eigenvalue (over the field in question). But the matrix is not nilpotent. > ************************************************* Point taken: then Santos surely meant alg. closed. And in order to make things clearer I supose it'd be better to say that A is nilpotent iff its char. pol if x^m, what would rule out the above example. Thanx Tonio === Subject: Re: ? nilpotent >> An extension which is algebraically closed, presumably. Anyway, I think there's no need to go out of the field we're working on: zero is the unique eigenvector of a given operator (matrix) iff x^m is the operator's char. >> Consider the matrix over the real numbers >> ( 0 -1 0) >> ( 1 0 0) >> ( 0 0 0) >> It has characteristic polynomial x(x^2+1), and therefore, 0 is the >> only eigenvalue (over the field in question). >> But the matrix is not nilpotent. > ************************************************* > Point taken: then Santos surely meant alg. closed. Indeed I did! Jose Carlos Santos === Subject: Re: ? nilpotent days. My association with the Department is that of an alumnus. ........................................................... >> An extension which is algebraically closed, presumably. >>Anyway, I think there's no need to go out of the field we're working >>on: zero is the unique eigenvector of a given operator (matrix) iff x^m >>is the operator's char. >> Consider the matrix over the real numbers >> ( 0 -1 0) >> ( 1 0 0) >> ( 0 0 0) >> It has characteristic polynomial x(x^2+1), and therefore, 0 is the >> only eigenvalue (over the field in question). >> But the matrix is not nilpotent. >************************************************* >Point taken: then Santos surely meant alg. closed. He meant what he said. If the only eigenvalue of the matrix is 0, ->over an algebraically closed extension of the field of definition<-, field of definition to be algebraically closed, but you need 0 to be the only root in ->any<- extension of the field of definition for your argument to apply. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Super size What is it ? how big they can be as you have seen ? === Subject: Re: Turing vs. Godel (Newbie Question) So, just what is the status of your claim that no-one before CB produced a >> system with axioms, rules of inference, formal queries to the system, >> proofs in the system and corresponding executable programs? None with any general capability. Your name is Smaill but your brain > is Small. :) I'll take this as CB's less than gracious way of admitting that yes, No, I said None. 1. Have you seen any examples of any programs that any system outside of CBL has ever produced? 2. Can you see anything wrong with the derivation of computer programs using CBL described in my ARXIV paper? The answers are: (1) No, you have no examples from any other systems, and (2) No, there is nothing wrong with the CBL derivation. The true conclusion is that CBL does what other systems do not. Agreed? Why not? Isn't the proof of this obvious (left as an exercised for the reader)? In contrast to the bullters, I will answer every question to any depth of detail, to show what CBL does, so you can see that it is for real. (The bullters use all sorts of tactics to hide the truth - that their system does nothing - by including non-standard terminology, unexplained expressions, violating the rules of the problem, refusing to answer, etc.) C-B > after all, he was not the first to produced a system with axioms, > rules of inference, formal queries to the system, proofs in the system > and corresponding executable programs. > > C-B > -- > Alan Smaill === Subject: Deep Thoughts # 28. The Meaning of Truth (Finally!) Deep Thoughts # 28. The Meaning of Truth (Finally!) Let's suppose that someone wants to get information from you and he may or may not know what questions you know the answer to. Every question is numbered and has a single answer, yes or no. One question is my secret. I want to answer my secret question falsely, to keep it a secret, but answer all other questions truthfully so as to not get in trouble as explained below, and in a way that he can't figure out what question is my secret. My answers may or may not depend on his past questions and my answers (which may contain randomness) after we start. It is also your responsibility to correct any mistakes made (with any consistent rules) in the rules given here. For example, questions include, 1. Is your secret Does 1 plus 1 equal 2?? 2. Is 3 the number of your secret question? 3. What is your answer to question # 3? 4. Do you always answer question # 3 correctly? 5. Is there a number whose question you answer falsely that is less than all numbers whose question you might not answer truthfully? 6. Is there an even number question that you answer falsely? For example, he might have a procedure that stops iff question # 3 is never answered by me correctly (# 4.) So he has a procedure for finding out information about procedures that I use to answer questions. He could also relate to my use of procedures to ask questions, especially in pursuit of a particular goal (e.g. something impossible but I don't know if it is impossible.) If I am ever inconsistent or do not answer within 1 second, then he knows I am lying and will expel me. All of my calculations take a total of .1 second, so after 1 second I can have answered the question correctly if I am able. I may guess but only if I can't tell the true answer. I never believe the wrong answer. I want to be consistent in order to continue in the game. I also do not want to tell him the truth when he asks me my secret question. How can I be consistent but not always tell him the truth? C-B Conjecture: The answer is a particular procedure that is strongly related to the original question above. C-B (Note that only a small subset of the above is needed to formalize conventional Mathematics e.g. theorem-proving using Peano Arithmetic or ZFC, while as written the above allows us to generate and discover other systems much more general than conventional Mathematics.) Find the function: (Current point in time) => (A,B,C,D,E,F,G) where: Each person has a number A that is assigned when he is born. Thus the current number B of A is always exactly the number of people who have ever been born. People also have a number C that indicates the total effort they will make throughout their life. While people are random, there is a formula D for what number C the Ath person will get, for any A. However, this formula may be replaced by a second formula E when formula F is true, where A is still the person number. So one or the other of the two formulas D or E gives the correct value. There is also a formula G for which formulas are to be used currently. There is always exactly one set of formulas that can be in use and work correctly There is also a total number that is the number of true statements answered by people (there is only one way to ask and answer questions, and we know what it is and who will be asking and who will be answering at each point in time or current person number.) Truth is the current value i.e. truth varies through time. C-B === Subject: Re: Deep Thoughts # 28. The Meaning of Truth (Finally!) Deep Thoughts # 28. The Meaning of Truth (Finally!) > POTM === Subject: Re: Deep Thoughts # 28. The Meaning of Truth (Finally!) Deep Thoughts # 28. The Meaning of Truth (Finally!) > POTM SLABTM === Subject: Re: integral of exp(-x^2) from 0 to infinity Does anyone know how to calculate exp(-x^2) from 0 to infinity? I can't use the FTC since exp(-x^2) does not have antiderivative in > terms of elem. functions. > Should I use FTC on the taylor series of the function? I know this is a classic example that I actually saw a prof do in a > past class, but now I am stumped. > === Subject: Re: integral of exp(-x^2) from 0 to infinity > Does anyone know how to calculate exp(-x^2) from 0 to infinity? > > I can't use the FTC since exp(-x^2) does not have antiderivative in > terms of elem. functions. > > > Should I use FTC on the taylor series of the function? > > I know this is a classic example that I actually saw a prof do in a > past class, but now I am stumped. The usual way to evdaluate I = Int_0^oo exp(-x^2) dx is a sneaky trick: I^2 = Int_0^oo Int0^oo exp(-x^2) exp( -y^2) dy dx = Int_0^p(pi/2) d(theta) Int_0^oo exp(-r^2) dr === Subject: Re: integral of exp(-x^2) from 0 to infinity Does anyone know how to calculate exp(-x^2) from 0 to infinity? I can't use the FTC since exp(-x^2) does not have antiderivative in > terms of elem. functions. > Should I use FTC on the taylor series of the function? I know this is a classic example that I actually saw a prof do in a > past class, but now I am stumped. Others posted the use of polar coordinates. I can avoid them as follows: The square A^2 of the integral A is again A^2 = int[x>0, y>0] exp(-x^2-y^2) dx dy but I calculate the half of it twice by A^2 = 2 * int[0(Az:zex<->zey) A2) ExAy ~(yex) A3) If |- (Ex)P(x), then |- ((Ax)(P(x) => x in x) or |- (Ax)(P(x) => x not in x)) were P(x) be a property with one free variable. Zuhair === Subject: Half integer Chebyshev solutions The differential equation named after Pafnuty Chebyshev is: (1-x^2) d^2T_n(x)/dx^2 - x dT_n(x)/dx + n^2 T_n(x) = 0 The integer solutions S_n(x) for n = 0,1,2,3 ... N are well known: S_n(x) = A T_n(x) + B sqrt(1-x^2) U_(n-1)(x) Where T_n(x) and U_n(x) are polynomials, defined by: T_0(x) = 1 U_0(x) = 1 T_1(x) = x U_1(x) = 2 x And the following recurrence relation, which is valid for both T_n(x) and U_n(x): S_(n+1)(x) = 2 x S_n(x) - S_(n-1)(x) But it seems that the general solution for n = 0 is less well known: S_0(x) = A + B arccos(x) And it's virtually impossible to find (on the web) that there also do exist _Half Integer Solutions_ of the Chebyshev equation, where n = 1/2,3/2,5/2,7/2 ... (2m+1)/2 ... They are: S_((2m+1)/2)(x) = A P_m(x).sqrt(1+x) + B Q_m(x).sqrt(1-x) Where P_m(x) and Q_m(x) are polynomials, defined by: P_0(x) = 1 Q_0(x) = 1 P_1(x) = 2x - 1 Q_1(x) = 2x + 1 And the following recurrence relation, which is valid for both P_m(x) and Q_m(x): S_(m+1)(x) = 2 x S_m(x) - S_(m-1)(x) Question: have these half integer solutions been found before? Or am I the first one who has discovered them? I consider the last possibility as highly improbable, but would like to know it for sure ... Oh, yeah. It's all in the following writeup about Chebyshev and stuff: http://hdebruijn.soo.dto.tudelft.nl/jaar2006/pafnuty.pdf Han de Bruijn === Subject: Vlcek - EINSTEIN comparison Calculation of the kinetic energy T /kin/ of a body moving at the velocity of v see on www.vlcek.euweb.cz/book.php?version=uk&page=3-2 Vlcek ?s theory and Einstein ?s theory v/c T /kin/ compare T /kin/compare 0.1 0.0050mc2 0.0050m0c2 0.2 0.0212mc2 0.0200m0c2 0.3 0.0517mc2 0.0480m0c2 0.4 0.1033mc2 0.0910m0c2 0.5 0.1895mc2 0.1550m0c2 0.6 0.3393mc2 0.2500m0c2 0.7 0.6233mc2 0.4010m0c2 0.8 1.2669mc2 0.6670m0c2 0.9 3.4327mc2 1.2930m0c2 0.99 47.294mc2 6.9200m0c2 1.0 infinite infinite L.Vlcek === Subject: Re: Proper class.Proper class ? Under some simple class theories, ALWAYS, because > EVERYthing is a class How about the classes that are not elements of themselves - That depends. Some theories have a foundation axiom that prevents > that. Other theories allow it. It is inconsistent to allow it, as it is inconsistent to say that > everything is a set or everything is a class. Hey, do you agree with me that every set is in itself. Then there is no empty set. Yes, that is the fact. There is not empty set. Every set is in itself. No set is not in itself. In fact what the word set mean is what is in itself. that's the true definition of set that eluded mathematicians and > logicians all of these years. Then mathematics would no longer work. How could I look for > {x|x^2+3*x-4=0}? It is meant to contain only one or two numbers, but > you say that it also contains itself. OF course it contains itself, no doubt. you are confusing determination > with membership. > yet it is true that this set is detemined by two numbers, but it > contain itself. Do you need to have an empty set? Should that be possible? Your idea is all ed up - it eliminates the empty set and all of the other familiar sets that we know and use. The idea of set axioms is to allow us to define sets as we have grown to use them. It is well known that the empty set is very useful (it is like 0 is in solving equations in algebra.) It is one of the two trivial sets, with special properties given by Rice. You idea actually makes no sense. That is rare in life. Usually a person has some sort of purpose and it makes sense to pursue one's purposes in life. However, what is the purpose here? Making every set contain itself does not serve any purpose that Mathematicians have ever had. It is pure, unadulterated academic (acanemic) bull. C-B > Zuhair Then we have no control over what is in a set. We can no longer just > give a condition and there is a set containing just its elements. Your Set Theory has little or nothing to do with how sets function. On the contrary have everything to do with how sets function. in > addition it has a universe that it can quantify on. it superseed > mathematics. Zuhair C-B the reason becaue they assumed the converse approach. a set that is not in itself is not a set, perhaps it is a ur-element. Zuhair > No, not by the normal (intuitive) > definition of a set. But regardless of whether sets can contain themselves or not, or > whether all do, saying that the set/class of sets/classes that don't > contain themselves is a set/class is inconsistent. Statements like: everything is a set > everything is a class > every set contains itself are just stupid. The fact of the matter is, a set can contain itself. There is nothing > logically wrong with a set containing itself. Have you ever > experienced a program that outputs only itself? There is no problem > there, and a set can likewise be defined so what matches the definition > happens to be a single definition, namely itself. Define f(x) = {x(x)}. Then f(f) = {f(f)} is a set that contains > itself. Look up QUINE ATOM on Google - see the first reference. A great deal of what is written on Logic, Set Theory and Computer > Science is stupid nonsense that professors get published because the > editors and referees are more stupid professors. It's true, but few > people seem to realize it. C-B Zuhair is that a class? There is a class of all SETS that are not elements of themselves. > No proper class is an element of ANY class, so, no, there is not a > class of all classes. That would be a hyperclass. Yes, this regress > is silly. Then how can you say that everything is a class? Likewise for those loons who say everything is a set - how > about the sets that are not elements of themself - is that a set? There is no set of all&only such sets; however, there is a class of > them. > There are no loons alleging that everything is a set. That is > alleged > by certain THEORETICAL TREATMENTS of this stuff (like ZFC). > People's opinions are NOT involved. They believe it but I know better because the sets that are not > elements of themselves is not a set, so it is inconsistent to say that > everything is a set, or everything is a class. That is loony. C-B C-B > (author of CBL which answers all of these questions formally and > automatically) Every individual set or class theory also answers them. > The problem is that different theories answer them in different ways. > Yours is just another raindrop in the hurricane. === Subject: Re: Proper class.Proper class ? Under some simple class theories, ALWAYS, because > EVERYthing is a class How about the classes that are not elements of themselves - That depends. Some theories have a foundation axiom that prevents > that. Other theories allow it. It is inconsistent to allow it, as it is inconsistent to say that > everything is a set or everything is a class. Hey, do you agree with me that every set is in itself. Then there is no empty set. Yes, that is the fact. There is not empty set. Every set is in itself. No set is not in itself. In fact what the word set mean is what is in itself. that's the true definition of set that eluded mathematicians and > logicians all of these years. Then mathematics would no longer work. How could I look for > {x|x^2+3*x-4=0}? It is meant to contain only one or two numbers, but > you say that it also contains itself. OF course it contains itself, no doubt. you are confusing determination > with membership. > yet it is true that this set is detemined by two numbers, but it > contain itself. Do you need to have an empty set? Should that be possible? no it cannot be possible you no that, since every set is a member of itself, then there cannot be an empty set, this is clear. Your idea is all ed up - it eliminates the empty set and all of the > other familiar sets that we know and use. True, but you can find equivalent sets to these sets you know, for example the empty set is somehow equivalent to the singlton set in this theory Q={Q}, since it can be regarded as empty of members other than itself. The idea of set axioms is to > allow us to define sets as we have grown to use them. It depends on the goal of the axioms, what if the set as we have grown to use them are not enough to explain universal sets like the set of all sets, the set of all ordinals, the set of all well founded sets,etc... It is well known > that the empty set is very useful (it is like 0 is in solving equations > in algebra.) It is one of the two trivial sets, with special > properties given by Rice. yes I know, but you can have its equivalent doing exactly the same uses as your empty set. You idea actually makes no sense. That is rare in life. Usually a > person has some sort of purpose and it makes sense to pursue one's > purposes in life. However, what is the purpose here? Making every set > contain itself does not serve any purpose that Mathematicians have ever > had. what are U saying, this is wrong. The purpose I made it very clear. my purpose it to understand universal sets, we cannot do that without saying that every set is in itself. You say why you don't go ask yourself, what raised Russell's paradox. You will see that what raised it is the idea that there can be sets that are not in themselfs,i.e all of their members are different from them. getting rid of this property will remove this paradox whatsoever, and enable us to traverse this unknown world of universal sets, that ZFC cannot deal with it without encountering Russell's paradox. It is difficult to construct a set theory that keep the sets we are used to know ( ie the sets that are not in themselfs) and at the same time have a universal set, without inflicting some restrictions on separation. To have a full theory in which separation and unrestriced comprehension exist and that can deal with these universal sets, then we should say that all sets are in themselfs. It is pure, unadulterated academic (acanemic) bull. You are speaking of you misunderstanding of the matter, not of my intentions. Zuhair === Subject: New Trends in Physics 1. VLCEK L.: New Trends in Physics, Slovak Academic Press, Bratislava 1996 ISBN 80-85665-64-6. 2. VLCEK L.: New Trends in Physics /book, elementes pictures, spheres in nuclei, forecasted nuclei, ZOO-3D editor for interactive inspecting of nuclei spheres/, Academic Electronic Press, Bratislava, 2000, CD- ROM, ISBN 80-88880-38-6. 3. Both book and CD - ROM are on http://www.vlcek.euweb.cz Vlcek L. === Subject: Vector spaces have bases implies AC? In ZF, does the hypothesis, Every vector space has a basis imply the Axiom of Choice? In ZF, does the hypothesis, Every real vector space has a basis imply the Axiom of Choice? Are the proofs easy to post? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Vector spaces have bases implies AC? > In ZF, does the hypothesis, Every vector space has a basis imply the > Axiom of Choice? > Yes, according to Infinite Ink, it's equivalent to AxC. > In ZF, does the hypothesis, Every real vector space has a basis imply > the Axiom of Choice? > Yes, according to Infinite Ink, it's equivalent to AxC. > Are the proofs easy to post? > None comes to mind. === Subject: Continuous functiions bounded implies compact? Suppose every continuous real-valued function on a topological space is bounded. Is the space necessarily compact? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Continuous functiions bounded implies compact? > Suppose every continuous real-valued function on a topological space is > bounded. Is the space necessarily compact? No. I think there's a name for that property, but I can't recall what it is. Anyway, the set of all countable ordinals with the order topology is a counterexample. === Subject: Re: Continuous functiions bounded implies compact? > Suppose every continuous real-valued function on a topological space is > bounded. Is the space necessarily compact? > No, it's called pseudo-compact and is strictly weaker than countably compact which is strictly weaker than compact. For example, every continuous function from omega_1 into R is eventually constant. === Subject: Re: Continuous functions bounded implies compact? >Suppose every continuous real-valued function on a topological space is >>bounded. Is the space necessarily compact? >> >No, it's called pseudo-compact and is strictly weaker than countably >compact which is strictly weaker than compact. For example, every continuous function from omega_1 into R >is eventually constant. > > Isn't the following function f: omega_1-> R continuous? f(omega * n + 1) = n; f(a) = 0 otherwise. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Continuous functions bounded implies compact? <456F1B21.2040905@netscape.net>Suppose every continuous real-valued function on a topological space is >>bounded. Is the space necessarily compact? >No, it's called pseudo-compact and is strictly weaker than countably >compact which is strictly weaker than compact. For example, every continuous function from omega_1 into R >is eventually constant. Isn't the following function f: omega_1-> R continuous? f(omega * n > + 1) = n; f(a) = 0 otherwise. No. Note that the sequence {omega * n + 1} converges to omega * omega. The function g(x,y) = x + y is continuous in y, but not in x; the successor function is discontinuous at limit ordinals! === Subject: Re: Cantor Confusion > >> We have to accept thart there are sets which are capable of >> growing, as Fraenkel et al. expess it. >> >> While there are several books by Fraenkel: >> >> Einleitung in die Mengenlehre 2nd ed. 1923, 3rd ed. 1946, >> Gesammelte Abhandlungen Cantor, Dedekind 1932 >> Das Leben Georg Cantors 1932 >> Abstract set theory 1961 >> Lebenskreise - Aus den Erinnerungen eines j.9fdischen Mathematikers 1967 >> >> perhaps there is only one book by Fraenkel et al.: >> Foundations of Set Theory 1958. Correct? >> >> Then we have finite sets without >> a largest element. >> >> Why not with a growing largest element? >> >> Eckard > > Growing elements would allow claims like: > 2 + 2 = 5 for large enough values of 2. Large enough is certainly not qualitatively different enough, infinity is the location where two parallel lines are thought to meet each other, and division by zero has been forbidden because it yields anything. Since I consider an infinite set just like a possibly useful fiction, I do not have any reason to speculate myself. I am just curious what monstrosity of split reasoning in best Cantorian tradition Fraenel et al. will offer. === Subject: Re: Cantor Confusion > > > Large enough is certainly not qualitatively different enough, infinity > is the location where two parallel lines are thought to meet each other, > and division by zero has been forbidden because it yields anything. Division by zero in a field yeilds a contradiction. That is why it is forbidden. 1/0 = x (for some x in the field) implies 1 = 0*x = 0. That simply will not do. Bob Kolker === Subject: Re: Cantor Confusion > In mathematics, claims unsupported by mathematically > valid proofs do not persuade, particularly when opposed by > mathematically valid proofs of their falsehood. Really? Fraenkel, 2nd ed. 1923, approved Cantor having managed not just to battle but also refute an assertion by Gauss. Is there any evidence for this proud claim? No. Whenever Cantor declared mathematicians like Aristotele, Locke, Descarte, or Leibniz proven wrong, he did it with evidence by assertion. His main argument DA2 convinced a lot of mathematicians because understanding of infinity requires more that mathematical education, and the attention focussed on the flawless evidence for non-countablilty. So Cantors invalid interpretation of DA2 persuaded. Dedekind admitted at the very beginning that he did not have a proof. === Subject: Re: Cantor Confusion > > Really? Fraenkel, 2nd ed. 1923, approved Cantor having managed not just > to battle but also refute an assertion by Gauss. > Is there any evidence for this proud claim? No. > Whenever Cantor declared mathematicians like Aristotele, Locke, Locke and Aristotle were NOT mathematicians. Aristotle was at most, a logician or an ethicist or a political scientist or a literarary critic. As a scientist he was a failure. Why? He didn't check. Bob Kolker === Subject: Re: Cantor Confusion hypotheses of the proof. So the set of transcendentals cannot be proved uncountable by this proof. Nor can it be proved uncountable by the diagonal argument. Do you have an idea why this is so? Could it be that Cantor's proof have not at all to do with countability? In my paper I stress the fact that his proof is equally vaild or invalid, when applied to the rational numbers alone as well as when applied to the transcenental numbers alone. What is wrong with this statement? === Subject: Re: Cantor Confusion > > > The set of transcendentals is not a connected set in R, so violates the > hypotheses of the proof. > > So the set of transcendentals cannot be proved uncountable by this > proof. Nor can it be proved uncountable by the diagonal argument. Do > you have an idea why this is so? Could it be that Cantor's proof have > not at all to do with countability? Since it CAN be shown that the set of transcendentals differs by a countable set from the set of all reals and the first proof applies to the set of reals, one can deduce that the set of trannscendentals is equally uncountable. > > In my paper I stress the fact that his proof is equally vaild or > invalid, when applied to the rational numbers alone as well as when > applied to the transcenental numbers alone. What is wrong with this > statement? Nothing. But see above for how to show the uncountability of the transcendentals. === Subject: Re: Cantor Confusion >> We have to accept thart there are sets which are capable of >> growing, as Fraenkel et al. expess it. >> While there are several books by Fraenkel: >> Einleitung in die Mengenlehre 2nd ed. 1923, 3rd ed. 1946, >> Gesammelte Abhandlungen Cantor, Dedekind 1932 >> Das Leben Georg Cantors 1932 >> Abstract set theory 1961 >> Lebenskreise - Aus den Erinnerungen eines j.9fdischen Mathematikers 1967 >> perhaps there is only one book by Fraenkel et al.: >> Foundations of Set Theory 1958. Correct? >> Then we have finite sets without >> a largest element. >> Why not with a growing largest element? > > Because Fraenkel, Bar-Hillel, and Levy said nothing that implies any > such thing as that there are finite sets without a largest element. WM > is just latching onto a particular quote(that even uses scare quotes) > out of context and with utter disregard to what the quote is meant to > actually summarize. I will look into the book myself. Everyone of us my be wrong in details. I guess, we cannot be cautious enough with respect to possibly changing meaning of words. === Subject: Re: Cantor Confusion are finite is to assume that there is no set of natural numbers > without a largest element. No. We have to accept thart there are sets which are capable of > growing, as Fraenkel et al. expess it. Then we have finite sets without > a largest element. Then we describe reality correctly. (Leave Fraenkel et al. out of it. Finite sets without a largest > element are your idea). I did not ascribe those sets to Fraenkel et al. I said, growing, as Fraenkel et al. express it. And, in fact, they used this expression. O.K. Let A be a finite set without a largest element. > What is the cardinal number of A. Clearly it cannot > be any finite cardinal number? It cannot have a cardinal number at all. We have no cardinals in potential infinity. Cantor knew that. Recall, you want to do more than just state that A does > not have a cardinal number. You want to show that assuming > that A has a cardinal number leads to a contradiction. Please do not mix up potential and actual infinity. Actual infinity is required, e.g., for the real numbers. They are not potentially infinite series, but actually infinite series. Consider just the paths in the infinite binary tree. In this case I have shown a contradiction. Here it is: The binary tree Consider a binary tree which has (no finite paths but only) infinite paths representing the real numbers between 0 and 1 as binary strings. The edges (like a, b, and c below) connect the nodes, i.e., the binary digits 0 or 1. 0. /a 0 1 /b c / 0 1 0 1 .......................... The set of edges is countable, because we can enumerate them. Now we set up a relation between paths and edges. Relate edge a to all paths which begin with 0.0. Relate edge b to all paths which begin with 0.00 and relate edge c to all paths which begin with 0.01. Half of edge a is inherited by all paths which begin with 0.00, the other half of edge a is inherited by all paths which begin with 0.01. Continuing in this manner in infinity, we see by the infinite recursion f(n+1) = 1 + f(n)/2 with f(1) = 1 that for n --> oo 1 + 1/2 + 1/ 4 + ... = 2 edges are related to every single infinite path which are not related to any other path. (By the way, the recursion would yield the limit value 2 for any starting value f(1).) The load of 2 edges is only related to infinite paths because any finite segment of a path with n edges will carry a load of (1 - 1/2^n)/(1 - 1/2) < 2 edges. The set of paths is uncountable, but as we have seen, it contains less elements than the set of edges. Cantor's diagonal argument does not apply in this case, because the tree contains all binary representations of real numbers within [0, 1], some of them even twice, like 1.000... and 0.111... . Therefore we have a contradiction: |IR| > |IN| || || |{paths}| => |{edges}| === Subject: Re: Cantor Confusion What is the cardinal number of A. Clearly it cannot > be any finite cardinal number? It cannot have a cardinal number at all. We have no cardinals in > potential infinity. Cantor knew that. Recall, you want to do more than just state that A does > not have a cardinal number. You want to show that assuming > that A has a cardinal number leads to a contradiction. Please do not mix up potential and actual infinity. Actual infinity is > required, e.g., for the real numbers. They are not potentially infinite > series, but actually infinite series. Consider just the paths in the > infinite binary tree. In this case I have shown a contradiction. Here > it is: The binary tree Consider a binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1 as binary strings. > The edges (like a, b, and c below) connect the nodes, i.e., the binary > digits 0 or 1. 0. > /a > 0 1 > /b c / > 0 1 0 1 > .......................... The set of edges is countable, because we can enumerate them. Now we > set up a relation between paths and edges. > Relate edge a to all paths > which begin with 0.0. Relate edge b to all paths which begin with 0.00 > and relate edge c to all paths which begin with 0.01. So your relation is supposed to be a function mapping edges -> sets of paths, correct? > Half of edge a is > inherited by all paths which begin with 0.00, the other half of edge a > is inherited by all paths which begin with 0.01. Why is only 1/2 inherited? Why not 1, or 1/3, or any other number? Isn't inheriting a different function than the original relation (which was presumably a function from edges to sets of paths)? How is it that now we have a function mapping (paths X edges) -> Q? > Continuing in this > manner in infinity, we see by the infinite recursion f(n+1) = 1 + f(n)/2 with f(1) = 1 that for n --> oo 1 + 1/2 + 1/ 4 + ... = 2 edges are related to every single infinite path which are not related > to any other path. Your notation is confusing. As you have described it, edges are not related to paths; they are related to /sets/ of paths. The best I can understand by edge e is related to path p is that p is in the set of paths related to e; but clearly for any path p of length greater than 2, there are more than 2 edges related to p by this interpretation. To clarify, can you indicate which 2 edges are related to the path which represents the real number 1/3, which are not related to any other path? === Subject: Re: Cantor Confusion are finite is to assume that there is no set of natural numbers > without a largest element. No. We have to accept thart there are sets which are capable of > growing, as Fraenkel et al. expess it. Then we have finite sets without > a largest element. Then we describe reality correctly. (Leave Fraenkel et al. out of it. Finite sets without a largest > element are your idea). I did not ascribe those sets to Fraenkel et al. I said, growing, as > Fraenkel et al. express it. And, in fact, they used this expression. O.K. Let A be a finite set without a largest element. > What is the cardinal number of A. Clearly it cannot > be any finite cardinal number? It cannot have a cardinal number at all. We have no cardinals in > potential infinity. Cantor knew that. Piffle. Extending the concept of bijection from sets to potentially infinite sets is trivial. Recall, you want to do more than just state that A does > not have a cardinal number. You want to show that assuming > that A has a cardinal number leads to a contradiction. The problem is although the term finite is used in the definition of A, A does not have many of the properties that are usually associated with finite sets. In particular it is not possible to produce A by induction. It is clear that there is no natural number n, such that there is a bijection between the set {1,2,3....,n} and A. Since the diagonal of your matrix has no largest element, it must be a finite set without largest element. It is clear that there is no bijection between one line and the diagonal. > Please do not mix up potential and actual infinity. Actual infinity is > required, e.g., for the real numbers. Piffle. The rational numbers form a potentially infinite set A. Let B and C be two potentially infinite sets such that: For any element a, that can be shown to exist in A, it is possible to show that a exists in exactly one of B and C. for any element, b, that can be shown to exist in B, for any element, c, that can be shown to exist in C and b < c. for any element b_1 that can be shown to exist in B, there is an element b_2 that can be shown to exist in B, and b_1 < b_2 The potentially infinite set of pairs (B,C) is the real numbers. - William Hughes === Subject: Re: Cantor Confusion one of > them leads to an identity map. You are confusing bijections within the model to bijections > outside of the model. And who told you that you were outside? And who told you that you were outside? You are noting that it is not possible to construct a bijection > in a nonstandard model and that it is > possible to construct a bijection in the standard model. This > is true, however, it is not a contradiction. In particular because there is neither a stadard model nor a > non-standard model of ZFC. You haven't proven that there is no model of ZFC. Mathematics exists nowhere except in the minds of mathematicians. At > present there is no mind with such a model. So, where should it exists? That such a model does not exist in WM's mind is not evidence that it > does not exist in minds competent to comprehend it. Do you know of such a mind? Why can't the model be written down? And in particular because the expression contradiction is not pat of > ZFC. In Z set theory, we can formulate definitions of 'S is a contradiction' > and 'T is inconsistent'. One could, but one wouldn't. Never! One has. But defining something does not instanciate it. One can define square circles and 4 sided triangles, too. Wrong definitions alltogether, the four-sided triangle as well as the possibility of set theory being proven inconsistent by one of its advocates. === Subject: Re: Cantor Confusion > > And in particular because the expression contradiction is not pat of > ZFC. In Z set theory, we can formulate definitions of 'S is a contradiction' > and 'T is inconsistent'. One could, but one wouldn't. Never! One has. But defining something does not instanciate it. One can define square circles and 4 sided triangles, too. > > Wrong definitions alltogether, the four-sided triangle as well as the > possibility of set theory being proven inconsistent by one of its > advocates. Definitions in mathematics, being merely abbreviations, cannot be right or wrong, they can only be more or less useful. The more useful ones tend to persist. Useless ones get discarded. === Subject: Re: Cantor Confusion bijection N <--> Q? The reason I believe it's a theorem of Z set theory that there is a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers is because I've studied at least one proof. It is believed to be a proof. But it isn't a proof. It is demonstarted > for few symbols. If it is not a proof, or rather if the bijection does not exist in ZF of > NBG or same other set theory, then WM should be able to produce a > counterexample. I did. Take the natural numbers between [e*10^10^100] and [pi*10^10^100]. You do not even know how many there are. But you insist to prove something for them all. Particularly when WM acknowledges himself not to be a mathematician, and > mathematicians claim otherwise? I am not a set theorist. WM does not get it. In mathematics, claims unsupported by mathematically > valid proofs do not persuade, particularly when opposed by > mathematically valid proofs of their falsehood. Unfortunately today mathematically valid is frequently mistaken with matheologically believed. === Subject: Re: Cantor Confusion > But why do you think that there is a > bijection N <--> Q? The reason I believe it's a theorem of Z set theory that there is a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers is because I've studied at least one proof. It is believed to be a proof. But it isn't a proof. It is demonstarted > for few symbols. If it is not a proof, or rather if the bijection does not exist in ZF of > NBG or same other set theory, then WM should be able to produce a > counterexample. > > I did. Take the natural numbers between [e*10^10^100] and > [pi*10^10^100]. You do not even know how many there are. But you insist > to prove something for them all. When I say that x <--> (x,x) is a bijection between members of any set S and the set {(x,x): x in S}, it is entirely immaterial what set S is. Particularly when WM acknowledges himself not to be a mathematician, and > mathematicians claim otherwise? > > I am not a set theorist. Nor a mathematician of any sort. WM does not get it. In mathematics, claims unsupported by mathematically > valid proofs do not persuade, particularly when opposed by > mathematically valid proofs of their falsehood. > > Unfortunately today mathematically valid is frequently mistaken with > matheologically believed. Not by the mathematically competent, of which WM is not one. === Subject: Re: Cantor Confusion In particular because there is neither a stadard model nor a > non-standard model of ZFC. You haven't proven that there is no model of ZFC. Mathematics exists nowhere except in the minds of mathematicians. At > present there is no mind with such a model. So, where should it exists? As I said, you've not proven that there is no model of ZFC. I did not say that I had proven that. I said In particular because there is neither a standard model nor a non-standard model of ZFC. Are you really incapable of understanding such simple sentences? And in particular because the expression contradiction is not pat of > ZFC. In Z set theory, we can formulate definitions of 'S is a contradiction' > and 'T is inconsistent'. One could, but one wouldn't. Never! WRONG. We DO. You're an ignoramus. Do you really believe that set theory becomes more popular if its adherents turn out to behave like undisciplined children without any self-control like you or Mr Bader? === Subject: Re: Cantor Confusion > In particular because there is neither a stadard model nor a > non-standard model of ZFC. You haven't proven that there is no model of ZFC. Mathematics exists nowhere except in the minds of mathematicians. At > present there is no mind with such a model. So, where should it exists? As I said, you've not proven that there is no model of ZFC. I did not say that I had proven that. I didn't say that you had said that you had proven that ZFC has no model. Rather, you just claim it to be true that ZFC has no model, and it is MY remark that you have not proven it. > I said In particular because > there is neither a standard model nor a non-standard model of ZFC. Are > you really incapable of understanding such simple sentences? Your posting there is neither a standard model nor a non-standard model of ZFC entails that you are claiming that there is no model of ZFC. And I have simply pointed out that you have not proven that claim. > And in particular because the expression contradiction is not pat of > ZFC. In Z set theory, we can formulate definitions of 'S is a contradiction' > and 'T is inconsistent'. One could, but one wouldn't. Never! WRONG. We DO. You're an ignoramus. Do you really believe that set theory becomes more popular if its > adherents turn out to behave like undisciplined children without any > self-control like you or Mr Bader? I've never taken a job as the public relations manager for Set Theory Inc. My goal is not to make set theory more popular, but rather to speak my mind. What is puerile is your entire approach, based on maintaining your ignorance, to discussing set theory, which is the most salient thing that comes through your postings, and is thus mine to comment upon. When you claim things that are patently false such as that in set theory one would never formulate definitions of 'contradiction' and 'inconsistent', or to deny that in set theory we prove that the set of naturals is equinumerous with the set of rationals, then you can pretty much count on being called for being the ignoramus that you are. MoeBlee === Subject: Re: Cantor Confusion growing, as Fraenkel et al. express it. Then we have finite sets without > a largest element. Then we describe reality correctly. You're a dishonest fool. If you continue to loose your self control in this way, then I will have to cease discussing with you. > I've already explained to you, as you should > have read for yourself in the original source, that what Fraenkel, > Bar-Hillel, and Levy mean by the universe of sets growing (THEIR > scare quotes) is that different axioms yield different universes of > sets, That is nonsense. I know that you cannot recognize it. Only for the lurkers: If *different* sets generated by different axiom systems were meant, then Fraenkel et al. would not only have to talk about growing but also about shrinking or simply about differing sets. But they don't. > not that any given set itself grows, let alone that there are > finite sets without a largest element (!). You misunderstand Fraenkel and you misunderstand me. Small wonder. I did not say that Frankel talks about these growing sets with the same We have to accept that there are sets which are capable of growing, *as Fraenkel et al. express it*. === Subject: Re: Cantor Confusion No. We have to accept that there are sets which are capable of > growing, as Fraenkel et al. express it. Then we have finite sets without > a largest element. Then we describe reality correctly. You're a dishonest fool. If you continue to loose your self control in this way, then I will > have to cease discussing with you. No loss of self-control. I comment as I see fit; and the most salient thing about your latest postings is that they reveal you to be a dishonest fool. Your choice to continue or cease posting in response to my posts is yours alone. > I've already explained to you, as you should > have read for yourself in the original source, that what Fraenkel, > Bar-Hillel, and Levy mean by the universe of sets growing (THEIR > scare quotes) is that different axioms yield different universes of > sets, > That is nonsense. I know that you cannot recognize it. Only for the > lurkers: If *different* sets generated by different axiom systems were > meant, then Fraenkel et al. would not only have to talk about growing > but also about shrinking or simply about differing sets. But they > don't. other hand, if what one does, as you do, is cruise the book looking for out of context quotes to misconstrue and misrepresent, then understanding is not required. > not that any given set itself grows, let alone that there are > finite sets without a largest element (!). You misunderstand Fraenkel and you misunderstand me. Small wonder. I > did not say that Frankel talks about these growing sets with the same > We have to accept that there are sets which are capable of growing, > *as Fraenkel et al. express it*. Nope I understand and represented accurately. Your notion of what Fraenkel, Bar-Hillel, and Levy say is terribly incorrect and the inferences you draw are ludicrous. MoeBlee === Subject: Re: Cantor Confusion The reason I believe it's a theorem of Z set theory that there is a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers is because I've studied at least one proof. It is believed to be a proof. But it isn't a proof. It is demonstarted > for few symbols. I don't know what you mean by 'demonstrated for few symbols'. Meanwhile, there does exist a proof in Z set theory that there exists a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers. Your proof concerns less than 10^100 elements. You are not even able to express some natural numbers of this domain, but you insist that there was a proof concerning all of them. A ridiculous self-overestimation. === Subject: Re: Cantor Confusion > The reason I believe it's a theorem of Z set theory that there is a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers is because I've studied at least one proof. It is believed to be a proof. But it isn't a proof. It is demonstarted > for few symbols. I don't know what you mean by 'demonstrated for few symbols'. Meanwhile, there does exist a proof in Z set theory that there exists a > bijection between the set of natural numbers and the set of ordered > pairs of natural numbers. Your proof concerns less than 10^100 elements. You are not even able > to express some natural numbers of this domain, but you insist that > there was a proof concerning all of them. A ridiculous > self-overestimation. You must be running a very high fever. In the proof I've mentioned, there's no mention nor dependency on any claims limited to 10^100 elements. MoeBlee === Subject: Re: Cantor Confusion > I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy How can you judge about that without the slightest idea of what they Quite a while ago I had read that whole book. Upon your mentioning the > quote, I re-read the quote in the book, re-read the section in which > the quote appears, and re-read the chapter in which the section > appears, and re-read several other parts of the book. And why did you not understand it? Only for the lurkers: Fraenkel et al. write: Platonistic point > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. the discussion leading to that quote to understand it. You don't. How can you judge without the least understanding? (I did not say that Fraenkel et al. support this point of view which they mention.) So a set (like the set of all sets) is not a > fixed entity. WRONG. You just keep saying that, but ignore the rest of Fraenkel, > Bar-Hillel, and Levy's discussion, and also you just keep skipping past > my own on-point replies to you on this matter. Meanwhile, you are ignorant of even such basic set theory as > proving the existence of an identity function. I think of a correct proof, not of blowing hot air. If you can't think of a proof that the identity function exists for any > set, then you are ignorant of basic set theory. Not ignorant of the theory but also not ignorant of its weaknessess and errors. If you state x = x for the elements x of an infinite set X, or talk about a subset of the Cartesion product XxX, then you have not proved anything. The information I require about the very basis of this discussion was > and which I mentioned yet again, but which you yet again ignore. I took > the effort to compose those questions for you at that time. If you want > me to consider your argument, then please respond to those posts. I'm > not inclined to recompose my questions for you again after I already > carefully composed them. Which posts? When posted? The ones Lester Zick responded to with his my way or the highway > remark. By copy and paste you should be able to > repeat those questions. I already took the time to compose the posts. I'm not inclined to use > up more of my time tracking them down for you after you've ignored > them. Find them on your own time, or don't, but then don't expect me to > address your tree argument until you do answer them. LOL! Why should I trust in your ability to prove the tree argument after you obviously turned out incapable to understand this argument I saw only the question whether I believed that > a proof in ZFC was possible. Then you didn't read the posts. Should this question decide whether such a > proof is really possible? Of course not. What a silly question. Why then did you ask me? === Subject: Re: Cantor Confusion > If you can't think of a proof that the identity function exists for any > set, then you are ignorant of basic set theory. > Not ignorant of the theory but also not ignorant of its weaknessess and > errors. If you state x = x for the elements x of an infinite set X, or > talk about a subset of the Cartesion product XxX, then you have not > proved anything. Using only first order logic with identity and the axioms of set theory, we prove that there exists an identity function on any set. > I already took the time to compose the posts. I'm not inclined to use > up more of my time tracking them down for you after you've ignored > them. Find them on your own time, or don't, but then don't expect me to > address your tree argument until you do answer them. LOL! Why should I trust in your ability to prove the tree argument > after you obviously turned out incapable to understand this argument YOU are the one who invited ME to try to put your argument into set theory. YOU are the one who keeps asking ME to comment on your argument. *I* never claimed to be able to prove the conclusion of your argument or to be able to put your argument into set theory. You really are a mixed up fool. > I saw only the question whether I believed that > a proof in ZFC was possible. Then you didn't read the posts. Should this question decide whether such a > proof is really possible? Of course not. What a silly question. Why then did you ask me? I never asked you such as question as whether your believing such a proof is possible decides whether such a proof is possible. You're completely mixed up. MoeBlee === Subject: Re: Cantor Confusion <456DC03A.1000206@et.uni-magdeburg.de We have to accept thart there are sets which are capable of > growing, as Fraenkel et al. expess it. While there are several books by Fraenkel: Einleitung in die Mengenlehre 2nd ed. 1923, 3rd ed. 1946, > Gesammelte Abhandlungen Cantor, Dedekind 1932 > Das Leben Georg Cantors 1932 > Abstract set theory 1961 > Lebenskreise - Aus den Erinnerungen eines j.9fdischen Mathematikers 1967 perhaps there is only one book by Fraenkel et al.: > Foundations of Set Theory 1958. Correct? Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: Foundations of Set Theory, 2nd edn., North Holland, Amsterdam (1984) Then we have finite sets without > a largest element. Why not with a growing largest element? That depends on definition. The largest element of a set of numbers today is not the same numbers as the largest element of the set tomorrow. But the object largest element considered as a variable, in fact would grow. Nevertheless this is not what Fraenkel et al. wish to express. They talk about the development of the set of all sets in a Platonic world view. === Subject: Re: Cantor Confusion > > The largest element of a set of numbers > today is not the same numbers as the largest element of the set > tomorrow. But the object largest element considered as a variable, in > fact would grow. Thatis fantasy, not mathematics. === Subject: Re: Cantor Confusion > I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. > > No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy > > How can you judge about that without the slightest idea of what they > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. So a set (like the set of all sets) is not a > fixed entity. There is nothing in that that shows that a set is not a fixed entity. You are able to produce bigger and bigger sets, but they are all different. When the universe has grown it allows bigger sets than where originally allowed, but the sets originally allowed are still sets and still the same and did not grow. How you conclude from the above statement that sets themselves are growing escapes me. -- 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: Cantor Confusion > I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. > No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy > > How can you judge about that without the slightest idea of what they > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. So a set (like the set of all sets) is not a > fixed entity. > > There is nothing in that that shows that a set is not a fixed entity. > You are able to produce bigger and bigger sets, but they are all > different. So far I recall the set IN of naturals is countable. I understood set theory means a hypothetical (alias fictitious) set off all natural numbers. Maybe, you are not aware of some shizophrenia in Cantor's concept. He actually imagines infinity like a quantity, like someting festes (concrete). At the same time he proofs by bijection, that the naturals are contable. The naturals are of course countable if considered one by one. However the entity of all naturals only exists like an uncountable fiction. === Subject: Re: Cantor Confusion ... > There is nothing in that that shows that a set is not a fixed entity. > You are able to produce bigger and bigger sets, but they are all > different. > > So far I recall the set IN of naturals is countable. > I understood set theory means a hypothetical (alias fictitious) set off > all natural numbers. Oh, for once, try to talk mathematics. By the axiom of infinity the set of all naturals is neither hypothetical nor fictitious. -- 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: Cantor Confusion I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. > No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy > How can you judge about that without the slightest idea of what they > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. So a set (like the set of all sets) is not a > fixed entity. There is nothing in that that shows that a set is not a fixed entity. The universe of all sets can grow. Define: The universe of all sets is called the set of all sets, and you see it. > You are able to produce bigger and bigger sets, but they are all > different. That is a matter of definition. If you consider a fixed set then it is fixed. Small wonder. If you consider a variable set then it is variable and perhaps changes its cardinal number. An easy example which should not escape you: The set of states of the EC has been growing and probably will continue to grow. > When the universe has grown it allows bigger sets than > where originally allowed, but the sets originally allowed are still > sets and still the same and did not grow. How you conclude from the > above statement that sets themselves are growing escapes me. It is simply a matter of definition. === Subject: Re: Cantor Confusion ... > How can you judge about that without the slightest idea of what they > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. So a set (like the set of all sets) is not a > fixed entity. > > There is nothing in that that shows that a set is not a fixed entity. > > The universe of all sets can grow. Define: The universe of all sets is > called the set of all sets, and you see it. And that is the confusion. If it is a set it cannot grow, but as Fraenkel et al. do not define it as a set it is allowed to grow. They do not state that a set can grow because they do not state that the universe is a set. > You are able to produce bigger and bigger sets, but they are all > different. > > That is a matter of definition. If you consider a fixed set then it is > fixed. Small wonder. If you consider a variable set then it is > variable and perhaps changes its cardinal number. An easy example which > should not escape you: The set of states of the EC has been growing and > probably will continue to grow. That is not the set of states. You can talk about the current set of states or about the set of states in 1957 or whatever. At least mathematically. In mathematics, by definition, a set can not grow. You are, of course, entitled to use another definition, but that will not clarify the discussion at all (and you are not using standard set theory). > When the universe has grown it allows bigger sets than > where originally allowed, but the sets originally allowed are still > sets and still the same and did not grow. How you conclude from the > above statement that sets themselves are growing escapes me. > > It is simply a matter of definition. Yes, with your definition a set can grow, but you put yourself outside set theory, and you must at first consider all results from set theory unproven theorems in your theorem, and you need to prove them (if possible). -- 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: Cantor Confusion I think if one swells to explosion about his knowledge of set theory, > he should at least know the very foundation. But I know, that you do > not even understand the simple texts of Fraenkel et al. > No, YOU radically MISunderstand what Fraenkel, Bar-Hillel, and Levy > How can you judge about that without the slightest idea of what they > of view is to look at the universe of all sets not as a fixed entity > but as an entity capable of growing, i.e., we are able to produce > bigger and bigger sets. So a set (like the set of all sets) is not a > fixed entity. There is nothing in that that shows that a set is not a fixed entity. The universe of all sets can grow. Define: The universe of all sets is > called the set of all sets, and you see it. the fact that different axioms yield different universes of sets. That is what they mean by the universe of sets growing (scare quotes in original text). > You are able to produce bigger and bigger sets, but they are all > different. That is a matter of definition. If you consider a fixed set then it is > fixed. Small wonder. If you consider a variable set then it is > variable and perhaps changes its cardinal number. You can go ahead and provide some theory in which there are such variable sets. Meanwhile, there is nothing like it in Z set theories. And Fraenkel, Bar-Hillel, and Levy's comments do not contradict this. > An easy example which > should not escape you: The set of states of the EC has been growing and > probably will continue to grow. Which you'll have a hard time proving to be a set in Z set theory. Of course no one denies that the everyday, non-mathematical, sense of 'set' includes the idea that many everyday conceived sets have changing membership. But that is distinct from the mathematics of Z set theory, which has an axiom that determines that the sense as applied to such theories is distinct from the everyday sense. No one is stopping anyone from formulating a mathematical theory in which sets have changing membership; but the option of doing this does not change that in Z set theories, sets do not have changing membership. > When the universe has grown it allows bigger sets than > where originally allowed, but the sets originally allowed are still > sets and still the same and did not grow. How you conclude from the > above statement that sets themselves are growing escapes me. It is simply a matter of definition. No, it's a matter of an axiom. MoeBlee === Subject: Re: Cantor Confusion uncountable by the same Cantor first proof argument, but disconnected > subsets may not. Cantor does not talk of connected or disconnected, but only of an arbitrary sequence beliebige Reihe reeller Zahlgr??en In addition, two sequences of transcendental numbers can converge to a > rational number, such that we have the same situation as described > above. The set of transcendentals is not a connected set in R, so violates the > hypotheses of the proof. There is no such hypothesis at all. Where did you think to read that? The *result* of the proof is applied to the transcendental numbers. kombiniert man die Inhalte dieser beiden Paragraphen, so ist damit ein neuer Beweis des zuerst von Liouville bewiesenen Satzes gegeben, da? es in jedem vorgegebenen Intervalle unendlich viele transzendente, d. h. nicht algebraische reelle Zahlen gibt. Therefore both sets, Q ???and T (transcendental numbers), have the same > status with respect to *this* uncountability proof. Both are totally disconnected as subsets of the reals (between any two > members there is a non-member) so that the hypotheses of the proof are > violated. There is no such hypothesis at all. The proof applied to these subsets cannot distinguish there cardinality. That is what I mentioned. And we are not > able, *based on this very proof*, to distinguish between them. Why should we, as the proof does not apply to either. As I mentioned, the proof yields the uncountability of the transcendental numbers, when the countability of the algebraic numbers is established. When also the countability of the nonalgebraic numbers is established, the proof shows a contradiction. On the other hand, the proof can show the uncountability of a countable > set. If, for instance, the alternating harmonic sequence > (-1)^n/ n --> 0 > is taken as sequence (1), yielding the intervals (-1 , 1/2), (-1/3 , > 1/4), ... we find that > its limit 0 does not belong to the sequence, although the set of > numbers involved is obviously > denumerable. That doesn't work it unless you show that NO sequence can be made > including all of the values of S = {0} union { (-1)^n/n: n in N} But consider the sequence 0, -1, 1/2, -1/3, 1/4, -1/5, ... which does > contain all of the members of S. Consider the sequence in Cantors proof which starts at position 0 with the common limit. > The alternating harmonic sequence does not, of course, contain all real > numbers, but this simple example demonstrates that Cantor's first proof > is not conclusive. > Wrong! The theorem speaks of ANY sequence, not just hand picked > sequences. Take any sequence of rational numbers selected from their order by magnitude. > Cantor's proof can be used to show the uncountability of the rational > numbers. In Q there are sequences which converge to rational numbers. But there are also sequences which do not, and it is the existence of > those sequences that do not which scuttles WM's misrepresentation. > Recall in the proof, that the analysis must hold for ANY sequence which > is alleged to count every member! if it fails for any, it fails entirely. Therefore Cantor's proof fails for those sequences which carry their limit as a first element. Therefore Cantor's second proof fails for the list 0.0 0.1 0.11 0.111 ... when replacing 0 by 1. Therefore the real numbers are countable, proven by the paths of the binary tree. > Ey in S Ax in S ((x != y) --> (y > x)) is false in every ordered set, > S, which does not have a largest member, such as N or Q or R. But it is not true in a set of elements of one line of the EIT which always has a largest member. === Subject: Re: Cantor Confusion leaves the plaine *being connected*. But it is not remarkable > that removing an uncountable set leaves the plaine being > connected? What a foolish assertion. > ... > Yes, but Cantor considered only point sets which are dense. > You did not state that above. > implied and need not be mentioned. let the examples where the inner portion of a circle were removed go, > while the apparently do *not* satisfy what Cantor considers! You cannot expect that I carefully read and answer every sentence of Mr. Bader. It is only by accident that I saw and answered his posting (because he seems to be very interested in my papers and to study them in great detail). I usually ignore him. > But under *those* conditions there there are still uncountable sets that > leave the plane connected and also sets that leave the plane disconnected. > Of course, but Cantor did not recognize it. And as far as I know nobody > recognized it before I did. At least nobody mentioned it. Of course nobody mentions it, because it is highly unremarkable! Well, after I showed it. Cantor's proof cannot be used to find an uncountable set which leaves the plane connected. > Remove > all irrational points and you still have a connected space. Remove in > addition a line and you have a disconnected space. What is remarkable > about that? What *is* remarkable is that whatever countable dense set > you remove, the result is still connected. And the property of being > dense is not even needed for both results. Just this shows that Cantor, who insists on it, was not aware of the fact that uncountable sets leave the plane connected. === Subject: Re: Cantor Confusion ... > Of course nobody mentions it, because it is highly unremarkable! > > Well, after I showed it. Cantor's proof cannot be used to find an > uncountable set which leaves the plane connected. Well, of course not, it was not targeted at that goal. And the fact that it is possible is totally unremarkable. > Remove > all irrational points and you still have a connected space. Remove in > addition a line and you have a disconnected space. What is remarkable > about that? What *is* remarkable is that whatever countable dense set > you remove, the result is still connected. And the property of being > dense is not even needed for both results. > > Just this shows that Cantor, who insists on it, was not aware of the > fact that uncountable sets leave the plane connected. Because that fact is false. Removing dense uncountable subsets can leave the plane connected or disconnected. Where did Cantor insist that removal of an uncountable set leaves the plane disconnected (as you assert he did insist on)? -- 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: Cantor Confusion > > ... > leaves the plaine *being connected*. But it is not remarkable > that removing an uncountable set leaves the plaine being > connected? What a foolish assertion. > ... > Yes, but Cantor considered only point sets which are dense. > You did not state that above. > implied and need not be mentioned. let the examples where the inner portion of a circle were removed go, > while the apparently do *not* satisfy what Cantor considers! > > You cannot expect that I carefully read and answer every sentence of > Mr. Bader. It is only by accident that I saw and answered his posting > (because he seems to be very interested in my papers and to study them > in great detail). I usually ignore him. > But under *those* conditions there there are still uncountable sets > that > leave the plane connected and also sets that leave the plane > disconnected. > Of course, but Cantor did not recognize it. And as far as I know nobody > recognized it before I did. At least nobody mentioned it. Of course nobody mentions it, because it is highly unremarkable! > > Well, after I showed it. Cantor's proof cannot be used to find an > uncountable set which leaves the plane connected. As Cantor's proof does not touch on connectivity in the plane, that is irrelevant. Besides which, it is quite easy to find uncountable subsets of the plane whose excision leaves the remainder not only connected but simply connected. === Subject: Re: Cantor Confusion Nothing more than opinion. You think there are no such representations, > so everything that states that there are such representations is wrong. > I proved that there are no such representations. If they were (the > paths in the infinite tree, for instance), then we had a contradiction. You are, with your tree, wrong. Please try to understand it first! Up to now you didn't. > In your tree (where the nodes represent > bits), it can be shown that all nodes represent a number: the collection > of bits found when going from the root to the current bit. So the nodes > represent all rational numbers in [0,1) where the denominator is a power > of two. You state that the paths represent numbers. Let us analyse that, > and begin with the finite paths that terminate at some finite edge. In my tree there are no finite (terminating) paths! >When > you consider such paths, you can assign numbers to each final edge: the > same number as the node were it comes from (nodes represent bits, so when > you come at an edge you can only add the bit of the last node you passed, > you can not look in the future). So your paths (whether finite or infinite) > all represent a number that is also represented by a node. In your terminology every digit 3 in 0.333... represents a number. The first 3 represents the number 0.3, the second 3 represents the number 0.33 and the 10th 3 represents the number 0.3333333333. What does this interpretation have to do with the existence of the decimal representations of the number 1/3, namely 0.333... in a list of real numbers? > In your infinite tree, all edges come fome a node > at finite distance, and so all edges together only represent the numbers > with a finite binary expansion. So all decimal representations represent finite decimal expansions? Or what is the difference? > Then 1/3 is not in Cantor's list. > It is. Can you imagine to represent the numbers in Cantor's list by paths? > Again, see above. As there is no node at an infinite distance, there is > also not an edge that comes from such a node, and so all edges represent > a number with a finite binary expansion. As there is no digit 3 at an infinite distance from the 0. in 0.333..., there is no infinite decimal expansion. What is the difference to a representations by paths? EIT: > Eh? N is the set of natural numbers. The quantifiers can be exchanged > when N is finite. Not when something completely different is finite. We consider only finite lines. Within every line the quantifiers can be exchanged. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > ... > Nothing more than opinion. You think there are no such representations, > so everything that states that there are such representations is wrong. > > I proved that there are no such representations. If they were (the > paths in the infinite tree, for instance), then we had a contradiction. > > You are, with your tree, wrong. > > Please try to understand it first! Up to now you didn't. I try to understand it, but you are not much help. > In your tree (where the nodes represent > bits), it can be shown that all nodes represent a number: the collection > of bits found when going from the root to the current bit. So the nodes > represent all rational numbers in [0,1) where the denominator is a power > of two. You state that the paths represent numbers. Let us analyse that, > and begin with the finite paths that terminate at some finite edge. > > In my tree there are no finite (terminating) paths! Sorry, as far as I see your tree has also finite paths. I see paths from the root to the first nodes below the roots. >When > you consider such paths, you can assign numbers to each final edge: the > same number as the node were it comes from (nodes represent bits, so when > you come at an edge you can only add the bit of the last node you passed, > you can not look in the future). So your paths (whether finite or > infinite) all represent a number that is also represented by a node. > > In your terminology every digit 3 in 0.333... represents a number. The > first 3 represents the number 0.3, the second 3 represents the number > 0.33 and the 10th 3 represents the number 0.3333333333. What does this > interpretation have to do with the existence of the decimal > representations of the number 1/3, namely 0.333... in a list of real > numbers? What is the tree? What are the nodes? You again switch to something different. But in a decimal tree, where each node is connected by edges to ten subnodes and where each of those subnodes is assigned a decimal digit, it can be shown that also each node represents a finite decimal number. And in that tree 1/3 is not represented by earlier. And, indeed, 0.333... does not represent a number until that notation has been defined. > In your infinite tree, all edges come fome a node > at finite distance, and so all edges together only represent the numbers > with a finite binary expansion. > > So all decimal representations represent finite decimal expansions? Or > what is the difference? The difference is that infinite decimal representations are not defined as given, there is a concept behind it: limit. > Then 1/3 is not in Cantor's list. > > It is. > > Can you imagine to represent the numbers in Cantor's list by paths? No, your paths represent finite expansions only. > Again, see above. As there is no node at an infinite distance, there is > also not an edge that comes from such a node, and so all edges represent > a number with a finite binary expansion. > > As there is no digit 3 at an infinite distance from the 0. in > 0.333..., there is no infinite decimal expansion. > What is the difference to a representations by paths? Limits. > EIT: giving proper references. > Eh? N is the set of natural numbers. The quantifiers can be exchanged > when N is finite. Not when something completely different is finite. > > We consider only finite lines. Within every line the quantifiers can be > exchanged. And due to your quoting all context has been lost. If I remember well it was about: forall{n in N} thereis(m in N} {etc.} and: thereis{m in N} forall{n in N} {etc.} The quantifiers are about N. And N is not finite, so you can not exchange them. What is stated in etc. is irrelevant. -- 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: Cantor Confusion In my tree there are no finite (terminating) paths! Sorry, as far as I see your tree has also finite paths. I see paths > from the root to the first nodes below the roots. That is an edge. But no path does stop there. Every path is continued, infinitely. What is the tree? What are the nodes? You again switch to something > different. I only want to make clear that decimal representations of real numbers and binary representations of real numbers (like the paths of the tree) are not different. > But in a decimal tree, where each node is connected by > edges to ten subnodes and where each of those subnodes is assigned a > decimal digit, it can be shown that also each node represents a > finite decimal number. And in that tree 1/3 is not represented by > earlier. And, indeed, 0.333... does not represent a number until > that notation has been defined. And I use exactly the same definition, adapted for the case of binary representations. Instead of writing a number like 0.11000... I write this same number 0. 1 1 0 0 0 ... And that is all the difference! I cannot understand how someone familiar with decimal representations of real numbers should have any difficulties with the binary tree. > In your infinite tree, all edges come fome a node > at finite distance, and so all edges together only represent the numbers > with a finite binary expansion. > So all decimal representations represent finite decimal expansions? Or > what is the difference? The difference is that infinite decimal representations are not defined > as given, there is a concept behind it: limit. Take simply the same concept for my tree: The limit. But whether there is a concept behind or not: Your mentioning of an infinite distance from the root to a node fails completely. In any case there is no digit of a real infinitely far from the decimal point. Why? Its position could not be indexed by a natural number. Its position, therefore, would be undefined. The same is true for the nodes of the binary tree. Its paths are infinitely long but no node has an infinite distance from the root. Do you see your error? > As there is no digit 3 at an infinite distance from the 0. in > 0.333..., there is no infinite decimal expansion. > What is the difference to a representations by paths? Limits. No. Paths are only another notation for the reals in usual representation and usual definition. === Subject: Re: Cantor Confusion >> In my tree there are no finite (terminating) paths! >> Sorry, as far as I see your tree has also finite paths. I see paths >> from the root to the first nodes below the roots. > That is an edge. But no path does stop there. Every path is continued, > infinitely. So you do not know what a path is, or a tree apparently. In a tree, there exists exactly one path between any two nodes. Stephen === Subject: Re: Cantor Confusion > > >> In my tree there are no finite (terminating) paths! >> Sorry, as far as I see your tree has also finite paths. I see paths >> from the root to the first nodes below the roots. > > That is an edge. But no path does stop there. Every path is continued, > infinitely. > > So you do not know what a path is, or a tree apparently. > In a tree, there exists exactly one path between any two nodes. > > Stephen There are those for whom what connects *adjacent* nodes is called an edge, and others call it a branch. Most of those whom I am aware, other than possibly WM, reserve path for the chain of connections starting at the root node and ending, if at all, in a leaf node. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > > > In my tree there are no finite (terminating) paths! > > Sorry, as far as I see your tree has also finite paths. I see paths > from the root to the first nodes below the roots. > > That is an edge. But no path does stop there. Every path is continued, > infinitely. I see paths from the root to the nodes below the nodes below the tree. > What is the tree? What are the nodes? You again switch to something > different. > > I only want to make clear that decimal representations of real numbers > and binary representations of real numbers (like the paths of the tree) > are not different. Indeed, they are not. Base 'pi' or base 'sqrt(2)' are also not inherently different. > But in a decimal tree, where each node is connected by > edges to ten subnodes and where each of those subnodes is assigned a > decimal digit, it can be shown that also each node represents a > finite decimal number. And in that tree 1/3 is not represented by > earlier. And, indeed, 0.333... does not represent a number until > that notation has been defined. > > And I use exactly the same definition, adapted for the case of binary > representations. Instead of writing a number like Ok. > And that is all the difference! I cannot understand how someone > familiar with decimal representations of real numbers should have any > difficulties with the binary tree. I have difficulty with the tree because your explanations are confused and sometimes contradictionary. > The difference is that infinite decimal representations are not defined > as given, there is a concept behind it: limit. > > Take simply the same concept for my tree: The limit. But whether there > is a concept behind or not: Your mentioning of an infinite distance > from the root to a node fails completely. In any case there is no > digit of a real infinitely far from the decimal point. Why? Its > position could not be indexed by a natural number. Its position, > therefore, would be undefined. The same is true for the nodes of the > binary tree. Its paths are infinitely long but no node has an infinite > distance from the root. Do you see your error? What has this to do with everything else? > As there is no digit 3 at an infinite distance from the 0. in > 0.333..., there is no infinite decimal expansion. > What is the difference to a representations by paths? > > Limits. > > No. Paths are only another notation for the reals in usual > representation and usual definition. You state that you are using limits with your infinite paths? -- 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: Cantor Confusion > In your terminology every digit 3 in 0.333... represents a number. The > first 3 represents the number 0.3, the second 3 represents the number > 0.33 In my number system it represents 0.03. > and the 10th 3 represents the number 0.3333333333. In my number system it represents 0.0000000003. > What does this > interpretation have to do with the existence of the decimal > representations of the number 1/3, namely 0.333... in a list of real > numbers? The way you have stated it, nothing. > As there is no digit 3 at an infinite distance from the 0. in > 0.333..., there is no infinite decimal expansion. There is that same confusion surfacing again. An endless sequence of single decimal digits creates an infinitely long expansion with no digit more that finitely far from the decimal point. Until WM learns this, his attempts to work with set theory will continue to fail. > EIT: > > Eh? N is the set of natural numbers. The quantifiers can be exchanged > when N is finite. Not when something completely different is finite. > > We consider only finite lines. Within every line the quantifiers can be > exchanged. But when what is being quantified are the lines themselves, that is not within a line, and cannot in general be 'exchanged. AxEy f(x,y) and EyAx f(x,y) are not in general equivalent. E.g. For every woman x there is (or at least once was) a woman y such that y is x's mother But is it the case that there is a woman y such that for every woman x, y is x's mother? Only in EB's family! === Subject: Re: Cantor Confusion In your terminology every digit 3 in 0.333... represents a number. The > first 3 represents the number 0.3, the second 3 represents the number > 0.33 In my number system it represents 0.03. and the 10th 3 represents the number 0.3333333333. In my number system it represents 0.0000000003. Read what Dik has written. What does this > interpretation have to do with the existence of the decimal > representations of the number 1/3, namely 0.333... in a list of real > numbers? The way you have stated it, nothing. As there is no digit 3 at an infinite distance from the 0. in > 0.333..., there is no infinite decimal expansion. There is that same confusion surfacing again. An endless sequence of single decimal digits creates an infinitely > long expansion with no digit more than finitely far from the decimal > point. That is my point. (I use this argument, although it is easily to recognize as matheological nonsense.) Until WM learns this, his attempts to work with set theory will continue > to fail. > It was Dik who assumed an infinite distance. Please try at least to understand what you read before you answer. === Subject: Re: infinite product Am 28.11.2006 00:30 schrieb Robert Israel: > ... and the answer is no. > > For example, consider the infinite product > > P = (1- 1/sqrt(2)) (1 + 1/sqrt(2)) (1 - 1/sqrt(3)) (1 + 1/sqrt(3)) ... > = product_{j=0}^infty a_j > where a_{2k} = 1 - 1/sqrt(k+2) and a_{2k+1} = 1 + 1/sqrt(k+2). > > Since (1 - 1/sqrt(k))(1 + 1/sqrt(k)) = 1 - 1/k and sum_k 1/k diverges, > P converges to 0. > > Now write P = P1 P2 > P1 = product_{j=0}^infty b_j > P2 = product_{j=0}^infty c_j > where for each j, either b_j = a_{2j} and c_j = a_{2j+1} or > b_j = a_{2j+1} and c_j = a_{2j}, chosen in the following way. > Consider the partial products > P1(k) = product_{j=0}^k b_j > P2(k) = product_{j=0}^k a_j > Start with b_j = a_{2j} and c_j = a_{2j+1} until P2(k) > 2 > then b_j = a_{2j+1} and c_j = a_{2j} until P1(k) > 2 > then again b_j = a_{2j} and c_j = a_{2j+1} until P2(k) > 2 > etc. > > Then P1 and P2 both diverge. > the other subthread. The problem which I see now were to classify that type of counter- arguments to have a field, where my initial intended statement is still useful. One of my example was the question related to the Euler-product. if f(x,s) = (1 - x^-s) and g(x,s) = (1 + x^-s) then f(x,s) = g(x,s)*g(x,s^2)*g(x,s^4)*... = g(x,s)*f(x,s^2) Then if, with r a nontrivial zeta-null zeta(r) = f(2,r)*f(3,r)* f(5,r)*.... = 0 = g(2,r)*f(2,r^2) * g(3,r)*f(3,r^2)*... and with p(r) = g(2,r)*g(3,r)* g(5,r)*.... q(r) = f(2,r^2)*f(3,r^2)* f(5,r^2)*.... = zeta(r^2) we had zeta(r) = p(r)* zeta(r^2) Now zeta(r^2)<>0 if zeta(r)=0 (no nulls when Re(s)<>1/2 under RH) thus zeta(r) = p(r)* zeta(r^2) = 0 ==> p(r) = 0 was my argument. (and my intention to extend this reasoning to partial-products of p(r) according to modular classes) Hmm. Gottfried Helms === Subject: Re: infinite product Am 28.11.2006 00:30 schrieb The World Wide Wade: > is again, in more conversational terms. Step 1: In the original > sequence go out as far as desired, and stop. Put all the 2's up > the stopping point in the first subsequence, and all the 1/2's in > the second subsequence. The partial product in the first subseq. > stopping point in the original sequence, we go out as far as we > want, then stop again. This time we put all the 1/2's we find > into the first subseq., and all the 2's into the second. If we > take the second stopping point far enough out, the partial > product in the first subseq. (of all terms up to the second > stopping point) is now very small, that of the second one is > huge. Continuing, we obtain the desired subsequences; in each > case the partial products oscillate between 0 and oo. Hi - I think, I've got an understanding of this, now. Please excuse the delay... What I can see (and I'm thinking about), and possibly could be a basis for defining a restriction on my initial statement: if one partial product pm_1 at index k1 is made>2, the companion product pn_1 <2, and at index k3, when again pm_2 >2 then pn_3 < pn_1, and at k4 we have, that pm_4 Does volume completely determine mass? Different measures measure > different things and need have any correlation. Since measure theory is mathematics, it should be in position to abstract from physics and have only one measure for let's say seven eleven. === Subject: Re: Galileo's Paradox > > Does volume completely determine mass? Different measures measure > different things and need have any correlation. > > Since measure theory is mathematics, it should be in position to > abstract from physics and have only one measure for let's say seven eleven. Nothing in measure theory requires that every quality of an object have the same standard of measure. For a 3D object, one can simultaneously and independently have measurements of surface area and volume. There are even such anomalies as Gabriel's horn of finite volume but infinite surface area: http://mathworld.wolfram.com/GabrielsHorn.html === Subject: Re: Galileo's Paradox > >> >> No. Infinite quantities include e.g. an infinite amount of points. >> Infinite means: The process of quantification has not been finished or >> cannot be finished at all. > > > A non-empty set is infinite if and only if it can be put in one to one > correspondence with a proper subset of itself. That is the standard > definition of infinite for sets. I consider Dedekind wrong, and he admitted to have no evidence in order to justify his basic idea. === Subject: Re: Galileo's Paradox > >> >> No. Infinite quantities include e.g. an infinite amount of points. >> Infinite means: The process of quantification has not been finished or >> cannot be finished at all. > > > A non-empty set is infinite if and only if it can be put in one to one > correspondence with a proper subset of itself. That is the standard > definition of infinite for sets. > > > I consider Dedekind wrong, and he admitted to have no evidence in order > to justify his basic idea. The fact that Dedekind's definition of infiniteness of sets has been widely adopted indicates that many others have found it to be a useful definition. And utility is the measure of the value of a definition. === Subject: Re: Galileo's Paradox > > > I consider Dedekind wrong, and he admitted to have no evidence in order > to justify his basic idea. What sort of evidence? Surely not empirical evidence. Mathematics done abstractly has no empirical content whatsoever. Bob Kolker > === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 07:32:58 -0500, Bob Kolker > >> >> I consider Dedekind wrong, and he admitted to have no evidence in order >> to justify his basic idea. What sort of evidence? Surely not empirical evidence. Mathematics done >abstractly has no empirical content whatsoever. Except apparently for axioms and definitions. ~v~~ === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <456AF6F8.5020307@et.uni-magdeburg.de> <456C5361.40706@et.uni-magdeburg.de> <456D7417.30000@et.uni-magdeburg.de> <4t64hoF1127iuU4@mid.individual.net> <456EBC4D.5080608@et.uni-magdeburg.de> <4t81bsF12prlrU2@mid.individual.net> I consider Dedekind wrong, and he admitted to have no evidence in order >> to justify his basic idea. What sort of evidence? Surely not empirical evidence. Mathematics done >abstractly has no empirical content whatsoever. Except apparently for axioms and definitions. ****************************************************************** What axioms of what part of maths have empirical evidence in the sense Eckard is trying to convey?!? For him, and for other trolls, Cantor not having evidence for his idea (what stupid this sounds!) means that he (cantor) never foiund an aleph_null under his bed, or that so far no one can buy aleph_beith apples out there. What empirical evidence are there in group theory's axioms? Or in Topology? Tonio === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <456AF6F8.5020307@et.uni-magdeburg.de> <456C5361.40706@et.uni-magdeburg.de> <456D7417.30000@et.uni-magdeburg.de> <4t64hoF1127iuU4@mid.individual.net> <456EBC4D.5080608@et.uni-magdeburg.de> <4t81bsF12prlrU2@mid.individual.net> I consider Dedekind wrong, and he admitted to have no evidence in order >> to justify his basic idea. What sort of evidence? Surely not empirical evidence. Mathematics done >abstractly has no empirical content whatsoever. Except apparently for axioms and definitions. ****************************************************************** What axioms of what part of maths have empirical evidence in the sense Eckard is tryuing to convey?!? For him, and for other trolls, Cantor not having evidence for his idea (what stupid this sounds!) means that he (cantor) never foiund an aleph_null under his bed, or that so far no one can buy aleph_beith apples out there. What empirical evidence are there in group theory's axioms? Or in Topology? Tonio === Subject: Re: Galileo's Paradox > On Thu, 30 Nov 2006 07:32:58 -0500, Bob Kolker I consider Dedekind wrong, and he admitted to have no evidence in order to justify his basic idea. >>What sort of evidence? Surely not empirical evidence. Mathematics done >>abstractly has no empirical content whatsoever. >> Except apparently for axioms and definitions. >****************************************************************** >What axioms of what part of maths have empirical evidence in the >sense Eckard is tryuing to convey?!? I don't know what sense Eckard is trying to convey. My response was in reply to Bob not Eckard. > For him, and for other trolls, >Cantor not having evidence for his idea (what stupid this sounds!) >means that he (cantor) never foiund an aleph_null under his bed, or >that so far no one can buy aleph_beith apples out there. Well empirical evidence would certainly be one criterion for the truth of what one claims. Otherwise one is forced to rely on analytical criteria for the the truth of infinities which no modern mathematikers appear willing to assert and demonstrate. >What empirical evidence are there in group theory's axioms? Or in >Topology? The axioms and definitions themselves are empirical. ~v~~ === Subject: Re: Galileo's Paradox >> >> Uncountable simply means requiring infinite strings to index the >> elements of the set. That doesn't mean the set is not linearly ordered, >> or that there exist any such strings which do not have a successor. > > Uncountable means infinite but not of the same cardinality as the > integers. For example the set of real numbers. It is an infinite set, > but it cannot be put into one to one correspondence with the set of > integers. Uncountable means: Counting is impossible. This property obviously belongs to fictitious elements of continuum. There is simply too much of them. So counting is not feasible. As long as one looks at a finite, just potentially infinite heap of single integers, one has to do with individuals. The set of all integers is something else. It is a fiction. It is to be thought constituted of an uncountable amount of non-elementary elements. Well this looks nonsensical. There is indeed a selfcontradiction within the notion of an infinite set. Non-elementary means not having a distinct numerical address. Element means exactly defined by an impossible task. Eckard Blumschein === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 12:08:03 +0100, Eckard Blumschein Uncountable simply means requiring infinite strings to index the elements of the set. That doesn't mean the set is not linearly ordered, or that there exist any such strings which do not have a successor. >> >> Uncountable means infinite but not of the same cardinality as the >> integers. For example the set of real numbers. It is an infinite set, >> but it cannot be put into one to one correspondence with the set of >> integers. Uncountable means: Counting is impossible. This property obviously >belongs to fictitious elements of continuum. There is simply too much of >them. So counting is not feasible. As long as one looks at a finite, >just potentially infinite heap of single integers, one has to do with >individuals. The set of all integers is something else. It is a fiction. >It is to be thought constituted of an uncountable amount of >non-elementary elements. Well this looks nonsensical. There is indeed a >selfcontradiction within the notion of an infinite set. >Non-elementary means not having a distinct numerical address. Element >means exactly defined by an impossible task. You make the same mistake of assuming infinite means larger than when it only means numerically undefined. Infinites are neither large nor small; they're only undefined. Consequently there are no numerical relations or operations possible between them and finites. The reason counting is not possible is not because infinites are huge or because they form a continuum but because there is no numeric metric defined for them and counting as well as every other arithmetic relation and operation requires some kind of numeric definitional metric. ~v~~ === Subject: Re: Galileo's Paradox > > Uncountable means: Counting is impossible. This property obviously > belongs to fictitious elements of continuum. There is simply too much of > them. So counting is not feasible. As long as one looks at a finite, > just potentially infinite heap of single integers, one has to do with > individuals. The set of all integers is something else. It is a fiction. > It is to be thought constituted of an uncountable amount of > non-elementary elements. Well this looks nonsensical. There is indeed a > selfcontradiction within the notion of an infinite set. > Non-elementary means not having a distinct numerical address. Element > means exactly defined by an impossible task. Have you forgotten to take your meds again? Bob Kolker === Subject: Re: Galileo's Paradox > >> >> There is no such thing as genuine for numbers in mathematics. >> >> Maybe it will exist in genuine mathematics. >> >> >> So that EB has just refused to accept all of Analysis, including >> calculus, which is based on just the sort of sets that EB denies exist. >> >> This is perhaps a lie. I feel well served by pre-Cantorian analysis and >> by modern mathematics which does not really rely on set theory. > > Since all of pre-Cantorian analysis is embeddable in Cantorian > analysis without loss, and with some gains (e.g., point-set topology > and measure theory), there is nothing to be gained by such retrograde > devolution. While colonies were changed when embedded in a Commonwealth, Cantorian distemper did effectively almost not at all affect mathematics. I just see some imperfections. I guess, point-set topology and measure theory do not require the claim of set theory to rule all mathematics. I wonder if they require aleph_2. === Subject: Re: Galileo's Paradox > >> >> There is no such thing as genuine for numbers in mathematics. >> >> Maybe it will exist in genuine mathematics. >> >> >> So that EB has just refused to accept all of Analysis, including >> calculus, which is based on just the sort of sets that EB denies exist. >> >> This is perhaps a lie. I feel well served by pre-Cantorian analysis and >> by modern mathematics which does not really rely on set theory. > > Since all of pre-Cantorian analysis is embeddable in Cantorian > analysis without loss, and with some gains (e.g., point-set topology > and measure theory), there is nothing to be gained by such retrograde > devolution. > > While colonies were changed when embedded in a Commonwealth, Cantorian > distemper did effectively almost not at all affect mathematics. EB's distempers will not have any effect on mathematics, though set theory has had effects. > I just see some imperfections. Get your eyes tested. > I guess, point-set topology and measure > theory do not require the claim of set theory to rule all mathematics. They cannot exist without a foundation of set theory. > I wonder if they require aleph_2. Try learning enough of them to find out. === Subject: Re: Galileo's Paradox > >> The relations smaller, equally large, and larger are invalid for infinite quantities. >> >> >> All one needs do is divorce the length from the number of points, >> which is probably what Galileo did, as being different sorts of measures >> (like weight versus volume), and the problem disappears. >> >> No. Infinite quantities include e.g. an infinite amount of points. >> Infinite means: The process of quantification has not been finished or >> cannot be finished at all. > > That may be your personal definition of infiniteness, but is not > everyones, and does not govern anyone but yourself. Cantor himself was the victim of his own stupid notion of infinity. I see it but I cannot believe it. Not more was correct and logical. === Subject: Re: Galileo's Paradox > Cantor himself was the victim of his own stupid notion of infinity. > I see it but I cannot believe it. Cantor defined his sense of more quite precisely, and according to If EB wishes to reject that definition, then he also rejects the right to comment on the validity of Cantor's statement. === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <456AF6F8.5020307@et.uni-magdeburg.de> <456C5361.40706@et.uni-magdeburg.de> <456D7417.30000@et.uni-magdeburg.de> <456EB22F.70703@et.uni-magdeburg.de The relations smaller, equally large, and larger are invalid for infinite quantities. >> All one needs do is divorce the length from the number of points, >> which is probably what Galileo did, as being different sorts of measures >> (like weight versus volume), and the problem disappears. >> No. Infinite quantities include e.g. an infinite amount of points. >> Infinite means: The process of quantification has not been finished or >> cannot be finished at all. That may be your personal definition of infiniteness, but is not > everyones, and does not govern anyone but yourself. Cantor himself was the victim of his own stupid notion of infinity. > I see it but I cannot believe it. ************************************************************** So..?? You cannot see beyond your own nose: what's new, then? are engineer. So clear now how it is possible that you know so little of maths! Of course, since most high school teachers are engineers, accountants, architects, etc., and in general people that doesn't have the faintest idea what maths REALLY can be (and not what it is, unfortunately, at HS level), no wonder they screwed you up a little too much in school. Unfortunately, somehow you gathered the nerve and chutzpah to try to deal with mathematical stuff that is way over your head...with mathematicians. And this when you're armed only with your....engineer maths!!!!!! Unbelievable, uh? So you cannot see it....hehe: what else is new? Tonio Ps Have you, and anyone else, noted how all the anticantorian cranks sci.math is an uncensored group, so anyone can offer his piece to all...and you know what? I think this is just fine. I'm convinced that also from the most stupid, dense and even annoying crank/troll we all can learn. === Subject: Re: Galileo's Paradox > Ps Have you, and anyone else, noted how all the anticantorian cranks > sci.math is an uncensored group, so anyone can offer his piece to > all...and you know what? I think this is just fine. I'm convinced that > also from the most stupid, dense and even annoying crank/troll we all > can learn. I too conduct my affairs in accord with this conviction. -- Michael Press === Subject: Re: Galileo's Paradox > > Ps Have you, and anyone else, noted how all the anticantorian cranks > sci.math is an uncensored group, so anyone can offer his piece to > all...and you know what? I think this is just fine. I'm convinced that > also from the most stupid, dense and even annoying crank/troll we all > can learn. > > I too conduct my affairs in accord with this conviction. Even the worst of them can serve as bad examples! === Subject: Re: Galileo's Paradox > Ps Have you, and anyone else, noted how all the anticantorian cranks > sci.math is an uncensored group, so anyone can offer his piece to > all...and you know what? I think this is just fine. I'm convinced that > also from the most stupid, dense and even annoying crank/troll we all > can learn. That does seem to be the case. Mathematicians stopped kicking about the Cantorian approach many decades ago. Giants, such as Hilbert, welcomed Cantorian mathematics. Bob Kolker === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 07:34:26 -0500, Bob Kolker > Ps Have you, and anyone else, noted how all the anticantorian cranks >> sci.math is an uncensored group, so anyone can offer his piece to >> all...and you know what? I think this is just fine. I'm convinced that >> also from the most stupid, dense and even annoying crank/troll we all >> can learn. That does seem to be the case. Mathematicians stopped kicking about the >Cantorian approach many decades ago. Giants, such as Hilbert, welcomed >Cantorian mathematics. Except the problem still remains as to whether that approach represents all of mathematics and whether set theory has any pre emptive prerogative to terms such as cardinality to the exclusion of other usages. I've yet to see anyone demonstrate or even claim the truth of any one approach over others and that includes those you mention. If you're going to insist your definitions of such terms as cardinality are mathematically correct and definitively used by mathematicians then you need to establish first that whatever theory they're defined within represents the paradigm for all mathematics. And so far all you've produced is problematic philosophical babbling. ~v~~ === Subject: Re: Galileo's Paradox >> Let's rather say in Cantor's illusion of allegedly being able to count >> the uncountable. >> > Uncountable simply means requiring infinite strings to index the > elements of the set. That doesn't mean the set is not linearly ordered, > or that there exist any such strings which do not have a successor. Infinite strings are a reasonable fiction. Requiring this fiction is effectively the same as requiring something impossible. We are guessing that in a line any point has to have a successor. Doing so, we see the piture of points with space in between. Weyl went a side-step further. He imagined the continuum a sauce with embedded single points. Actually the continuum is a concept that complements the concept of discrete numbers and complements it at a time. A genuine continuum cannot at all be resolved into single points. What about existence, I got aware of an appealing idea: Existence means common propereties. Reals, as indirectly defined with DA2, differ from genuine numbers in being uncountable. So they only exist like a fiction. === Subject: Re: Galileo's Paradox > Infinite strings are a reasonable fiction. Well that's a matter of personal opinion, what's reasonable and what's not reasonable, but I basically agree, there's no such thing as an infinite string in the real universe, but it's reasonable to sometimes hypothesize them so you can have automata generate them (potentially, if allowed to run forever) etc. > Requiring this fiction is effectively the same as requiring > something impossible. This is where you go astray, attacking a straw man. Nobody requires anybody to work with infinite strings, or talk about infinite strings, or perform derivations/proofs that deal with infinite strings. You (Eckard Blumschein) are perfectly free to ignore them and never mention them again. It's your *choice* that you in fact do talk about them here in this newsgroup. Why do you do that??? > We are guessing that in a line any point has to have a successor. Who is we??? I'm not aware of anyone in the whole world, except maybe you (Eckard Blumschein) who has ever guessed that. In the first place, from a single point, a line goes both directions, equally, and neither of those directions is the direction in preference to the other, so even if you allowed only integer points along a line, still there'd be no successor, merely two (2) different neighbors, one on each side. But moreso, even if you mean directed line instead of line, still there's a really elementary property of classical geometry, that any two different points define a line segment between them, and that line segment has at least one interior point, it doesn't consist *solely* of its two endpoints. Accordingly, if you assume two points are neighbors on a line, or one is the successor of the other on a directed line, you get a contradiction, a third point between the two. So who, other than you (Eckard Blumschein), would be so careless to make such a stupid guess as you cited above?? > Doing so, we see the piture [sic] of points with space in between. Again, who other than you (Eckard Blumschein) would see a picture of a line like that? Have you been looking at your dot-matrix computer-display so long you mistakenly believe the row of discrete dots on the screen *is* an actual geometrical line? > Actually the continuum is a concept that complements the concept of > discrete numbers and complements it at a time. A genuine continuum > cannot at all be resolved into single points. What do you mean by resolved? > Existence means common propereties. Hmm, per science that sorta makes sense. Nothing is demonstrated true except by experiments that can be repeated by different people to yield consistent results. Anything not sharable in that way can be disregarded as mere halucination, not actual existence. > Reals, as indirectly defined with DA2, If by DA2 you mean Cantor's second diagonal argument, you are quite mistaken. DA2 doesn't define anything. DA2 merely shows that it's impossible to enumerate a particular set. Any supposed enumeration of that set results in contradiction (some number that the supposed enumeration overlooked, failed to include). > differ from genuine numbers in being uncountable. I'm not familiar with the term genuine numbers in mathematics. I've seen you use the term several times, so are they your own personal definition? If so, please present your definition for us to read, either a consistent set of axioms for genuine numbers, or a way to construct genuine numbers from some well-known mathematical system. If all you real mean are rational numbers, or constructable numbers, we already have a name for them, so your synonym is of no value. So you'd better hope your genuine numbers are a brand-new system you invented that nobody else invented (and named) before you. === Subject: Re: Galileo's Paradox where one set contains all the elements of another, plus more, it can rightfully be considered a larger set. >> >> All of oo? >> >> > > Yes. All of the naturals are integers. Only half of all the integers are > naturals. > > All of the points in (0,1] are in (0,2], but only half of all of the > points in (0,2] are in (0,1]. You are equating two quite different notions: smaller and half as large. In case of two finite heaps of size a and b of numbers, a=b/2 implies a gallons you've certainly measured the gas. Or if you ask how much > space and get the answer two inches you've certainly measured the > space. Common sense terminology tends to be nearly as logical as the mathematical one but more subtle. The question how much gas refers to the insight that the amount of gas is not directly countable. Nowadays, almost any continuous measurement yields in the end a numerical result because processing and transmission of digital signals is superior to processing and transmission of analogous signals. Eckard Blumschein === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 10:09:10 +0100, Eckard Blumschein > I mean if you ask how much gas and get the answer two >> gallons you've certainly measured the gas. Or if you ask how much >> space and get the answer two inches you've certainly measured the >> space. Common sense terminology tends to be nearly as logical as the >mathematical one but more subtle. >The question how much gas refers to the insight that the amount of gas >is not directly countable. The amount of gas or even space is as directly countable as anything else. All you have to do is superimpose a metric.The fact that certain things come neatly prepackaged suggests a metric nothing more. > Nowadays, almost any continuous measurement >yields in the end a numerical result because processing and transmission >of digital signals is superior to processing and transmission of >analogous signals. That kind of efficiency has nothing to do with the issue of a metric and its application. If you want to count you need some metric to do it. It's that simple and the metric is what establishes cardinality in mathematics. ~v~~ === Subject: Re: Galileo's Paradox >> >> It has the same cardinality perhaps, but where one set contains all the >> elements of another, plus more, it can rightfully be considered a larger >> set. > > Not necessarily so, if it is an infinite set. > > Bob Kolker This time I agree with Bob. === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 09:52:49 +0100, Eckard Blumschein It has the same cardinality perhaps, but where one set contains all the elements of another, plus more, it can rightfully be considered a larger set. Tony, you know we've been over this previously. All infinite means is lack of definition for a particular predicate such as numerical size. And when you add numerical finites to numerical infinites the result is still infinite. This problem mainly arises I suspect because mathematikers insist on portraying infinites as larger than naturals and somehow coming beyond the range of naturals such as George Gamow's famous 1, 2, 3, . . . 00. Then mathematkers try to establish certain numerical properties for infinities by comparative numerical analysis and mapping with numerically defined finites. However one cannot do comparative numerical analysis and numerical analysis with numerically undefined infinites anymore than one can do arithmetic. Infinites are neither large nor small; they're just numerically undefined. >> Not necessarily so, if it is an infinite set. >> >> Bob Kolker This time I agree with Bob. ~v~~ === Subject: Re: Galileo's Paradox >I was addressing your claim that there was no extension from the >finite case. In the finite case, two sets have the same number >of elements if and only if there exists a one to one correspondence >between them. This very simple idea has been extended to the >infinite case. Incidentally, can you point me please to convincing evidence for all pertaining conclusions to be justified? OK. The idea of a 1:1 corresondence is indeed a simple idea. The idea of infinity is not. I see it the other way round. >>That depends on what 'idea of infinity' of you are talking about. >>The mathematical definition of 'infinite' is as simple as the >>idea of a 1:1 correspondence. >> The mathematical definition of infinity may be simple, but is it > unproblematic? It seems to me that infinity is a sublte and difficult > concept. It is indeed a fiction. >What concept of infinity? Note, I said 'infinite', not 'infinity'. You have been talking about Cantor and one-to-one correspondences, so you have been talking about set theory. The word 'infinity' is generally not used in set theory. It has been implicitely used if we directly or indirectly say there is an (infinite) set. The axiom of extensionality claims: A set is unambiguously set by its elements. In other words: There are infinitly many elements being united into an allegedly existing set. The infinite set is a fiction. So it is questionable whether or not it actually is in the sense one can attribute the same properties to it like to a finite set. It has no formal definition. Archimede's definition of a number is the same as the definition of potential infinity. What about actual infinity, one could write oo=a/0. Do not worry. Division by zero is strictly forbidden, yes. However, belief in the existence of an infinite set is pretty much the same. 'infinite' is used to describe sets, Systems of numbers are potentially infinite. If the original meaning of a set was not somewhat contrasting, there would not be any reason to deal with sets instead of opem ended series of numbers. The peculiarity of a set is: It has already been set. In other words, it is something perfect, something static. An infinite set is an intentional selfcontradiction. and it has a very simple definition. What definition do you refer to? Cantor's definition of an (infinite) set has been proven untennable and withdrawn without substitute by Fraenkel. > >> I'm talking about mathematical meaning. Specifically I'm talking >> about How many?, more or less etc.. > > How many is not a technical term. Ironically, Cantorian mathematics cannot accept a consequent distinction between countable (many) and uncountable (much) although it was Cantor himself who demonstrated that rationals are contable while reals are uncountable. The reason for this stunning fact is Cantor's misinterpretation of uncountable as more than countable (ueberabzaehlbar). > Cardinality corresponds to our > notion of how many in the finite case, In the finite case there is no reason to ask how much. Numbers are countable. According to Dedekind, ... A set is infinite if there exists a bijection between the set and a proper subset of itself. Because an infinite set is a fiction, it is questionable whether or not it is justified to attribute any subset to it. An infinite set would have indefinitely much of subsets, definitely too much as to focus on a single one. That is what mathematicians mean when they say a set is infinite. There are other equivalent definitions. > >> I know already. I too, and I understood they are imprecise altogether. > So what are you asking? That is the definition of 'infinite set'. > It means mathematically exactly what it says. It is based on lacking understanding of what it means to have no limit. It ignores that an infinite set is the same as the fiction actual infinity. Dedekind's definition contradicts Euclid who stated that the whole is always larger than its parts. Such contradiction is only allowed within the realm of mathematical fiction. Do you have the same problem with prime numbers? Or even numbers? The words 'prime' and 'even' have meanings outside of mathematics. Do you feel obligated to drag those meanings into a discussion of prime or even numbers? Telltale terminology unveils which errors led to the present mess in an allegedly fundamental part of mathematics. > I accept that. The contradiction comes about if the one notion > suggests equality of size and the other notion suggests inequality. Which > they do, so there is a prima facie paradox. >The problem is that you are using a word 'size' that you have not defined. > >> True. I took it that people knew what I meant. And I think they do. > > No. I do not know what it means when applied to a set. Does > it mean cardinality? Fraenkel indeed explained cardinality like generalized size. > If so then we would not be having this discussion. > If it does not mean cardinality, what does it mean? Can you give > me a mathematical definition of size? The size of a number is its value. If there are no infinite (no transfinite) numbers, then there is no reason to distinguish cardinal and ordinal nonsense. >> Certainly I write things in the heat of the moment which I later >> regret. But this wasn't meant as a cheap jibe. I've already conceded that >> following Cantor might in some deep way be right, Deep? You might even be prone to join scientology. Cantor himself claimed immediate contact to Him. Cantor's theory has been called naive for decades. It has been replaced by an aximatic method being not really correct but sucessfully designed in a slick manner. Look into the book Zahlen by Ebbinghaus et al. in order to find that an obvious error led to something valuable. if it comes down to >> following productive branches and forsaking dead ends. I is overdue to forsake the dead end set theory. Ongoing big fuss about CH and AC, R*, etc. did not led to anything tangible. There is not even a single application of aleph_2 or higher alephs. Nobody needs aleph_0 and aleph_1 because the two qualities infinite and uncountable do not need a misinterpretation. >> There is an intuition that there are less squares (even numbers, >> primes, whatever) than naturals. This is not an intuition but correct reasoning for any given number of numbers. It merely lacks justification for open ended strings. >> We are talking here precisely of intuitions about infinite sets. Ebbinghaus called Cantor's way of thinking highly intuitive. What did he mean? He meant trusting in the guess of analogy. What is valid for numbers should be valid for transfinite numbers, too. >> It is not good enough to say: You're >> getting mixed up with finite sets, or: You can't rely on common sense >> intuitions in maths. Ironically, it's the proponents of what I call the Dedekind Cantor illusion who rely on half-religious half-common sense intuitions in maths. It was Galileo Galilei who resolved the paradox in a purely rational manner of thinking. >> So if there are less squares than naturals, squares as well as naturals do not have an upper limit. Therefore, we may conclude that there are neiter more nor equally many nor less. They are simply not comparable with each other. > Again, your problem is insisting that cardinality match some vague notion of 'how many' > that you have not defined. The basic problem is: He lacks the insight that cardinality is a cardinal mistake, something that has proven unfounded as well as useless. Eckard Blumschein === Subject: Re: Galileo's Paradox > GALILEO'S PARADOX 1 2 3 4 5 ................. > 1 4 9 16 25 ............... There is a paradox because the 1:1 Correspondence suggests the sets > are equal in size, by extension from the finite case, and yet clearly the > second set is contained in the first set. That an infinite set can be put > into 1:1 C with a proper subset is not by itself paradoxical. That is only > the beginning, the facts of the case. The paradox is that the squares seem > to be both smaller than N and the same size as N. I want to suggest there are only two sensible ways to resolve the > paradox: 1) So- called denumerable sets may be of different size. 2) It makes no sense to compare infinite sets for size, neither to say one > is bigger than the other, nor to say one is the same size as another. The > infinite is just infinite. Yea, a quite negative approach. But it is not without intuitive backround. Intuitivelly speaking the idea that an infinite set has no fixed size comes to ones mind. That idea that infinity makes all infinite sets equal in size is also beautiful, and I think it was the idea before Cantor showed that there can be infinite sets of different sizes, the alephes and the powers are different in size, though infinite. If you want to change the definition of infinity to a one like saying, infinity is that quality which cause all sets that possess it to be equal in size, instead of the current definition of an infinite set, that is a set injectable to some proper subset of it, then you are free to do that,provided you bring a new definition of set size, other than cardinality. But this definition that looks to be their in your mind, is a negative one, I mean it canceal the chance of having meaningful comparisons of sizes of sets when they are infinite. If you bring a more positive claim, for example a method by which you can detect that there can exist difference is size of infinite sets that are currently considered to have equal size, then this idea would be somewhat chanllenging, but as I said you should bring a different rule of size comparison than cardinality. People here desire infinity to be determined by sets and desire the size of an infinite set to be also solelly determined by sets, i.e. knowledge of the members in a set is enough for you to know that they are infinite and let you know their set size, once apon a time I suggested the idea of generational size, which seems to be a measure of the generational size of sets as they are generated from themselfs or from other sets, a quality that is determined by the generational function from one set to the other. However even generationally speaking there are some types of generational size comparison that is solelly determined by sets only without the need to know the generational function of it from the other set. Example a set and its power set, whatever generational function that generated P(x) from x, then this function is strictly serjective from P(x) to x. and accordingly P(x) has always a bigger generational size than x. Anyhow this idea of generationl size was not apealing to the majority people in this forum, and it is certainly in the opposite direction to what you are suggesting here. However, two ideas are strong when intuitivelly speaking of infinite sets size, the first is that there nothing called infinite set. i.e to state that they are contradictive since bijection to a proper subset somewhat seems unapealing intuitivelly. the other idea is that if infinite sets exists then they should be equal in size. Both of these ideas though negative, yet can be true. I will discuss the idea of generational size again in a separate thread. Zuhair > My line of thought is that the 1:1C is a sacred cow. That there is > no extension from the finite case. If we want to compare the two sets for size we would write, not the > above, but: 1 2 3 4 5 6 7 8 9 ............... > 1 2 3 4 5 6 7 8 9................ > ^ ^ ^ (The intention here is to highlight the squares in the second row of > integers.) Then we would notice that the relative size of the squares set > becomes ever smaller as n increases, that increasingly large numbers of > integers are missed out. In fact, if we wanted to find a plausible > candidate for a set eual in size to N, then we would choose not the > squares, but the non-squares. The contrived nature of the 1:1C becomes more obvious when we > compare with N, sets that appear to be larger than N. The clearest example > is Z. We have: 1 2 3 4 5 6 7 ......... > 1 -1 2 -2 3 -3 4 ......... which is mildly clever, but again if we wanted to compare the two sets for > size we would write: 0 1 2 3 4.... > ...-4 -3 -2 -1 0 1 2 3 4.... with a perfect 1:2 Correspondence. Here one would like to say, since not only is there 1:1C between Z > and a proper subset, but an identity (1 2 3 4 ......), that however you > define infinity there has got to be more in Z than in N. Not, of course, if > you make all countable sets equal in size by definition. But for me, that > doesn't relieve the paradox at all. On the contrary it builds it into the > foundation of the mathematics. > I would like to suggest that the existence of 1:1C between the two > sets is a CONSEQUENCE of the fact that they are both infinite. The > infinities are what gives one room to manoeuvre, to manufacture a 1:1C. It > has no bearing on their relative size. Can one make sense of Z = 2N, of Q = N^2, etc.? (Incidentally the > number of squares would be sq.rt. of N, since after n^2 integers there are > n squares.) Maybe it's complete rubbish, but my argument is that the > alternative is the ineffable infinity. If it does make sense, there is no > place for a diagonal argument, or a power set argument, since it would > already be conceded that 10^N > N, that 2^N > N, or in general that k^N N, just as Z > N and Q > N. There remains of course Cantor's proof that R cannot be put into a > 1:1C with N, which is very interesting. But what does it mean? > Maybe something like this: So-called denumerable sets can be represented on a > finite-dimensional lattice, so that a self-avoiding walk can be shown to > systematically cover the entire line, are, volume or hyper-volume. For R > understood as a set of decimals (to choose that -- perfectly good -- > representation), by contrast, every decimal place can be construed as an > axis. In any case what I don't understand is how this affects the simple > paradox with which we began. > However, it may very well be that my insufficiently tutored brain > has flown its coop again, in which case I would be very grateful for any > illumination. > > > Six Letters 24/11/06 === Subject: Re: Galileo's Paradox > GALILEO'S PARADOX >> 1 2 3 4 5 ................. >> 1 4 9 16 25 ............... >> There is a paradox because the 1:1 Correspondence suggests the sets >> are equal in size, by extension from the finite case, and yet clearly the >> second set is contained in the first set. That an infinite set can be put >> into 1:1 C with a proper subset is not by itself paradoxical. That is only >> the beginning, the facts of the case. The paradox is that the squares seem >> to be both smaller than N and the same size as N. >> I want to suggest there are only two sensible ways to resolve the >> paradox: >> 1) So- called denumerable sets may be of different size. >> 2) It makes no sense to compare infinite sets for size, neither to say one >> is bigger than the other, nor to say one is the same size as another. The >> infinite is just infinite. Yea, a quite negative approach. But it is not without intuitive >backround. Intuitivelly speaking the idea that an infinite set has no >fixed size comes to ones mind. That idea that infinity makes all >infinite sets equal in size is also beautiful, and I think it was the >idea before Cantor showed that there can be infinite sets of different >sizes, the alephes and the powers are different in size, though >infinite. If you want to change the definition of infinity to a one >like saying, infinity is that quality which cause all sets that possess >it to be equal in size, instead of the current definition of an >infinite set, that is a set injectable to some proper subset of it, >then you are free to do that,provided you bring a new definition of set >size, other than cardinality. But this definition that looks to be >their in your mind, is a negative one, I mean it canceal the chance of >having meaningful comparisons of sizes of sets when they are infinite. >If you bring a more positive claim, for example a method by which you >can detect that there can exist difference is size of infinite sets >that are currently considered to have equal size, then this idea would >be somewhat chanllenging, but as I said you should bring a different >rule of size comparison than cardinality. People here desire infinity to be determined by sets and desire the >size of an infinite set to be also solelly determined by sets, i.e. >knowledge of the members in a set is enough for you to know that they >are infinite and let you know their set size, once apon a time I >suggested the idea of generational size, which seems to be a measure of >the generational size of sets as they are generated from themselfs or >from other sets, a quality that is determined by the generational >function from one set to the other. However even generationally >speaking there are some types of generational size comparison that is >solelly determined by sets only without the need to know the >generational function of it from the other set. Example a set and its >power set, whatever generational function that generated P(x) from x, >then this function is strictly serjective from P(x) to x. and >accordingly P(x) has always a bigger generational size than x. Anyhow this idea of generationl size was not apealing to the majority >people in this forum, and it is certainly in the opposite direction to >what you are suggesting here. However, two ideas are strong when intuitivelly speaking of infinite >sets size, the first is that there nothing called infinite set. i.e to >state that they are contradictive since bijection to a proper subset >somewhat seems unapealing intuitivelly. the other idea is that if >infinite sets exists then they should be equal in size. Both of these ideas though negative, yet can be true. I will discuss the idea of generational size again in a separate >thread. Zuhair > Lots of interesting stuff there which I need time to digest. I look forward to your post on generational size. Much appreciated, Six Letters === Subject: Re: Galileo's Paradox >> 2) It makes no sense to compare infinite sets for size, neither to say one >> is bigger than the other, nor to say one is the same size as another. The >> infinite is just infinite. > > Yea, a quite negative approach. But it is not without intuitive > backround. Intuitivelly speaking the idea that an infinite set has no > fixed size comes to ones mind. That idea that infinity makes all > infinite sets equal in size is also beautiful, Really? The OP did not write equal in size. Infinite is just infinite but no size. > and I think it was the > idea before Cantor showed that there can be infinite sets of different > sizes, He did not! Not! Not! He just misinterpreted uncountable as mor than countable. Is incorrect more than correct? > the alephes and the powers are different in size, though > infinite. If you want to change the definition of infinity to a one > like saying, infinity is that quality which cause all sets that possess > it to be equal in size, instead of the current definition of an > infinite set, that is a set injectable to some proper subset of it, > then you are free to do that,provided you bring a new definition of set > size, other than cardinality. But this definition that looks to be > their in your mind, is a negative one, I mean it canceal the chance of > having meaningful comparisons of sizes of sets when they are infinite. The idea of considering sets was born in order to make the illusion by Dedekind and Cantor less obviously wrong than it was with numbers. > If you bring a more positive claim, The argument Cantors transfinite numbers are somthing positive something progressive is old and has proven wrong. Not even aleph_2 has found an application. === Subject: Re: Galileo's Paradox > > The argument Cantors transfinite numbers are somthing positive something > progressive is old and has proven wrong. Not even aleph_2 has found an > application. So what? The criterion for goodness in pure mathematics is consistency, not usability. After that aesthetic issues dominate. Are the systems interesting. Do they have a kind of beauty? etc. etc. Bob Kolker === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 11:13:38 -0500, Bob Kolker >> The argument Cantors transfinite numbers are somthing positive something >> progressive is old and has proven wrong. Not even aleph_2 has found an >> application. So what? The criterion for goodness in pure mathematics is consistency, >not usability. Truth would be an even better criterion. > After that aesthetic issues dominate. Are the systems >interesting. Do they have a kind of beauty? etc. etc. Yes, yes, truth is beauty and beauty truth but I still haven't gotten anyone to answer whether set theory represents all of mathematics and if not why the term cardinality cannot have other equally valid mathematical definitions than used in set theory? ~v~~ === Subject: Re: Galileo's Paradox <456EEEBB.7070007@et.uni-magdeburg.de> <4t8e9lF12uobhU2@mid.individual.net> anyone to answer whether set theory represents all of mathematics No, it doesn't. One can have mathematical theories other than set theory. What is usually said is that all of the usual theorems of classical mathematics can be expressed and proven in certain set theories (for example Z set theory with dependent choice). But that is not to claim that set theories, especially any given set theory, is the only mathematical theory, or only possible foundational theory, or that it exhausts all of mathematics. > and if not why the term cardinality cannot have other equally valid > mathematical definitions than used in set theory? I wouldn't use the word 'valid', but, for my own view, 'cardinality' is an English language nickname I use to talk about a defined symbol of certain theories. I don't demand that the word 'cardinality' may not be used in a different sense for different theories. In fact, even among different set theories, 'cardinality' is used in somewhat different ways. But if a conversation about mathematics is to be coherent, then if we use the word 'cardinality', we should be clear as to which definition we are working with at any given point in the conversation. MoeBlee === Subject: Re: Galileo's Paradox > Cardinality is generalized from the simple count of finite sets to the infinite case. In the finite case, the cardinality of a set is exactly a natural number, a quantity. In the infinite case, cardinality becomes something more ephemeral, >> Epheremal means shortlived. We have a saying: Lies live short. >> but it still has its roots in the count of a set. >> Let's rather say in Cantor's illusion of allegedly being able to count >> the uncountable. > What about when there is more than one type of measure that can be > applied to a set, or none at all? What happens then? >> Then perhaps a red light will indicate logical error. > > Does volume completely determine mass? Different measures measure > different things and need have any correlation. Given a density of the material, indeed, volume determines mass. Given a set density, value range determines count. === Subject: Re: Galileo's Paradox > Given a > set density, value range determines count. Compare the set densities of the set of naturals, the set of rationals, the set of algebraics, the set of transcendentals, the set of constructibles, and the set of reals. === Subject: Re: Galileo's Paradox > > >> Given a >> set density, value range determines count. > > Compare the set densities of the set of naturals, the set of > rationals, the set of algebraics, the set of transcendentals, the set of > constructibles, and the set of reals. Rather difficultt o formulate relations between those in standard theory. In the name of IST, I'll avoid any criteria including the notion of standard and state the following. The size of the set of hypernaturals is the square root of the size of the set of hyperreals. The set of hyperrationals corresponds to the square of the set of hypernaturals, minus all those pairs that are redundant, such as 2/4 or 6/18. That number of the hyperreals are the hyperirrationals. I am not sure how to relatively quantify transcendentals, constrictibles, or algebraics. Those are probably considered all countable by you, which doesn't say much about their relative sizes. === Subject: Re: Galileo's Paradox > > >> Given a >> set density, value range determines count. > > Compare the set densities of the set of naturals, the set of > rationals, the set of algebraics, the set of transcendentals, the set of > constructibles, and the set of reals. > > Rather difficultt o formulate relations between those in standard > theory. In the name of IST, I'll avoid any criteria including the notion > of standard and state the following. The size of the set of > hypernaturals is the square root of the size of the set of hyperreals. > The set of hyperrationals corresponds to the square of the set of > hypernaturals, minus all those pairs that are redundant, such as 2/4 or > 6/18. That number of the hyperreals are the hyperirrationals. I am not > sure how to relatively quantify transcendentals, constrictibles, or > algebraics. Those are probably considered all countable by you, which > doesn't say much about their relative sizes. When challenged to support his fool theories, TO resorts to nonsense. === Subject: Re: Galileo's Paradox > > >> Given a >> set density, value range determines count. > > Compare the set densities of the set of naturals, the set of > rationals, the set of algebraics, the set of transcendentals, the set of > constructibles, and the set of reals. Either discrete or continuous. Nothing in between. === Subject: Re: Galileo's Paradox > > >> Given a >> set density, value range determines count. > > Compare the set densities of the set of naturals, the set of > rationals, the set of algebraics, the set of transcendentals, the set of > constructibles, and the set of reals. > > Either discrete or continuous. Nothing in between. The natural ordering of the set of rationals is neither continuous not discrete, at least in the mathematical meanings of those words. It is dense, but not complete, and both density and completeness are required for the set to have a continuous ordering. So of the sets mentioned above, the reals and only the reals are continuous in amy mathematically acceptable sense. === Subject: Re: Galileo's Paradox > > > Either discrete or continuous. Nothing in between. You obviously have no knowledge of fractal dimension or Hausdorf dimension. For example the Peano space filling curves have a dimension between 1 and 2. Bob Kolker === Subject: Re: Galileo's Paradox On Thu, 30 Nov 2006 11:15:09 -0500, Bob Kolker >> >> Either discrete or continuous. Nothing in between. You obviously have no knowledge of fractal dimension or Hausdorf >dimension. For example the Peano space filling curves have a dimension >between 1 and 2. Don't forget about fractional dimensions and fractional cardinality. ~v~~ === Subject: Re: Galileo's Paradox >> Uncountable simply means requiring infinite strings to index the >> elements of the set. That doesn't mean the set is not linearly >> ordered, or that there exist any such strings which do not have a >> successor. > > Uncountable means infinite but not of the same cardinality as the > integers. For example the set of real numbers. It is an infinite set, > but it cannot be put into one to one correspondence with the set of > integers. > > Bob Kolker Yes, Bob, I know that. It boils down to the same thing. Any uncountable set with element indexes expressed as digital numbers will require infinitely long indexes for most elements, such as is the case for the reals in any nonzero interval. Standard integers each require only a finite number of digits. Tony === Subject: Re: Galileo's Paradox > Any uncountable > set with element indexes expressed as digital numbers will require > infinitely long indexes for most elements, such as is the case for the > reals in any nonzero interval. There is nothing in being a set, including being a set of reals, that requires its members to be indexed at all. === Subject: Re: Galileo's Paradox > >> Any uncountable >> set with element indexes expressed as digital numbers will require >> infinitely long indexes for most elements, such as is the case for the >> reals in any nonzero interval. > > There is nothing in being a set, including being a set of reals, that > requires its members to be indexed at all. To establish an explicit bijection between infinite sets does require an ordering on the sets, at least in general. === Subject: Re: Galileo's Paradox > >> Any uncountable >> set with element indexes expressed as digital numbers will require >> infinitely long indexes for most elements, such as is the case for the >> reals in any nonzero interval. > > There is nothing in being a set, including being a set of reals, that > requires its members to be indexed at all. > > To establish an explicit bijection between infinite sets does require an > ordering on the sets, at least in general. It requires a function between the sets, but neither set need be ordered. For example, let P = R^2 be the Cartesian plane, then for any reals a and b with a^2 + b^2 > 0, (x,y) |--> (a*x + b*y, -b*x + a*y) bijects P to itself, but P is not an ordered set. === Subject: Re: Galileo's Paradox > > > To establish an explicit bijection between infinite sets does require an > ordering on the sets, at least in general. Not true. One can map the disk of radius one one onto the disk of radius two without ordering points in either disk. Hint: Use a cone. Or if you like vectors map the vector V of unit length into 2*V which has length 2. No ordering in sight. So in genaral one does not require an ordering. Bob Kolker === Subject: Re: Galileo's Paradox > >> To establish an explicit bijection between infinite sets does require >> an ordering on the sets, at least in general. > > > Not true. One can map the disk of radius one one onto the disk of radius > two without ordering points in either disk. Hint: Use a cone. Or if you > like vectors map the vector V of unit length into 2*V which has length > 2. No ordering in sight. Oops. Map the vector V of length <= 1 to vector 2*V which has length <= 2. Sorry about that. Bob Kolker === Subject: Re: Galileo's Paradox > > If you think you have some wonderful notion of integrated count and measure that applies to all sets then even if you're correct (you're NOT) then cardinality is still a valid definition and still as useful as it is now. >> Yes, and the square wheel will always be as useful as it ever has. > > Any squareness is all in TO's wheels. > What about when there is more than one type of measure that can be applied to a set, or none at all? What happens then? >> Where count can be calculated from either of two measures, then one has >> a choice in that matter. Hopefully, one gets the same result either way. >> Do you have an example you'd like to explore? > > Outer measure of sets in R^n, defined as the LUB of the content of a > covering by open intervals, for one. Where standard measure is the same, there still may be an infinitesimal difference, such as between (0,1) and [0,1], if that's what you mean. === Subject: Re: Galileo's Paradox > > If you think you have some wonderful notion of integrated count and measure that applies to all sets then even if you're correct (you're NOT) then cardinality is still a valid definition and still as useful as it is now. >> Yes, and the square wheel will always be as useful as it ever has. > > Any squareness is all in TO's wheels. > What about when there is more than one type of measure that can be applied to a set, or none at all? What happens then? >> Where count can be calculated from either of two measures, then one has >> a choice in that matter. Hopefully, one gets the same result either way. >> Do you have an example you'd like to explore? > > Outer measure of sets in R^n, defined as the LUB of the content of a > covering by open intervals, for one. > > Where standard measure is the same, there still may be an infinitesimal > difference, such as between (0,1) and [0,1], if that's what you mean. The outer measure of those two sets is exactly the same. === Subject: Re: Galileo's Paradox > If you think you have some wonderful notion of integrated count and >> measure that applies to all sets then even if you're correct (you're >> NOT) then cardinality is still a valid definition and still as useful >> as it is now. > Yes, and the square wheel will always be as useful as it ever has. Any squareness is all in TO's wheels. What about when there is more than one type of measure that can be >> applied to a set, or none at all? What happens then? > Where count can be calculated from either of two measures, then one has > a choice in that matter. Hopefully, one gets the same result either way. > Do you have an example you'd like to explore? Outer measure of sets in R^n, defined as the LUB of the content of a covering by open intervals, for one. >> Where standard measure is the same, there still may be an infinitesimal >> difference, such as between (0,1) and [0,1], if that's what you mean. > > The outer measure of those two sets is exactly the same. Right, and yet, the second is missing two elements, and is therefore infinitesimally smaller in measure. That doesn't show up on the standard ruler. 0.999...=1. === Subject: Re: Galileo's Paradox >> Where standard measure is the same, there still may be an infinitesimal >> difference, such as between (0,1) and [0,1], if that's what you mean. > > The outer measure of those two sets is exactly the same. > > Right, and yet, the second is missing two elements, and is therefore > infinitesimally smaller in measure. Except that in outer measure there are no infinitesimals, and the outer measure of the difference set, {0,1} is precisely and exactly zero. === Subject: Re: Galileo's Paradox On Wed, 29 Nov 2006 14:09:19 -0500, Bob Kolker Actually not because infinity a well defined mathematical concept >> whereas cardinality is only an ambiguously defined concept >> mathematically restricted to undemonstrable set analytical techniques. Nonsense. Nonsense that infinity is a well defined mathematical concept or that the definition of cardinality is an ambigously defined concept restricted to set analytical techniques? I mean unless you have some as yet undemonstrated idea that parochial set analytical techniques represent some kind of paradigm for mathematics as a whole I see no necessity to restrict the concept of cardinality in one for the other. > Two sets have the same cardinality if and only if there exists >a one to one onto mapping from one to the other. That is a plain >definition. Same cardinality produces and equivalence relation defined >on sets. The cardinal number of a set is the equivalence class of sets >with the same cardinality as the the given set. Yeah, yeah, yeah, Bob. You seem to confuse set analtyical techniques and definitions with mathematics. >Don't give up your day job Lester. But this is my day job, Bob. I mean why else would anyone pander to a bunch of congential idiots trying to make sense out of set theory? ~v~~ === Subject: Re: Galileo's Paradox On Wed, 29 Nov 2006 14:11:57 -0500, Bob Kolker Just out of curiosity, Bob, why is cardinality in set theory not a >> measure? I mean if you ask how much gas and get the answer two >> gallons you've certainly measured the gas. Or if you ask how much >> space and get the answer two inches you've certainly measured the >> space. It seems to me that you can obviously superimpose cardinality >> on questions like how much without having to count or match things. Consider two measure sets, disjoint. The measure of the union is the sum >of the measures of each. So two inches is not the measure of the space in question? >Now consider two sets of the same cardinality, disjoint. The cardinality >of the union equals the cardinality of either. But not the cardinality of both? >In short cardinality does not add like measure. Well you say so, Bob, but I can't decipher the reason why. If you add the cardinality of two inches to the cardinality of two inches don't you get the cardinality of four inches? ~v~~ === Subject: Re: Galileo's Paradox I want to suggest there are only two sensible ways to resolve the > paradox: >> 1) So- called denumerable sets may be of different size. >> 2) It makes no sense to compare infinite sets for size, neither to say one > is bigger than the other, nor to say one is the same size as another. The > infinite is just infinite. >> > My line of thought is that the 1:1C is a sacred cow. That there is > no extension from the finite case. >What do you mean by that? The one-to-one correspondence works perfectly in the finite case. That is the entire idea behind counting. Given any two finite sets, such as { q, x, z, r} and { #, %, * @ }, there exists a one-to-one correspondence between them if and only if they have the same number of elements. This is the idea that let humans count sheep using rocks long before they had names for the numbers. > I love this quaint, homely picture of the origin of arithmetic. I >> am sure that evolutionary arithmetic will soon be taught in universities, >> if it is not already. Disregarding the anthropology, however, you have said >> absolutely nothing about whether !:!C is adequate for the infinite case. I was addressing your claim that there was no extension from the >finite case. In the finite case, two sets have the same number >of elements if and only if there exists a one to one correspondence >between them. This very simple idea has been extended to the >infinite case. OK. The idea of a 1:1 corresondence is indeed a simple idea. The idea of infinity is not. >>That depends on what 'idea of infinity' of you are talking about. >>The mathematical definition of 'infinite' is as simple as the >>idea of a 1:1 correspondence. >> The mathematical definition of infinity may be simple, but is it > unproblematic? It seems to me that infinity is a sublte and difficult > concept. >What concept of infinity? Note, I said 'infinite', not 'infinity'. You have been talking about Cantor and one-to-one correspondences, so you have been talking about set theory. The word 'infinity' is generally not used in set theory. It has no formal definition. 'infinite' is used to describe sets, and it has a very simple definition. > I'm talking about mathematical meaning. Specifically I'm talking >> about How many?, more or less etc.. How many is not a technical term. Cardinality corresponds to our >notion of how many in the finite case, and that is likely what people >will think of when you ask how many. I know that later on you complain >about the term cardinality, but I will respond to that later. > And that we are entitled to ask how well the simple mathematical > defintion captures what we mean by it, not necessarily in all its wilder > philosphical nuances, but what we mean by it mathematically, or if you > like, proto- mathematically. >A set is infinite if there exists a bijection between the set and a proper subset of itself. That is what mathematicians mean when they say a set is infinite. There are other equivalent definitions. > I know already. So what are you asking? Is it a good definition? > That is the definition of 'infinite set'. >It means mathematically exactly what it says. > There is no point in dragging >>philosophical baggage into a mathematical discussion. >> In my opinion the philsosopy is already there, and it impoverishes > mathematics to pretend otherwise. >Do you have the same problem with prime numbers? Or even numbers? The words 'prime' and 'even' have meanings outside of mathematics. Do you feel obligated to drag those meanings into a discussion of prime or even numbers? > See above I do not see an answer to the question above. couldn't help it. I responded to your imputation that I was smuggling in extraneous philosophical material well enough, I thought, that this rather facetious question of yours did not require an additional answer. >> I accept that. The contradiction comes about if the one notion > suggests equality of size and the other notion suggests inequality. Which > they do, so there is a prima facie paradox. >The problem is that you are using a word 'size' that you have not defined. > True. I took it that people knew what I meant. And I think they do. No. I do not know what it means when applied to a set. Does >it mean cardinality? If so then we would not be having this discussion. >If it does not mean cardinality, what does it mean? Can you give >me a mathematical definition of size? Certainly I write things in the heat of the moment which I later >> regret. But this wasn't meant as a cheap jibe. I've already conceded that >> following Cantor might in some deep way be right, if it comes down to >> following productive branches and forsaking dead ends. > Look at what you've written. It consists of repeating things I >> already know (definitions etc.) coupled with the suggestion that I'm mixing >> up different notions of size. Saying that people are confusing two >> different notions of X is a classic manoeuvre of 20th century philosophy in >> the moribund analytic movement, and in every case, I'd venture to say, it >> sells the argument short. As if anybody that disagreed with your point of >> view was a complete idiot. You seem to be taking this all far too personally. You miss my point, I think. I was not suggesting that you were calling me an idiot. I was trying to typify your style of argument. >You have not provided >a definition of 'size'. You are using a vaguely defined word, which >is always going to get you into trouble in mathematics. My God, it's a wonder mathematics ever got started! > There is an intuition that there are less squares (even numbers, >> primes, whatever) than naturals. We are talking here precisely of >> intuitions about infinite sets. It is not good enough to say: You're >> getting mixed up with finite sets, or: You can't rely on common sense >> intuitions in maths. > So if there are less squares than naturals, then since they have >> the same cardinality, how can cardinality have anything to do with size >> (how many)? Why not just say there's a bijection and forget about >> cardinality. Why not just say 'having no factors other than itself and one' instead of >'prime'? Whe not just say 'divisible by 2' instead of even? Cardinality >has a very precise definition. Yes, we could replace the word 'cardinality' >with its definition. It would not change anything. Again, your problem is insisting that cardinality match some vague notion of 'how many' >that you have not defined. Until you come up with a precise definition of 'how many', >any questions about 'how many' elements are in a set simply cannot be answered. > You suggested I conduct my argument without using the term >> 'infinity'. I am quite happy to do that. I suggest you conduct the rest of >> your argument without using the term 'cardinality'. Why? Cardinality has a definition in set theory. 'infinity' does not have >a definition. Do you really think that the two words are on an equal footing? Stephen If I'm questioning the fitness of a definition, it hardly makes sense to keep bashing me over the head with it. There seems both to be as many squares as naturals (because of correspondence) and less squares than naturals (because of containment). I don't see how anything could be clearer than that. I was tempted to prefix this with 'in exactly the same sense of how many '. But there aren't multiple meanings of 'how many', not at least until mathematicians get to work on it. It's not the layman that has the problem here, it's the mathematician. It is quite in order for me to question the mathematical response to this paradox. It is quite in order for you to defend it. Please begin. I suspect though that there may be no proof of the matter either way, as has been hinted at in other parts of this thread. It may come down to this: that someone who wants to take a different, but still very reasonable (maybe more reasonable) appoach to this paradox, would need to demonstrate that some interesting and viable mathematics can result from it. This would certainly involve having precise defintions and so forth; it is just that they would be more or less different ones. I certainly do not have the wherewithall to even begin such a task. But it's interesting to speculate. And it's good to keep an open mind. === Subject: Re: Galileo's Paradox >> OK. The idea of a 1:1 corresondence is indeed a simple idea. The idea of > infinity is not. >That depends on what 'idea of infinity' of you are talking about. The mathematical definition of 'infinite' is as simple as the idea of a 1:1 correspondence. > The mathematical definition of infinity may be simple, but is it >> unproblematic? It seems to me that infinity is a sublte and difficult >> concept. What concept of infinity? Note, I said 'infinite', not 'infinity'. >You have been talking about Cantor and one-to-one correspondences, >so you have been talking about set theory. The word 'infinity' >is generally not used in set theory. It has no formal definition. >'infinite' is used to describe sets, and it has a very simple >definition. I'm talking about mathematical meaning. Specifically I'm talking about How many?, more or less etc.. >>How many is not a technical term. Cardinality corresponds to our >>notion of how many in the finite case, and that is likely what people >>will think of when you ask how many. I know that later on you complain >>about the term cardinality, but I will respond to that later. >> And that we are entitled to ask how well the simple mathematical >> defintion captures what we mean by it, not necessarily in all its wilder >> philosphical nuances, but what we mean by it mathematically, or if you >> like, proto- mathematically. A set is infinite if there exists a bijection between the set and >a proper subset of itself. That is what mathematicians mean when >they say a set is infinite. There are other equivalent definitions. I know already. >>So what are you asking? > Is it a good definition? What do you mean by a 'good definition'? What makes something a good definition, as opposed to a bad definition. Is 'a number is prime if it is only divisible by itself and 1' a good definition. This is a serious question. You seem to find the definition of 'infinite' somehow questionable. Is this unique to 'infinite', or does it apply to mathematical definitions in general, such as 'even', 'prime', 'odd', etc.? The definition is good in the sense that there exist objects that satisfy it. For example the set of natural numbers satisfies the definition. It is a good definition because you can determine when something meets the definition. It is also a good definition that if you were to try to list the elements of an 'infinite' set one at a time, you would never reach the end of them, and this corresponds nicely with one of the common definitions of the word 'infinite'. So why do you think it might not be a good definition? >> That is the definition of 'infinite set'. >>It means mathematically exactly what it says. There is no point in dragging philosophical baggage into a mathematical discussion. > In my opinion the philsosopy is already there, and it impoverishes >> mathematics to pretend otherwise. Do you have the same problem with prime numbers? Or even numbers? >The words 'prime' and 'even' have meanings outside of mathematics. >Do you feel obligated to drag those meanings into a discussion >of prime or even numbers? See above >>I do not see an answer to the question above. > couldn't help it. which to me implies that the answer to the question was in the text above. > I responded to your imputation that I was smuggling in extraneous > philosophical material well enough, I thought, that this rather facetious > question of yours did not require an additional answer. It is not a facetious question. You seem to think there is something wrong with the definition of 'infinite'. I am trying to determine if your problem is soley with the word 'infinite', or with mathematical definitions in general. If your objection is soley to the word 'infinite', then I think you are making the mistake of worrying about philosophical implications of the word that are irrelevant to mathematics. > Look at what you've written. It consists of repeating things I already know (definitions etc.) coupled with the suggestion that I'm mixing up different notions of size. Saying that people are confusing two different notions of X is a classic manoeuvre of 20th century philosophy in the moribund analytic movement, and in every case, I'd venture to say, it sells the argument short. As if anybody that disagreed with your point of view was a complete idiot. >>You seem to be taking this all far too personally. > You miss my point, I think. I was not suggesting that you were > calling me an idiot. I was trying to typify your style of argument. I am just trying to get you to answer some questions. You have still not provided a definition of 'size' or 'how many'. >>You have not provided >>a definition of 'size'. You are using a vaguely defined word, which >>is always going to get you into trouble in mathematics. > My God, it's a wonder mathematics ever got started! There is an intuition that there are less squares (even numbers, primes, whatever) than naturals. We are talking here precisely of intuitions about infinite sets. It is not good enough to say: You're getting mixed up with finite sets, or: You can't rely on common sense intuitions in maths. So if there are less squares than naturals, then since they have the same cardinality, how can cardinality have anything to do with size (how many)? Why not just say there's a bijection and forget about cardinality. >>Why not just say 'having no factors other than itself and one' instead of >>'prime'? Whe not just say 'divisible by 2' instead of even? Cardinality >>has a very precise definition. Yes, we could replace the word 'cardinality' >>with its definition. It would not change anything. >>Again, your problem is insisting that cardinality match some vague notion of 'how many' >>that you have not defined. Until you come up with a precise definition of 'how many', >>any questions about 'how many' elements are in a set simply cannot be answered. You suggested I conduct my argument without using the term 'infinity'. I am quite happy to do that. I suggest you conduct the rest of your argument without using the term 'cardinality'. >>Why? Cardinality has a definition in set theory. 'infinity' does not have >>a definition. Do you really think that the two words are on an equal footing? >>Stephen > If I'm questioning the fitness of a definition, it hardly makes > sense to keep bashing me over the head with it. What definition are you questioning? 'infinite'? 'cardinality'? What about those definitions are you questioning? > There seems both to be as many squares as naturals (because of > correspondence) and less squares than naturals (because of containment). > I don't see how anything could be clearer than that. I was tempted > to prefix this with 'in exactly the same sense of how many '. But there > aren't multiple meanings of 'how many', not at least until mathematicians > get to work on it. Can you provide me with that single meaning of 'how many'? Until you actually define what you mean by the phrase 'how many', then it is impossible to answer any question about 'how many'. > It's not the layman that has the problem here, it's the > mathematician. It is quite in order for me to question the mathematical > response to this paradox. It is quite in order for you to defend it. Please > begin. You seem to be arguing against a position I have not taken. I cannot defend anything involving 'how many' until you define what 'how many' means, especially with regard to infinite sets. You have repeatedly refused to do this. My point is that the paradox is a result of thinking that 'size' and 'how many' have a common sense definition that applies to infinite sets. As far as I know, they do not. If you think otherwise, say what the definition is. > I suspect though that there may be no proof of the matter either > way, as has been hinted at in other parts of this thread. It may come down > to this: that someone who wants to take a different, but still very > reasonable (maybe more reasonable) appoach to this paradox, would need to > demonstrate that some interesting and viable mathematics can result from > it. This would certainly involve having precise defintions and so forth; it > is just that they would be more or less different ones. I certainly do not > have the wherewithall to even begin such a task. But it's interesting to > speculate. And it's good to keep an open mind. You are more than welcome to come up with a definition of 'how many' that you like. If other people like it, they may use it. It is not going to change the fact that the naturals and the squares have the same cardinality, nor the fact that the squares are a proper subset of the naturals. Stephen === Subject: Re: Galileo's Paradox [. . .] >What do you mean by a 'good definition'? What makes something a good >definition, as opposed to a bad definition. It helps if the definition is true. ~v~~ === Subject: Re: Galileo's Paradox OK. The idea of a 1:1 corresondence is indeed a simple idea. The idea of >> infinity is not. That depends on what 'idea of infinity' of you are talking about. >The mathematical definition of 'infinite' is as simple as the >idea of a 1:1 correspondence. The mathematical definition of infinity may be simple, but is it unproblematic? It seems to me that infinity is a sublte and difficult concept. >>What concept of infinity? Note, I said 'infinite', not 'infinity'. >>You have been talking about Cantor and one-to-one correspondences, >>so you have been talking about set theory. The word 'infinity' >>is generally not used in set theory. It has no formal definition. >>'infinite' is used to describe sets, and it has a very simple >>definition. >> I'm talking about mathematical meaning. Specifically I'm talking > about How many?, more or less etc.. >How many is not a technical term. Cardinality corresponds to our notion of how many in the finite case, and that is likely what people will think of when you ask how many. I know that later on you complain about the term cardinality, but I will respond to that later. >> And that we are entitled to ask how well the simple mathematical defintion captures what we mean by it, not necessarily in all its wilder philosphical nuances, but what we mean by it mathematically, or if you like, proto- mathematically. >>A set is infinite if there exists a bijection between the set and >>a proper subset of itself. That is what mathematicians mean when >>they say a set is infinite. There are other equivalent definitions. >> I know already. >So what are you asking? > Is it a good definition? What do you mean by a 'good definition'? What makes something a good >definition, as opposed to a bad definition. Is 'a number is prime if it is only divisible by itself and 1' a good >definition. This is a serious question. You seem to find the definition >of 'infinite' somehow questionable. Is this unique to 'infinite', or >does it apply to mathematical definitions in general, such as 'even', >'prime', 'odd', etc.? The definition is good in the sense that there exist objects that >satisfy it. For example the set of natural numbers satisfies the >definition. It is a good definition because you can determine when >something meets the definition. It is also a good definition that >if you were to try to list the elements of an 'infinite' set one at >a time, you would never reach the end of them, and this corresponds >nicely with one of the common definitions of the word 'infinite'. So why do you think it might not be a good definition? >> That is the definition of 'infinite set'. It means mathematically exactly what it says. >> There is no point in dragging >philosophical baggage into a mathematical discussion. In my opinion the philsosopy is already there, and it impoverishes mathematics to pretend otherwise. >>Do you have the same problem with prime numbers? Or even numbers? >>The words 'prime' and 'even' have meanings outside of mathematics. >>Do you feel obligated to drag those meanings into a discussion >>of prime or even numbers? >> See above >I do not see an answer to the question above. > couldn't help it. which to me implies that the answer to the question was in the text above. There was a silly, harmless joke there. >> I responded to your imputation that I was smuggling in extraneous >> philosophical material well enough, I thought, that this rather facetious >> question of yours did not require an additional answer. It is not a facetious question. You seem to think there is something >wrong with the definition of 'infinite'. I am trying to determine if >your problem is soley with the word 'infinite', or with mathematical >definitions in general. If your objection is soley to the word 'infinite', >then I think you are making the mistake of worrying about philosophical >implications of the word that are irrelevant to mathematics. Look at what you've written. It consists of repeating things I > already know (definitions etc.) coupled with the suggestion that I'm mixing > up different notions of size. Saying that people are confusing two > different notions of X is a classic manoeuvre of 20th century philosophy in > the moribund analytic movement, and in every case, I'd venture to say, it > sells the argument short. As if anybody that disagreed with your point of > view was a complete idiot. >You seem to be taking this all far too personally. > You miss my point, I think. I was not suggesting that you were >> calling me an idiot. I was trying to typify your style of argument. I am just trying to get you to answer some questions. You have >still not provided a definition of 'size' or 'how many'. >>You have not provided a definition of 'size'. You are using a vaguely defined word, which is always going to get you into trouble in mathematics. > My God, it's a wonder mathematics ever got started! > There is an intuition that there are less squares (even numbers, > primes, whatever) than naturals. We are talking here precisely of > intuitions about infinite sets. It is not good enough to say: You're > getting mixed up with finite sets, or: You can't rely on common sense > intuitions in maths. >> So if there are less squares than naturals, then since they have > the same cardinality, how can cardinality have anything to do with size > (how many)? Why not just say there's a bijection and forget about > cardinality. >Why not just say 'having no factors other than itself and one' instead of 'prime'? Whe not just say 'divisible by 2' instead of even? Cardinality has a very precise definition. Yes, we could replace the word 'cardinality' with its definition. It would not change anything. >Again, your problem is insisting that cardinality match some vague notion of 'how many' that you have not defined. Until you come up with a precise definition of 'how many', any questions about 'how many' elements are in a set simply cannot be answered. >> You suggested I conduct my argument without using the term > 'infinity'. I am quite happy to do that. I suggest you conduct the rest of > your argument without using the term 'cardinality'. >Why? Cardinality has a definition in set theory. 'infinity' does not have a definition. Do you really think that the two words are on an equal footing? >Stephen > If I'm questioning the fitness of a definition, it hardly makes >> sense to keep bashing me over the head with it. What definition are you questioning? 'infinite'? 'cardinality'? >What about those definitions are you questioning? > There seems both to be as many squares as naturals (because of >> correspondence) and less squares than naturals (because of containment). >> I don't see how anything could be clearer than that. I was tempted >> to prefix this with 'in exactly the same sense of how many '. But there >> aren't multiple meanings of 'how many', not at least until mathematicians >> get to work on it. Can you provide me with that single meaning of 'how many'? Until you >actually define what you mean by the phrase 'how many', then it is >impossible to answer any question about 'how many'. > It's not the layman that has the problem here, it's the >> mathematician. It is quite in order for me to question the mathematical >> response to this paradox. It is quite in order for you to defend it. Please >> begin. You seem to be arguing against a position I have not taken. >I cannot defend anything involving 'how many' until you define what 'how >many' means, especially with regard to infinite sets. You have repeatedly >refused to do this. My point is that the paradox is a result of >thinking that 'size' and 'how many' have a common sense definition that >applies to infinite sets. As far as I know, they do not. If you think >otherwise, say what the definition is. >> I suspect though that there may be no proof of the matter either >> way, as has been hinted at in other parts of this thread. It may come down >> to this: that someone who wants to take a different, but still very >> reasonable (maybe more reasonable) appoach to this paradox, would need to >> demonstrate that some interesting and viable mathematics can result from >> it. This would certainly involve having precise defintions and so forth; it >> is just that they would be more or less different ones. I certainly do not >> have the wherewithall to even begin such a task. But it's interesting to >> speculate. And it's good to keep an open mind. You are more than welcome to come up with a definition of 'how many' >that you like. If other people like it, they may use it. It is not >going to change the fact that the naturals and the squares have the >same cardinality, nor the fact that the squares are a proper subset >of the naturals. > That is exactly what it is going to change. Your insistence that I define my terms in advance is back to front. There is a primitive intuition that naturals exceed squares and that they are equinumerous. There is no simple confusion here. Though undoubtedly there is something to be learned. It seems to me that your understanding of the unclarity about how many is a phantom product, a reflection of the set theory you accept so uncritically. You want to pretend the ambiguity about how many pre-exists, whereas in fact it is a creation of the mathematics. Conventional set theory has no unique. proprietary rights over this paradox. and it seemed to me that, whatever else it does and no doubt does very well, it deals with this paradox rather poorly. === Subject: Re: Galileo's Paradox > I suspect though that there may be no proof of the matter either way, as has been hinted at in other parts of this thread. It may come down to this: that someone who wants to take a different, but still very reasonable (maybe more reasonable) appoach to this paradox, would need to demonstrate that some interesting and viable mathematics can result from it. This would certainly involve having precise defintions and so forth; it is just that they would be more or less different ones. I certainly do not have the wherewithall to even begin such a task. But it's interesting to speculate. And it's good to keep an open mind. >>You are more than welcome to come up with a definition of 'how many' >>that you like. If other people like it, they may use it. It is not >>going to change the fact that the naturals and the squares have the >>same cardinality, nor the fact that the squares are a proper subset >>of the naturals. > That is exactly what it is going to change. How can that possibly change? There exists a bijection between the naturals and the squares, therefore the two sets have the same cardinality. Every square is a natural, and some naturals are not squares, therefore the squares are a proper subset of the naturals. Those two simple facts cannot change no matter what definition of 'how many' you come up with. > Your insistence that I define my terms in advance is back to front. > There is a primitive intuition that naturals exceed squares and that they > are equinumerous. There is no simple confusion here. Though undoubtedly > there is something to be learned. If you are not going to define your terms, then what you say is meaningless. You say 'how many' but you refuse to say what you mean by that. How is anyone supposed to understand you? > It seems to me that your understanding of the unclarity about how > many is a phantom product, a reflection of the set theory you accept so > uncritically. You want to pretend the ambiguity about how many pre-exists, > whereas in fact it is a creation of the mathematics. Conventional set > theory has no unique. proprietary rights over this paradox. and it seemed > to me that, whatever else it does and no doubt does very well, it deals > with this paradox rather poorly. I am not claiming any unique proprietary rights over this paradox. I am simply claiming that it is only a paradox because you are relying on vague notions of 'how many' and 'size'. There is absolutely nothing paradoxical about the statement Every element of S is an element of N, but N contains elements not in S, and There exists a bijection between S and N. That is all we are saying when we say that S is a proper subset of N, and S has the same cardinality as N. That is all set theory says on the subject. The only problem is that you have some vague notion of 'size' or 'how many' that you are applying. But you apparently refuse to even look at your notion of 'size' or 'how many', and instead would rather complain about others being close minded. Stephen === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <381sm25eu8nq5aivngnp4hre823nn45r28@4ax.com>>A set is infinite if there exists a bijection between the set and a proper subset of itself. That is what mathematicians mean when they say a set is infinite. There are other equivalent definitions. > Is it a good definition? x is infinite <-> ~ x is finite is, in my opinion, a good definition (depending on what you mean by a 'good definition'). Es(s is a proper subset of x & x and s are 1-1) is equivalent to ~ x is finite with the axiom of choice added to Z set theory. > There seems both to be as many squares as naturals (because of > correspondence) and less squares than naturals (because of containment). Sure, if you use 'as many' in two different ways. > I don't see how anything could be clearer than that. I was tempted > to prefix this with 'in exactly the same sense of how many '. But there > aren't multiple meanings of 'how many', not at least until mathematicians > get to work on it. It's better that there are NOT multiple definitions of the same terminology. > It's not the layman that has the problem here, it's the > mathematician. You mean the problem of non-mathematicians not understanding certain mathematical defintions? Yes, I suppose this is a public relations problem. > It is quite in order for me to question the mathematical > response to this paradox. It is quite in order for you to defend it. Please > begin. It's paradox only in a broad sense of the word 'paradox'. It is not a contradiction in any usually used mathematical theory. MoeBlee === Subject: Re: Galileo's Paradox >A set is infinite if there exists a bijection between the set and >a proper subset of itself. That is what mathematicians mean when >they say a set is infinite. There are other equivalent definitions. > Is it a good definition? x is infinite <-> ~ x is finite is, in my opinion, a good definition (depending on what you mean by a >'good definition'). Yes, I like that bit. >Es(s is a proper subset of x & x and s are 1-1) is equivalent to ~ x >is finite with the axiom of choice added to Z set theory. But that's the bit that worries me. >> There seems both to be as many squares as naturals (because of >> correspondence) and less squares than naturals (because of containment). Sure, if you use 'as many' in two different ways. So, give me an explication of the relevant, multiple meanings which does not already assume ZF and all its paraphernalia. >> I don't see how anything could be clearer than that. I was tempted >> to prefix this with 'in exactly the same sense of how many '. But there >> aren't multiple meanings of 'how many', not at least until mathematicians >> get to work on it. It's better that there are NOT multiple definitions of the same >terminology. Oh, agreed. >> It's not the layman that has the problem here, it's the >> mathematician. You mean the problem of non-mathematicians not understanding certain >mathematical defintions? Yes, I suppose this is a public relations >problem. Public relations? Looks like you could usefully work on it anyway. >> It is quite in order for me to question the mathematical >> response to this paradox. It is quite in order for you to defend it. Please >> begin. It's paradox only in a broad sense of the word 'paradox'. Oh thank God. I can sleep at night again > It is not a >contradiction in any usually used mathematical theory. It is better that there are not contradictions in accepted theories. My, some of you mathematicians are not only piss-poor at conducting an argument, but arrogant dickheads to boot. Six Letters === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <381sm25eu8nq5aivngnp4hre823nn45r28@4ax.com> <5gium2588j35ua14b8g3kbmqmmvamtni8c@4ax.comEs(s is a proper subset of x & x and s are 1-1) is equivalent to ~ x >is finite with the axiom of choice added to Z set theory. But that's the bit that worries me. I don't know why that worries you. >> There seems both to be as many squares as naturals (because of >> correspondence) and less squares than naturals (because of containment). Sure, if you use 'as many' in two different ways. So, give me an explication of the relevant, multiple meanings which > does not already assume ZF and all its paraphernalia. I'd just as soon use the time to change the oil in my car. > It is not a >contradiction in any usually used mathematical theory. It is better that there are not contradictions in accepted > theories. I don't know of a theory that has demonstrated a contradiction yet remains accepted. My, some of you mathematicians are not only piss-poor at conducting > an argument, but arrogant dickheads to boot. I'm not a mathematician. As to being an arrogant dickhead, you've made your call. As to an argument, I don't know what argument of mine you find fault. I'm not an expert, judge, or jury. But you are welcome for my post. MoeBlee === Subject: Re: Galileo's Paradox > Cardinality of set S is denoted | S | . Cardinality of its power set is > denoted 2 | S | . Shouldn't that be 2 ^ | S | ? === Subject: Re: Galileo's Paradox > > > >>Cardinality of set S is denoted | S | . Cardinality of its power set is >>denoted 2 | S | . > > > Shouldn't that be 2 ^ | S | ? Bob Kolker === Subject: Re: Galileo's Paradox >>The cardinal number of a set is the equivalence class of sets >>with the same cardinality as the the given set. > In what theory is this? Standard set theory. > > Wrong. > > The usual definition is that the cardinality of a set (or the cardinal > number of a set) is the least ordinal equinumerous with the set. That > is not the class of sets with the same cardinality as the given set. In > Z set theories, the class of sets having the same cardinality as the > given set is not even itself a set, let alone a cardinal number. > > MoeBlee There are set theories in which equivalence classes need not be proper classes or cause other problems, and the equivalence class definition is commonly used in them. In ZF or NBG, the least ordinal definition works nicely. === Subject: Re: Galileo's Paradox <9ttjm2pv5lmmkqnptsgc8fqp6ooo9n2klq@4ax.com> <64irm29mh0c6ctui50e5mpg6ophft5pbn9@4ax.com> <4t646uF1127iuU1@mid.individual.net> <4t64pqF1127iuU6@mid.individual.net> >The cardinal number of a set is the equivalence class of sets >>with the same cardinality as the the given set. > In what theory is this? Standard set theory. Wrong. The usual definition is that the cardinality of a set (or the cardinal > number of a set) is the least ordinal equinumerous with the set. That > is not the class of sets with the same cardinality as the given set. In > Z set theories, the class of sets having the same cardinality as the > given set is not even itself a set, let alone a cardinal number. MoeBlee There are set theories in which equivalence classes need not be proper > classes or cause other problems, and the equivalence class definition is > commonly used in them. Other than type theories such as PM or a stratification theory such as NF, what is an example of a set theory in which the class of all sets equinumerous with a given nonempty set is a set? (I don't deny that such a thing is possible; I'm just curious as to an example.) > In ZF or NBG, the least ordinal definition works nicely. Right, and they don't use the class of all sets equinumerous with a given nonempty set to serve as the cardinality of the given set. MoeBlee === Subject: Re: Galileo's Paradox >The cardinal number of a set is the equivalence class of sets >>with the same cardinality as the the given set. > In what theory is this? Standard set theory. Wrong. The usual definition is that the cardinality of a set (or the cardinal > number of a set) is the least ordinal equinumerous with the set. That > is not the class of sets with the same cardinality as the given set. In > Z set theories, the class of sets having the same cardinality as the > given set is not even itself a set, let alone a cardinal number. MoeBlee There are set theories in which equivalence classes need not be proper > classes or cause other problems, and the equivalence class definition is > commonly used in them. > > Other than type theories such as PM or a stratification theory such as > NF, what is an example of a set theory in which the class of all sets > equinumerous with a given nonempty set is a set? (I don't deny that > such a thing is possible; I'm just curious as to an example.) I was thinking of NF or variations on that theme. > > In ZF or NBG, the least ordinal definition works nicely. > > Right, and they don't use the class of all sets equinumerous with a > given nonempty set to serve as the cardinality of the given set. > > MoeBlee === Subject: Re: Galileo's Paradox > > > Just out of curiosity, Bob, why is cardinality in set theory not a > measure? I mean if you ask how much gas and get the answer two > gallons you've certainly measured the gas. Or if you ask how much > space and get the answer two inches you've certainly measured the > space. It seems to me that you can obviously superimpose cardinality > on questions like how much without having to count or match things. > > Consider two measure sets, disjoint. The measure of the union is the sum > of the measures of each. > > Now consider two sets of the same cardinality, disjoint. The cardinality > of the union equals the cardinality of either. This only holds for sets whose cardinality is either 0 or not finite. > > In short cardinality does not add like measure. > > Bob Kolker === Subject: Re: Galileo's Paradox > > > This only holds for sets whose cardinality is either 0 or not finite. Precisely. However you can take the measure of the union of two disjoint infinite (in cardinality) sets each with finite measure the the measure of the union is the sum of the measures of the constituent set. In short, measure is additive in the algebraic sense and cardiality (for infinite sets) is not. That is one of the reasons that how much and how many are different questions. Bob Kolker === Subject: Re: Galileo's Paradox > > > This only holds for sets whose cardinality is either 0 or not finite. > > Precisely. However you can take the measure of the union of two disjoint > infinite (in cardinality) sets each with finite measure the the measure > of the union is the sum of the measures of the constituent set. In > short, measure is additive in the algebraic sense and cardiality (for > infinite sets) is not. > > That is one of the reasons that how much and how many are different > questions. > > Bob Kolker Right! === Subject: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? Hi. I'm going to attempt to open a study group for non-mathematicians that will mimick the way geometry was taught when that was the entire corpus of mathematical knowledge. But, instead of just going through the Elements page-by-page, I would prefer to create the course in a way where I first ask a question, let the group try to give possible solutions, and after all the ideas were exhausted, go to Euclid and show Euclid's solution, which then would lead to other questions. I thought the best way to start would be with what is probably the most important theorem of pre-Euclidean geometry, the Pythagorean Theorem. I read the vast commentary by Heath on I.47 who tried to explain how this proof is a new one not based on similarity of triangles, and thus not based on Eudoxus' theory of proportion, but I can't seem to find an explanation why the proof based on 'moving triangles around' (Proof #9 in http://www.cut-the-knot.org/pythagoras/index.shtml) isn't more simple and intuitive. I tried to figure out which assumptions are implicit in that proof, and could not find any. Does anyone notice any such assumptions I've missed? To make this harder, I am sure it was not a new idea to prove things in similar ways, since in Plato's Meno we see a very similar solution to a specific case of I.47. So I would think Plato would have figured out that this method is extendable to the general case, and would have created a proof of the same spirit of Meno's proof which even a slave-boy would understand (or recollect, as Plato explains). Avital. === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? > I thought the best way to start would be with what is probably the most > important theorem of pre-Euclidean geometry, the Pythagorean Theorem. I > read the vast commentary by Heath on I.47 who tried to explain how this > proof is a new one not based on similarity of triangles, and thus not > based on Eudoxus' theory of proportion, but I can't seem to find an > explanation why the proof based on 'moving triangles around' (Proof #9 > in http://www.cut-the-knot.org/pythagoras/index.shtml) isn't more > simple and intuitive. It requires area, which is done in Euclid only much later. > I tried to figure out which assumptions are > implicit in that proof, and could not find any. Does anyone notice any > such assumptions I've missed? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? <301120060741213691%edgar@math.ohio-state.edu.invalid> Euclid's proof of I.47 also assumes area (see http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html). It assumes facts like I.41 (http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI41.html) - the area of a triangle is half the area of a parallelogram surrounding it (equivelant to S=bh/2) and other related arguments about area. And, in essence, all proofs of the Pythagorean theorem must involve area, since the actual theorem says that the areas on the two squares combined are equal to the area of the third square. The thorem is not only not provable without area, but it is not even stateable. On Nov 30, 7:41 am, G. A. Edgar I thought the best way to start would be with what is probably the most > important theorem of pre-Euclidean geometry, the Pythagorean Theorem. I > read the vast commentary by Heath on I.47 who tried to explain how this > proof is a new one not based on similarity of triangles, and thus not > based on Eudoxus' theory of proportion, but I can't seem to find an > explanation why the proof based on 'moving triangles around' (Proof #9 > inhttp://www.cut-the-knot.org/pythagoras/index.shtml) isn't more > simple and intuitive.It requires area, which is done in Euclid only much later. I tried to figure out which assumptions are > implicit in that proof, and could not find any. Does anyone notice any > such assumptions I've missed?-- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? <301120060741213691%edgar@math.ohio-state.edu.invalid> I thought about it and realized that there are several implicit assumptions in the nice visual proof: a) One can construct a square on any length - this assumes the parallel postulate and a bunch of other constructions b) The quadrilateral constructed inside the square must be proven to be a square - this assumes certain propeties of right angles c) Possibly in order to prove rigorously the possibility of moving around triangles you must assume properties of parallel lines [I'm not sure about this one] But it appears that the rigorous version of the visual proof would indeed need to assume practially all that is assumed in Euclid's I.47. > Euclid's proof of I.47 also assumes area (seehttp://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html). It > assumes facts like I.41 > (http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI41.html) - > the area of a triangle is half the area of a parallelogram surrounding > it (equivelant to S=bh/2) and other related arguments about area. And, > in essence, all proofs of the Pythagorean theorem must involve area, > since the actual theorem says that the areas on the two squares > combined are equal to the area of the third square. The thorem is not > only not provable without area, but it is not even stateable. On Nov 30, 7:41 am, G. A. Edgar important theorem of pre-Euclidean geometry, the Pythagorean Theorem. I > read the vast commentary by Heath on I.47 who tried to explain how this > proof is a new one not based on similarity of triangles, and thus not > based on Eudoxus' theory of proportion, but I can't seem to find an > explanation why the proof based on 'moving triangles around' (Proof #9 > inhttp://www.cut-the-knot.org/pythagoras/index.shtml) isn't more > simple and intuitive.It requires area, which is done in Euclid only much later. I tried to figure out which assumptions are > implicit in that proof, and could not find any. Does anyone notice any > such assumptions I've missed?-- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? >Euclid's proof of I.47 also assumes area (see >http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html). It >assumes facts like I.41 >(http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI41.html) - >the area of a triangle is half the area of a parallelogram surrounding >it (equivelant to S=bh/2) and other related arguments about area. And, >in essence, all proofs of the Pythagorean theorem must involve area, >since the actual theorem says that the areas on the two squares >combined are equal to the area of the third square. The thorem is not >only not provable without area, but it is not even stateable. That's not strictly true. You need *multiplication* to state the theorem, and for some proofs (e.g. by similar triangles), that's sufficient to prove it. Mike Guy === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? <301120060741213691%edgar@math.ohio-state.edu.invalid> What *is* multiplication of two real numbers? If you think about it, it's either defined using something complex like Dedekind cuts (which are equivelant to Eudoxus' theory of proportion, given in Euclid V) or its defined as the area of a rectangle. The latter case has the problem of it being sometimes hard to prove certain products are equal. >Euclid's proof of I.47 also assumes area (see >http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html). It >assumes facts like I.41 >(http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI41.html) - >the area of a triangle is half the area of a parallelogram surrounding >it (equivelant to S=bh/2) and other related arguments about area. And, >in essence, all proofs of the Pythagorean theorem must involve area, >since the actual theorem says that the areas on the two squares >combined are equal to the area of the third square. The thorem is not >only not provable without area, but it is not even stateable.That's not strictly true. You need *multiplication* to state the > theorem, and for some proofs (e.g. by similar triangles), that's > sufficient to prove it. > > Mike Guy === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? >>Euclid's proof of I.47 also assumes area (see >>http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html). >It >>assumes facts like I.41 >>(http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI41.html) - >>the area of a triangle is half the area of a parallelogram >surrounding >>it (equivelant to S=bh/2) and other related arguments about >area. And, >>in essence, all proofs of the Pythagorean theorem must involve area, >>since the actual theorem says that the areas on the two squares >>combined are equal to the area of the third square. The thorem is not >>only not provable without area, but it is not even stateable. That's not strictly true. You need *multiplication* to state the >theorem, and for some proofs (e.g. by similar triangles), that's >sufficient to prove it. I guess the question is, whose version of the theorem? For the Greeks, the theorem is about areas. As far as I am aware, they did not have the concept of multiplication of real numbers except for the idea of the area of a rectangle with sides of certain lengths or the volume of a solid with base of a given area and height a given length. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? <301120060741213691%edgar@math.ohio-state.edu.invalid> As far as I am aware, they > did not have the concept of multiplication of real numbers > except for the idea of the area of a rectangle with sides of > certain lengths or the volume of a solid with base of a given > area and height a given length. I guess a lot of math students would have wanted to be back in the days of the Greeks ;>) But seriously, though, it is difficult to imagine a world of an advanced society like the Greeks had where multiplication was not really used. I knew that they did not have a concept of a variable, but how could they do simple things like carry on commerce and accounting without multiplication? Was it all addition to them? Matt B === Subject: Re: Is most of Euclid I necessary for Euclid I.47 (Pythagorean Theorem)? <301120060741213691%edgar@math.ohio-state.edu.invalid> They had multiplication of integers, but not of real numbers. As to fractions, it seems that earlier Greek mathematicians did not multiply fractions (in Euclid there is no trace of this), as to later on, where they do (e.g. Diophantus). > As far as I am aware, they did not have the concept of multiplication of real numbers > except for the idea of the area of a rectangle with sides of > certain lengths or the volume of a solid with base of a given > area and height a given length.I guess a lot of math students would have wanted to be back in the days > of the Greeks ;>) But seriously, though, it is difficult to imagine a world of an > advanced society like the Greeks had where multiplication was not > really used. I knew that they did not have a concept of a variable, but > how could they do simple things like carry on commerce and accounting > without multiplication? Was it all addition to them? > > Matt B === Subject: Re: Calculate Gibbs phenomenon (how to?) >The mathematical term calculate might be wrong, I am not sure, and >it might be that my mathematical skills are not good enough to >communicate my problem. First of all: My goal is to prove that there is a limit for Gibbs' >phenomenon - at 9% - by math. I have managed to do this by using >graphs, and derivate the fourier series in order to find the correct >maximum point. Still, I would like to do more than just proving the 9% >by presenting a graph and derivating. There's a proof in Folland Real Analysis. >Now: as my solution has evolved, so has my problem. I am currently >trying to find (what sosmath.com calls) the trigonometrical identities >for my fourier series. This is something I have always struggled with. An example of a fourier series that I am trying to prove Gibbs' 9% with >is: >-x^2-4x, -4 < x < 0 >x, 0 < x < 4 This for instance gives me e.g. an a0 = 14/3, >and I can calculate that the it has a max discontinuity close to 4,36 >(which is the 9%). Maybe I finally see what you've been talking about when you ask about trig identities. For certain simple functions one can use a trig identity to find a simple form for the n-th partial sum of the Fourier series. But you shouldn't expect that to happen for all functions - for most functions trig identities are no help in simplifying the n-th partial sum. The basis for proving the Gibb's phenomenon happens in general is that the n-th partial sum for f is given by the convolution of f with the Dirichlet kernel. If you don't know what that means you need to read a book on Fourier series (or the relevant chapter in Folland...) >I am not satisfied proving it with graphs only, and am therefore >searching for a more theoretical approach. Hopefull we may use the >example in order to communicate better. >David C. Ullrich skrev: >>Yes, I guess my writing was inadequate and the question did not appear >>as it should. >>My questions were all mixed up, as I tried to sum up my problem in one >>question. Firstly I have now understood that the percentage is not used >>together with the actual f(x), but the difference in height between >>Also, I have understood that the partial sum of the Fourier series >>isn't difficult at all: It is simply a partial sum where e.g. n=50 or >>(as the example at the website) 2n-1. >>My real problem now is about the trigonometric identities, which I have >>always found hard. Since the function I am trying to calculate the >>Gibbs' phenomenon on >> Exactly what are you trying to do? Calculate the Gibb's phenomenon >> doesn't make much sense - you calculate things like numbers. >> is more complicated than the example, and the >>partiell sum consist of Cos and Sin elements, I am having a real >>struggle converting them to their trigonometric identities. >>Any tips for my trigonometric idendities will be highly appreciated. >> Again, exactly what are you trying to do? There are lots of trig >> identities. >>David C. Ullrich skrev: >I am actually trying to solve a (somewhat) complicated school asignment, which I only have some days to finish. This is of course just before the final exams and everything is quite chaotic at the moment :) >Although I now understand the findings of this particular example, I am still needing some more general info on the subject. The question that I am still having trouble with is: * How may I derivate the partial sum of my fourier series? >The Riemann sum and Taylor polynomial I still hope to solve on my own, but I am strugling with the derivative of my partial sum. >If you would like, I could publish my findings so far. It is actually a quite simple function that I need to compute, but the derivation part has set me back some hours already... > It's not at all clear what question you really mean to be asking. Because the answer to the question you actually _asked_ is so simple that it seems you must really mean to be asking something else: > If > S(t) = sum_{n=-N}^N c_n exp(in t) > is the N-th partial sum of a Fourier series then the derivative is > S'(t) = sum_{n=N}^N in c_n exp(in t). >Hope for more advices. Badger skrev: I noted that they only calculated for the right side of origo, since >the function is odd. Does this give them .18 instead of .09? The web site you referenced in your first post, > , > uses a particular function f(x) as an example. The height of that f(x) at the discontinuity is 2. .09 * 2 = ... >>Still, my main concern is about the the partial fourier sum. How do >they calculate this partial sum, and especially concidering the 4/pi >factor (which I have seen many times at other web sites reading about >Fourier)? The factor 4/pi arises when finding the Fourier series for the square > wave being used in the example. >>Also, may you give me some advice on how to derivate the Fourier >series? I don't know if you are asking about the theory of Fourier series in > general, or how to find the Fourier series of a square wave. But in > either case, you can use Google to easily find a number of web sites > that provide the information. >David C. Ullrich skrev: >>Hi. >>I am trying to calculate Gibbs phenomenon and have after searching the >>internet found some help. Still I am somewhat unsure of how I may use >>my Fourier series to calculate the maximum overshoot around the >>discontinuity point(s). >>This website might be a n indication to what I am trying to achieve. >>http://www.sosmath.com/fourier/fourier3/gibbs.html >>The page calculates the overshoot to be 1,18 (18%), while Gibbs himself >>found this to be 9%. Why is there a difference? The difference is >>actually noted on the webpage, but not entirely explained. >> No, there is no discrepancy noted on the web page. Both number >> .18 and .09 do appear there, but there's no inconsistency here >> at all. You just have to read more carefully: >> Taking x_0 = 0, it's stated that in general the bump has size >> approximately >> .09(f(0+) - f(0-)). >> For the specific example at the start of the page what is >> f(0+) and what is f(0-)? >>Looking forward to your answers. >> ************************ >> David C. Ullrich >> ************************ David C. Ullrich >> >> >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Calculate Gibbs phenomenon (how to?) On Thu, 30 Nov 2006 06:43:42 -0600, David C. Ullrich >>An example of a fourier series that I am trying to prove Gibbs' 9% with >>is: >>-x^2-4x, -4 < x < 0 >>x, 0 < x < 4 >>This for instance gives me e.g. an a0 = 14/3, >>and I can calculate that the it has a max discontinuity close to 4,36 >>(which is the 9%). Maybe I finally see what you've been talking about when you >ask about trig identities. For certain simple functions one >can use a trig identity to find a simple form for the n-th >partial sum of the Fourier series. But you shouldn't expect >that to happen for all functions - for most functions >trig identities are no help in simplifying the n-th >partial sum. Hopefully the post by David Ullrich will serve to point you in the right direction. I do not know where you got the function you gave as an example, or what you hope to show with it, but here's what I got for the coefficients (which I believe are correct, but be warned that I might have gone wrong somewhere): > -x^2-4x, -4 < x < 0 > x, 0 < x < 4 a 0 = 14/3 a n = -4/(n^2 pi^2) [ 3 cos(n pi) + 5 ] b n = -4/(n^3 pi^3) [ (n^2 pi^2 - 8) cos(n pi) + 8 ] Where you go from here I don't know, but good luck! === Subject: Re: Lebesgue measure of a curve On 29 Nov 2006 11:45:37 -0800, Dave L. Renfro > Denote Lambda_2 the Lebesgue measure on R^2 >> Define G a nondecreasing function >> Define the curve C={(x,G(x)), x in R} >> Show that Lambda_2(C)=0 Well, it's immediate from the Lebesgue density theorem, How is that? >if you can use that. There's also a very short and elementary proof for >rectifiable curves in the following paper: Casper Goffman, Rectifiable curves are of zero content, >Mathematics Magazine 44 (1971), 179-180. Note: Any nondecreasing function on a compact interval > is rectifiable, What's your definition of rectifiable here? (I've only seen the word applied to _curves_, and in particular continuous curves). >and passing to a countable union > takes care of your case. Dave L. Renfro ************************ David C. Ullrich === Subject: Re: Lebesgue measure of a curve Well, it's [Lebesgue planar measure of the graph of >> a nondecreasing function is zero] immediate from the >> Lebesgue density theorem, [...] > How is that? Not only does the graph fail to have any points of density (points where the planar Lebesgue density is equal to 1), but at each point p on the graph and for each R > 0, the measure of the intersection of the ball B(p,R) with the graph is at most 1/2 the measure of the ball B(p,R). Indeed, the interiors of the 2'nd and 4'th quadrant sectors at each point p on the graph are disjoint from the graph. Think about a set satisfying a uniform exterior cone condition, which is something I think comes up in your field sometimes. >> Casper Goffman, Rectifiable curves are of zero content, >> Mathematics Magazine 44 (1971), 179-180. >> Note: Any nondecreasing function on a compact interval >> is rectifiable, [...] > What's your definition of rectifiable here? (I've only > seen the word applied to _curves_, and in particular > continuous curves). A function f:[a,b] --> R is rectifiable if it has finite length, where length is the usual least upper bound over polygonal paths whose (finitely many) vertices lie on the graph. So basically, I'm just talking about a function of bounded variation. I think there's a more general notion of rectifiable that applies to any subset of the plane (any set for which there exists a rigid motion that takes it to the graph of a rectifiable curve), but I think the only role it plays is to get countable unions of such sets, which apparently are important in geometric measure theory. Dave L. Renfro === Subject: Re: Lebesgue measure of a curve >> Denote Lambda_2 the Lebesgue measure on R^2 >> Define G a nondecreasing function >> Define the curve C={(x,G(x)), x in R} >> Show that Lambda_2(C)=0 >With Fubini, easy. >Without Fubini, cover by finitely many rectangles, and >see how small you can make the sum of their areas. I can not use Fubini's theorem. Nor can I use Fubini's density theorem. I can see that since the width of the curve is null, the Lebesgue measure of the curve will be zero. However, I am not able to cover the curve by finitely many rectangles, and minimize the covering so to obtain the value of zero. === Subject: Re: Lebesgue measure of a curve <31130758.1164851049018.JavaMail.jakarta@nitrogen.mathforum.org>, > I can see that since the width of the curve is null, the Lebesgue measure > of the curve will be zero. However, I am not able to cover the curve by > finitely many rectangles, and minimize the covering so to obtain the value of > zero. Consider the function restricted to [0,1]. Partition the y axis into intervals of length epsilon. The corresponding sets on the x-axis are intervals (because the function is monotone). The sum of the areas of all those rectangles is epsilon. === Subject: Re: Lebesgue measure of a curve <31130758.1164851049018.JavaMail.jakarta@nitrogen.mathforum.org>, > >> Denote Lambda_2 the Lebesgue measure on R^2 >> Define G a nondecreasing function >> Define the curve C={(x,G(x)), x in R} >> Show that Lambda_2(C)=0 > >With Fubini, easy. >Without Fubini, cover by finitely many rectangles, and >see how small you can make the sum of their areas. > > > I can not use Fubini's theorem. Right, if you could use Fubini, nondecreasing would have been replaced by measurable. > Nor can I use Fubini's density theorem. That would be Lebesgue's density theorem. I'm not sure why that was suggested. > I can see that since the width of the curve is null, the Lebesgue measure > of the curve will be zero. However, I am not able to cover the curve by > finitely many rectangles, and minimize the covering so to obtain the value of > zero. > Hint: It suffices to consider a nondecreasing f : [a,b] -> R (why?). Partition [f(a),f(b)] into n disjoint intervals of length < eps. The inverse images of the these intervals under f partition [a,b] into n disjoint intervals. Proceed ... === Subject: Re: Do reals really have to be genuine numbers? > >> >> Then there cannot be a whole line with points in it, unless there are >> other things in it which are not points and do not, at some iteration, >> contain points. >> >> EB is trying to create a geometry totally distinct from that of both >> Euclid and Descartes. > > Also totally disjoint from Hilbert's perfection of Euclid's geometry. I did recall Euclid like a witness twice. Hibert is to me the Staatsmacht (ruler of the state). > EB is very busy avoiding any mathematical sense. Not avoiding sense but avoiding nonsense. === Subject: Re: Do reals really have to be genuine numbers? > >> >> Then there cannot be a whole line with points in it, unless there are >> other things in it which are not points and do not, at some iteration, >> contain points. >> >> EB is trying to create a geometry totally distinct from that of both >> Euclid and Descartes. > > Also totally disjoint from Hilbert's perfection of Euclid's geometry. > > I did recall Euclid like a witness twice. > Hibert is to me the Staatsmacht (ruler of the state). > > > EB is very busy avoiding any mathematical sense. > > Not avoiding sense but avoiding nonsense. As far as the mathematics goes, EB has them backwards. His sense is mathematically nonsense and mathematical sense is to EB nonsense. === Subject: Re: Do reals really have to be genuine numbers? > > Not avoiding sense but avoiding nonsense. You entire position is nonsense, as I have demonstrated on several occassions. Bob Kolker === Subject: Re: Do reals really have to be genuine numbers? > what > difference does the very minor one-point assymetry at 0 make in > engineering? The asymmetry is not limited to zero and in case of genuine numbers quite understandable. I consider IR to be continuous. Continua must not behave asymmetrical. I started wondering when I got aware that mathematicians did not have a unique answer how to deal with the nil between IR+ and IR-. For some surrounding questions cf. http://iesk.et.uni-magdeburg.de/~blumsche/M283.html >> The ancient Greeks saw the need for real numbers when working on basic >> geometric problems and moved to compass and ruler constructions since >> it was better able to describe them then their number system. I'm sure >> you're welcome to discuss the usefulness of a circle but for most folk >> there is nothing objectionable about a perfectly round intellectual >> object to model real word round objects that we can only measure with >> imperfect precision. >> >> Being an engineer myself I agree. > > Stick to engineering. I will return as soon as possible after clarification. >> Personally, I consider all mathematical objects to >> be fictitious in this sense including those that stem from finite >> counting processes and I feel any opinion on the subject is merely a >> matter of taste as long as the system maintains logical consistency >> >> The larger system (not Cantor's little paradise) is inconsistent. > > EB keeps alleging inconsistency but provides no mathematically valid > arguments (1) > in support of it In mathematics, only mathematically valid arguments (2) > need be persuasive to mathematicians. mathematically invalid hand waving, (3) Three times the same non-factual declamation. The end of discussion seems to approach. === Subject: Re: Do reals really have to be genuine numbers? > what difference does the very minor one-point assymetry at 0 make > in engineering? > > The asymmetry is not limited to zero and in case of genuine numbers > quite understandable. I consider IR to be continuous. Continua must > not behave asymmetrical. Whyever not? The geometry of a line has the same sort of asymmetry, in that when any point bisects a line into disjoint pieces, the point will be in one and only one of those pieces. How is that necessary geometrical asymmetry any different from the real numbers numerical asymmetry? >> Personally, I consider all mathematical objects to be >> fictitious in this sense including those that stem from finite >> counting processes and I feel any opinion on the subject is >> merely a matter of taste as long as the system maintains logical >> consistency >> >> The larger system (not Cantor's little paradise) is inconsistent. > > EB keeps alleging inconsistency but provides no mathematically > valid arguments in support of it In mathematics, only > mathematically valid arguments need be persuasive to > mathematicians. Mathematically invalid hand waving, such as EB > indulges in, is irrelevant. > Three times the same non-factual declamation. The end of discussion > seems to approach. EB's provides lots of arguments, but none of them are *mathematically* valid. Someone will keep repeating this straightforward fact as long as EB keeps denying it. === Subject: Re: Do reals really have to be genuine numbers? >> >> What does all points of a line mean? > > Every bit of that line. Unless you are referring to binary digit (in this case you were wrong) you should have an extraordinary acute bit and endless time for biting. population of the line with points. >> If we agree, that a line cannot be resolved into single points > > Why should we agree to anything so silly? Reason 1: Else you have to quantify the sufficient number convincingly. Reason 2: More intelligent people like Spinoza came to the same result. Reason 3: A point is something that has no parts. A continuum is something every part of which has parts. Reason 4: Homeomorphism is insensitive against any zoom. Reason 5: your turn reason 6: .... > >> regardless how many, we do not deny the possibility to imagine the >> fiction that actually infinitely much of points do the job. >> Consequently, all points of a line implies a fictitious plurality of >> points which cannot be separated from each other. > > Then there cannot be a whole line with points in it, unless there are > other things in it which are not points and do not, at some iteration, > contain points. Do not utter wild and unfounded guesses. Try to understand 0*oo instead. > EB is trying to create a geometry totally distinct from that of both > Euclid and Descartes. Not at all. === Subject: Re: Do reals really have to be genuine numbers? > >> >> What does all points of a line mean? > > Every bit of that line. > > Unless you are referring to binary digit (in this case you were wrong) In English every bit does not necessarily refer to binary bits, but is logically equivalent to every jot and tittle. > you should have an extraordinary acute bit and endless time for biting. > population of the line with points. > > >> If we agree, that a line cannot be resolved into single points > > Why should we agree to anything so silly? > > Reason 1: Else you have to quantify the sufficient number convincingly. Euclid resolved lines into points, in the sense that one cannot pass across a line in a plane without passing through a point of that line, and geometers have been doing it without let or hindrance ever since. The same thing occurs in Cartesian geometry. > Reason 2: More intelligent people like Spinoza came to the same result. Spinoza was not doing mathematics, and we have no evidence that he was sufficiently mathematically competent to be mathematically authoritative, anyway. > Reason 3: A point is something that has no parts. Improper as a definition. See Hilbert on the subject of Euclids foundations for geometry. > A continuum is something every part of which has parts. The reals are topologically an open interval, each open (sub)interval of which has open (sub)intervals. So topologically, the reals fit that definition of a continuum precisely and completely. > Reason 4: Homeomorphism is insensitive against any zoom. So are the reals. For any non-zero real number (zoom factor) z, the mapping x |--> z*x is a homeomorphism of the reals. > Reason 5: your turn See above. All objections overcome, so that, at least in the mathematical world, the reals are as much a continuum as anything mathematical. > reason 6: .... EB seems to have run out of steam before making any relevant points. > > >> regardless how many, we do not deny the possibility to imagine the >> fiction that actually infinitely much of points do the job. >> Consequently, all points of a line implies a fictitious plurality of >> points which cannot be separated from each other. > > Then there cannot be a whole line with points in it, unless there are > other things in it which are not points and do not, at some iteration, > contain points. > > Do not utter wild and unfounded guesses. Try to understand 0*oo instead. If 0 represents EB's navel and oo his eyes, 0*oo could represent EB contemplating his navel. > > EB is trying to create a geometry totally distinct from that of both > Euclid and Descartes. > > Not at all. That EB does not realize what he is dong is part of his problem. === Subject: Re: Do reals really have to be genuine numbers? > > Reason 1: Else you have to quantify the sufficient number convincingly. > Reason 2: More intelligent people like Spinoza came to the same result. > Reason 3: A point is something that has no parts. > A continuum is something every part of which has parts. A fractal set has parts at every scale, but is not a continuum. Fractal sets have holes.. Don't quit your day job. Bob Kolker === Subject: Re: Do reals really have to be genuine numbers? > > >> But clearly a world in which the real number line is complete is a >> simpler world to work in for purposes of solving mathematical problems. >> >> However, while one may indeed imagine a line >> completely constituted by an actually infinite amount of points, this >> completion must not be extended to numbers because systems of numbers >> have to have empty spaces in between. > > That is a mental image that in no way reflects necessity. No. Restriction to discreteness has illogical consequences. For instance, a continuous transition from nothing to something (Aristoteles: rest --> beginning movement) is paradoxical with this restriction. The perhaps first approach was fluentism, then Fermat, Newton, and Leibniz invented calculus. >> The concept of counting numbers >> requires single, distinct from each other, elements. >> >> Why should everyone who solves a differential equation have to >> apologize for using procedures which may involve fictitious numbers? There is no reason for that because fictitious does not mean American science fiction. >> >> There is no reason for that. > > There is if one uses EB's model. No more than to apologize for hosting criminals when eating a Hamburger. >> As I pour milk into a glass, it is true that the number of atoms in the >> glass is an integer... >> >> Do not disdain the mathematical concept continuity just because it does >> not ideally fit some applications. > > Do not disdain numerical continuity just because it does not fit your > prejudices. Google for numerical continuity. You will not get many hits because there is at best the possibility to numerically approximate continuity. Jumps and continuity exclude each other. >> Our issue is fundamentals of mathematics, not physics. > > But EB applies physical imaging in determining his version of > mathematics, Never. I always refer to either the mathematical concept of continuity or the also mathematical concept of numbers. so is hardly in a position to criticize others for doing > the same. === Subject: Re: Do reals really have to be genuine numbers? >> >> Do not disdain the mathematical concept continuity just because it does >> not ideally fit some applications. > > Do not disdain numerical continuity just because it does not fit your > prejudices. > > Google for numerical continuity. You will not get many hits because > there is at best the possibility to numerically approximate continuity. > Jumps and continuity exclude each other. 5 million hits, and about a quarter million for real numbers and continuum. What does EB consider not many hits? > > > >> Our issue is fundamentals of mathematics, not physics. > > But EB applies physical imaging in determining his version of > mathematics, > > Never. I always refer to either the mathematical concept of continuity > or the also mathematical concept of numbers. What mathematical concept of continuity does EB refer to? I cannot find any such notion anywhere in mathematics that excludes the real number system. And I do not find that mathematics concepts of numbers anywhere conform to EB's non-mathematical concepts of them. === Subject: Re: Do reals really have to be genuine numbers? > A line is the shortest connection between two points. >> >> That definition, however, doesn't fully describe the points a line is >> made out of. >> >> Unless you are ready for border crossing from numbers to fictitious >> numbers, be cautious with the idea that a line is made out of points. > > Be even more cautious with the idea that a line is anything more than a > particular set of points. Consider, for example, Cartesian geometry. Not more but different. Analytic geometry is bound to approximations e.g. of pi. === Subject: Re: Do reals really have to be genuine numbers? > A line is the shortest connection between two points. >> >> That definition, however, doesn't fully describe the points a line is >> made out of. >> >> Unless you are ready for border crossing from numbers to fictitious >> numbers, be cautious with the idea that a line is made out of points. > > Be even more cautious with the idea that a line is anything more than a > particular set of points. Consider, for example, Cartesian geometry. > > Not more but different. > > Analytic geometry is bound to approximations e.g. of pi. All of Euclidean geometry is included in Cartesian geometry. If EB disagrees, let him present even one theorem of Euclidean geometry not also true in Cartesian geometry. === Subject: Re: Do reals really have to be genuine numbers? >> >> Aren't compact spaces pretty close to finite sets while the naturals are >> open ended? > > A subset of a topological space is compact if and only if given any open > covering of set (open with respect to the topology of the set) a finite > sub set of the covering exists which covers the set. I imagine this like the skin covering the body. > > Compactness is a topological property. It is definied topologically and > is preserved under homeomorphism. If I recall correctly, homeomorphism includes: Zoom does not matter as long as it is finite. While all tangible natural numbers and even all sets of natural numbers are finite, the naturals are not finite, and therefore the set of all naturals is an uncountable fiction if seen as an entity of all naturals. In other words I consider: all finite sets {1, 2, 3,...,n} with n-->oo countable and closed: ...,n] the set {1, 2, 3, ...} if seen as an entity infinite uncountable and clopen Eckard Blumschein === Subject: Re: Do reals really have to be genuine numbers? >> >> Aren't compact spaces pretty close to finite sets while the naturals are >> open ended? > > A subset of a topological space is compact if and only if given any open > covering of set (open with respect to the topology of the set) a finite > sub set of the covering exists which covers the set. > > I imagine this like the skin covering the body. That particular analogy leaves out too much of what topology is all about. > > > Compactness is a topological property. It is definied topologically and > is preserved under homeomorphism. > > If I recall correctly, homeomorphism includes: Zoom does not matter as > long as it is finite. Then you do not recall correctly. Homeomorphism essentially bijects topologies as well as the underlying spaces. For topologies on finite spaces scale must be preserved. > > While all tangible natural numbers and even all sets of natural numbers > are finite, the naturals are not finite, and therefore the set of all > naturals is an uncountable fiction if seen as an entity of all naturals. > > In other words I consider: EB's considerations are irrelevant to any serious mathematics. He merely plays games with words. === Subject: Re: Do reals really have to be genuine numbers? <455C1C1D.2020500@et.uni-magdeburg.de> <455C5076.8040005@et.uni-magdeburg.de> <4561845F.2070708@et.uni-magdeburg.de> <45645276.4020804@et.uni-magdeburg.de> <4565D2A2.5040606@et.uni-magdeburg.de> <4sm5mbF10bd5gU9@mid.individual.net> <4567495A.2030706@et.uni-magdeburg.de> <456ABA27.4050307@et.uni-magdeburg.de> <456C3393.1020008@et.uni-magdeburg.de> <4t2umlF11usi8U1@mid.individual.net > Dedekind's cut is based on an illusion. It is seemingly easy to decide a > number has to be either smaller, or equally large, or larger than let's > say five. The other way round, one cannot quantify a successor of five. > Correspondingly, Dedekind was unable to find any unknown real number. Dedikind's cut is based on the division of the rationals into two > mutually exclusively sets. All the elements of one set being less than > any of the elements of the other. What is illusory about that? > It's possible to view cuts geometrically. Consider the (integral) lattice points in R^2. Think of them as rational numbers. Now ask: Is it possible to draw a line thru (0,0) that misses all other points in this lattice? The answer is clearly yes, *if* one assumes the usual axioms. EB apparently does not accept all of the usual axioms. Fair enough, I guess. Others, accept the axioms as true and happily explore the consequences that follow from these assumptions: i.e. they do some mathematics. This might be considered a rather silly pursuit except that every now and then mathematics turns out to be damn useful: Robert Israel's recent post on where to most effectively plant a bush being a good example! EB's position is neither useful nor interesting. Rich === Subject: Re: Do reals really have to be genuine numbers? >> Dedekind's cut is based on an illusion. It is seemingly easy to decide a >> number has to be either smaller, or equally large, or larger than let's >> say five. The other way round, one cannot quantify a successor of five. >> Correspondingly, Dedekind was unable to find any unknown real number. >> Dedikind's cut is based on the division of the rationals into two >> mutually exclusively sets. All the elements of one set being less than >> any of the elements of the other. What is illusory about that? > > It's possible to view cuts geometrically. Consider the (integral) > lattice points in R^2. Think of them as rational numbers. Now ask: Is > it possible to draw a line thru (0,0) that misses all other points in > this lattice? The answer is clearly yes, *if* one assumes the usual > axioms. EB apparently does not accept all of the usual axioms. On the contrary. I accept the axioms by Archimede and by Euclid. I abstain from a summary judgement on the axioms presently en vogue. Maybe, they are clever. However Dedekind as well as Cantor did not have any sound basis for their idea that numbers and continuum are quantitatively comparable and all of the great thinker before were declared wrong. When Hilbert admitted: The axioms maintain the possibility to believe in certain dogmas (Cantor's naive set theory), this says a lot. > Fair enough, I guess. ??? > > Others, accept the axioms as true and happily explore the consequences > that follow from these assumptions: i.e. they do some mathematics. Happily? Maybe. But where is any progress due to set theory after 125 years of useless quarrel and confusion with CH, AC, etc.? > This might be considered a rather silly pursuit except that every now > and then mathematics turns out to be damn useful: Robert Israel's > recent post on where to most effectively plant a bush being a good > example! Such warm recommendation will cost me time. I recall Goodstein and Dehornoy. Is Israel presenting something new and more convincing? If so, please provide the link. > EB's position is neither useful nor interesting. Doesn't mind as long as it is not disproven. > > Rich Poor Eckard === Subject: Re: Do reals really have to be genuine numbers? > sqrt(2)*sqrt(5) =3D sqrt(10). Look - I just *directly* performed an > operation involving irrational numbers! Pretty cool how it works. That's not interesting compared to computing this: sqrt(3 + 2*sqrt(2)) Now that's *really* cool! === Subject: Re: Do reals really have to be genuine numbers? > sqrt(2)*sqrt(5) =3D sqrt(10). Look - I just *directly* performed an > operation involving irrational numbers! Pretty cool how it works. > > That's not interesting compared to computing this: > sqrt(3 + 2*sqrt(2)) > Now that's *really* cool! sqrt(3 + 2*sqrt(2)) = +/-(1 + sqrt(2)) since (1 + sqrt(2))^2 = 3 + 2*sqrt(2) === Subject: Re: Do reals really have to be genuine numbers? > It's also trivial to see that there is an uncountable > number of what I might call unspecifiable numbers. > A number is specifiable if one can describe it; > said description by necessity cannot be infinite. If one can describe a Turing machine that generates the digits of a number, then is that sufficient to satisfy one can describe it as you use the phrase above? Do you allow that somebody might describe a number without knowing for sure that he has in fact described a number, for example if one cites some random Turing machine and doesn't know whether it halts or not, but if in fact it *does* produce an infinite sequence of digits without halting the would you say that person has in fact described a number? Or does your requirement that one can describe it require the person not only to cite the Turing machine but *prove* that machine really does generate an infinite sequence of output digits? Depending on your answers, I'm going to claim that the set of specifiable numbers is ill-defined (hence unspecifiable numbers is also ill-defined) or the set of specifiable is *not* enumerable, so it's nothing new if the set of unspecifiable numbers is also not eumerable. The set of computable numbers is what I call sub-countable, a subset of an enumeration, but not an enumeration itself because there's no decision procedure for whether an element is a member of that subset or not. (The two definitions, decision procedure for membership and enumeration of sub-set, are almost equivalent. If you have a decision procedure, then you can run the full enumeration, testing each output for membership, keep the members, discard the rest, and viola you have an enumeration of the subset. If you have an enumeration of the subset, you have a green-light machine for testing elements for membership in the subset, just run the sub-enumeration until it hits the desired element, but you don't have a red-light machine for being sure some non-member really is a non-member, no matter how long you run the sub-enumeration you can't be sure the desired element won't turn up later. Of course if you have a sub-enumeration that generates in the same ordering as the full enumeration, then of course you can run the two in parallel, pausing the sub-enumeration whenever it has skipped past something in the full enumeration, until the full enumeration reaches the desired element, so you have a full decision procedure for any element in/notin subset.) > An old joke involves finding the most boring number; 42. > If I am correct, then I hope for some minor corrections including the > possibility to symmetrically cut IR into IR+ and IR-. > I have no idea what you mean by IR here. I've seen that notation in this newsgroup before, so let me guess what it means again: That's a sort of ASCII-ART image of a capital R with an extra bold leading stroke, standard mathematical notation for the system (set with arithmetic and metric) of real numbers. If IR+ and IR- denote the positive and negative real numbers, the cut is a destructive cut, removing the singleton number 0, leaving two disconnected subsets, one on each side of the cut. === Subject: Re: Do reals really have to be genuine numbers? > Are you familiar with the construction of the real numbers from > the rationals by either Dedekind cuts or Cauchy sequences? I'm not the person you asked, but I'm jumping in here to make a couple side comments of my own: In college I was told that Dedekind cuts were the most common way to construct the reals from the rationals, but we were going to be different and learn Cauchy sequences instead, so we did. It was only many years later that I finally saw the full definition of Dedekind cuts written out so I could stop guessing what they were. But nowadays I prefer nested intervals instead of either. > To what part of this construction do you object? One possible nitpick is that Dedekind cuts beg the question what subsets of the rationals are allowed. If you have a set, and a subset of it, then presumably there's a decision procedure as to whether some element of the set is or is not a member of that particular subset, right? But the question is begged what sorts of decision procedures are allowed for defining subsets of the rationals. And even if you have a decision procedure for a subset of the rationals, we need another decision procedure for whether that subset has an upper bound or not, as to whether its convex hull is or is not a Dedekind cut. That's a second begged question. Of course Cauchy sequences suffer similar question-begging. (And nested intervals likewise.) No fair saying the decision procedure is whether the element is in the subset or not, because that's a circular definition of any particular subset. === Subject: Re: Do reals really have to be genuine numbers? > I'm not the person you asked, but I'm jumping in here to make a > couple side comments of my own: Me too. we need another decision procedure for whether > that subset has an upper bound or not, as to whether its convex > hull is or is not a Dedekind cut. That's a second begged question. > > Of course Cauchy sequences suffer similar question-begging. > (And nested intervals likewise.) > > No fair saying the decision procedure is whether the element is in > the subset or not, because that's a circular definition of any > particular subset. You are possibly at the beginning of insight. Fictitious Cauchy sequences do the job. Nested intervals, Dedekind cuts and the like are equally unfit. However if you arrive at the same real numbers as assumed with DA2, then you will get aware: They are fictitious. This means, they do not have the same properties as ordinary numbers (rational ones). In particular, the are uncountable. Eckard Blumschein === Subject: Re: Do reals really have to be genuine numbers? > Fictitious Cauchy sequences do the job. That's too ambiguous for me to understand. What do you mean by fictitious? What specific job do they do?? > Nested intervals, Dedekind cuts and the like are equally unfit. Unfit for what purpose? > However if you arrive at the same real numbers as assumed with > DA2, then you will get aware: They are fictitious. What do you mean by DA2? I've see you mention that several times, but never define it. Do you mean Cantor's second diagonal argument, the one based on decimal representation of each real number in a sequence? Cantor's diagonal arguments don't show that real numbers are ficticious. They merely show the set of *all* real numbers can't be enumerated (put into a sequence that includes all of them without missing anything and without including anything else and without duplicating anything). Are you saying the individual real numbers are each ficticious, or that the entire set of them all is a ficticious set? > This means, they do not have the same properties as ordinary > numbers (rational ones). In particular, the [sic] are uncountable. Countable/uncountable is a property of sets, not an individual property of the elements within such a set. The individual real numbers have most of the same kinds of properties that rational numbers have. Any one of them can be incremented. Any two of them can be added or multiplied or subtracted, and if the second one isn't zero they can be divided. Any two can be compared to see whether they are the same, and if not the same then one is less than the other. The only property that all rationals have that not all reals have is that they are solutions to equations of the form a*x=b where a and b are integers. Among the real numbers, there are some that have that property and some that don't. And each real number satisfies the Archimedes property (either it's zero already, or there's some positive integer such that the number divided the integer is between -1 and 1, or equivalently the absolute value of the number is less than the absolute value of the integer). The *system* of real numbers has a nice property that the *system* of rational numbers doesn't have: Every bounded-above subset has a LUB. The *set* of rational numbers has a nice property that the *set* of real numbers doesn't have: They can be enumerated. But both of those are properties of entire sets or algebraic systems, not of individual elements. Other than the triviality that some reals aren't solutions to linear integer-coefficient equations, what other property do you claim each individual rational satisfies but not every real satisfies? === Subject: Re: Do reals really have to be genuine numbers? subset of it, then presumably there's a decision procedure as to > whether some element of the set is or is not a member of that > particular subset, right? As far as I know, in usual set theory. there is no such presumption of such a decision procedure. Maybe in some theory other than Z set theory and its most prominent variations. Moeblee === Subject: Re: Do reals really have to be genuine numbers? > there is perhaps no benefit from the utopic view that reals are > indeed numbers like the rationals. We can likewise operate with > the reals like symbols that merely denote the task to perform an > operation which cannot be exactly performed directly in numerical > terms. The rational number 2/3 can not be exactly calculated in binary notation within finite time. If that fraction really means the equivalance class consisting of all pairs [x,y] such that 3*x = 2*y, that set can't be completely expressed in finite time. If we accept only binary fractions, then 2/3 can be defined as LUB(set of all x such that 3*x < 2). The irrational number LUB(set of all x such that x*2 < 2) can be expressed as SQRT(2), but it can't be calculated either in binary notation nor the set completely expressed in finite time. So what's the difference between the two?? Why grant 2/3, but not SQRT(2), the right to be called a number? The non-negative integers are the only true numbers if you want to get picky about it. For any non-negative integer N, you can have N billiard balls on a table (possibly requiring larger-than-standard sized table if N is large), and you can express N exactly in finite time in binary or decimal or any other integer base you want, even in fractal complex bases such as -1+i. Now for any well-defined set of construction tools, you can define the set of constructable numbers. For example you might start with zero, use Peano's axioms to construct the non-negative integers, then use division to construct the rationals, then use solutions to polynomial equations to construct the algebraics, then use limits of series and areas under curves to construct various trigonometric and stastical numbers, and maybe using LUB on various bounded sets would extend that even further I'm not sure. I think if you accept output from all Turing machines (of a class that actually generates output in a meaningful way) you'll get them all, but again I'm not sure. Are there any other construction operators that would go even further, generating constructable numbers not already included above? Caveat: Given some random Turing machine, it's not possible to decide whether it does or does not generate meaningful output (an infinite sequence of digits in some base). So the *set* of constructable numbers is not itself constructable. You can enumerate all the Turing machines, and define a mapping from those which produce meaningful output to the number thereby represented by that output, but you can't decide which sub-sequence of Turing machines generate meaningful output, so you can't enumerate the numbers expressed by the output of that sub-sequence of Turing machines. Now I get to ask a question: Does anybody know of a Turing machine, or other algorithm, which generates an infinite sequence of digits (i.e. there's a proof the sequence keeps going on and on, never stops), but where nobody knows whether the real number expressed by that sequence of digits is algebraic or transcendental? Long ago somebody came up with 0.123456789101112131415161718192021... but then somebody else proved that was transcendental. Is there any number like that whose current alg/trans status isn't yet known? === Subject: Re: Do reals really have to be genuine numbers? >> there is perhaps no benefit from the utopic view that reals are >> indeed numbers like the rationals. We can likewise operate with >> the reals like symbols that merely denote the task to perform an >> operation which cannot be exactly performed directly in numerical >> terms. > > The rational number 2/3 can not be exactly calculated in binary > notation within finite time. If that fraction really means the > equivalance class consisting of all pairs [x,y] such that 3*x = > 2*y, that set can't be completely expressed in finite time. > If we accept only binary fractions, then 2/3 can be defined as > LUB(set of all x such that 3*x < 2). > > The irrational number LUB(set of all x such that x*2 < 2) can be > expressed as SQRT(2), but it can't be calculated either in binary > notation nor the set completely expressed in finite time. > > So what's the difference between the two?? Why grant 2/3, but not > SQRT(2), the right to be called a number? I was pondering over such questions for quite a while. My first idea was: Each selected system of rational numbers does not cover the same locations as do other ones. One could consider all imaginable bases altogether. While this would yield the best cover possible, it did not provide decisive advantages and it was not used with DA2. It was once again DA2 which paved the way when I asked how to characterize real numbers. Whether or not a number is taken for real does not depend on its possible termination into a period. Real numbers have to be considered a sauce altogether. So it makes no difference at all whether they are irrational of just embedded. They all have in common that they are fictitious, i.e. impossible to immediately represent numerically. > The non-negative integers You meant the positive ones. Zero is more reasonably the result of subtraction. > are the only true numbers if you want > to get picky about it. Negative numbers and imaginary numbers derive from IN by multiplication with -1 or sqrt(-1), respectively. > For any non-negative integer N, you can have > N billiard balls on a table (possibly requiring > larger-than-standard sized table if N is large), and you can > express N exactly in finite time in binary or decimal or any other > integer base you want, even in fractal complex bases such as -1+i. > > Now for any well-defined set of construction tools, you can define > the set of constructable numbers. For example you might start > with zero, use Peano's axioms to construct the non-negative > integers, then use division to construct the rationals, then use > solutions to polynomial equations to construct the algebraics, then > use limits of series and areas under curves to construct various > trigonometric and stastical numbers, and maybe using LUB on various > bounded sets would extend that even further I'm not sure. I think > if you accept output from all Turing machines (of a class that > actually generates output in a meaningful way) you'll get them all, > but again I'm not sure. Are there any other construction operators > that would go even further, generating constructable numbers not > already included above? > > Caveat: Given some random Turing machine, it's not possible to > decide whether it does or does not generate meaningful output (an > infinite sequence of digits in some base). So the *set* of > constructable numbers is not itself constructable. You can > enumerate all the Turing machines, and define a mapping from > those which produce meaningful output to the number thereby > represented by that output, but you can't decide which sub-sequence > of Turing machines generate meaningful output, so you can't > enumerate the numbers expressed by the output of that > sub-sequence of Turing machines. > > Now I get to ask a question: Does anybody know of a Turing machine, > or other algorithm, which generates an infinite sequence of digits > (i.e. there's a proof the sequence keeps going on and on, never > stops), but where nobody knows whether the real number expressed > by that sequence of digits is algebraic or transcendental? > Long ago somebody came up with 0.123456789101112131415161718192021... > but then somebody else proved that was transcendental. Is there any > number like that whose current alg/trans status isn't yet known? upthe word Turing. . So my comment is a shot in the dark: Turing machines belong to genuine (i.e. rational) numbers. Eckard Blumschein === Subject: Re: Do reals really have to be genuine numbers? > > > Then there cannot be a whole line with points in it, unless there are > other things in it which are not points and do not, at some iteration, > contain points. > > EB is trying to create a geometry totally distinct from that of both > Euclid and Descartes. > > Also totally disjoint from Hilbert's perfection of Euclid's geometry. > > EB is very busy avoiding any mathematical sense. > > Bob Kolker Right on! === Subject: Simple way to multiply big numbers (video) This counting-based technique lets ANYONE multiply big numbers together without a Calculator. It uses lines with Representative multitudes of the value of the digits to criss-cross, forming a plaid-like pattern, one then adds the number of intersections in certain crossing points to get the product of two numbers, no matter how big...F'ing awesome! Video: http://huderon.blogspot.com/2006/11/simple-way-to-multiply-big-numbers.html === Subject: Re: Simple way to multiply big numbers (video) What happens if there is a 0? > This counting-based technique lets ANYONE multiply big numbers together > without a Calculator. It uses lines with Representative multitudes of > the value of the digits to criss-cross, forming a plaid-like pattern, > one then adds the number of intersections in certain crossing points to > get the product of two numbers, no matter how big...F'ing awesome! Video: > http://huderon.blogspot.com/2006/11/simple-way-to-multiply-big-numbers.html > === Subject: Re: Simple way to multiply big numbers (video) Xadai > This counting-based technique lets ANYONE multiply big numbers together > without a Calculator. It uses lines with Representative multitudes of > the value of the digits to criss-cross, forming a plaid-like pattern, > one then adds the number of intersections in certain crossing points to > get the product of two numbers, no matter how big...F'ing awesome! Video: > http://huderon.blogspot.com/2006/11/simple-way-to-multiply-big-numbers.html The video is quite neatly done. Cute. But the algorithm is just a re-do of the usual one. And it gets pretty awkward when you multiply numbers whose _digits_ are relatively big. Try 96*87. This is basically what's involved: sum (a_n 10^n) sum (b_n 10^n) = sum_n (sum_{x+y=n}a_x b_y) 10^n The condition x+y=n means in some sense that a _diagonal_ is participating. The algo in the video does show this diagonal clearly. === Subject: Calculating mixed high-order derivatives Let f be a function of N complex parameters and a complex result (sorry for my English). Everyone knows that a derivative df/dx can be approximately calculated like this: df/dx = (f(x+epsilon) - f(x-epsilon)) / (2*epsilon) So it's easy to calculate an N-th order mixed derivative (e.g. d^4f/dx dy dz dw for N=4), calculating f in 2^N points. The question is, could we use less calculations than 2^N? I encountered this question in my research, and answering it surprisingly appears to be not so simple. Using the brute force approach, it's easy to prove that no combination of 3 different calculations of function f can approximate its second order mixed derivative d^2/dx dy, so for N=2 we need at least 4. However, for N=3 such a proof already requires solving a non-linear system of 20 or some such equations, and it get worse when N increases. I also tried searching for an answer using Google, mathworld.worfram.com, citeseer and similar resources, but failed. So, I'm asking for help. If you know a proof (or a counterexample), or === Subject: Re: virgil crankheit A line is the shortest connection between two points. >> That definition, however, doesn't fully describe the points a line is >> made out of. >> Unless you are ready for border crossing from numbers to fictitious >> numbers, be cautious with the idea that a line is made out of points. >> Be even more cautious with the idea that a line is anything more than a >> particular set of points. Consider, for example, Cartesian geometry. > > be even even more cautious > about being wrong > when you so delight in the error of others > > take a point > > what's its derivative? Most likely, engineering does not really need the distribution theory by Laurent Schwartz. In my set of singularity functions in IR+ cf. http://iesk.et.uni-magdeburg.de/~blumsche/M277.html integration of a impulse (alias point) yields a step. Integrate once again and get a falling (not inconvergent) ramp. > > okay > now take a couple of points more > take 20 > take a million > just keep collecting them > for infinity many times > in whatever way you count infinity Sorry, infinity cannot be reached by counting. It is a different quality, not a quantity. Sometimes mathematicians shold not forget that there are conceptual basics of their notions. > have you found a derivative? > do you have a cospace? Do not imagine me as ill as Hawking. > screw cartesian geometry > take algebraic geometry or differential geometry > since you belittle the aulden arguments > confront the modern > > are varieties collections of points? > are manifolds? > > a line has much more to it than points > it has continuity That's I am about. === Subject: Problem of the Week 2 This week the problems of the week at http://www.mymathforum.com are very classical. I am still looking for an elementary solution to the problem of the previous week, which was about the coverings of a unit square (it appeared in an Iberoamerican olympiad). Enjoy, JS. === Subject: Re: Baby stat question > Strictly speaking, the difference between of the two sample > means is either 11 or more, or it isn't. So the correct answer > is, 0 or 1. This reminds me of the Car vs. Goats problem. For the player of the game, the odds favor switching, but for a guy who walks into the room after Monty opens a door, the odds are even. Two guys standing in the same room facing the same problem have different probabilities. This is because the player has some knowledge that the newcomer doesn't. > This somewhat illustrates problems in interpreting p-values > and frequentist statistics. A better way to phrase the question > might be the following. Given the two sample standard deviations, > under the null hypothesis that the samples are taken from the same > population, under repeated sampling, in what proportion of > the repetitions would you expect to observe an absolute difference > between the two means of at least 11 pounds? standpoint, they have some knowledge (sample sizes and SD's) but not other knowledge (sample means) so we can ask them the probability from that viewpoint, eh? Your proposed wording has quite a few multisyllable words in it; that's not necessarily better for a freshman stats test. Bart -- The man without a .sig === Subject: Re: Baby stat question > >This is incorrect. The degrees of freedom is 18 (= 10 + 10 - 2). This >>gives an answer of about 7.55%, from the Excel function TDIST. >> Not so much in our textbook. The only way that n1+n2-2 comes up >is in the context of what most(?) books call Option 1. I'm >avoiding software as much as possible, so we're using only >Option 2, which _estimates_ (underestimates) the df by always >using min(n1-1,n2-2). (If the samples happen to be the same size, >then the complicated Option 1 formula boils down to n+n-2.) I think. > I am unfamiliar with that estimate, nor have I seen thesereferrred to as option 1 and option 2 before. However, the degress of freedom I gave can be derived rigorously. Thus, option 2 does not provide a very good estimate here. The fact that the normal approximation is closer to the to option 1 is telling. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Baby stat question > I am unfamiliar with that estimate, nor have I seen thesereferrred to as > option 1 and option 2 before. However, the degress of freedom I > gave can be derived rigorously. Thus, option 2 does not provide a > very good estimate here. The fact that the normal approximation is > closer to the to option 1 is telling. The text's tact is that option 2 is an underestimate that is easier to compute. It's not really meant to be accurate so much as quick. I'd have to read the text more closely (something I'm trying hard not to do) to find out whether it says much more about df in such cases. Bottom line: I believe you, but I think, as far as our text goes, the underestimate is acceptable, since it errs on the side of giving too wide a C.I., etc. Bart -- The man without a .sig === Subject: Re: How many chess games are there?...and who wins? > How many chess games are there?...and who wins? > http://members.iinet.net.au/~ray/Chessgames.htm ray@iinet.com.au > www.iinet.com.au/~ray See links from relevant sequences here: http://www.research.att.com/~njas/sequences/Sindx_Ch.html#chess Rob Pratt === Subject: Is an invariant measure a statistical distribution Is it correct to say that the invariant measure of the logistic map (x(t+1)=4*x(t)*(1-x(t))) is the statistical distribution of the typical trajectories of the logistic map? And in general, the invariant measure of a discrete ergodic dynamical system corresponds to the statistical distribution of its typical trajectories? Paul === Subject: Re: Is an invariant measure a statistical distribution > > Is it correct to say that the invariant measure of the logistic map > (x(t+1)=4*x(t)*(1-x(t))) is the statistical distribution of the typical > trajectories of the logistic map? > > And in general, the invariant measure of a discrete ergodic dynamical > system corresponds to the statistical distribution of its typical > trajectories? > > > Paul > compare to Elton's ergodic theorem: Elton, John H. An ergodic theorem for iterated maps. Ergodic Theory Dynam. Systems 7 (1987), no. 4, 481--488 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: vector fields and coordinate transforms <1164829327_2021@sicinfo3.epfl.ch [Wikipedia on vector fields] > [...] >> (*) V_y := @x/@y V >> is the expression for the same vector field in the new >> coordinates. > [...] >> Reference: >> [1] (accessed 29.11.06, >> 10:06 CET). > [...] > [lapiz ] >We have (@_x1,@_x2, ... ,@_xn) (@y/@x) = (@_y1,@_y2, ... ,@_yn) . Sorry, but I don't think I get it. Why is this not (@_x1,...,@_xn) (@x/@y) = (@_y1,...,@_yn) or, equivalently, (@./@x) (@x/@y) = (@./@y) ? This latter equation results in the same conflict. [lapiz ] >So, V_y := @x/@y V means a representation of V_y by (@_x1,@_x2, ... >,@_xn). >I think your conflict is Vy_i = V yi. >Vy_i (yi component of V) is not V yi (V orientation differential of >yi). Why not? What is wrong with the following derivation? V yi = (sum_j Vy_j @_yj) yi = Vy_i. Sorry, I had many mistakes. Once again. Just as you say, (@_x1,...,@_xn) (@x/@y) = (@_y1,...,@_yn). Then V = (sum_j Vy_j @_yj) = (@_y1,...,@_yn) t[Vy_1,Vy_2,...,Vy_n] = (@_x1,...,@_xn) (@x/@y) t[Vy_1,Vy_2,...,Vy_n], where t[ ] is a column vector. This shows that @x/@y converts a components w.r.t. a basis (@_y1,...,@_yn) to a components w.r.t a basis (@_x1,...,@_xn). I think Similarly, @y/@x converts a components w.r.t. a basis (@_x1,...,@_xn) to a components w.r.t a basis (@_y1,...,@_yn). This is the same as your case : > So, given another coordinate system y, I would write the i-th > component of V in the new coordinate system as Vy_i = V yi = sum_j Vx_j @yi/@xj I agree there is a mistake in the Wikipedia. === Subject: Re: vector fields and coordinate transforms Originator: pouya@localhost [lapiz ] [...] >I agree there is a mistake in the Wikipedia. -- Pouya D. Tafti p dot d dot tafti at ieee dot org === Subject: Re: Volume of finite body I got it I think The Qurqirish Dragon skrev: Calculate the volume of the finite body, in the first quadrant, that is > limited by the surfaces > y^2 -x^2 -z^2 =2 > and > y = 4 > Try making a sketch of the solid. If you are not good at drawings, then > simply draw the level curves for various fixed y-values. If you recall your geometric forms, this is a hyperboloid in 2 sheets > (although you only care about the one with vertex (0,sqrt(2),0), since > you are looking in the (+x,+y,+z) octant - I assume that is what you > meant by first quadrant). You may have some rules about volumes for > basic shapes (such as this)- you can use those to check your answer, > but I would advise trying to do the problem without the canned result. Once you have the basic shape, you should see how a double-integral can > get the volume. > Depending on your class, a conversion into cylindrical coordinates (of > the form (y,r,theta))may also help, as in rectangular coordinates the > integrand is not particularly nice. Another possibility (and if you haven't taken multivariable calculus > yet, the above won't help) is to rewrite this as a volume of revolution > (revolved about the y-axis), and note that this volume is 1/4 the full > volume of revolution (since you are in only one ocatant). This will get > you the same integral as the polar form above, if you use cylindrical > shells. Or a different integral (but the same answer, of course) if you > use discs. === Subject: Re: isomorphic subgroups of a finite group <12mqpmsqvqppr9c@corp.supernews.com On 29-Nov-2006, Derek Holt H is maximal, is it true that H' is also maximal? No. Looking through the ATLAS of Finite Groups, one finds that the > Mathieu group M12 has a maximal subgroup isomorphic to PSL(2,11), and > also a non-maximal subgroup which is contained in M11. Darn, you beat me to it. :) I don't suppose you know of a general way to approach the > problem, maybe even to characterize all such groups G? No, but here is another easily described class of examples, brought to my attention by Avinoam Mann. Let S be any nonabelian simple group, and let G = S x S. Then the diagonal subgroup { (g,g) | g in S } is maximal in G and isomorphic to S, whereas { (g,1) | g in G } is isomorphic to S but non maximal in G. > In particular, I couldn't think of any small solvable examples, but > neither could I find a quick way to prove there aren't any. I'm > wondering if Phillip Hall's Theorem about pi-subgroups might be > used as a starting point to show there's no solvable G (if indeed > that's true), but haven't had time to think about it much yet, > beyond the obvious result that all primes dividing |G| must also > divide |H|. I am going to make so bold as to conjecture that there are no solvable examples. I can sort of almost prove this, but it is getting messy, so I am not completely confident. I will try to find out whether this is known. Derek Holt. === Subject: Re: interpolation/approximation functions > All, There are many methods to perform interpolation/approximation. The > main ones that spring to mind are using Trigonometric functions > Rational Functions > Polynomial Functions > Splines I'm looking for a book/reference material that will give me an idea of > which particular functions fit a particular interpolation/approximation > problem. For example, I know that if i have a system/model/function > f(t) that is approximately linear, then i should probably avoid using > trigonometric functions as a basis for my approximation. On the other > hand, if f(t) is periodic, then i probably should consider > trigonometric functions as my first port of call. Do you know of a > source that discusses when and where to use the multitude of > functions/polynomials that are out there? Or can you direct me to a > more appropriate newsgroup? see: http://en.wikipedia.org/wiki/Levenberg-Marquardt_algorithm === Subject: Re: interpolation/approximation functions > All, >> There are many methods to perform interpolation/approximation. The >> main ones that spring to mind are using >> Trigonometric functions >> Rational Functions >> Polynomial Functions >> Splines >> I'm looking for a book/reference material that will give me an idea of >> which particular functions fit a particular interpolation/approximation >> problem. For example, I know that if i have a system/model/function >> f(t) that is approximately linear, then i should probably avoid using >> trigonometric functions as a basis for my approximation. On the other >> hand, if f(t) is periodic, then i probably should consider >> trigonometric functions as my first port of call. Do you know of a >> source that discusses when and where to use the multitude of >> functions/polynomials that are out there? Or can you direct me to a >> more appropriate newsgroup? see: >http://en.wikipedia.org/wiki/Levenberg-Marquardt_algorithm Excuse, please, but what relevance does that have to the question being asked? === Subject: any difference between a conditional Fubini theorem vs. a normal Fubini theorem? Hi all, For conditional Fubini theorem, say, A1=E[ Integrate[ f(s), integrating w.r.t. s from t to +infinity] | given F_t ] where F_t is the filtration; May I interchange the order of E(.) and Integrate(.), getting: A2=Integrate[ E[ f(s) | given F_t ] , integrating w.r.t. s from t to +infinity] ??? The final result is a random variable A2, if this A2 is a random variable that exists, i.e. finite almost surely, can I equate A1 = A2 to obtain A1? If it doesn't work, could you please tell me under what technical conditions will the conditional Fubini theorem work? I didn't find it in any book. === Subject: Re: computers of future having a science key to write science fluently > >> Do you mean that all the symbols you create will be listed in a table >> and accessed by clicking? Where is this table going to be displayed? >> Will the symbols be listed in some kind of order, so that they can be >> located once you have a few hundred? > > I mean that this figure 8 with diagonals or two figure 8's with > diagonals back to back can produce all the known science and math > symbols we now commonly use and more. > > Build that special digital key into every new computer. And it allows > the user to custom make any symbol he/she wants. Once a symbol is > constructed, hit the preserve key of a folder that saves all formed > symbols from this custom make your own symbol key. > > physics symbol or a physics equation that has alot of symbols to it. So > you draw up your symbol folder and you want a partial derivative symbol > for the next slot. You simply click on to your partial derivative > symbol and it fills that slot just as easy as you typing in a letter of > the alphabet. > > So in this manner, of a special key, we custom make a large list of > symbols we use or plan to use. And when we want that symbol in a > sentence or line of the page, we click on it and there it is. this already exists! I use MathType in word- brings up a page (or text-box, or in-line equation space). I select and click on whatever symbol I want. I have the common ones(for me) programmed as shortcut keys. I can use any combination I like of letters (from keyboard), subscript/superscripts(keyboard shortcut commands), any greek letters (dropdown menus), function indicators, fraction bars, etc. etc.... So how is this different? I have this functionality (yes it's software, but you can't write a document with formatting completely without software, either). It's only different in that you have a built-in new bunch of keys on your keyboard- you still need some kind of software to word-process. In fact, there is a (limited functionality) equation / symbol editor right *inside* most word-processing software packages. My office 2000 has MS equation editior, symbol selection, etc. (though it's less than convenient). This is just (IMO) not a hardware issue. It's a software issue. As someone said earlier- how easy has it been to get everyone (US) to use the metric system, when we ALL use weights and measures? So, considering that, how easy is it going to be to introduce and promote use of a system that, in reality, most people who use word-processing functions of computers will rarely, if ever, use? -k wallace > > The benefit and superiority of a science-key is that, unlike software > such as LaTex, the science key is integral to the alphabet keys. They > are one and the same software. LaTex is a different software from the > language you write, and websites like Mathematica or many of > Wikipedia's is a hodge podge of different softwares. So that when you > cut and paste, it comes out unreadable. > > With the science-key, since every computer of the future has the same > key built in, then language is one and the same as science symbols, all > integrated, and only one software needed. > > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies > === The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. He and his family later sued the U.S. government for damages. We lived in 1984, Mayfield told reporters Wednesday. I'm talking about the George Orwell, frightening brave new world in which Big Brother is constantly watching you. (Watch Mayfield discuss the case Video) I, myself, have dark memories of stifling paranoia, of being monitored, followed, watched, tracked, he said, choking back emotion. I've been surveilled, followed, targeted primarily because I've been an outspoken critic of this administration and doing my job to defend others who can't defend themselves, to give them their day in court, and mostly for being a Muslim. The government refused, he said, to tell him where they put their cameras and surveillance devices, leaving his family wondering if their private conversations and intimate moments were on display. The days and weeks and months following my arrest were some of the hardest and darkest that myself and my family have ever had to endure, he said. And all because of this government's ill-conceived war on terror. ... What I really want is for this not to happen to anyone else. Wednesday's settlement includes not only a $2 million payment and an apology, but also an agreement by the government to destroy communications intercepts conducted by the FBI against Mayfield's home and office during the investigation. The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. A Justice Department statement released Wednesday said Mayfield was not targeted because of his Muslim faith and that the FBI had taken steps to improve its fingerprint identification process to ensure that what happened to Mr. Mayfield does not happen again. Mr. Mayfield and his family felt it was in their best interest to get on with their lives, said Mayfield's attorney, Elden Rosenthal. No amount of money can compensate Mr. Mayfield for being held as a prisoner and being told he faced the death penalty [for the Madrid bombings]. Mayfield said his suit was not about money. It's about regaining our civil rights, our freedom and most important, our privacy, he said. He and his attorneys said the settlement will allow him to continue the portion of his lawsuit challenging the constitutionality of the Patriot Act. Mayfield contends that his home was searched under provisions of the Patriot Act. Criminal, Rogue and RACIST elements have penetrated all the branches of USA and world governments. We know that on 911 FBI bastards ran all over in the neighborhood of Pentagon trying to confiscate EVIDENCE of the 911 crime that no Boeing Passenger Airliner hit Pentagon. They have stil not released the evidence. 911 was inside job. FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, Mossad, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED THE ONLY ONLY REASON FBI DID THIS IS BECAUSE THEIR CRIME WAS UNCONCEALABLE. UNDER THE PRETEXT OF SELF-DEFENSE, MANY OF THESE AGENCY OFFICIALS HAVE COVERED THEIR CRIMES AND DESTROYED EVIDENCE OF THEIR CRIMES. FBI COVERED UP A CHILD PEDOPHILIA RING THAT WAS LINED TO REAGAN AND === Subject: *** ARE The FBI Bastards are making movies of YOUR PRIVATE MARITAL SEXUAL MOMENTS *** The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. He and his family later sued the U.S. government for damages. We lived in 1984, Mayfield told reporters Wednesday. I'm talking about the George Orwell, frightening brave new world in which Big Brother is constantly watching you. (Watch Mayfield discuss the case Video) I, myself, have dark memories of stifling paranoia, of being monitored, followed, watched, tracked, he said, choking back emotion. I've been surveilled, followed, targeted primarily because I've been an outspoken critic of this administration and doing my job to defend others who can't defend themselves, to give them their day in court, and mostly for being a Muslim. The government refused, he said, to tell him where they put their cameras and surveillance devices, leaving his family wondering if their private conversations and intimate moments were on display. The days and weeks and months following my arrest were some of the hardest and darkest that myself and my family have ever had to endure, he said. And all because of this government's ill-conceived war on terror. ... What I really want is for this not to happen to anyone else. Wednesday's settlement includes not only a $2 million payment and an apology, but also an agreement by the government to destroy communications intercepts conducted by the FBI against Mayfield's home and office during the investigation. The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. A Justice Department statement released Wednesday said Mayfield was not targeted because of his Muslim faith and that the FBI had taken steps to improve its fingerprint identification process to ensure that what happened to Mr. Mayfield does not happen again. Mr. Mayfield and his family felt it was in their best interest to get on with their lives, said Mayfield's attorney, Elden Rosenthal. No amount of money can compensate Mr. Mayfield for being held as a prisoner and being told he faced the death penalty [for the Madrid bombings]. Mayfield said his suit was not about money. It's about regaining our civil rights, our freedom and most important, our privacy, he said. He and his attorneys said the settlement will allow him to continue the portion of his lawsuit challenging the constitutionality of the Patriot Act. Mayfield contends that his home was searched under provisions of the Patriot Act. Criminal, Rogue and RACIST elements have penetrated all the branches of USA and world governments. We know that on 911 FBI bastards ran all over in the neighborhood of Pentagon trying to confiscate EVIDENCE of the 911 crime that no Boeing Passenger Airliner hit Pentagon. They have stil not released the evidence. 911 was inside job. FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, Mossad, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED THE ONLY ONLY REASON FBI DID THIS IS BECAUSE THEIR CRIME WAS UNCONCEALABLE. UNDER THE PRETEXT OF SELF-DEFENSE, MANY OF THESE AGENCY OFFICIALS HAVE COVERED THEIR CRIMES AND DESTROYED EVIDENCE OF THEIR CRIMES. FBI COVERED UP A CHILD PEDOPHILIA RING THAT WAS LINED TO REAGAN AND === Subject: *** DONT HESITATE TO SUE THE OFFICIAL BASTARDS IN THE GOVERNMENT, FEDERAL, STATE OR CITY *** The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. He and his family later sued the U.S. government for damages. We lived in 1984, Mayfield told reporters Wednesday. I'm talking about the George Orwell, frightening brave new world in which Big Brother is constantly watching you. (Watch Mayfield discuss the case Video) I, myself, have dark memories of stifling paranoia, of being monitored, followed, watched, tracked, he said, choking back emotion. I've been surveilled, followed, targeted primarily because I've been an outspoken critic of this administration and doing my job to defend others who can't defend themselves, to give them their day in court, and mostly for being a Muslim. The government refused, he said, to tell him where they put their cameras and surveillance devices, leaving his family wondering if their private conversations and intimate moments were on display. The days and weeks and months following my arrest were some of the hardest and darkest that myself and my family have ever had to endure, he said. And all because of this government's ill-conceived war on terror. ... What I really want is for this not to happen to anyone else. Wednesday's settlement includes not only a $2 million payment and an apology, but also an agreement by the government to destroy communications intercepts conducted by the FBI against Mayfield's home and office during the investigation. The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. A Justice Department statement released Wednesday said Mayfield was not targeted because of his Muslim faith and that the FBI had taken steps to improve its fingerprint identification process to ensure that what happened to Mr. Mayfield does not happen again. Mr. Mayfield and his family felt it was in their best interest to get on with their lives, said Mayfield's attorney, Elden Rosenthal. No amount of money can compensate Mr. Mayfield for being held as a prisoner and being told he faced the death penalty [for the Madrid bombings]. Mayfield said his suit was not about money. It's about regaining our civil rights, our freedom and most important, our privacy, he said. He and his attorneys said the settlement will allow him to continue the portion of his lawsuit challenging the constitutionality of the Patriot Act. Mayfield contends that his home was searched under provisions of the Patriot Act. Criminal, Rogue and RACIST elements have penetrated all the branches of USA and world governments. We know that on 911 FBI bastards ran all over in the neighborhood of Pentagon trying to confiscate EVIDENCE of the 911 crime that no Boeing Passenger Airliner hit Pentagon. They have stil not released the evidence. 911 was inside job. FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, Mossad, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED THE ONLY ONLY REASON FBI DID THIS IS BECAUSE THEIR CRIME WAS UNCONCEALABLE. UNDER THE PRETEXT OF SELF-DEFENSE, MANY OF THESE AGENCY OFFICIALS HAVE COVERED THEIR CRIMES AND DESTROYED EVIDENCE OF THEIR CRIMES. FBI COVERED UP A CHILD PEDOPHILIA RING THAT WAS LINED TO REAGAN AND === Subject: Binary Division Problem Help I'm trying to work out a binary division problem 1100 / 101010101010111 Here is what I have so far but I'm not sure if I'm doing it correctly and I'm suppose to continue the division until there is only a remainder left 110 ______________________ 1100 [ 101010101010111 1100 _______________ 1101 1100 _______________ 01010 1100 _____________ 110 I'm not sure if I'm doing this correctly and I don't really have a way of checking if I'm doing it right. Does anyone know of a simple web calculator that I can check against my answer or if someone could run through this example with me, that would be great. thank you === Subject: Re: Binary Division Problem Help I'm trying to work out a binary division problem 1100 / 101010101010111 Here is what I have so far but I'm not sure if I'm doing it > correctly > and I'm suppose to continue the division until there is only a > remainder left 110 > ______________________ > 1100 [ 101010101010111 > 1100 > _______________ > 1101 > 1100 > _______________ > 01010 > 1100 > _____________ > 110 I'm not sure if I'm doing this correctly and I don't really have a > way > of checking if I'm doing it right. Does anyone know of a simple web > calculator that I can check against my answer or if someone could > run > through this example with me, that would be great. thank you Try it again in a fixed width font, Courier New, for example. [Read and write all Usenet stuff in a fixed width font.] So the sum would begin 1100 into 1, I cannot 1100 into 10, I cannot 1100 into 101, I cannot 1100 into 1010, I cannot 1100 into 10101, is 1 etc | 1 | ---------------- | 1100 ) 101010101010111 | 1100 | ----- | 10010 Now 1100 into 10010, is 1 etc -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: Re: Is it possible for a probability to be unknown? > I present you with a bag, with the following properties: 1. You cannot see inside it. > 2. It contains at least one marble, and this marble is red. > 3. It may or may not contain more marbles. > 4. If it does contain more marbles, they may or may not be red. Is it valid to say that the probability of picking a red marble is unknown > here? Or must we commit to 50%? Maybe the right question to ask is: if you were forced to make a bet on whether a marble taken out of the bag at random would be red, what odds would you offer? Your decision is primarily subjective, since you have very little information, and if you are a Bayesian, you don't hesitate to take into account everything you know. E.g., someone has told you that the bag has at least one red marble. Do you trust that person? Did they look you in the eye when they talked to you, or did they look sneaky? [Bayesians should make good poker players ... I am not sure that in real life they actually do.] Or are they hoping you will bet on red, knowing that there are 12 marbles in the bag, and only one is red? Or are they just giving you a neutral fact? Or a hint? You cannot see inside the bag, but you can see how large it is. If it is so small that it could contain only one marble, you know that marble is red. A person betting against you who does not make use of this information, who just estimates the probability with no basis, should be at a disadvantage. I.e., as some have argued, the Bayesian approach is 'coherent' or rational. If you are right in your subjective guesses, you should have an advantage in betting. Marcus. === Subject: Re: Is it possible for a probability to be unknown? > For example, suppose the question were: What is the probability that > life exists in the Alpha Centauri solar system? We might be tempted > to say, Of the choice 'life exists there' and 'no life exists there', > we do not have a preference for one over the other. Therefore, with > two equiprobable possible outcomes, the probability must thus be 0.5. > But to say that there is a 50/50 chance for the existence of life in > Alpha Centauri, based exclusively on the fact that we don't know, would > be ridiculous. I can kind of see what you're saying, but I'm still not entirely sure about this. Probability, to me, seems to be simply a measure of what we don't know. Let's say that Jane just tossed a coin and observed it land on tails. Bob hasn't observed this. If Jane then asks Bob the probability that it landed on heads, Bob might say 50%. But to *Jane*, the probability that it landed on heads is 0%. Does that mean that Bob was somehow wrong to commit to 50%? It seems to me that the there are two choices; that's all we know thing is very similar to this coin example -- we have incomplete knowledge, and we state a probability anyway. Let's say that we want to determine whether or not X exists (where X is any arbitrary phenomenon). We currently have no evidence either way. But then, a few weeks later, we make a discovery that in effect slightly doubts that X exists. Does the existence of X now have a probability? Would it now be only (something like) 40% probable? If so, wasn't it 50% to begin with? === Subject: Re: Is it possible for a probability to be unknown? <1gd2379tbyt20$.dlg@tjb.invalid.invalid It seems to me that the there are two choices; that's all we know thing > is very similar to this coin example -- we have incomplete knowledge, and > we state a probability anyway. That there are two choices is fine. The assumption that those choices are equi-probaable is the problem. With the coin, it's a reasonable assumption. With the life on Jupiter problem, on what basis can you make the assumption that those choices are equi-probable? > Let's say that we want to determine whether or not X exists (where X is any > arbitrary phenomenon). We currently have no evidence either way. But > then, a few weeks later, we make a discovery that in effect slightly doubts > that X exists. Does the existence of X now have a probability? Would it > now be only (something like) 40% probable? If so, wasn't it 50% to begin > with? No. If you don't know either way, you can't assume 50%. You can assume this only if you have reason to believe the choices are equi-probable. When you say slightly doubts, do you mean believed to be less probable than before? If so, that means that you had some idea before what the probability was to think it's less now. With the life on Jupiter scenario, more doubtful evidence would likely take the figure down much more than 40%, because it would be the *only* evidence. Give me an example of a problem in which we have no idea which of two choices is correct (nor do we have reason to believe the choices to be equiprobable), and some discovery causes a slight doubt in our otherwise complete ignorance. === Subject: Re: Is it possible for a probability to be unknown? > I present you with a bag, with the following properties: 1. You cannot see inside it. > 2. It contains at least one marble, and this marble is red. > 3. It may or may not contain more marbles. > 4. If it does contain more marbles, they may or may not be red. Is it valid to say that the probability of picking a red marble is unknown > here? Or must we commit to 50%? No! Never 50% - you should go for 65% if you're selling, 35% if you're buying. oo-._.-._.-._.-._! Actually this 50% is uncannily like the Orlovian Big'un: Bill enters a competition. What is the probability he will win? Why, 50%. And the probability he will come third? Why, 50%. And the probability he will come in the top three? Why, 50%. But once we formulaically declare that one of these 50%s is the Real Who'nos, the others can be calculated. Suppose the 50% for top 3 is the real Who'nos, then the probability of coming top is Who'nos/3. Well, given any specific enough question, I'm sure I could make up a story to answer it. (Which of course is TO's MO) Brian Chandler http://imaginatorium.org === Subject: Re: Is it possible for a probability to be unknown? > Bayesian probability is an estimate of our degree of belief in > different outcomes. A Bayesian would say you can assign > probabilities to events about which you have no knowledge > whatsoever, such as the existence of life on Mars. Ah. I take it then that Bayesian probability would put the probability at 50% in the example in the original post in this thread. This may well be what I'm thinking of. Well, maybe I should give some context by stating why I made this thread in the first place. Basically, a book I'm reading quotes a definition of agnosticism (lack of belief in God, but also lack of disbelief; i.e., fence-sitting), and the book suggests that this quoted definition implies that God's existence and non-existence are equiprobable. However, it's not actually stated within the quotation that they are equiprobable, and I couldn't quite determine how it was *implied* within it either. My theory was that by implying that there is no evidence either way, the quoted definition also implied that the probability was 50-50 (with the author using the logic that if there is no evidence, it is 50-50). I wonder if the author was using Bayesian probability then. Hmm. === Subject: Re: Is it possible for a probability to be unknown? <4t5k72F12cqibU1@mid.individual.net> <1n5b9hrfb0jhu.dlg@tjb.invalid.invalid> <97u5shvwhpij$.dlg@tjb.invalid.invalid Bayesian probability is an estimate of our degree of belief in > different outcomes. A Bayesian would say you can assign > probabilities to events about which you have no knowledge > whatsoever, such as the existence of life on Mars. Ah. I take it then that Bayesian probability would put the probability at > 50% in the example in the original post in this thread. No. There is no such thing as a single, uniquely-defined Bayesian probability. A Bayesian named Jones might put p = 0.5, Smith might use p = 0.4 and Schwartz might use p = 0.0001. Much depends on the extra information (valid or not) they bring to the problem. For example, are they allowed to look at the bag? They can judge from the bag's size that it likely will not hold millions of balls, marbles, or whatever. They might differ in their judgement of exactly how many colours exist in the world (that could be alternatives to red), etc. Indeed, the true Bayesian would assign a probability distribution to the value of p (so that p is truly uncertain). They might then use a summary statistic, such as the expectation of the distribution, as a point estimate of p, but that would not mean that they regarded a single value as true---just most likely, perhaps. If a Bayesian has a non-informative prior (a uniform distribution) the expectation of p would be 0.5, but still, any value from 0 to 1 would be equally likely. I think you might benefit by looking at the book by E. Jaynes, Probability Theory (the Logic of Science). Jaynes rejects the standard Kolmogorov formulation of probability, and opts instead for a formulation based on plausible reasoning. His treatment is refreshing, but sometimes a bit too extreme and rebellious for my taste. He rejects notions of randomness completely, so for him, probability is all about degrees of knowledge. It sounds like his approach is in synch with what you want. But you would still come away realizing that your simple p = 0.5 estimate is baseless. R.G. Vickson This may well be what I'm thinking of. Well, maybe I should give some > context by stating why I made this thread in the first place. Basically, a book I'm reading quotes a definition of agnosticism (lack of > belief in God, but also lack of disbelief; i.e., fence-sitting), and the > book suggests that this quoted definition implies that God's existence and > non-existence are equiprobable. However, it's not actually stated within > the quotation that they are equiprobable, and I couldn't quite determine > how it was *implied* within it either. My theory was that by implying that there is no evidence either way, the > quoted definition also implied that the probability was 50-50 (with the > author using the logic that if there is no evidence, it is 50-50). I > wonder if the author was using Bayesian probability then. Hmm. === Subject: Re: Is it possible for a probability to be unknown? <4t5k72F12cqibU1@mid.individual.net> <1n5b9hrfb0jhu.dlg@tjb.invalid.invalid> <97u5shvwhpij$.dlg@tjb.invalid.invalid Bayesian probability is an estimate of our degree of belief in > different outcomes. A Bayesian would say you can assign > probabilities to events about which you have no knowledge > whatsoever, such as the existence of life on Mars. Ah. I take it then that Bayesian probability would put the probability at > 50% in the example in the original post in this thread. In Bayesian language, the initial belief is called the a priori or prior distribution. There isn't one Bayesian probability. I think you could make a case for distributions with p(red) = 50%, p(red) = 100%, or p(red) = any nonzero value. > This may well be what I'm thinking of. Well, maybe I should give some > context by stating why I made this thread in the first place. Basically, a book I'm reading quotes a definition of agnosticism (lack of > belief in God, but also lack of disbelief; i.e., fence-sitting), and the > book suggests that this quoted definition implies that God's existence and > non-existence are equiprobable. However, it's not actually stated within > the quotation that they are equiprobable, and I couldn't quite determine > how it was *implied* within it either. You are correct that the author has made a mistake, thinking that two possibilities implies two equally-likely possibilities. I think a Bayesian agnostic could again put the a priori p(God) as any value between 0 and 100%, exclusive. (If he/she chose 0 or 100%, that would indicate firm atheism or firm theism). > My theory was that by implying that there is no evidence either way, the > quoted definition also implied that the probability was 50-50 (with the > author using the logic that if there is no evidence, it is 50-50). I > wonder if the author was using Bayesian probability then. Hmm. I think the author is following a common misconception about probability. - Randy === Subject: Re: Is it possible for a probability to be unknown? <4t5k72F12cqibU1@mid.individual.net> <1n5b9hrfb0jhu.dlg@tjb.invalid.invalid> <97u5shvwhpij$.dlg@tjb.invalid.invalid Bayesian probability is an estimate of our degree of belief in > different outcomes. A Bayesian would say you can assign > probabilities to events about which you have no knowledge > whatsoever, such as the existence of life on Mars. Ah. I take it then that Bayesian probability would put the probability at > 50% in the example in the original post in this thread. In Bayesian language, the initial belief is called the a priori > or prior distribution. There isn't one Bayesian probability. > I think you could make a case for distributions with p(red) = 50%, > p(red) = 100%, or p(red) = any nonzero value. This may well be what I'm thinking of. Well, maybe I should give some > context by stating why I made this thread in the first place. Basically, a book I'm reading quotes a definition of agnosticism (lack of > belief in God, but also lack of disbelief; i.e., fence-sitting), and the > book suggests that this quoted definition implies that God's existence and > non-existence are equiprobable. However, it's not actually stated within > the quotation that they are equiprobable, and I couldn't quite determine > how it was *implied* within it either. You are correct that the author has made a mistake, thinking > that two possibilities implies two equally-likely possibilities. He doesn't mention what the book is, but a (wild) guess might be that it is The Probability of God by Stephen Unwin. This book explicitly brings a Bayesian approach to the question of religious belief. The book (though not the cover - curse those marketing types) is fairly modest in that its main point seems to be that Bayesian methods have the potential to clarify a person's subjective beliefs and it doesn't claim to have once and for all settled the existence (or nonexistence) of God. His conclusion is that it is more likely than not that God exists, though like a true Bayesian he acknowledges that other people with other prior beliefs might come to different conclusions. At once ground) Bayesian analysts would generally agree that the choice of prior probabilties must be considered and justified on a case by case basis ... Here, I think that the expression of complete ignorance [of whether or not God exists] is a good case for the 50-50 argument (pg 57). In the next couple of pages he tries to give some argumentation to back this up, so he is not simply unaware of the issue or naively assuming that uncertainty implies probability 50%. Somewhat humorously, he has an appendix for setting up an Excel spreadsheet to play around with his calculations in which the initial prior probability is a parameter that anyone could modify if they see fit. If this isn't the book, I would be interested in hearing from the OP what it was. > I think a Bayesian agnostic could again put the a priori p(God) > as any value between 0 and 100%, exclusive. (If he/she chose > 0 or 100%, that would indicate firm atheism or firm theism). My theory was that by implying that there is no evidence either way, the > quoted definition also implied that the probability was 50-50 (with the > author using the logic that if there is no evidence, it is 50-50). I > wonder if the author was using Bayesian probability then. Hmm. I think the author is following a common misconception about > probability. > > - Randy === Subject: Re: Is it possible for a probability to be unknown? <4t5k72F12cqibU1@mid.individual.net> <1n5b9hrfb0jhu.dlg@tjb.invalid.invalid> <97u5shvwhpij$.dlg@tjb.invalid.invalid Bayesian probability is an estimate of our degree of belief in > different outcomes. A Bayesian would say you can assign > probabilities to events about which you have no knowledge > whatsoever, such as the existence of life on Mars. Ah. I take it then that Bayesian probability would put the probability at > 50% in the example in the original post in this thread. This may well be what I'm thinking of. Well, maybe I should give some > context by stating why I made this thread in the first place. Basically, a book I'm reading quotes a definition of agnosticism (lack of > belief in God, but also lack of disbelief; i.e., fence-sitting), and the > book suggests that this quoted definition implies that God's existence and > non-existence are equiprobable. However, it's not actually stated within > the quotation that they are equiprobable, and I couldn't quite determine > how it was *implied* within it either. My theory was that by implying that there is no evidence either way, the > quoted definition also implied that the probability was 50-50 (with the > author using the logic that if there is no evidence, it is 50-50). I > wonder if the author was using Bayesian probability then. Hmm. Bob enters a competition. What is the probability that he wins? 50% -- mike === Subject: Re: Is it possible for a probability to be unknown? <4t5k72F12cqibU1@mid.individual.net> <1n5b9hrfb0jhu.dlg@tjb.invalid.invalid> <97u5shvwhpij$.dlg@tjb.invalid.invalid> Mike Kelly ha scritto: > Bob enters a competition. What is the probability that he wins? 50% > I agree. In my opinion a probability can never be unknown. It reflects our knowledge - or lack of. In complete lack of knowledge the probability of exclusive and complete events should be equally distributed. We could reformulate your example as: An experiment can give only one of the two results A and B. What is the probability of getting result A? With no knowledge of what the experiment is, nor what the results are, but just that they are mutually exclusive and that one of the two _will_ take place, I think we must say 0.5 (the letters A and B could be exchanged without modifying what we know). To apply this to the original example, if we accept its non-realistic formulation (no limit is specified to the number of marbles) then we should say 0.5. To keep it simple, if there was a limit of total 2 marbles then: Probability that there is just one marble: 0.5 (and it _is_red, we know). Probability that there are two marbles: 0.5 Probability that the second marble, if present, is red: 0.5. Probability of picking a red marble: 0.5 * 1 + 0.5 * (0.5 * 1 + 0.5 * 0.5) = 0.875 Increasing the limit, this approaches 0.5 === Subject: Re: How To Get Rich Quick. === > Subject: How To Get Rich Quick. > > It is really simple if you live with your parents. Just get a full > time job for 2 or 3 years, and save every last penny you make. You > should have about $25,000. And you can make that kind of money in > high school. Now once you have that cash saved up, don't spend it on > college, or your bills, or weed, or anything else you need money for. > Just start playing the stock market, and it will be like you have a > second job. First of all, nobody who ever made money in the market ever called it playing... > If you can spot lows in things like Oil, the US Dollar, > ot a solid company that just has a good buying opportunity, then you > stand to make about 10% every month. Of course, every professional and amateur trader in the other 23 time zones will be watching each market segment and individual stock in real time with much higher level computer access and information sources than you are going to get not having spent any of the $25,000 it apparently took you three years to make working at McDonalds... > At least I seem to. And 10% of > $25,000 is about $2,500. Cool right? Pathetic really, because if you parents were willing to pick up absolutely all of your expenses you should have been able to cut out of your low level job after only a few months, invest the money in yourself, and leverage your way up to a much higher paying opportunity during the remainder of the three year period, WAAAAAaaaaay outperforming a 10 percent return on minimum wage. > Then you just take your > earnings out of the market and you use them to help you get through > school, while you work your other jobs. It is great! Why do people who've never made real money give advice on how easy it is to do so??? > The real secret > is just knowing how to invest conservatively, and you can take it from > me you want to diversify. Half the time you should be keeping your > money liquid and looking for deals in the stock market. Uh huh. Brace yourself...here it comes...any second now... > And the other > half of the time you should be thinking about where the peaks are that > you want to sell. YES! Told ya! That sophomoric tautology you just expounded is called Market Timing. It's the original Mother of All Bad Ideas that is as old as the market itself. Read A Random Walk Down Wall Street which was written way back in 1973 to understand that you haven't just been a little stupid - but rather you've dipped yourself in nitro and decided to go try your hand at walking on hot coals. > Just don't trust anyone who says a stock is going > to double or tripple, because it probably never will. Except for all those hundreds and thousands that have done way more than that. > Look for good 5% > or 10% increases in the stock price, cause those are the sweet spots. > And play lots of texas no limit poker. Actually YOU should play Powerball. At least you'll only be losing a dollar at a time. > Because the stock market isn't > much different than gambling. Unless, of course, you actually have money and you have even some clue what you are doing. That is called investing, not playing. > There is just less risk because you are > only trying to make a fraction of what you are gambling with. So go > out there and get rich! Or not! > And if you can keep putting money back into > your stocks without spending it from your real job, you will be a > millionaire by the time you are 40. Or you could get a REAL job, do something useful, and actually make a million at any damn age. Nevermore (Always short sell the stupid) === Subject: Re: How To Get Rich Quick. <20061130135511384-0500@news1.news.adelphia.net> === > Subject: How To Get Rich Quick. It is really simple if you live with your parents. Just get a full > time job for 2 or 3 years, and save every last penny you make. You > should have about $25,000. And you can make that kind of money in > high school. Now once you have that cash saved up, don't spend it on > college, or your bills, or weed, or anything else you need money for. > Just start playing the stock market, and it will be like you have a > second job. First of all, nobody who ever made money in the market ever called it > playing... > If you can spot lows in things like Oil, the US Dollar, > ot a solid company that just has a good buying opportunity, then you > stand to make about 10% every month. Of course, every professional and amateur trader in the other 23 time > zones will be watching each market segment and individual stock in real > time with much higher level computer access and information sources than > you are going to get not having spent any of the $25,000 it apparently > took you three years to make working at McDonalds... At least I seem to. And 10% of > $25,000 is about $2,500. Cool right? Pathetic really, because if you parents were willing to pick up > absolutely all of your expenses you should have been able to cut out of > your low level job after only a few months, invest the money in yourself, > and leverage your way up to a much higher paying opportunity during the > remainder of the three year period, WAAAAAaaaaay outperforming a 10 > percent return on minimum wage. Then you just take your > earnings out of the market and you use them to help you get through > school, while you work your other jobs. It is great! Why do people who've never made real money give advice on how easy it is > to do so??? Because the only people who have EVER made REAL money is AT&T Who are well know to be US Governement subsidized morons. The real secret > is just knowing how to invest conservatively, and you can take it from > me you want to diversify. Half the time you should be keeping your > money liquid and looking for deals in the stock market. Uh huh. Brace yourself...here it comes...any second now... And the other > half of the time you should be thinking about where the peaks are that > you want to sell. YES! Told ya! That sophomoric tautology you just expounded is called > Market Timing. It's the original Mother of All Bad Ideas that is as > old as the market itself. Read A Random Walk Down Wall Street which was written way back in 1973 > to understand that you haven't just been a little stupid - but rather > you've dipped yourself in nitro and decided to go try your hand at > walking on hot coals. Just don't trust anyone who says a stock is going > to double or tripple, because it probably never will. Except for all those hundreds and thousands that have done way more than > that. Look for good 5% > or 10% increases in the stock price, cause those are the sweet spots. > And play lots of texas no limit poker. Actually YOU should play Powerball. At least you'll only be losing a > dollar at a time. Because the stock market isn't > much different than gambling. Unless, of course, you actually have money and you have even some clue > what you are doing. That is called investing, not playing. There is just less risk because you are > only trying to make a fraction of what you are gambling with. So go > out there and get rich! Or not! And if you can keep putting money back into > your stocks without spending it from your real job, you will be a > millionaire by the time you are 40. Or you could get a REAL job, do something useful, and actually make a > million at any damn age. > > Nevermore (Always short sell the stupid) === Subject: Re: How To Get Rich Quick. > === >>Subject: How To Get Rich Quick. >>It is really simple if you live with your parents. Just get a full >>time job for 2 or 3 years, and save every last penny you make. You >>should have about $25,000. And you can make that kind of money in >>high school. Now once you have that cash saved up, don't spend it on >>college, or your bills, or weed, or anything else you need money for. >>Just start playing the stock market, and it will be like you have a >>second job. > > > First of all, nobody who ever made money in the market ever called it > playing... > Please pardon the piggy-back, Nevermore, I didn't read the original. ;) Second of all, sponging off your parents while you save every penny you make is no better than sucking up welfare when you could be supporting yourself. Unless, of course, you are both agreed that you will support *them* fully when it comes to their retirement time and *then follow through*. That doesn't seem to be part of the game-plan though, does it? Just sponge off your parents for as long as they let you, and use the money you save on bills to fund your own wealth-plan. pfst. What a pile of crap, and indicative of the greedy, selfish, me-me-me mentality all too common in this society. Carry on, Nevermore. Rip away. ;) L. > > > Or you could get a REAL job, do something useful, and actually make a > million at any damn age. > > Nevermore (Always short sell the stupid) -- Lilith Dragonswife, Yin Bitch ~ Better to be an enemy than a slave. ~ === Subject: Re: How To Get Rich Quick. <20061130135511384-0500@news1.news.adelphia.net> <9gGbh.11404$dX4.10455@clgrps13> In what kind of world do you live, Corey? The dollar is plummeting. The stock markets are shaky as a result. In your philosophy, should we all buy now? If you would want to invest at all, silver and/or gold is all you should go for - at least that's what the experts say. But congratulations with the $25,000. I would love to hear how you earn 10% a MONTH. === Subject: Re: How To Get Rich Quick. > It is really simple if you live with your parents. Just get a full > time job for 2 or 3 years, and save every last penny you make. You > should have about $25,000. And you can make that kind of money in high > school. Now once you have that cash saved up, don't spend it on > college, or your bills, or weed, or anything else you need money for. > Just start playing the stock market, and it will be like you have a > second job. If you can spot lows in things like Oil, the US Dollar, ot > a solid company that just has a good buying opportunity, then you stand > to make about 10% every month. At least I seem to. And 10% of $25,000 > is about $2,500. Cool right? Then you just take your earnings out of > the market and you use them to help you get through school, while you > work your other jobs. It is great! The real secret is just knowing > how to invest conservatively, and you can take it from me you want to > diversify. Half the time you should be keeping your money liquid and > looking for deals in the stock market. And the other half of the time > you should be thinking about where the peaks are that you want to sell. > Just don't trust anyone who says a stock is going to double or > tripple, because it probably never will. Look for good 5% or 10% > increases in the stock price, cause those are the sweet spots. And > play lots of texas no limit poker. Because the stock market isn't much > different than gambling. There is just less risk because you are only > trying to make a fraction of what you are gambling with. So go out > there and get rich! And if you can keep putting money back into your > stocks without spending it from your real job, you will be a > millionaire by the time you are 40. post I have ever read that actually makes any sense. If all youngsters would take this advice we could get rid of welfare in this country. BTW, if you only save $500,000 by age 40 or even half of that you will still be a success. Ivan Ivan === Subject: Re: How To Get Rich Quick. > post I have ever read that actually makes any sense. That probably says more about WebTV than the Internet. > If all youngsters would take this advice we could get rid of welfare > in this > country. Or we could just raise the minimum wage to reflect what employees really contribute to fast food restaurants and then we wouldn't have to force them to live rent free with their parents for three years and try to beat the market in order to finally have real money some 24 years after they've entered the workforce... > BTW, if you only save $500,000 by age 40 or even half of that > you will still be a success. Ivan Ivan So, by your definition, somebody with $240,000 in the bank would be a failure? Even if they had a real computer and not a WebTV? Nevermore === Subject: Re: How To Get Rich Quick. <20061130140031766-0500@news1.news.adelphia.net>: > >> post I have ever read that actually makes any sense. > > That probably says more about WebTV than the Internet. Bwahahahahaha!!! OMG.. whew. < gryn > That's gotta be the best one liner I've seen in months. Kudos. -- blu*goddess.of.groundhogs*juju blu 3=3 master of irrelevance Cancel my subscription to the resurrection. -Jim Morrison http://blu05.port5.com/cantmakeit.htm === Subject: Re: How To Get Rich Quick. > It is really simple if you live with your parents. Just get a full > time job for 2 or 3 years, and save every last penny you make. You > should have about $25,000. And you can make that kind of money in high > school. Now once you have that cash saved up, don't spend it on > college, or your bills, or weed, or anything else you need money for. > Just start playing the stock market, and it will be like you have a > second job. If you can spot lows in things like Oil, the US Dollar, ot > a solid company that just has a good buying opportunity, then you stand > to make about 10% every month. At least I seem to. And 10% of $25,000 > is about $2,500. Cool right? Then you just take your earnings out of > the market and you use them to help you get through school, while you > work your other jobs. It is great! The real secret is just knowing > how to invest conservatively, and you can take it from me you want to > diversify. Half the time you should be keeping your money liquid and > looking for deals in the stock market. And the other half of the time > you should be thinking about where the peaks are that you want to sell. > Just don't trust anyone who says a stock is going to double or > tripple, because it probably never will. Look for good 5% or 10% > increases in the stock price, cause those are the sweet spots. And > play lots of texas no limit poker. Because the stock market isn't much > different than gambling. There is just less risk because you are only > trying to make a fraction of what you are gambling with. So go out > there and get rich! And if you can keep putting money back into your > stocks without spending it from your real job, you will be a > millionaire by the time you are 40. > post I have ever read that actually makes any sense. If all youngsters > would take this advice we could get rid of welfare in this country. > BTW, if you only save $500,000 by age 40 or even half of that you will > still be a success. Ivan > Ivan ******************************************************* Hi: Nonsense: people loves fun, and kids love fun the most. Want to make them into boring, rather weird cheap cuckoos playing with actions and not spending a dime and thus being damn parasytes? So that theyll 'be rich by the time they're 40 but most probably without friends, family, etc.? Bitter rich old men 40 years old....great future! Tonio === Subject: Re: Has bad mathematics lead to engineering disasters? <87bqmq1vfs.fsf@nonospaz.fatphil.org > Sorry,it is not so as we gather.As Progenoskes also observed,Tacoma > Narrows Bridge was a new physics/mechanics situation as a strange > unsteady state aerodynamics phenomenon that had not been encountered in > engineering situations up till that point of time...naturally not > mathematically modeled to be handed down to practising design > engineers.The failure analysis commitee also cleared them of any > charges of negligence. Even in late nineties a Boeing aircraft > empennage was destroyed at Japan Norita airport due to such > causes.However the degree of goofitude in Hyatt Hotel is an order of > magnitude higher. The Hyatt Hotel walkway was not built as designed, so it is hard to > fault the designer. Not built as *originally* designed. They changed the designs because > they were not practical from the manufacturing standpoint. It was > built in accordance with the new designs. Designers somewhere, and > the people who reviewed and signed off those designs, are to blame. And you *can* fault the original designer too for designing something > for which the components and construction were practically impossible. Phil > -- If you build 10 stories above the foundation, the foundation has to take the weight of 10 stories in compression.If you hang 10 stories down from the ceiling,the ceiling has to take the weight of 10 stories in tension.What prevented adoption of floorwise tapering I-Beams or welded taper tubes? It is the speed needing to ignore blind spots.Granted things are more clear in hindsight.However insistence on routine elementary but mandatory structural checks by a second party in large constructions e.g., by finite element analysis would have routinely eliminated such loopholes when elemental modeling would be considered.This is not maths but managerial practice of safety concern in engineering. ButSafety does not sell. === Subject: Re: Has bad mathematics lead to engineering disasters? <87bqmq1vfs.fsf@nonospaz.fatphil.org > Sorry,it is not so as we gather.As Progenoskes also observed,Tacoma > Narrows Bridge was a new physics/mechanics situation as a strange > unsteady state aerodynamics phenomenon that had not been encountered in > engineering situations up till that point of time...naturally not > mathematically modeled to be handed down to practising design > engineers.The failure analysis commitee also cleared them of any > charges of negligence. Even in late nineties a Boeing aircraft > empennage was destroyed at Japan Norita airport due to such > causes.However the degree of goofitude in Hyatt Hotel is an order of > magnitude higher. The Hyatt Hotel walkway was not built as designed, so it is hard to > fault the designer. Not built as *originally* designed. They changed the designs because > they were not practical from the manufacturing standpoint. It was > built in accordance with the new designs. Designers somewhere, and > the people who reviewed and signed off those designs, are to blame. And you *can* fault the original designer too for designing something > for which the components and construction were practically impossible. Phil If you build 10 stories above the foundation, the foundation has to take the weight of 10 stories in compression.If you hang 10 stories down from the ceiling,the ceiling has to take the weight of 10 stories in tension.What prevented adoption of floorwise tapering I-Beams or a welded taper tubes? It is the speed needing to ignore blind spots.Granted things are more clear in hindsight.However insistence on routine elementary but mandatory structural checks by a second party in large constructions e.g., by finite element analysis would have routinely eliminated such loopholes when elemental modeling would be considered.This is not maths but managerial practice of safety concern in engineering. Narasimham === Subject: Re: Has bad mathematics lead to engineering disasters? > I'd like to know of some specific examples where mathematics was > misused in the engineering sciences leading to disasters or at least > bad results. Thinking of this in terms of questions on other threads makes me see > that this should be a rarity. Errors in arithmetical calculations, or using an empirical formula in > areas where it does not apply (as mentioned in this thread), could > cause engineering disasters. But someone using bad mathematics? It's highly unlikely that we will > ever see a bridge fall down because someone used a formula that > mistakenly assumed the truth of the Riemann conjecture! > I wonder if the Ambiguous Case in oblique triangles has ever bitten > an engineer or designer in the ass? Where, of course, by ass, you refer to angle-side-side. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Has bad mathematics lead to engineering disasters? > > The engineers at Morton Thiokol, who built the solid rocket boosters >> containing the infmaous o-rings, recommended against launch at cold >> temperatures. They wer over-ruled by their management and NASA's >> management. I've seen the blame (for the exact scenario you describe) put at > the feet of the visualisation of the data. i.e. it was misleading > charts in the powerpoint slides that caused the management to hold > the views that they did. (A disease which the engineers were immune > to.) > > I heard it was not enough animation. And the powerpoint sound effects > weren't in surround sound. The powerpoints were _too_ persuasive, those 'improvements' would, oh, yeah, I get it. Phil -- Home taping is killing big business profits. We left this side blank so you can help. -- Dead Kennedys, written upon the B-side of tapes of /In God We Trust, Inc./. === Subject: help deriving formula for stock-price random changes in a stock's price relative to the stock price: dS/S = r dt + (.83[CapitalEth] .81 .8bdt) .83Ì Now, from that, the professor got: d(ln S) = (r [CapitalEth] 1ò2.83[CapitalEth]^2)dt + (.83[CapitalEth] .81 .8bdt) .83Ì And then: S(t+.83¢t) = S(t)e[(r [CapitalEth] 1ò2.83[CapitalEth]^2) .83¢t + (.83[CapitalEth] .81 .8b.83¢t) .83Ì] Can anyone whow the steps on how to get there? step by step? Also, I'm confused about the result of this formula. Why is it (r - 0.5.83[CapitalEth]^2)? Isn't that systematically going to result in returns lower than the actual average rate of return on the stock, r? E.g., assume epsilon, the error, is 0. The expected rate of return is 10%, and the standard deviation is 20%. Thus, if you have a given return with no deviation from the mean, you would expect the return to be 10%. However, if you use this formula, the actual return is going to be lower than 10%...it'll be 8% (10% - 0.5 * 20%^2 = 8%). === Subject: Recurrent Equation Hello everybody I am trying to solve this : x( i ) = ( 3/7 )*x( i-1 ) + ( 4/7 )*x( i+1 ) 0 < i < N ( integers ) ( I ) ( 4/7 )*x( 0 )+( 4/7 )*x( 1 )=x( 0 ) ( II ) ( 3/7 )*x( N-1) + ( 3/7 )*x( N ) = x(N) ( III ) with Sum ( x( i ) , i = 0..N ) = 1 I get this : )^(i) 0 < i < N c1,c2 constants and x( N-1 ) = ( 4/3 )*x( N ) Then, c1 + c2*( 3/4 ) = (3/4)*x( 0 ) and c1 + c2*( 3/4 )^(N-1) = ( 4/3 )*x( N ) So, x( 0 ) = ( 4/3 )*c1 + c2 and x( N ) = (3/4)*c1 + c2*( 3/4 )^(N) Now, I supose that I have to use Sum ( x( i ) , i = 0..N ) = 1 but I don 't see how to get every x( i ) and the values c1 and c2. Any suggestions, please ? === Subject: Re: Recurrent Equation > Hello everybody I am trying to solve this: I never thought I'd say this to anyone, but ... Your formulas have too many spaces in them. The word-wrap is out of control and is giving me a headache. I've fixed them below so that I can comment. > x(i) = (3/7) x(i-1) + (4/7) x(i+1) 0 < i < N ( integers ) (I) (4/7) x(0) + (4/7) x(1) = x(0) (II) (3/7) x(N-1) + (3/7) x(N) = x(N) (III) with Sum (x(i), i = 0..N) = 1 (IV) The number of variables is (N+1), and the number of equations is (N-1)+3 = N+2 > I get this: Solving (I) , x(i) = c1 + c2 (3/4)^(i) 0 < i < N c1,c2 constants > Then, c1 + c2 (3/4) = (3/4) x(0) and c1 + c2(3/4)^(N-1) = (4/3) x(N) > So, x(0) = (4/3) c1 + c2 and x(N) = (3/4) c1 + c2 (3/4)^(N) Now, I supose that I have to use Sum (x(i) , i = 0..N) = 1 (IV) > but I don 't see how to get every x(i) > and the values c1 and c2. i.e., it doesn't appear that you can solve uniquely for c1 and c2. I think trying to fit x(0) and x(N) is what caused the trouble. What you had originally was a system of linear equations with one more equation than variable. If this system is to have a solution, then some equation among (I)-(IV) is redundant. You should find out which one it is and then use Cramer's Rule (or other linear algebra techniques) to get an explicit solution. --- Christopher Heckman ((For those interested, the rest of the derivation is: (IV) states x(0) + [x(1) + x(2) + ... + x(N-1)] + x(N) = 1 x(0) + x(N) + [c1 + c2 (3/4) + c1 + c2 (3/4)^2 + ... + c1 + c2 (3/4)^(N-1)] = 1 x(0) + x(N) + (N-1) c1 + c2 (3/4) * ((3/4)^N - 1) / (3/4 - 1) = 1 (from a + a r + a r^2 + a r^3 + ... + a r^(N-1) = a (r^N - 1)/(r - 1).) which simplifies to x(0) + x(N) + (N-1) c1 + 3 c2 (1 - (3/4)^N) = 1 Using the formulas for x(0) and x(N) above, you get [(4/3) c1 + c2] + [(3/4) c1 + c2 (3/4)^N] + (N-1) c1 + 3 c2 (1 - (3/4)^N) = 1.)) === Subject: Notation question: a^b, avb I'm reading (OK, that's WAY too generous an assessment of my progress, more like stumbling through) Tom Kurtz's notes on stochastic analysis, and have encountered the notations a^b (actually, the ^ is more of a kind of inline wedge, not looking so much like a superscript as ^ might imply), and avb, where the 'v' is not the letter 'v', but the upside- down version of the mystery symbol ^. Here is a good example where these symbols occur. It is in the introduction to stopping times: If t1, t2, ... are stopping times and c geq 0 is a constant, then 1) t1 v t2 and t1 ^ t2 are stopping times. 2) t1 + c, t1 ^ c, and t1 v c are stopping times. There are a couple of other things that turn out to be stopping times as well, but they don't use the mystery symbols, so aren't relevant to my question. I have been assuming this: a v b : the greater of the two quantities a,b a ^ b : the lesser of the two quantities a,b. It makes sense, since I seem only to have run into the notation wrt real-valued quantities, but I'd really like someone to confirm these guesses (or more importantly, to set me straight with the correct version in the event that I'm just plain wrong). I can surely go along with my pretend version of what these symbols mean, but it's kind of like reading The Brothers Karamazov, and replacing every Slavic surname with Smith or Jones. Dale. === Subject: Re: Notation question: a^b, avb The use of the non-seriffed v and upside-down v to respectively represent min and max is really not that uncommon. Come to think of it, though, I may have seen these symobls (usually without definition) mainly in the context of stopping times. See, for example, Karlin and Taylor or Karatzas and Shreve. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Notation question: a^b, avb > I'm reading (OK, that's WAY too generous an assessment of > my progress, more like stumbling through) Tom Kurtz's notes > on stochastic analysis, and have encountered the notations > a^b (actually, the ^ is more of a kind of inline wedge, not > looking so much like a superscript as ^ might imply), and > avb, where the 'v' is not the letter 'v', but the upside- > down version of the mystery symbol ^. The symbols you specified are usually reserved for meet and join. http://en.wikipedia.org/wiki/Meet_%28mathematics%29 http://en.wikipedia.org/wiki/Join_(mathematics) In a practical sense, the ^ can usually be read as AND and the other one as OR. I would say more, but Wikipedia (URLs above) has said it all, in color, with pretty pictures. Dave. === Subject: Re: Notation question: a^b, avb [W. Dale Hall] > I'm reading (OK, that's WAY too generous an assessment of > my progress, more like stumbling through) Tom Kurtz's notes > on stochastic analysis, and have encountered the notations > a^b (actually, the ^ is more of a kind of inline wedge, not > looking so much like a superscript as ^ might imply), and > avb, where the 'v' is not the letter 'v', but the upside- > down version of the mystery symbol ^. Here is a good example where these symbols occur. It > is in the introduction to stopping times: If t1, t2, ... are stopping times and c geq 0 > is a constant, then 1) t1 v t2 and t1 ^ t2 are stopping times. > 2) t1 + c, t1 ^ c, and t1 v c are stopping times. There are a couple of other things that turn out to be > stopping times as well, but they don't use the mystery > symbols, so aren't relevant to my question. I have been assuming this: a v b : the greater of the two quantities a,b > a ^ b : the lesser of the two quantities a,b. That interpretation is explictly confirmed by a different author, page 7: http://home.uchicago.edu/~nstokey/chap1.pdf It seems a bit strange there too, though. I suppose it's related to that a v b is traditionally the least upper bound of {a, b} in lattice theory, and that a ^ b is traditionally the greatest lower bound of {a, b}. But when 'join' and 'meet' are plain old max and min ... > ... === Subject: Re: Notation question: a^b, avb > [W. Dale Hall] ... stuff deleted ... > I have been assuming this: >> a v b : the greater of the two quantities a,b >> a ^ b : the lesser of the two quantities a,b. > > That interpretation is explictly confirmed by a different author, page 7: > > http://home.uchicago.edu/~nstokey/chap1.pdf > > It seems a bit strange there too, though. I suppose it's related to that a > v b is traditionally the least upper bound of {a, b} in lattice theory, and > that a ^ b is traditionally the greatest lower bound of {a, b}. But when > 'join' and 'meet' are plain old max and min ... > How about that. I had thought it might be related to lattices, but then the source for more notes for my further edification. Dale === Subject: Re: Notation question: a^b, avb [W. Dale Hall] ... stuff deleted ... > I have been assuming this: > a v b : the greater of the two quantities a,b a ^ b : the lesser of the two quantities a,b. [Tim Peters] >> That interpretation is explictly confirmed by a different author, >> page 7: >> http://home.uchicago.edu/~nstokey/chap1.pdf >> It seems a bit strange there too, though. I suppose it's related to >> that a v b is traditionally the least upper bound of {a, b} in lattice >> theory, and that a ^ b is traditionally the greatest lower bound of {a, >> b}. But when 'join' and 'meet' are plain old max and min ... [W. Dale Hall] > How about that. I had thought it might be related to lattices, but then > as the source for more notes for my further edification. I don't know enough about the field to do better than guess. My /suspicion/ is that, while you may have seen only real numbers used for stopping time in examples so far, they'll eventually generalize the notion to, e.g., stopping time functions where lattice terminology may make a lot more natural sense. In any case, you're certain to find out for sure before I do ;-) === Subject: Re: Notation question: a^b, avb I'm reading (OK, that's WAY too generous an assessment of >my progress, more like stumbling through) Tom Kurtz's notes >on stochastic analysis, and have encountered the notations >a^b (actually, the ^ is more of a kind of inline wedge, not >looking so much like a superscript as ^ might imply), and >avb, where the 'v' is not the letter 'v', but the upside- >down version of the mystery symbol ^. Look at this link and you should see your V and upside-down V. === Subject: Re: Notation question: a^b, avb > >> I'm reading (OK, that's WAY too generous an assessment of >> my progress, more like stumbling through) Tom Kurtz's notes >> on stochastic analysis, and have encountered the notations >> a^b (actually, the ^ is more of a kind of inline wedge, not >> looking so much like a superscript as ^ might imply), and >> avb, where the 'v' is not the letter 'v', but the upside- >> down version of the mystery symbol ^. > > Look at this link and you should see your V and upside-down V. > > I'm reading (OK, that's WAY too generous an > assessment of > my progress, more like stumbling through) Tom Kurtz's > notes > on stochastic analysis, and have encountered the > notations > a^b (actually, the ^ is more of a kind of inline > wedge, not > looking so much like a superscript as ^ might imply), > and > avb, where the 'v' is not the letter 'v', but the > upside- > down version of the mystery symbol ^. > > Here is a good example where these symbols occur. It > is in the introduction to stopping times: > > If t1, t2, ... are stopping times and c geq 0 > is a constant, then > > 1) t1 v t2 and t1 ^ t2 are stopping times. > 2) t1 + c, t1 ^ c, and t1 v c are stopping times. > > There are a couple of other things that turn out to > be > stopping times as well, but they don't use the > mystery > symbols, so aren't relevant to my question. > > I have been assuming this: > > a v b : the greater of the two quantities a,b > a ^ b : the lesser of the two quantities a,b. > > It makes sense, since I seem only to have run into > the > notation wrt real-valued quantities, but I'd really > like > someone to confirm these guesses (or more > importantly, > to set me straight with the correct version in the > event > that I'm just plain wrong). > > I can surely go along with my pretend version of what > these symbols mean, but it's kind of like reading The > Brothers Karamazov, and replacing every Slavic > surname > with Smith or Jones. > > > Dale. Do the logic symbols, / = and; / = or fit your context? Just taking a stab in the dark. Tom === Subject: Re: Notation question: a^b, avb >> I'm reading (OK, that's WAY too generous an >> assessment of >> my progress, more like stumbling through) Tom Kurtz's >> notes >> on stochastic analysis, and have encountered the >> notations >> a^b (actually, the ^ is more of a kind of inline >> wedge, not >> looking so much like a superscript as ^ might imply), >> and >> avb, where the 'v' is not the letter 'v', but the >> upside- >> down version of the mystery symbol ^. >> Here is a good example where these symbols occur. It >> is in the introduction to stopping times: >> If t1, t2, ... are stopping times and c geq 0 >> is a constant, then >> 1) t1 v t2 and t1 ^ t2 are stopping times. >> 2) t1 + c, t1 ^ c, and t1 v c are stopping times. >> There are a couple of other things that turn out to >> be >> stopping times as well, but they don't use the >> mystery >> symbols, so aren't relevant to my question. >> I have been assuming this: >> a v b : the greater of the two quantities a,b >> a ^ b : the lesser of the two quantities a,b. >> It makes sense, since I seem only to have run into >> the >> notation wrt real-valued quantities, but I'd really >> like >> someone to confirm these guesses (or more >> importantly, >> to set me straight with the correct version in the >> event >> that I'm just plain wrong). >> I can surely go along with my pretend version of what >> these symbols mean, but it's kind of like reading The >> Brothers Karamazov, and replacing every Slavic >> surname >> with Smith or Jones. >> Dale. > > > > Do the logic symbols, / = and; / = or fit > your context? Just taking a stab in the dark. > > Tom That's hard to say, since the variables in the formulas are all real-valued (whether constants or [samples of] random variables). The response by Badger, directing me to this table at Wikipedia: http://en.wikipedia.org/wiki/Table_of_mathematical_symbols actually did the trick. Dale. === Subject: Re: Notation question: a^b, avb I have been assuming this: a v b : the greater of the two quantities a,b > a ^ b : the lesser of the two quantities a,b. I have read a fair amount of mathematics and statistics in my year or two 's past experience but i've NEVER seen the above usage. The ^ has been used in these groups mostly for the FORTRAN ** for the power exponent such as 3^2 for 3**2. I've seen the v symbol for logical expressions. > I can surely go along with my pretend version of what > these symbols mean, but it's kind of like reading The > Brothers Karamazov, and replacing every Slavic surname > with Smith or Jones. > Dale. It's hard to believe Tom Kurtz or anyone can write a book withoug explaining the symbols used in it. Surely there must be a glossary of symbols so that you would not have to make wild guesses on what the symbols mean! -- Reef Fish Bob. === Subject: Re: Notation question: a^b, avb >> I have been assuming this: >> a v b : the greater of the two quantities a,b >> a ^ b : the lesser of the two quantities a,b. > > I have read a fair amount of mathematics and statistics in > my year or two 's past experience but i've NEVER seen > the above usage. > > The ^ has been used in these groups mostly for the > FORTRAN ** for the power exponent such as 3^2 for > 3**2. > Right, that's pretty much universal. > I've seen the v symbol for logical expressions. > Right, as or. In the same context, one frequently sees ^ as and. One also encounters them in lattice theory as meet and join but I can never remember which is which. Good thing that's not my field! > >> I can surely go along with my pretend version of what >> these symbols mean, but it's kind of like reading The >> Brothers Karamazov, and replacing every Slavic surname >> with Smith or Jones. >> Dale. > > It's hard to believe Tom Kurtz or anyone can write a book > withoug explaining the symbols used in it. Surely there must > be a glossary of symbols so that you would not have to make > wild guesses on what the symbols mean! > > -- Reef Fish Bob. > I only have a .pdf file, and hadn't done a keyword search for any of the words glossary,table,symbol. Having just done that, it came up empty. These are somewhat informal course notes, so it seems reasonable that they might not have all details required for a more formal publication. precise I had imagined the notation was standard for the specialty (Stochastic Differential Equations), but had no clue as to how narrowly one had to focus to find the specific application. I did luck out by Googling optional sampling theorem and came up with a set of lecture notes from Stanford with the ^ notation in which it was explicitly defined as T^n = min(n,T). Given that, I'm comfortable with the corresponding provisional definition for v. I suppose I've discovered (not to anyone's surprise) that the notation is a bit narrow in its application. Dale === Subject: Re: Notation question: a^b, avb Just a little posthumous editing. a buncha stuff, with a hanging word that shoulda hit the crapper: > > I only have a .pdf file, and hadn't done a keyword search for > any of the words glossary,table,symbol. Having just done > that, it came up empty. These are somewhat informal course > notes, so it seems reasonable that they might not have all > details required for a more formal publication. Here it floats: > precise ....flush.... > I had imagined the notation was standard for the specialty (Stochastic > Differential Equations), but had no clue as to how narrowly one had > to focus to find the specific application. and so forth ... Dale. === Subject: problem with sum function How can one optimize a discrete function given by a sum? For example, suppose s is the variable (positive integer), and one has: sum(something, from 0 to s) Suppose that the function is convex. I don`t know how to define that concept in the discrete area, but I try to mean that the function has an U shape, so it has a minimum (and only one). Is there a way to find the minimun without setting an interval, and trying all the s?? === Subject: Re: problem with sum function >How can one optimize a discrete function given by a sum? For >example, suppose s is the variable (positive integer), and >one has: sum(something, from 0 to s) Suppose that the function is convex. I don`t know how to >define that concept in the discrete area, but I try to mean >that the function has an U shape, so it has a minimum (and >only one). >Is there a way to find the minimun without setting an >interval, and trying all the s?? Find where the something changes from negative to positive. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Miquel's Six Circle Theorem - a new (?) 3d extension Hi! I have found an extension of Miquel's Six Circle Theorem in three dimensions, and I'd like to know if it is publishable. The only relevant web reference I have found is: http://11011110.livejournal.com/42096.html My theorem uses the same configuration of circles, but both the premise and the conclusion are different! === Subject: Re: number of cycles in digraphs with maximum density > Consider adigraphwith n vertices and (n choose 2) directed edges > (maximal density). There are a total of 2^(n choose 2) unique such > digraphs. We believe that n! of these graphs are acyclic (contain no directed > cycle), and 2^(n choose 2) - n! are cyclic (contain a directed cycle). Can anyone suggest how might we prove this (or disprove, if we're > wrong)?Induction? THEOREM. If D is a (finite)digraphwhere every vertex has in-degree >= > 1, then D contains a directed cycle. COROLLARY. If D is an acyclicdigraph, then D contains a vertex v such > that v has indegree 0. There are n ways to choose this vertex v (for adigraphwith n > vertices), and you cannot have two vertices with this property if you > have a completedigraph. There are exactly (n-1)! ways to order the > edges of Dv so that Dv is acyclic, by induction. By taking Dv union > the edges from v to all vertices of Dv, you get exactly n*(n-1)! = n! > digraphs with nocycles. You could also do it non-inductively, by repeatedly choosing the vertex > v that satisfies the corollary above, and thus set up a bijection > between the permutations of V(D) and the acyclic complete digraphs. --- Christopher Heckman Does a proof already exist for this problem?Probably. In most graduate-level Graph Theory texts, in fact. > > > -Dave === Subject: polynomials with rational coefficients. let f and h be polynomials with rational coefficients, and let g be a polynomial with complex coefficients. If f=gh then, is g also a polynomial with rational coefficients? Is there a theorem that says that? === Subject: Re: polynomials with rational coefficients. days. My association with the Department is that of an alumnus. >let f and h be polynomials with rational coefficients, and let g be a >polynomial with complex coefficients. >If f=gh then, is g also a polynomial with rational coefficients? Yes. > Is there a >theorem that says that? The division algorithm. If f = gh, then h | f in C[x]. If you divide f by h in Q[x], you get f(x) = q(x)h(x) + r(x) with q(x) and r(x) polynomials with rational coefficients, r(x)=0 or deg(r)http://www.mediamuslim.info/index.php?option=com_content&task=view&id=35 4&Itemid=23 Merasa Bernasib Sial atau Meramal Nasib Buruk >http://www.mediamuslim.info/index.php?option=com_content&task=view&id=35 5&Itemid=23 Berprilaku Baik melalui Ucapan, Perbuatan dan Al-Ma'ruf http://www.mediamuslim.info/index.php?option=com_content&task=view&id=355 &Itemid=23 Konsentrasi Untuk Menghadapi Hari Inihttp://www.mediamuslim.info/index.php?option=com_content&task=view&id= 355&Itemid=23 Pandanglah Kebawah, Akan Terlihat Besarnya Nikmat Allohhttp://www.mediamuslim.info/index.php?option=com_content&task=view&i d=355&Itemid=23 === Subject: Re: Moebius Band is not homeomorphic with a Torus Here is what I think you are saying: > g: {(t,k,m)| 0 <=3D t <=3D 4 Pi, 1 <=3D k <=3D 2m, 1 <=3D m <=3D oo} -> R= > ^3 > g(t,k,m) :=3D ( ( cos(t/2 + k Pi/m) + 3 ) cos(t), ( cos(t/2 + k Pi/m) + 3= > ) sin(t), > sin(t/2 + k Pi/m ) ), f_m(t,k) :=3D g(t,k,m). (***) Then for any m, 1 <= m <= oo, the image of f_m is a surface. And, f_1 > is a MB, and f_oo is a torus. That is fine, only now we are beginning to talk for the first time on > addressable terms. This is not a homeomorphism between a MB and a torus. If we restrict m > to be in some finite interval, then we get a homotopy. For example, g > gives a homotopy between the maps f_1 and f_2. OK, there is homotopy between an MB and that I now call the square > tube S. However, two surfaces can be homotopic without being homeomorphic. You have to give some more examples (of course other than the present > one) and its logic, if there is something that can be brought in here. Do you mean you want an example of surfaces that are homotopic, but not > homeomorphic? I'll give a simpler example, which should give the idea. > Consider f: {(x,t)| 0 <= x <= 2 pi, 0 <= t <= 1} -> R^2, > f(x,t) := (t cos(x), t sin(x) ), > g_t(x) := f(x,t). Then the image of g_0 is a point and the image of g_1 is a circle. f is > a homotopy between a point and a circle. However a point and a circle > are not homeomorphic, since a circle has more than one point. > > In this example, there is a singularity at origin, in which size > shrinking is associated with annihilation of direction. When radius t > shrinks to zero at pole, normal vector direction of (x-t) field is > indeterminate. Likewise, exactly at the north/south poles the direction > of (lat, long) lines is lost when using spherical coordinates. > Quaternions are employed to have a definitive direction at all points. So? It is still a homotopy. > In the helicoidal 3D surface ( t cos(x), t sin(x), x ) has a definite > direction of surface normal along z-axis but its projection considered > as 2D disc (t cos(x), t sin(x)) the center would not have any > determinate direction. Does not orientation suffer here while > considering homeomorphism? Could we consider such an example while > considering homemorphism? A homotopy is not a homeomorphism. If you want to show two things are homeomorphic, you should exhibit the homeomorphism. > The function is bounded, finite intervals in its > range and domain m. The points occupy a slightly different position in > R^3, that is all. We can let m be 6, no? I am satisfied with m=6 or f_6 > homotopy with torus m=oo or f_6, you too will surely agree to this. It > is like accepting a situation in which the sum of external angles of a > hexagon (m=6) and circle (m=oo) is the same total angle 2 pi, we are > quite indifferent to polygon smoothness in topology or the difference > between a hexagon and circle. M values 6 -> 5 -> 4 -> 3 -> 2 between > and among the maps f_2, f_3, f_4, f_5,and f_6 gives a homotopy. Then > why do you bar holds suddenly in front of 2 -> 1 map between f_2 and > f_1? It makes no difference with inclusion of the third argument of > parameter domain, hope you can see it in Mathematica animation for the > entire torus set: Do[ ParametricPlot3D[{x[th,k,m],y[th,k,m],z[th,k,m] },{th,0,4 > Pi},{k,1,2 m},PlotPoints ->{101,2m}, > Boxed=EF=82=AEFalse] ,{m,1,6,1}]: > Do[ ParametricPlot3D[{x[th,k,m],y[th,k,m],z[th,k,m] },{th,0,4 > Pi},{k,1,2 m},PlotPoints ->{101,2m} ] A homeomorphism is different. Is different how? Formally, the definitions are different. Conceptually, two things that > are homeomorphic have identical topological properties. Two things that > are homotopic can be continuously deformed into each other, but can have > very different topological properties. > > That states the present position again; with no new inputs I cannot > understand why it should be so, appears still to me as not so > consequentially established facts stated a priori. What are your definitions of the words homotopy and homeomorphism? > > Let M be a subset of R^3 that is a MB. > Let T be a subset of R^3 that is a torus. Alright, in one statement, (M,T) are subsets of R^3 as separate > embedments. Then a homeomorphism is a map h from M to T such that h is bijective, > continuous, and its inverse is continuous. Yes this holds, T -> S -> M homeomorphism. don't know what you mean. > S is the square tube. It is beneficial to recognize its good vantage > topological position standing between MB and torus, close to MB. The > square tube S is not disoriented, has the same orientation as MB. As > the thickness disappears so the orientation disappears, that is all. > The map T -> S -> M or the reversed M -> S -> T homotopy is allowed > all the way down as a deformation. The part of it S -> T is also > homotopically OK, but S -> T is not being a homeomorphism seems to be > the present stumbling block, its nature as yet to me incomprehensible. > > If you consider torus f_oo is too long drawn out, consider only the > square tube f_2 of S. Why is not a map s a homeomorphism from S to M such that s is > bijective, continuous, and its inverse is continuous? You can't demonstrate two things are homeomorphic by writing down a > homotopy. What is your map between the two surfaces? Please write the > formulas. It should be a map from (a subset of) R^3 to (a subset of) > R^3. > The formula is already mentioned in (***) above, map between f_2 to f_1 > is S -> T. The images are mapped as they belong to the same general > parametric domain. That's not what a homeomorphism is. If you insist on confusing two different things, then naturally you will be confused. > m=1 for image of f_1 (MB) -> m=2 for image of f_2 (S) > > ((th,0,4 Pi ), (k,1,2)) -> ((th,0,4 Pi ), (k,1,4)) > > Between f_2 and f_1? You can have a homotopy between f_2 and f_1, but the homeomorphism has > to be between the *image* of f_2 and the *image* of f_1. A homotopy is > between maps. A homeomorphism is between sets. > In the example above, if we take the range of variation of deformed > points in 3D as homotopy, and the generator parameter space or the > generating image space as above, then, is not homeomorphism sufficient > and complete with definitions of identified domain as image and range > put and considered together? No. > I am going through the Staford University lecture notes. But much seems > to be hidden. I would like to see many more concrete examples on: > (homotopic, not homeomorphic). I gave an example to show a circle is homotopic to a point. But, a circle is not homeomorphic to a point. Doesn't this convince you that the two are different? -- David Marcus === Subject: Re: Moebius Band is not homeomorphic with a Torus > Is f_6 to f_2 homotopy ok? Yes. > If so,why so and if not,why not? It satisfies the definition of homotopy. > Is f_6 to f_2 homeomorphism ok? That's not a homeomorphism. > If so,why so and if not,why not? Because a homotopy is not a homeomorphism. > Is f_2 to f_1 > homotopy ok? If so,why so and if not,why not? Is f_2 to f_1 > homeomorphism ok? If so,why so and if not,why not? -- David Marcus === Subject: Re: Moebius Band is not homeomorphic with a Torus > Moebius Band is not homeomorphic with a Torus or even orientable like > it. Mathematica formula and images of the unifying set of tori in toroidal > co-ordinate system I have indicated in: http://img139.imageshack.us/img139/3489/mbnoorientci0.jpg I made a square section steel ring of m = 2 configuration by > cut/twist/weld and feel that the surface morphology of all these > surfaces has a unifying character by its fiber bundle.Of course the > standard Moebius Band is made by cut/twist/paste of a rectangular sheet > having two sides, small thickness of the sheet is conveniently > overlooked to result in one side only! There are actually four sides > in all and two tracks. By varying m or Aspect Ratio of a rectangular section of a tube (or > solid section) of torus one can morph/animate images among all tori > including the Moebius Band. I am not comfortable with the Moebius Band singled out topologically > either in terms of orientability or homeomorphability. What exactly am I missing? Narasimham Time to close the thread.I thank David Marcus,Lee Rudolph and others who participated. To be considered for acceptance are: 1) If a torus T fully encloses/is enclosed by an MB, corresponding points being identified by a common tori set formula given, there is homotopy but no homeomorphism between an MB and T. 2) If a square tube S fully encloses /is enclosed by an MB,-Ditto-. 3) The above is valid, not withstanding the fact that the topological invariant, Euler Characteristic X for MB, T and even the KB are the same, equal to zero. 4) In a square tube there are two tracks. If one track ( homeomorphic to a disc) is removed, the other track is not homeomorphic to an MB. When googling this topic, I was very surprised to see a large Moebius sculpure in stone like the square tube S. Keizo Ushimo starts with a massive granite ring having a hole width equal to the thickness of the ring. He then drills into the granite to slice it longitudinally, not the way you would normally slice a bagel to get two halves, but with a cut that makes a 180-degree twist during its travel around the ring.In effect, such a cut creates a space that can be considered a M?bius strip . It is topologically same as what I had made in steel and rubber. In the following link from which the above is quoted,in effect we should see no (topological) difference between the square tube, M?bius strip and the torus. Narasimham === Subject: Re: Moebius Band is not homeomorphic with a Torus > Moebius Band is not homeomorphic with a Torus or even orientable like > it. Mathematica formula and images of the unifying set of tori in toroidal > co-ordinate system I have indicated in: http://img139.imageshack.us/img139/3489/mbnoorientci0.jpg I made a square section steel ring of m = 2 configuration by > cut/twist/weld and feel that the surface morphology of all these > surfaces has a unifying character by its fiber bundle.Of course the > standard Moebius Band is made by cut/twist/paste of a rectangular sheet > having two sides, small thickness of the sheet is conveniently > overlooked to result in one side only! There are actually four sides > in all and two tracks. By varying m or Aspect Ratio of a rectangular section of a tube (or > solid section) of torus one can morph/animate images among all tori > including the Moebius Band. I am not comfortable with the Moebius Band singled out topologically > either in terms of orientability or homeomorphability. What exactly am I missing? Narasimham Time to close the thread.I thank David Marcus,Lee Rudolph and others who participated. Consider these for acceptance: 1) If a torus T fully encloses/is enclosed by an MB, corresponding points being identified by a common tori set formula given, there is homotopy but no homeomorphism between an MB and T. 2) If a square tube S fully encloses /is enclosed by an MB,-Ditto-. 3) The above is valid, not withstanding the fact that the topological invariant, Euler Characteristic X for MB, T and even the KB are the same, equal to zero. 4) In a square tube there are two tracks. If one track ( homeomorphic to a disc) is removed, the other track is not homeomorphic to an MB. When googling this topic, I was very surprised to see a large Moebius sculpure in stone like the square tube S. Keizo Ushimo starts with a massive granite ring having a hole width equal to the thickness of the ring. He then drills into the granite to slice it longitudinally, not the way you would normally slice a bagel to get two halves, but with a cut that makes a 180-degree twist during its travel around the ring.In effect, such a cut creates a space that can be considered a M?bius strip . It is topologically same as what I had made in steel and rubber. In the following link from which the above is quoted,in effect he sees no (topological) difference between the square tube, M?bius strip and the torus. Narasimham === Subject: *** 911 WAS AN INSIDE JOB. YOU WILL GET ALL SCIENTIFIC EVIDENCE ON st911.org *** 911 WAS AN INSIDE JOB. YOU WILL GET ALL SCIENTIFIC EVIDENCE ON st911.org The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. He and his family later sued the U.S. government for damages. We lived in 1984, Mayfield told reporters Wednesday. I'm talking about the George Orwell, frightening brave new world in which Big Brother is constantly watching you. (Watch Mayfield discuss the case Video) I, myself, have dark memories of stifling paranoia, of being monitored, followed, watched, tracked, he said, choking back emotion. I've been surveilled, followed, targeted primarily because I've been an outspoken critic of this administration and doing my job to defend others who can't defend themselves, to give them their day in court, and mostly for being a Muslim. The government refused, he said, to tell him where they put their cameras and surveillance devices, leaving his family wondering if their private conversations and intimate moments were on display. The days and weeks and months following my arrest were some of the hardest and darkest that myself and my family have ever had to endure, he said. And all because of this government's ill-conceived war on terror. ... What I really want is for this not to happen to anyone else. Wednesday's settlement includes not only a $2 million payment and an apology, but also an agreement by the government to destroy communications intercepts conducted by the FBI against Mayfield's home and office during the investigation. The written apology reads: The United States of America apologizes to Mr. Brandon Mayfield and his family for the suffering caused by the FBI's misidentification of Mr. Mayfield's fingerprint and the resulting investigation of Mr. Mayfield, including his arrest as a material witness in connection with other court orders in the Mayfield family home and in Mr. Mayfield's law office. A Justice Department statement released Wednesday said Mayfield was not targeted because of his Muslim faith and that the FBI had taken steps to improve its fingerprint identification process to ensure that what happened to Mr. Mayfield does not happen again. Mr. Mayfield and his family felt it was in their best interest to get on with their lives, said Mayfield's attorney, Elden Rosenthal. No amount of money can compensate Mr. Mayfield for being held as a prisoner and being told he faced the death penalty [for the Madrid bombings]. Mayfield said his suit was not about money. It's about regaining our civil rights, our freedom and most important, our privacy, he said. He and his attorneys said the settlement will allow him to continue the portion of his lawsuit challenging the constitutionality of the Patriot Act. Mayfield contends that his home was searched under provisions of the Patriot Act. Criminal, Rogue and RACIST elements have penetrated all the branches of USA and world governments. We know that on 911 FBI bastards ran all over in the neighborhood of Pentagon trying to confiscate EVIDENCE of the 911 crime that no Boeing Passenger Airliner hit Pentagon. They have stil not released the evidence. 911 was inside job. FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, Mossad, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED FBI is the BEST of the criminal branches ... THERE ARE MORE EVIL AND BIGGER ROGUES IN CIA, NSA, PENTAGON AND THE WHITE HOUSE. Dont forget that BASTARD, NIXON and JOHNSON who sank USS LIBERTY and KILLED THE ONLY ONLY REASON FBI DID THIS IS BECAUSE THEIR CRIME WAS UNCONCEALABLE. UNDER THE PRETEXT OF SELF-DEFENSE, MANY OF THESE AGENCY OFFICIALS HAVE COVERED THEIR CRIMES AND DESTROYED EVIDENCE OF THEIR CRIMES. FBI COVERED UP A CHILD PEDOPHILIA RING THAT WAS LINED TO REAGAN AND www.nkusa.org Name: www.nkusa.org Nickname: www.st911.org Location: WTC Building 7 Title: Pentagon Videos Industry: Non-Profit Email address: s...@rock.com Website or Blog: www.infowars.com www.prisonplanet.org www.counterpunch.org Quote: On 911, the only buildings that were demolished were the ones that belonged to Larry Silverstein. Building 7 Committed suicide even though no plane had hit it. Within minutes, FBI bastards were running around confiscating the pentagon videos to coverup the fact that there was no plane that hit the pentagon. Every bastard in the plot spoke in passive language: Rudy Guiliani said ... we were told ... WHO TOLD YOU!!! The physics shows that the building were demolished by synchronized explosions. Residue of Thermate = Thermite + Sulfur have been found in the molten metal pools. The time of fall is equal to free fall, debunking the pancake theory. Uncontrolled burning of Jet fuel does not create molten metal pools. View the videos with your own eyeballs. On the next day the uncivilized mad dogs of europe stood up and barked that an attack on one country is on all of them, becoming infact the most effective enablers of the actual criminals of 911. Instead of a dispassionate investigation they barked for war. This led to invasion of Afghanistan, the false cooked evidence of WMDs on Iraq. The use of white phosphorus, war crime. The naked human towers. The mad dogs of europe in the govt are all quiet on this. Only humane civilians of europe are vocal and making their efforts at individual levels. === Subject: Re: How to explain this to someone (highschooler)? >How would I explain this to a highschooler? Say highschool does this: 2x = 1 >3x = 1 Thus 2x = 3x because both equal same thing = 1 > 3x-2x = 0 > x = 0 Wrong! Check back in the 2 equations and one finds that this solution >doesn't work. I can explain why doing non-reversible operations to both sides of a >*single* equation can produce nonsense at the end, but can't explain >this type of thing. Of course, the best way to try and solve this >system is to use gaussian elimination and then one will notice that >something of the form c=0 (where c <>0) pops up. But what's the >explanation I could use to tell someone NOT to use this transitive rule >to solve such things (assume it's a highschooler who has no idea what >Linear Algebra or Gaussian Elimination are)? How would I explain this? The problem is that there are no words before 2x = 1 3x = 1 about the context. It should perhaps be Suppose x is a real number such that 2x = 1 and 3x = 1. Then ... When you get the contradiction, you realize the suppose was a mistake, and no such x exists (in the context which existed when x was introduced). The part about is a real number might be omitted in a first algebra course, but you still need Suppose x satisfies both 2x = 1 and 3x = 1. Then ... -- 44 months after Japan attacked Pearl Harbor, Japan surrendered. 44 months after US attacked Iraq, it's time for the US to surrender. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Lin algebra Q about orthonormal complex vectors Is the following true (no proof expected): If U is an nxn matrix with orthonormal columns, then the conjugate of it's rows are also orthonormal. Yes or No answer is fine. I am just trying to generalize all the theorems in this book I have (that works over Reals). === Subject: Re: Parametric solutions of Euler Brick? It's certainly true that g*(E) need not be Lebesgue measurable > for Lebesgue measurable E. That's not a problem - we're just > trying to construct a _Borel_ measure. (In fact, come to think > of it, if g*(E) _were_ Lebesgue measurable for every Lebesgue > measurable E then _that_ would show the argument must be > wrong, because it would imply that every Borel measure > extends to a measure on the Lebesgue-measurable sets, > which is not so.) > What is an example of a Borel measure that cannot be so extended? === Subject: Re: A Rudin's approach >>Let f be a nondecreasing, left continuous, real function on [a,b]. Then there exists a Borel measure mu on [a,b] such that f(x)-f(a)=mu([a,x)) for all x in [a,b]. The usual proof of this theorem is using Caratheodory extension theorem, extending the measure from an algebra to a sigma-algebra. But this method is long. In Rudin's Real and Complex Analysis, he seems to suggest another approach which is not clear to me, seems like you consider g(x)=x+f(x) and go from there. Does anyone know this Rudin's approach, >> I was puzzled by the question at first, but looking at the book I see >> that this is not proved explicitly, it's left as an exercise. But >> there's a simple proof using the idea you mention, implicit in the >> proof of Theorem 7.18 (at least that's the number in the third >> edition). >> As you say, let g(x) = x + f(x). The point to that is that g is >> _strictly_ increasing. Now, as in the proof of Theorem 7.18, >> define >> nu(E) = m(g(E)), >> where m is Lebesgue measure. Since g is stictly increasing it's >> easy to verify that nu is a positive measure. Since g is >> left-continuous it follows that >> nu([x,y)) = g(y) - g(x). This is not right: Let f(x) be the Heaviside function (0 for x<=0, 1 >for >x>0). Then >nu([-1,1))=2, while g(1)-g(-1)=3. > Hmm. Good point, sorry. > Let's see. What I said about using the Stieltjes integral and the Riesz Representation Theorem certainly works; that's a standard thing, the way I've always thought of this. Of course that can't be what Rudin had in mind for the exercise, since I doubt he's assuming the reader knows about the Stieltjes integral. > Maybe what he had in mind was something inspired by the proof of that theorem in the book, but not such a literal copy. For example, perhaps we could say that for any x, S_x is the closed interval [g(x-), g(x+)], and define nu(E) to be m(g*(E)), where g*(E) is the union of S_x for x in E? That's the same as above if f is continuous... > I bet that works. If x <> y then S_x and S_y are disjoint, and that means that E -> g*(E) preserves unions and intersections. >>A problem with this approach is that g*(E) may not be measurable. Even >>if >>f is continuous, f(E) may not be measurable, unless f is absolutely >>continuous. It's certainly true that g*(E) need not be Lebesgue measurable > for Lebesgue measurable E. That's not a problem - we're just > trying to construct a _Borel_ measure. (In fact, come to think > of it, if g*(E) _were_ Lebesgue measurable for every Lebesgue > measurable E then _that_ would show the argument must be > wrong, because it would imply that every Borel measure > extends to a measure on the Lebesgue-measurable sets, > which is not so.) I gave a hint how to show that g*(E) is Borel for every Borel > set E. In a little more detail: Let A be the class of all E such that g*(E) is Borel. > Show that every interval is in A and that A is a > sigma-algebra. > I think you are right. === Subject: Re: The Unity Problem > Clean sweep. I preapologize. This whole model is premised on the physical space analysis that has > been going on here for me for some time. > I have already rejected the real numbers as lacking physical > correspondence. > continuum. > We find that by instantiating a distance we recover most of the > properties of the old representation. << We observe the peculiar nature of this physical continuum. OK I see this: http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html > We keep refining our reference distance down to a smaller and smaller > value. No... we don't have that long. We can divide by 2 forever. This is not a method that we practically use. Pricision becomes > problemmatic. > This theory makes a supposition. It is entirely based upon that > supposition. > Your suggestion of dividing by two breaks that supposition. You didn't dispute that ISL is valid ~property~ of empty space. > But you didn't offer any other properties either. > Are you saying ISL is invalid because I can't measure it? I don't think this breaks /r/r (ISL) behavior. > I am trying to consider general dimension continuum as well. > Let's not forget there is no electromagnetism even yet in this > construction. > Just delta's which are so far very small drab things. > They are like photons but more primitive. > It's true that we can't measure the fundamental frequency, at least not > yet and not directly. > Only when it coalesces into lower frequency phenomena do we see it as > light. > These things happen naturally because of the thermodynamic effect of > you consider rotational frames. > There is polarization to worry about also. This has to be accounted > for. Can deltas be little bars of legnth 'a' hurling and rotating? > That's a lot of geometry from a primitive. I can't fill all of this out > right now. Just have to take it as it comes. Skepticism requires that > we break this model efficiently. The energy scenario is the worst part. > Still, others have already stepped on this threshhold. This theory is > almost perfectly in line with > http://chaos.fullerton.edu/~jimw/general/inertia/index.htm > Woodward's perspective is almost exactly what I am talking about here. > The concern over radiation I have regenerated under a new principle. > Inertia may be called a side-effect of this system. If so... it is 377 ohms for the space we live in. Currently the meter is a discrete count of an atomic radiation pattern. You lost me the eleventh CGPM defined the metre in the new SI system as equal to > 1,650,763.73 wavelengths of the orange-red emission line in the > electromagnetic spectrum of the krypton-86 atom in a vacuum. That is metrology, not physics. Can you dispute ISL (1/r^2) > Can you dispute 377 ohms. > Can you think of any other properties of empty space? Note: 377 ohms is based on the observation that EM > signals never come back to us. 1/2 of the matter we > need to launch an EM signal is not part of the emitter > but dispersed out in our universe. I had to introduce some matter to say anything beyond > 1/r^2 about empty space so it isn't really empty anymore. > I called it space we live in to make a distinction. > Free-space seems a popular term. http://en.wikipedia.org/wiki/Free_space > http://en.wikipedia.org/wiki/Wave_impedance Again...Can you think of any other properties > of *empty* space other than istropic attenuation (1/r^2) ? I'm not seeing it right now. I am considering these oscillators to go > into higher dimensions though. That could corrupt a simplistic 3D > Just a delta source and so completely generic. This can be simpler. Just think of light as pushing a distant > electron while pulling a distant proton: the net force is zero. Electro-induction, changes the spatial relation of a distant > electron and proton, so the net force is attractive. Both effect exist as components of the same ~light~. > I still don't actually have charges in the model, but if we plop them in and a piece of radiation hits an atom as I reflect your first suggestion above we would see the energy gain in the atom as a dual reaction requiring both electron and proton and taking them as a unit. That's pretty good. Though you would not necessarily have gravity that wasn't really the initial goal of all of this. To get your second part won't we have to address coherence and isotropic gravity? Doesn't this neeed some transient analysis? You can try to just call this a residue, but you will also have the repulsive electrostatic residuals right? How these can cancel to get your 10^-40 factor net affinity is not a done deal is it? The supposition that a smallest distance exists alleviates the infinitessimal and supports a finite existence. As I read over http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/elescat.html#c1 it really seems that the nuclear model holds, though a little bit of math can go a long way by turning things inside out. Anyhow the critical distance is about 1x10^-15 m, the same as the max photon frequency that I found earlier. There are models predicting Tev gamma rays, but the information that I find on actual wavelength measurements here is very thin. Everything becomes approximate so I can take this as a loose form of support. This data so far indicates 'a' to be 1x10^-15 meters; the fermi. Perhaps the electron is the photon's sister in this model. Please check out http://www.newton.dep.anl.gov/askasci/phy00/phy00466.htm are being generated mysteriously. This fits with the static delta radiator oscillator, where a mode of no change is really a mode of constant acceleration. Notice how neatly they defer any maximum known measured photon wavelength from this decisive statement. You'd think a serious response would at least include that figure. Something goes amiss up at this high frequency end. The cosmic rays he speaks of are not photons, though he loosely couples them linguistically. If a photon were detected at 10^-16 meters then the nicety pointed out above goes away. I'm having a hard time finding this data. This is your chance to break my theory!!! We've been ignoring blue-shifting and red-shifting and all of the relativistic possibilities. I wonder if there is a neat short argument for a gamma ray burst. The dynamics of oscillating modes is daunting. The easiest thing that I can coherence, but getting to that state would be highly unlikely in free experience a pressure and an attenuation of thier motion. Just two right simultaneously. They accelerate left then right, but as the other stifled. The pair have become glued together and are emmitting as if Is their inertial mass halved as well? Their motion does appear to be attenuated from what it would be as independents. They seem to be quantitatively less than what they are individually. Normally we think of a coherent superposition as being a reinforced state that effectively doubles everything. This is not the case here since 'a' must be maintained. Maybe the simplest disproof of a simplistic electrostatic gravitation model is a three body system: three equal masses in free space at zero velocity and colinear with symmetrical seperation: A-----B-----C If B is to pull A toward it due to an electrostatic residual then B's influence on C must be an inverse electrostatic residual. This is a symmetry argument. Since the construction is perfectly symmetrical the result ought to be too. Please assign the differential charge to these bodies that will explain their gravitational behavior. -Tim > Combining this model with an n-pole seems the obvious thing to do. > If n-poles are the right mix then the light electron may still be a > 1-pole; it has no receptors. > This junction will need some work. It doesn't immediately look clean. > Simply calling electrons 0D makes more sense, then photons 1D, then > protons 2D. I'm not settled on this portion. There is plenty of room to > adapt the new principle of radiation to many contexts. Physical > correspondence will be the test. Sue... > -Tim minimum matter we can introduce to explain 377 ohms aren't adding anything to my view of the problem you perceive. Sue... > - from Wikepedia Sorry, it's not quite discrete count, but awfully close. I guess I was > thinking cesium and the time base: This frequency is the natural resonance frequency of the cesium atom > (9,192,631,770 Hz), or the frequency used to define the second. > - from http://tf.nist.gov/cesium/fountain.htm Why the two are not unified at this point? I suppose to preserve the > meter. > Sue... Observing this and seeing the necessity of a reference distance to > obtain a finite measure we make a suppositional leap: that a smallest > deistance exists in this continuum. This means that the concept of a > zero distance is nonexistent. When we take pieces of matter and move > them about we find that they will get no closer than 'a'. When we > radiate energy we find that that there is no wavelength less than 'a'. > When we look in time we find that there is no time less than 'a'. The > duration of a delta of length 'a' is 'a'. In effect we have created a > basis out of thin air from this suppositional 'a'. Under this construction the notion of instantaneous is nonexistent. The > can get is 'a'. The shortest wavelength of radiation is 'a'. This wavelength of radiation( going along the lines that delta is a > photon ) is the shortest possible and has the highest energy possible. momentum we will have to balance this radiation. It will have to > balance perfectly. So the radiational oscillator is born. > > -Tim