mm-1048 === Subject: Re: Attacking my own math proof, fun >But first I need you to acknowledge some *basic* facts. If you >disagree, or if anyone else disagrees, they need to step forward. And yes, from that very interesting assumption that you can force a >variable dependency on a constant factor of a polynomial I can >hopefully show you that it has no mathematical basis. After all, you can multiply a polynomial by *any* constant factor you >choose. How does your action force variable dependency on something >independent, like what if I multiply by 398234? How could that affect the *factorization*? >>There you go, I didn't step forward. But how is a_3 proven to be coprime to f ? >> > It's proven to be coprime by checking at m=0, as the constant f^2 > provably divides off the same way without regard to m. > a_3(0) is coprime to f^2... ok. Since that's obvious enough, you must *believe* that possibly it > divides off differently dependent on the value of m, which is a false > assumption. > Why? Others have shown quite clearly that it does. Well then, if that is true, then wouldn't it *still* be true if f=3? However, it turns out that with f=3 NONE of the a's are ever coprime to f, for any m, not even m=0. If you're an ethical mathematician, then it's over. If you're a liar, or not even a mathematician, you may wish to continue to debate. For readers, yes, that's mathematics. The poster is checkmated that quickly. However, because it's Usenet, the poster may simply post as if it didn't happen. That assumption is provably false by letting f=3, as then ALL of the > a's have at least a factor of 3 that is 3^{2/3} without regard to m, > so they have that factor at m=0. > Special cases do NOT prove the general case. That proves the general case, unless you expect readers to believe that algebraic equations first check a variable value, like to see if it's 3 or not, before deciding to be functions. Hmmm...maybe you do expect readers to believe such a thing. However, when f does not equal p, a_3 is coprime to f, when m=0, as > I've shown. > Special cases do NOT prove the general case. Obviously the math equations do not decide to be functions or > variables dependent on m after checking the value of f. Either they are dependent on m or they are not. You can't have it both > ways. It's a *constant* so why would it be dependent on m? And in fact it's not, as is easily proven. However, in retrospect, I can see how unethical posters with some math knowledge could confuse people on the issue. James Harris === Subject: Re: Attacking my own math proof, fun But by introducing b_1, b_2 you put yourself in that field. That is the false assumption onto which you've grabbed hold. It's easily proven false. See below... > It seems to me that you've seized on one possibility and now are > holding on to it though it doesn't follow from the math argument. > The actual math argument gives an answer, and not just a guess. a's at m=0, right? So then, if f^2 divides off of > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > without regard to m, then my result applies for all m. > Do you agree? > I think I do. I'm actually though a believer in things like the commutative > associative and distributive laws, so you can actually divide out f^2 any which > way you want. The fact that > it looks nicer when you do it one particular way when m=0, so you want to do it > the same way for all other m, is fine by me. I don't argue with you there. I do > disagree that you have to do it one way and not another, but I think that's a > slightly different topic than the one here. Well, it turns out that each of the a's does have a constant factor in common with the constant factor f^2, where for one of them it's a unit factor. The point is that the expression f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) is NOT a polynomial, and I call it an uber-polynomial. Given a polynomial and a constant factor, like 4(x^2 + 3x + 2) the factorization is not constrained. But with my uber-polynomial and the factorization f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) there IS a constraint on how the a's must divide out, which may seem odd. For those readers who might not understand the ambiguity with a normal polynomial consider the factorizations 4(x^2 + 3x + 2) = (2x + 2)(2x+4) = (4x+4)(x+2) which are just a couple out of an infinite number. Splitting up 4 as prime integers shortens the number but leaves ambiguity. > For example, > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m), > so > b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m), > might let you mistakenly think that a_3 must be coprime to f. > This doesn't follow. > It might not seem to follow to you if you *believe* that the f^2 > divides off of a_1 a_2 a_3 one way for some m and a different way for > different m. > See above. I'd say we have our wires crossed at this point. Let's try to get > something out of > it though. My guess is that you're holding on to an intuitive feel from polynomials. they're non zero, > you leave the world of algebraic integers behind. So I'm really not clear with > how f^2 divides off. But as I said, as far as I can see, you can divide it off > how you prefer. Actually, no, as there's *provably* only one way f^2 divides off. I've shown. > Let's see - at m=0 a_3=3 . And f is an integer coprime to 3. Yep, you're correct > when m=0 Yes. So now you can see what one of the a's is, which shows you how the f^2 *does* divide off, once you consider the constant term P(0) of P(m). That should be it, unless you believe that somehow the dividing off of the *constant* f^2 has a dependency on m. That's the point that I've been making for several posts now. > Obviously the math equations do not decide to be functions or > variables dependent on m after checking the value of f. > James Harris > I wouldn't mind doing a sanity check, then restate my problem if you don't mind. Fine. > We have > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > so it's really a cubic polynomial in x. f is an integer coprime to 3. m is any > integer coprime to f. Not it's NOT really a cubic polynomial in x, as it's not a polynomial at all. A polynomial has constant coefficients while it has variable ones. You can consider it to be a polynomial with respect to various factors, which implies that certain variables are constant; however, your assumption is not a mathematical constraint. That's an important point. After all, how could it be? Is the math supposed to read your mind and figure out which point of view you have? Whether or not you're looking at a polynomial with respect to x or m, or even f? > Not sure about u. It's an integer that usually ends up being set to 1 in most > discussions. > So you can pick values for f, m and u, constrained only by their coprimality > requirements, so > you're actually considering an infinite number of poynomials of x. Nope. You keep trying to force the expression into a box, when it can't be so forced. While you can consider it as containing an infinite number of polynomials of x, it also has an infinite number of polynomials of m, or other variables. That's why I call it the uber-polynomial. It's a special construction for a particular purpose. > Doing some algebra, by expanding the right-hand side and equating like values of > x, among > other things you can discover: > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m) > Hmmm. Too much cutting and pasting. > I think you actually get > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) > Never mind, no arguments have changed. > and the a's are the solutions to: > a^3 + 3 (-1 + mf^2)a^2 -f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0 > So they are indeed algebraic integers. Nice construction. At m=0, the solutions > are 0,0,3 Which tells you how f^2 divides off, if you also consider that with P(m) at m=0, you have P(0) = u^2(3x + uf), which shows that a_3 is coprime to f, since f is coprime to 3 and x. That's important, as if f is not coprime to 3 and x, you get a different result, so it matters to check the constant term P(0). > Now considering Q(m) = P(m)/f^2 > Q(m) = m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > = (b_1 x + u)(b_2 x + u)(a_3x + uf) > and you get > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m > and the b's are two of the solutions to: > fb^3 + 3 ( -1 + mf^2)b^2 -m^3 f^4 - 3m^2 f^2 + 3m = 0 > In general then, the b's are not algebraic integers as m and f are coprime. You're making leaps. If you let f=sqrt(2), m=1, you'll see that two of the b's are algebraic integers, while one is not for your cubic. > Is this sanity check correct? Please fix it if not. The cubics are actually > derived quite easily using your substitution of v = -1 + m^2 They look ok to me. > As a side issue, the 2 b's you want are the 2 that are equal to 0 at m=0. > Probably if you solved these cubics, these 2 b's might be evident. As side point > to consider, what the hell is the 3rd b and where did it come from? One of the b's is a_3. > Now onto the disagreement we have: > You claim that because > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m > and RHS is coprime to f (which it is as m and f are coprime) > then a_3 is coprime to f > Note that b_1 and b_2 are NOT algebraic integers. > Not by any theorem of mathematics is your claim correct. You're making several false assumptions, and it's easy to refute your position by having you consider f=sqrt(2), m=1. Clearly, you've seized on one idea, and you keep holding on to it, despite my efforts to get you to follow the math. > Except at m=0 when b_1 and b_2 are =0 and are algebraic integers. Well, try m=1 with f=sqrt(2), and welcome to a more complicated mathematical world than you might have realized. James Harris === Subject: Re: Big Bertha Thing blogs Big Bertha Thing progress Cosmic Ray Series Possible Real World System Constructs http://web.onetel.com/~tonylance/progress.html Access page to 6K Web page Astrophysics net ring access site Newsgroup Reviews including uk.transport The Progress of Discontent by T. Warton (Written at Oxford in 1746) Half-hours With the Best Authors The Chandos Classics Edited by Charles Knight Published by Frederick Warne and Co. 1890 Big Bertha Thing besides Some people think that Big Bertha is less suitable for the Have they heard the long name for CP Conf? Perspective. They think that is a better place to put Big Bertha. They must be besides themselves, which proves the thing. (C) Copyright Tony Lance 1998 To comply with my copyright, please distribute complete and free of charge. Tony Lance pobox47@big-bertha-thing.com Big Bertha Thing NASA II Hi James, Further to your above link, you have missed one or two sub-plots:- 1. Case documentation has been published on my web site. 2. My nbci web site has been closed. 3. You did not do this. 4. The opposition did it. (3rd Battle of Cyberspace) 5. You do not believe the opposition exists. 6. All Big Bertha Thing postings on that Saturday had the same prefix. 7. NASA would tell you to shut up. 9. There is no problem my end. 10. A rocket scientist counts double. 11. John Prescott, Minister of the Crown counts double. 12. Ken Livingstone, Mayor of Greater London counts double. 13. Both have appointed directors to LPFA. (see mayor) http://web.onetel.com/~tonylance/mayor.html 14. Pensions Ombudsman's judgements cannot be enforced on these directors. 15. Magna Carta does not apply; they are above the law. 16. It is feudal personal political patronage; all opticians to a man. Tony Lance pobox47@big-bertha-thing.com http://users.mweb.co.za/d/da/dalen/tclockex.htm Keep site owners from using too much of your computer systems resources, with free software. Systems info:- dd/MM/yy h':'m'Sy'S'CP'C'MM'L When Sys is equal to zero, then windows crashes due to shortage of resources. Use Ctrl Alt Del to restart. === Subject: Re: Constant factors and polynomials This is actually a reply to a post by William Hale, who was replying to my post: >[cut] >> Your statement that >> ... there's no reason to believe it varies >> with m, or that it cares if the polynomial is irreducible >> over Q >> is one of the weakest you have ever put forward. First, >> saying there's no reason to believe it is one of that lamest >> non-proofs I have seen. What you are really saying is, >> '*I* can't think of any reason to believe it.' Second, justifying a >> conclusion on the basis that it cares whether a polynomial is >> irreducible - this is truly new methodology. You are basing an >> argument on the psychology of some utterly fictional entity. In fact >> REDUCIBILITY MATTERS - that is, conclusions regarding irreducible >> polynomials are *qualitatively different* from conclusions >> regarding reducible ones. >Here's is my point of view of his statement you quoted. >James Harris want to prove a statement about the factors of >a polynomial of a particular general form. Offhand, one >might suppose that the statement is true for some such >polynomials and false for other such polynomials. It could >happen that the statement is false for random particular >polynomials. Or, it may happen that the statement is >false for certain polynomials that satisfy some particular >property (like, being irreducible or having prime degree or >something else). This latter case does not exclude the >statement being false for other random particular polynomials >or being false for some other particular property. Agreed. >James wants to prove that his statement is true for all >such polynomials. Until I have agreed that his proof >is correct, I can't rule out that his statement cares >whether the polynomial is irreducicble or whether its >degree is prime or whether some other property holds >true. I think when he talks about something that cares about such, he means that the math is what cares or doesn't care. But your interpretation could be right too. >In fact, some additional property on such >polynomials might be enough for me to show that >his statement is false (under the additional assumed >property, like irreducible). >If his statement can't care whether the polynomials have >additional properties, then of course his statement that >he is trying to prove would be true. But, I don't know >that a priori. Right. >In fact, if I doubt his statement is true, I might try >to find a counterexample. One way is to randomly >try various polynomials to see if his statement is >false. Of course, this could take a long time. It won't. 'Most' polynomials with integer coefficients are irreducible. Here I am being informal about the probability statement 'Most'. If, for example, you randomly generated degree 3 polynomials with integer coefficients randomly selected between 1 and 1000, the majority will be irreducible. Computation of factors of the roots could be tedious however. >A better >way is to try to analyse the statement into various >cases (eg, irreducible or reducible; or prime degree >not prime degree). Indeed, with the case of irreducible, >I have enough additional properties to show that >his statement is always false for polynomials satisfying >that property (of being irreducible). Yes - several proofs of that have now been given. >Note, that this >does not say that the statement is true for all other >such polynomials. For reducible polynomials of such >a form, the statement still may be true always, or >may be false always, or may be sometimes true or >sometimes false. Since I am only interested in finding >at least one counterexample, I need not worry about >analysing the reducible case. I agree. The case he is interested in for his FLT proof is the irreducible case. Our counterproofs explicitly mention 'irreducible' as an assumption. He keeps trying to say that the proofs are wrong because he can find *reducible* polynomials where the conclusion is not true. To which the right response is, Duh? Why do you think we included irreducible as an assumption? Here is the germ of his thinking. He notes that we say, in the irreducible case, that each root has algebraic integer factors in common with each prime divisor of the constant term. For polynomials of 3rd degree or higher, this is hard to see explicitly. He cannot follow our arguments and cannot do the computations for himself, so he assumes we are lying or wrong. He CAN do the computations in some reducible cases, and they do not agree with what we say in the irreducible cases. He then says, 'How can the math care whether a given polynomial is irreducible?' and concludes (using faulty intuition) that it cannot, and therefore all the more, we must be lying. So I gave quadratic examples where he might possibly be able to do the computations: one reducible and one irreducible, and (as expected!) they have very different behavior with respect to factors of the roots. He will either ignore this or deny it, or say the quadratic case is irrelevant (even though his arguments if valid would apply equally well there). He knows he is at the point now where he cannot give an inch on this, or he loses everything. He is remarkably creative in throwing out bogus and irrelevant arguments. He makes you realize that for most problems, there are only a very few right solutions, but infinitely many wrong ones. His mathematical intuition is actually quite bad, and he comes up with wrong ideas in an incredibly high percentage of cases. But eventually he is going to run out of even remotely plausible options on this. Nora B. >-- Bill Hale === Subject: Re: Constant factors and polynomials >It occurred to me that some of you may be hampered in understanding >certain math arguments of mine because > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) >has that constant factor of f^2. >Normally when considering factorizations, you separate off constant >factors, as otherwise you don't have a unique factorization even with >polynomial factors. >For instance > > 4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4) >along with an infinity of other factorizations, but typically you'd >just have > 4(x^2 + 2x + 1) = 4(x+1)(x+1). >Besides all that the expression I use is rather imposing, and it has a >lot of symbols, so I thought I'd remind you of a few things. >1. You *can* look at an actual example with m=1, f=sqrt(2), as then >all that complexity drops away and you have > P(1) = 2x^3 - 3x + 1 >which actually does reduce over Q. >Some of you may have realized that you can consider m=1(mod sqrt(2)) >to blow apart several assertions made by some posters. >2. A requirement I give is that f be coprime to 3, but letting f=3, >you get that *each* of the a's in the factorization > 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - > 3(-1+m3^2 )x u^2 + u^3 3) = > (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u) >has a non-unit factor in common with 3, which is a radical factor of >3, and there's no reason to believe it varies with m, or that it cares >if the polynomial is irreducible over Q. >That actually destroys >several claims made about using Galois Theory where reducibility over >rationals is an issue. > Not even close. First, in this case, f factors out of your > polynomial 3 times, not 2 times as in the cases you were > considering (f <> 3). This is a special case of no interest > to you or me. It is irrelevant. The Galois argument does not > apply here. No claims based on that argument are 'destroyed'. > Second, only one of the proofs that you are wrong in the cases > where f <> 3 is dependent on Galois Theory. The other proofs are > based on an elementary theorem from algebraic number theory, > which you have previously accepted. All of the proofs *do* > require irreducibility of P(x)/f^2. Well, it IS the case that for f=sqrt(2) only *two* of the a's have a factor that is sqrt(2), so your claim that it is otherwise is false. Readers can easily verify by considering P(1) = 2x^3 -3xy^2 + y^3, where y=uf, using f=sqrt(2), m=1, with P(m) given above, as it is reducible. Then consider that using m=1(mod sqrt(2)) then destroys any claims that reducibility is the issue, as, for instance, m=3 is NOT reducible. Now if Nora Baron were an ethical mathematician, there would be no debate. But it's not even clear who Nora Baron is, as was pointed out by another poster the name itself is a palindrome. That is, it's the same reversed as Nora B-aron. The actual poster may not even be female. > Your statement that > ... there's no reason to believe it varies > with m, or that it cares if the polynomial is irreducible > over Q > is one of the weakest you have ever put forward. First, > saying there's no reason to believe it is one of that lamest > non-proofs I have seen. What you are really saying is, > '*I* can't think of any reason to believe it.' Second, justifying a > conclusion on the basis that it cares whether a polynomial is > irreducible - this is truly new methodology. You are basing an > argument on the psychology of some utterly fictional entity. In fact > REDUCIBILITY MATTERS - that is, conclusions regarding irreducible > polynomials are *qualitatively different* from conclusions > regarding reducible ones. > Here is an example: > x^2 -5*x + 6 = 0 > has constant term 6 = 2*3. The roots are 2 and 3. One root is > coprime to 3 and the other is coprime to 2. > But > x^2 -3*x + 6 = 0 > also has two roots. They are: > r1 = (3 + sqrt(-15))/2 and > r2 = (3 - sqrt(-15))/2. > You can check (by computing a 4th degree polynomial) > that: > NEITHER r1 nor r2 is coprime to 2, and > NEITHER r1 nor r2 is coprime to 3. > Thus: REDUCIBILITY MATTERS: note that: > x^2 - 5*x + 6 is reducible over Q, and > x^2 - 3*x + 6 is irreducible over Q. > It's not a coincidence. That was hand-waving. In fact, your position requires that people believe that given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) that some f^2 divides off as some function of m, or variable dependent on m. What I noted is the oddity of believing that a constant factor of a polynomial, which f^2 is, could factor off as a function. It seems unlikely to me that a math expert would make that mistake. And in fact in this case it's rather easy to prove that belief is false by letting f=3, as then ALL of the a's have a factor of 3 that is 3^{2/3}. >What I want you to understand is that for trained mathematicians, >these are not issues. However, when it comes to confusing people >about even relatively basic mathematics, who would be better at it >than mathematicians? >They need to confuse you here for *social* reasons. > This is not true. All we want is that the truth > emerges. Social consequences are irrelevant and > unimportant. You, in contrast, are desperately seeking > social recognition for your proof. As has happened > in the past, we are dragging you kicking and screaming > into the realization that your proof is incorrect. Then why are you ignoring the actual math argument that I repeatedly give? Besides, why would you care? There are any number of posters all over Usenet with various notions. It makes more sense that realizing my work is correct, and fearing the social consequences of acknowledging the truth, people like yourself Nora Baron bank on your ability to confuse others about the mathematics. The more I simplify, the more you desperately work to confuse, which I think explains your recent postings. > Although social reasons - recognition, especially - are > paramount for you, you are losing the battle. That is > why you desperately keep starting new threads and you refuse > to answer rigorous arguments which prove you are wrong. Then why bother continuing to reply? In actuality, you are lying. You have not given valid math arguments but instead have given bogus ones. However, you have relied on using math where you can hide the fact that it is invalid by various tricks, mostly banking on people not checking thoroughly. Faced with your efforts, and the support you get from other poster helping you to try and obscure the truth, I keep bringing focus back to the actual math argument I've presented. >Notice that with f=3, the constant term P(0) = u^2(3x + 3u) = >3u^2(x+u), so it *still* has a factor that is 3, and that's why I >always have the condition that f be coprime to 3. >Here, however, I'm hoping it'll help to point out why that requirement >is there, and what happens if you ignore it. >Well then, what are some posters trying to convince you about > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f)? >They're trying to convince you that there is a mathematical limitation >based on reducibility over Q that determines how f^2 can divide >through when you have the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >and the first question that should come to you is, how could a >constant factor be constrained by reducibility over rationals? > See above: REDUCIBILITY MATTERS greatly and is central to > the disagreement here. It is reasonable to ask, as you do, how > reducibility might constrain factoring. It is NOT reasonable to assume > with no hint of proof that it doesn't. Our counterproofs need either > a basic theorem from Galois theory (which *requires* irreducibility), or > a basic theorem from algebraic number theory (which *also requires* > irreducibility). If you really want to understand why, you need to > study the proofs of those theorems. However the quadratic examples I > gave above should be sufficient to show exactly how reducibility can > affect the form of factors of roots. Yet I can give f=sqrt(2), with m=1, and get a reducible polynomial, which you already made a false statement about in this post, where only two of the a's have a factor that is sqrt(2). Then I can simply use m=1(mod sqrt(2)) for a family of results where most are irreducible. Now your claim would require that suddenly, against mathematical logic, the single one of the a's that doesn't have a non-unit factor in common with sqrt(2) gets one. Faced with the destruction of your argument, you side-stepped it, which diligent readers can see at the start of this post. >Now that question is resolvable, but I know the answer is in my favor, >so mathematicians are avoiding even letting you know that IS the >question, and instead those who post work to confuse. >Now given that I know I have a short proof of Fermat's Last Theorem, >and that mathematicians have been avoiding dealing with reality, while >some posters have gotten away with *deliberately* confusing people, >why would I quit talking about my proof of FLT? >If you'd found a short proof of Fermat's Last Theorem, would you quit >talking about it? > No: *** If ***. But if other people had presented rigorous > proofs that I was wrong, I would try very hard to > dismantle those proofs. You have made no effort to do that, > preferring instead to bring up irrelevant issues as in the > present thread. The bottom line here is, whether you like it > or not, reducibility DOES matter. If a ridiculous argument based > on whether it (math) doesn't care about reducibility is all > you have, you had better give up mathematics. > Nora B. So now even though I have a proof of my own, you claim that I should have to work to show falsity in alternate proofs given by people who have proven a need to deceive, who have a long record of deception, who may be actual mathematicians, i.e. math experts, with a willingness to use their knowledge to confuse others? Here readers should consider that Nora Baron is probably a pseudonym that the poster believes is clever because it's a palindrome, who may not even be a woman. They can also consider the clear dodge of an issue brought up against the poster's claims with f=sqrt(2), m=1. But best of all, readers can consider that *mathematics* gives them their best way to figure out what's going on, which is to consider the argument Nora Baron keeps working to obscure. Consider P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f. Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, as f is coprime to 3 from before, which leaves b_3 = 3. Essentially objections now come down to claiming that the w's are dependent on m, but consider that w_1 w_2 = 1, when m=0, here where f is coprime to 3. But that was an arbitrary choice *I* made, so let f=3. Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO m. So those posters who try to convince you that the w's are actually dependent on m, like being functions of m, must now also convince you that the w's make a decision, first looking to see if f=3 or have some non-unit factor in common with 3, and THEN they decide if they're dependent on m. People can waffle trying to figure out who they are, but mathematics is logical, which is why I've emphasized that posters are acting on *social* not mathematical reasons. James Harris === Subject: constructing an ovoid from N given points It is well known how to construct a circle, given three distinct points. Let's generalize this. Say that we have N number of points such that: N >= 3, the points are coplanar, no three points are colinear, and a polygon whose vertices are these points is CONVEX. Now suppose that we connect the points with a set of smooth curves (instead of line segments), thus forming an ovoid. I'm guessing that the ovoid could be an ellipse for N = 4 or maybe 5, but not necessarily so for N >= 6. (That is: given any four points that meet the conditions above, those points determine an ellipse.) There is probably be more than one way to come up with a smooth curve, of course, and not all of them lead to ellipses. If we manage to abide by a consistent method, our ovoids will all share certain features. Thus, we would have a generalization that begins something like this: circle --> ellipse --> ??? --> ... or circle --> other shape --> ??? --> ... Can someone point me in a good direction for more about this? I don't care whether the study is in the direction of geometry or the algebra equations, or something else I hadn't thought of. (A similar question appeared on this discussion group some time ago, but Google doesn't show any replies to it. If you replied to the Ted Shoemaker shoemakerted@yahoo.com === Subject: converse of the theorem of lagrange?? if G is abelian group, converse of the theorem of lagrange is true. this is true ? false ? please, reply to the problem. sir === Subject: Re: converse of the theorem of lagrange?? I guess this means that the theorem, which states that the order of a subgroup divides the order of a finite group, is true the other way around for abelian groups only (ie a finite abelian group of order n has a subgroup of order d, where d divides n. This is NOT true of groups in general (ie non-abelian groups). hth Guy > if G is abelian group, > converse of the theorem of lagrange is true. > this is true ? false ? > please, reply to the problem. sir === Subject: Re: converse of the theorem of lagrange?? > if G is Abelian group, > converse of the theorem of Lagrange is true. Likely you're referring to H subgroup G implies ord H | ord G > this is true ? false ? > please, reply to the problem. sir Arrgg, vaguery. First off clarify finite groups only. Now just what do you mean by converse? If G Abelian group then ord H | ord G implies H subgroup G ? Naw, false. If G Abelian group and H group then ord H | ord G implies H subgroup G ? Naw, still false. If G Abelian group and H Abelian group then ord H | ord G implies H subgroup G ? Naw, still false. If ord H = ord G, would H be subgroup G ? If G Abelian group and H Abelian group then ord H | ord G implies H embeds G ? By embeds I mean an isomorphic image of H is a subgroup of G. But you didn't ask that question, or did you? Naw, still false. Look for a counterexample. === Subject: Re: converse of the theorem of lagrange?? Although it is not immediately obvious what `the converse of Lagrange's Theorem' might mean, the expression has acquired the following meaning among group theorists. A finite group G satisfies the converse of Lagrange's Theorem if and only if, for each divisor n of |G|, G has a subgroup of order n. That statement is indeed true for abelian groups. Since the OP did not request a proof, I will not supply one! >I guess this means that the theorem, which states that the order of a >subgroup divides the order of a finite group, is true the other way around >for abelian groups only (ie a finite abelian group of order n has a >subgroup of order d, where d divides n. You appear to have said that the converse of Lagrange's theorem is true for abelian groups only. Maybe you did not mean that, but it is not true, anyway. For example the nonabelian group of order 6, and all finite groups of prime power order satisfy the converse of Lagrange's Theorem. Derek Holt. > This is NOT true of groups in >general (ie non-abelian groups). >hth >Guy >> if G is abelian group, >> converse of the theorem of lagrange is true. >> this is true ? false ? >> please, reply to the problem. sir === Subject: Re: converse of the theorem of lagrange?? > if G is abelian group, > converse of the theorem of lagrange is true. > this is true ? false ? > please, reply to the problem. sir Technically the converse of Lagrange would read something like |H| divides |G| imples H < G ... If quantified over all sets then this is not true in general for abelian groups. Thus in texts you often see them say things like Sylow's 1st theorem provides something like a converse to Lagrange. The converse in this case looks like if n divides |G| then there exists a subgroup of order n. This version of the converse is true if G is Abelian. One way to prove it would be to use the fundamental theorem of abelian groups. Hugh === Subject: Re: converse of the theorem of lagrange?? |Technically the converse of Lagrange would read something |like |H| divides |G| imples H < G ... If quantified over |all sets then this is not true in general for abelian |groups. The converse of a statement of the form A->B is B->A, but there are multiple ways of writing Lagrange's theorem as A->B, which give you different converses. Saying that if H is a subgroup of G then the order of H divides the order of G, is equivalent to saying that if there exists a subgroup of G of order n, then n divides the order of G. People discuss whichever converse seems most natural, and I seem to remember hearing conversations where more than one converse of a given result was considered appropriate. Keith Ramsay === Subject: Re: converse of the theorem of lagrange?? > The converse of a statement of the form A->B is B->A, but there > are multiple ways of writing Lagrange's theorem as A->B, which > give you different converses. Saying that if H is a subgroup of > G then the order of H divides the order of G, is equivalent to saying > that if there exists a subgroup of G of order n, then n divides the > order of G. People discuss whichever converse seems most natural, > and I seem to remember hearing conversations where more than > one converse of a given result was considered appropriate. Admittedly, when I said that one sometimes sees texts say, Sylow's 1st theorem provides something like a converse to Lagrange., I was basing this on my (possibly faulty) memory. Looking at Dummitt and Foote, they actually say it provides a partial converse (ie the converse is true if you add some hypotheses). Of course, the converse they refer to is based on the version of Lagrange you gave. Hugh === Subject: Re: Counterexample needed >> In my algebra textbook, the product of two ideals I,J is defined as >> { sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J } >> Now it is rather easy to prove that IJ is an ideal in R. The last >> question of the exercise is: >> Is A = { ab | a in I, b in J } an ideal of R. >> Now the preceding questions strongly suggest that the answer in >> general is no, but I can't find a counterexample. Clearly, (since it >> is understood that R is commutative), if one of I or J is a principal >> ideal, the set A is an ideal, so a counterexample has to consist of a >> non-PID and two ideals generated by at least two elements each [...] > HINT: Find proper ideals whose product contains an irreducible element, > e.g. p in (p,a)(p,b) if (a,b) = (1) > Examples abound. Domains where ideals multiply simply as IJ = { ij : i in I, j in J }, are called condensed domains. Below are reviews of related papers. ---------------------------------------------------------------------------- -- 84a:13019 13F99 Anderson, David F.; Dobbs, David E. On the product of ideals. Canad. Math. Bull. 26 (1983), no. 1, 106-114. ---------------------------------------------------------------------------- -- In this paper the authors define an integral domain R to be a condensed domain provided IJ = {ij: i in I, j in J} for all ideals I and J of R. Bezout domains are condensed domains. The main results of the paper characterize condensed domains within some large class of domains. For example, it is shown that a GCD-domain is condensed if and only if it is a Bezout domain, and a Krull domain is condensed if and only if it is a principal ideal domain. For a Noetherian domain R to be condensed it is necessary that dim R <= 1 and that the integral closure of R be a principal ideal. Reviewed by J. T. Arnold ---------------------------------------------------------------------------- -- 86h:13017 13F05 (13B20 13G05) Anderson, David F.(1-TN); Arnold, Jimmy T.(1-VAPI); Dobbs, David E.(1-TN) Integrally closed condensed domains are Bezout. Canad. Math. Bull. 28 (1985), no. 1, 98-102. ---------------------------------------------------------------------------- -- An integral domain R is termed quasicondensed if I^n = {i_1 i_2...i_n : i_j in I for 1 <= j <= n} for each positive integer n and each two-generated ideal I = (a,b) of R. R is said to be condensed if IJ = {ij: i in I, j in J} for all ideals I and J of R. The main theorem shows that an integral domain is a Bezout domain if and only if it is integrally closed and condensed. An example (a D+M construction) is given of an integrally closed quasicondensed domain which is not a Bezout domain. Reviewed by Anne Grams ---------------------------------------------------------------------------- -- 90e:13019 13F30 (13B20 13G05) Gottlieb, Christian (S-STOC) On condensed Noetherian domains whose integral closures are discrete valuation rings. Canad. Math. Bull. 32 (1989), no. 2, 166-168. ---------------------------------------------------------------------------- -- Following D. F. Anderson and the reviewer [same journal 26 (1983), no. 1, 106-114; MR 84a:13019] an integral domain R is said to be condensed in case IJ = {ij : i in I, j in J} for all ideals I,J of R. The author defines an integral domain R to be strongly condensed if for every pair I,J of ideals of R, either IJ = aJ for some a in I or IJ = Ib for some b in J. Suppose henceforth that R is a Noetherian integral domain whose integral closure R' is a discrete valuation ring. It is proved that if R is condensed, then R contains an element of value 2 (in the associated discrete rank 1 valuation). It is not known whether the converse holds, nor whether all condensed domains are strongly condensed. As a partial converse, it is proved that R is strongly condensed under the following conditions: (R',M') is a finitely generated R-module, R'/M' is isomorphic to R/M and R contains an element of value 2. Reviewed by David E. Dobbs ---------------------------------------------------------------------------- -- 1 955 608 13A15 (13Bxx) Anderson, D. D.; Dumitrescu, Tiberiu Condensed domains. http://journals.cms.math.ca/cgi-bin/vault/view/anderson8107 ---------------------------------------------------------------------------- -- Abstract: An integral domain D with identity is condensed (resp., strongly condensed) if for each pair of ideals I,J of D, IJ = {ij : i in I, j in J} (resp., IJ = iJ for some i in I or IJ = Ij for some j in J). We show that for a Noetherian domain D, D is condensed if and only if Pic(D) = 0 and D is locally condensed, while a local domain is strongly condensed if and only if it has the two-generator property. An integrally closed domain D is strongly condensed if and only if D is a Bezout generalized Dedekind domain with at most one maximal ideal of height greater than one. We give a number of equivalencies for a local domain with finite integral closure to be strongly condensed. Finally, we show that for a field extension k < K, the domain D = k + XK[[X]] is condensed if and only if [K:k] <= 2 or [K:k] = 3 and each degree-two polynomial in k[X] splits over k, while D is strongly condensed if and only if [K:k] <= 2. === Subject: Counting holes of the union of two polygons Situation: ---------- Let P and Q be simple polygons. Assume, they intersect in a non-degenerate way, i.e. no vertex of one polygon is on the border of the other. Idea: ----- I'm interested in a lower bound on the number of holes in P cup Q. I want to get it by counting holes coming from parts of the borders of P and Q. Formally: --------- Split the border of P into disjoint polylines p_i in a way that all p_i are either convex or concave and the total change of direction when walking along each p_i is less than pi. Split the border of Q into q_j the same way. (I don't know whether all these conditions on p_i and q_j are necessary. But this is what I have.) Let n_{ij} be the number of enclosed regions between p_i and q_j that are on the outer side (concerning P or Q, respectively) of the bounding parts of p_i and q_j Question: --------- Prove or disprove P cup Q has at least sum_{ij} n_{ij} holes. I know: ------- Every hole counting for n_{ij} contains a hole of P cup Q. (That's easy because a point close to an intersection of p_i and q_j cannot be covered by P cap Q.) However the same hole of P cup Q might be enclosed by several (p_i,q_j). I hope that it's possible to understand my question though several exact defintions are missing. I will ask more precisely and/or modify the question if necessary. Any help will be appreciated. === Subject: Re: Counting holes of the union of two polygons >(That's easy because a point close to an intersection of p_i >and q_j cannot be covered by P cap Q.) This should be P cup Q. Sorry. === Subject: Re: Criss-crossed Dodecagon: (better?) puzzle > ...start with a dodecagon (12-gon) [...] make a path that moves > from vertex to vertex [...] visiting each vertex exactly one time. > And the path returns to the starting-point. But in this puzzle, > consecutive vertexes MAY be connected by a segment. > So, I give a list of nonnegative integers below. As the path is drawn > (as opposed to after the path is completed), the n_th segment crosses > a(n) previously drawn segments, where a(n) is the n_th term of the > integer-list. > The path starts at 12. And the first segment goes from 12 to 8. > The list {a(n)}: 0, 0, 1, 0, 2, 2, 0, 3, 2, 2, 5, 2 ... > I tried (but not hard) to find an alternative path using this list. > But to do so seemed somewhat difficult. So this might be an > interesting puzzle. ... 17 of the 10! paths that start off {12, 8} match that crossing- count sequence, so solutions seem fairly rare - about 1 per 213000 paths - but on the other hand, there are perhaps O((n-3)!) possible crossing-count sequences, so 17 could instead be an unusually high prevalence, and this case no more rare than thousands of others. I plan to look at this more and find out, next weekend. -jiw -- See even more 17's at http://www.vinc17.org/yp17_eng.html === Subject: Re: CURVATURE OF EARTH SURFACE PER MILE?? In sci.math, ChrisCoaster circle, but beyond that I don't know how to go about it. > Edia = 7,926(miles) or 41,849,280(feet) > Ecirc = 24,900(miles) or 131,472,000(feet) > Eradius = 1/2 dia = 3963miles or 20,924,640feet. > What is formula or procedure to find the answer? I'm learning > disabled so please explain it as to a 10 year old. > Chris If you understand trigonometry, you'll understand this explanation reasonably enough, because that's what's needed to solve this particular problem. Take the Earth's radius to be 6.378 * 10^6 m, or 6,378 km. (Scientists and I prefer metric, for various reasons. This is slightly higher than your value. Feel free to rework the computations if you want to; they work well enough in miles as it turns out but remember to convert the result at the end to feet or inches.) We're assuming a perfectly circular Earth for the purpose of this argument; obviously that's not the case but the errors won't affect the amount much. This means that the circumference of the Earth will be 2 * pi * 6.378 * 10^6 = 4.007 * 10^7 m, or 40,070 km. One mile on the Earth's surface therefore subtends 1.609344 / 6378 radian = 0.0002523 radian. (1 mile = 5280 feet x 12 inches/ft x 0.0254 m/in x 1/1000 km/m. Bear in mind the circumference is 2 pi radians. Mathematicians love radians as 1 radian along a circle's circumference is the same distance as the radius. :-) It turns out a radian is 180 / pi degrees, or about 57.3.) Cut that triangle in half lengthwise and you get 2 very thin right triangles. The rise of the surface is the distance between the base of each right triangle and the arc. The base is the cosine of half the angle times the radius, or 6.378 * 10^6 * cos(0.0002523) = 6377999.796959 m. The difference is then about 0.203 m or 20.3 cm -- less than a foot. This means that you'll be able to see your friend standing a mile away although you may not be able to see his shoetops, assuming sufficiently powerful binoculars and a nice flat surface, such as the ocean. :-) ) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: CURVATURE OF EARTH SURFACE PER MILE?? In sci.math, ChrisCoaster In sci.math, ChrisCoaster >> I know it involves determining the height of a chord through a given >> circle, but beyond that I don't know how to go about it. >> >> Edia = 7,926(miles) or 41,849,280(feet) >> Ecirc = 24,900(miles) or 131,472,000(feet) >> >> Eradius = 1/2 dia = 3963miles or 20,924,640feet. >> >> What is formula or procedure to find the answer? I'm learning >> disabled so please explain it as to a 10 year old. >> >> >> Chris >> If you understand trigonometry, you'll understand this explanation >> reasonably enough, because that's what's needed to solve >> this particular problem. >> Take the Earth's radius to be 6.378 * 10^6 m, or 6,378 km. >> (Scientists and I prefer metric, for various reasons. >> This is slightly higher than your value. Feel free to >> rework the computations if you want to; they work well >> enough in miles as it turns out but remember to convert >> the result at the end to feet or inches.) >> We're assuming a perfectly circular Earth for the purpose >> of this argument; obviously that's not the case but the >> errors won't affect the amount much. >> This means that the circumference of the Earth will be >> 2 * pi * 6.378 * 10^6 = 4.007 * 10^7 m, or 40,070 km. >> One mile on the Earth's surface therefore subtends >> 1.609344 / 6378 radian = 0.0002523 radian. (1 mile = >> 5280 feet x 12 inches/ft x 0.0254 m/in x 1/1000 km/m. >> Bear in mind the circumference is 2 pi radians. Mathematicians >> love radians as 1 radian along a circle's circumference is the >> same distance as the radius. :-) It turns out a radian >> is 180 / pi degrees, or about 57.3.) >> Cut that triangle in half lengthwise and you get 2 very thin >> right triangles. The rise of the surface is the distance >> between the base of each right triangle and the arc. >> The base is the cosine of half the angle times the radius, >> or 6.378 * 10^6 * cos(0.0002523) = 6377999.796959 m. >> The difference is then about 0.203 m or 20.3 cm -- less >> than a foot. This means that you'll be able to see your >> friend standing a mile away although you may not be able to >> see his shoetops, assuming sufficiently powerful binoculars >> and a nice flat surface, such as the ocean. :-) ) > ______________________________ > closer to 3 feet(~90cm) per mile because I remember reading that as a > kid, but I just wanted a more scientific way to be sure. One problem is that if one doubles, triples, etc. the angle, the amount is not linearly doubled, tripled, etc., but quadrupled, multiplied by 4, etc. (for small angles, anyway). 2 miles translates into 81.2 cm. 3 miles translates into 1.83 m. 4 miles translates into 3.25 m. > I live close to Long Island sound about 40 miles from NYC. So 40 x 8 > inches(20.3cm) would approximate 320 inches in rise, or about 26 feet > if my mental math isn't completely comotose. Your math is unfortunately a bit off; it's more like 40 x 40 x 8 inches, or more than a thousand feet. (The actual amount is more like 324.8 m, if one uses trigonometry similar to the above.) One reason is simply that cos(x) = 1 - x^2/2! + x^4/4! - ...; for very small x the value 1 - cos(x) is approximately equal to x^2/2. > In reality, when I look > at the Empire State Building from that distance, it seemed that the > entire lower 1/4th of the building (all those setbacks) were missing > due to the earth's curve. Now 40 miles x 3feet does seem to explain > what I'm seeing: 120 feet of building obscured by the horizon. 26 > feet at 40 miles? Nothing would appear to be missing. The Empire State Building's height is 102 stories or 1,250 feet, although that apparently doesn't include the spire. http://www.pbs.org/wgbh/buildingbig/wonder/structure/empire_state.html 1/4 of the building would be about 312 feet or 95 m. Assuming a perfectly flat Earth (which is unlikely) one could deduce that your distance away is about acos((6.378 * 10^6 - 95) / 6.378 * 10^6) / 0.0002523 = 21.6 miles. (This odd calculation is mostly because we already have the equation above for 1 mile; I'm basically working backwards.) > -CC -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: CURVATURE OF EARTH SURFACE PER MILE?? > The difference is then about 0.203 m or 20.3 cm -- less > than a foot. This means that you'll be able to see your > friend standing a mile away although you may not be able to > see his shoetops, assuming sufficiently powerful binoculars > and a nice flat surface, such as the ocean. :-) ) But this assumes that I'm viewing my friend with my eyes at the ground, instead of about 5.5 feet above the surface, right? So I think I could see his shoes. === Subject: Delta explained? I'm having a basic calc problem and I thought I'd go straight to the source to find the answer. I am writing a little vba program for land surveyors and in it I need to find the azimuth of a line. I have the northings and eastings of two points on the line and now need to use that to find the azimuth. I am trying to convert a formula that someone give me which finds an angle given the northing and easting of a point. Here is the formula angle = tan^-1 ((delta northing) / (delta easting)) I'm embarresed to say but I don't remember what delta means, if I ever knew. Can someone demonstrate to me what delta means? Can I use this formula to find the azimuth of a line? Any help would be great! === Subject: Re: Delta explained? > I'm having a basic calc problem and I thought I'd go straight to the > source to find the answer. > I am writing a little vba program for land surveyors and in it I need > to find the azimuth of a line. I have the northings and eastings of > two points on the line and now need to use that to find the azimuth. > I am trying to convert a formula that someone give me which finds an > angle given the northing and easting of a point. > Here is the formula > angle = tan^-1 ((delta northing) / (delta easting)) > I'm embarresed to say but I don't remember what delta means, if I ever > knew. > Can someone demonstrate to me what delta means? > Can I use this formula to find the azimuth of a line? > Any help would be great! delta often means the change or difference. So when it says delta northing, it means the northing of B - northing of A, where A and B are the two points. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: Delta explained? > I'm having a basic calc problem and I thought I'd go straight to the > source to find the answer. > I am writing a little vba program for land surveyors and in it I need > to find the azimuth of a line. I have the northings and eastings of > two points on the line and now need to use that to find the azimuth. > I am trying to convert a formula that someone give me which finds an > angle given the northing and easting of a point. > Here is the formula > angle = tan^-1 ((delta northing) / (delta easting)) > I'm embarresed to say but I don't remember what delta means, if I ever > knew. Usually it means change in or difference in. That's the case here. If your two points are at (N1, E1) and (N2, E2), then by using delta northing = N2-N1 delta easting = E2-E1 in your formula, you will get the angle of the line from point 1 to point 2. This angle will be measured counterclockwise relative to due east. That is, 0 means east, 90 degrees means north, -90 degrees = south. Is that the angle you want? One problem with formulas like this is that there's an ambiguity. You'll never get an angle outside the range of -90 to +90 degrees, because -135 degrees (or +225 degrees) has the same tangent as 45 degrees. For this reason computers and many calculators usually have a two argument version of arctan, called something like atan2. You give it the y coordinate (delta-northing) and the x coordinate (delta-easting) separately. If your calculator doesn't include this function, you'll need to write a little if..then logic to handle this and to handle the cases where delta-easting = 0 (so you don't divide by 0). - Randy === Subject: Re: Delta explained? > I am writing a little vba program for land surveyors and in it I need > to find the azimuth of a line. I have the northings and eastings of > two points on the line and now need to use that to find the azimuth. > angle = tan^-1 ((delta northing) / (delta easting)) > Can someone demonstrate to me what delta means? Delta means difference in or change in. You said you have two points. (delta northing) would be the difference in their northings. Make sure to subtract in the same order in both the numerator and denominator. > Can I use this formula to find the azimuth of a line? Yes, if by azimuth you mean the angle that the line makes with East. The symbol tan^-1 will probably be accessed in your computer language as atan or arctan. === Subject: Re: Delta explained? > I am writing a little vba program for land surveyors and in it I need > to find the azimuth of a line. I have the northings and eastings of > two points on the line and now need to use that to find the azimuth. angle = tan^-1 ((delta northing) / (delta easting)) Can someone demonstrate to me what delta means? > Delta means difference in or change in. You said you have two > points. (delta northing) would be the difference in their northings. > Make sure to subtract in the same order in both the numerator and > denominator. > > Can I use this formula to find the azimuth of a line? > Yes, if by azimuth you mean the angle that the line makes with East. > The symbol tan^-1 will probably be accessed in your computer language > as atan or arctan. === Subject: Delta patterns related to the primes. This is a little off the wall, but what the heck here goes --- Given this sequence where a prime (p) has exactly 2 of the same (p) and (p)count of other unique primes between them. 2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61,67,71,73,17, 79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149,151,31,... Having the sequence I then produced an algorithm to find the deltas. I am just showing the first few here. 2, 3, 5, 2, 7, 3, 11, 13, 5, 17, 19, 23, 7, 29, 31, 37,.. 1 2 3 5 4 8 2 8 12 2 4 16 22 2 6,... 1 1 2 1 4 6 6 4 10 2 12 6 20 4,... 0 1 1 3 2 0 2 6 8 10 6 14 16,... 1 0 2 1 2 2 4 2 2 4 8 2,... 1 2 1 1 0 2 2 0 2 4 6,... 1 1 0 1 2 0 2 2 2 2,... 0 1 1 1 2 2 0 0 0,... 1 0 0 1 0 2 0 0,... 1 0 1 1 2 2 0,... 1 1 0 1 0 2,... 0 1 1 1 2,... 1 0 0 1,... 1 0 1,... 1 1,... 0,... Starting with the first diagonal on the left and starting with the 4th delta the repeating sequence of ---- 1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,... The next diagonal delta on the left starting with the 6th delta the repeating sequence of --- 1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,... The third diagonal on the left starting with the fifth delta the repeating sequence of --- 1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,... Finally the fourth diagonal from the left and starting with the fourth delta the repeating sequence of --- 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,.. . The reason I believe this important about the primes is if a mistake is made for any primes where not enough other primes are placed between the pair of same primes in the sequence, at some point in conjunction with the error this will break the 4 repeating diagonal delta patterns. Also there appears to be no left delta diagonals that have any patterns when using the composites that have the same rules for constructing a sequence. Will these patterns continue where p ----->oo? I have only done 400 terms in this sequence that include only 66 same pair of primes where the above repeating sequences hold. I also discovered an error where I did not place enough primes between a pair of same primes, which in turn broke the pattern of the repeating sequences of the deltas. The error was one other prime short between a pair of same primes! The largest prime I have tested too in completing the pair is 317, which has 317 unique primes between the two. This = 400 terms in the sequence and after correcting 1 error the 4 repeating patterns of the left delta diagonals hold. These 4 left diagonal repeating delta sequences from their starting points have about 396 terms that repeat. I don't know if this is trivial, but I found it interesting. How could the calculations for the number of terms in this sequence be made where the last completed pair of same primes where say p= 6221? Which is the 810th prime which would need 6,221 unique primes between them. ;-) Dan === Subject: Re: Delta patterns related to the primes. >This is a little off the wall, but what the heck here goes --- >Given this sequence where a prime (p) has exactly 2 of the same (p) >and (p)count of other unique primes between them. >2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61,67,71,73,17 ,79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149,151,31,... I had a hard time figuring out what you meant here. I think it's this. Enumerate the primes as p_i, i=1,2,3,.... Define your sequence a_j as follows by the following process. Start with all a_j undefined. After the (k-1)'th round, let a_i be the first entry that hasn't yet been defined, take a_i = p_k and a_{i+p_k+1} = p_k. >Having the sequence I then produced an algorithm to find the deltas. >I am just showing the first few here. >2, 3, 5, 2, 7, 3, 11, 13, 5, 17, 19, 23, 7, 29, 31, 37,.. > 1 2 3 5 4 8 2 8 12 2 4 16 22 2 6,... > 1 1 2 1 4 6 6 4 10 2 12 6 20 4,... i.e. let A_{i,1} = a_i and A_{i,j+1} = |A_{i,j} - A_{i+1,j}| for j >= 1. >Starting with the first diagonal on the left and starting with the 4th >delta the repeating sequence of ---- >1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,... >The next diagonal delta on the left starting with the 6th delta the >repeating sequence of --- >1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,... >The third diagonal on the left starting with the fifth delta the >repeating sequence of --- >1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,... >Finally the fourth diagonal from the left and starting with the fourth >delta the repeating sequence of --- >1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,. .. Wow! This certainly looks impressive. It looks like these patterns will continue... Note that as long as A[5,j] is 0 or 2, starting with A[i,7] = 1,1,0,1,..., we will have these patterns continuing from then on. Now for fixed j, A[i,j] can't increase too rapidly as i increases: A[i,1] is less than the i'th prime, which is O(i ln(i)), and of course is odd for i > 4 (so A[i,j] is even for i >= 5 and j >= 1). A[i,2] is usually the gap between two consecutive primes, which is on the average about ln(i), but sometimes the gap between two far-apart primes, about i ln(i). The differencing process tends to make large entries for one j become smaller as they propagate down the table (increasing j and decreasing i), and it looks like none of them (or at least none we've seen so far) make it all the way to i=5. For example, A[61,45] = 40 is unusually large. But A[60,46] = |2 - 40| = 38, A[59,47] = |2 - 38| = 36, and so on, decreasing each time you take a difference with a nonzero entry, until A[16,90] = 2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Delta patterns related to the primes. >This is a little off the wall, but what the heck here goes --- >Given this sequence where a prime (p) has exactly 2 of the same (p) >and (p)count of other unique primes between them. > >2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61,67,71,73,17 , 79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149,151,31,... > I had a hard time figuring out what you meant here. I think it's this. > Enumerate the primes as p_i, i=1,2,3,.... Define your sequence a_j as > follows by the following process. Start with all a_j undefined. > After the (k-1)'th round, let a_i be the first entry that hasn't yet > been defined, take a_i = p_k and a_{i+p_k+1} = p_k. >Having the sequence I then produced an algorithm to find the deltas. >I am just showing the first few here. >2, 3, 5, 2, 7, 3, 11, 13, 5, 17, 19, 23, 7, 29, 31, 37,.. > 1 2 3 5 4 8 2 8 12 2 4 16 22 2 6,... > 1 1 2 1 4 6 6 4 10 2 12 6 20 4,... > i.e. let A_{i,1} = a_i and A_{i,j+1} = |A_{i,j} - A_{i+1,j}| for j >= 1. >Starting with the first diagonal on the left and starting with the 4th >delta the repeating sequence of ---- >1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,... >The next diagonal delta on the left starting with the 6th delta the >repeating sequence of --- >1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,... >The third diagonal on the left starting with the fifth delta the >repeating sequence of --- >1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,... >Finally the fourth diagonal from the left and starting with the fourth >delta the repeating sequence of --- > >1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,. . . > Wow! This certainly looks impressive. It looks like these patterns > will continue... > Note that as long as A[5,j] is 0 or 2, starting with > A[i,7] = 1,1,0,1,..., we will have these patterns continuing from then > on. Now for fixed j, A[i,j] can't increase too rapidly as i increases: > A[i,1] is less than the i'th prime, which is O(i ln(i)), and of course > is odd for i > 4 (so A[i,j] is even for i >= 5 and j >= 1). > A[i,2] is usually the gap between two consecutive primes, which is on > the average about ln(i), but sometimes the gap between two far-apart > primes, about i ln(i). The differencing process tends to make large > entries for one j become smaller as they propagate down the table > (increasing j and decreasing i), and it looks like none of them (or at > least none we've seen so far) make it all the way to i=5. For example, > A[61,45] = 40 is unusually large. But A[60,46] = |2 - 40| = 38, A[59,47] > = |2 - 38| = 36, and so on, decreasing each time you take a difference > with a nonzero entry, until A[16,90] = 2. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Robert, the deltas. I had to study your reply to understand what you were explaining about how the sequence is generated and what happens with the deltas derived from this sequence. Is this sequence in EIS? The reason I ask is, my new broadband provider will not connect me to any ATT sites and it doesnët matter what browser I use! Dan === Subject: Re: Delta patterns related to the primes. >This is a little off the wall, but what the heck here goes --- >Given this sequence where a prime (p) has exactly 2 of the same (p) >and (p)count of other unique primes between them. > >2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61,67,71,73,17 , 79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149,151,31,... > I had a hard time figuring out what you meant here. I think it's this. > Enumerate the primes as p_i, i=1,2,3,.... Define your sequence a_j as > follows by the following process. Start with all a_j undefined. > After the (k-1)'th round, let a_i be the first entry that hasn't yet > been defined, take a_i = p_k and a_{i+p_k+1} = p_k. I admit I didn't read this very carefully, but it reminds me of something I've seen a few years ago and as far as I know is an open problem. It also looks a bit more natural than the OP's definition. Take the difference sequence of the primes, and iteratively take the difference sequence of the previous sequence, but make sure you take all differences in absolute value. 2 3 5 7 11 13 15 17 19 . . . . 1 2 2 4 2 2 2 2 . . . . 1 0 2 2 0 0 0 . . . . 1 2 0 2 0 0 . . . . 1 2 2 2 0 . . . . The claim is that the leftmost diagnoal is always 1 (except for the initial 2). This seems to be some sort of delicate statement about the distribution of primes. Has anyone seen this claim? Just thought I'd mention it, - EM === Subject: Re: Delta patterns related to the primes. >>2,3,5,2,7,3,11,13,5,17,19,23,7,29,31,37,41,43,11,47,53,13,59,61,67,71,73,1 7,79,83,19,89,97,101,103,23,107,109,113,127,131,137,139,29,149,151,31,... >Is this sequence in EIS? No, it isn't. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Delta patterns related to the primes. >Take the difference sequence of the primes, and iteratively take the >difference sequence of the previous sequence, but make sure you take >all differences in absolute value. >2 3 5 7 11 13 15 17 19 . . . . 15? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Delta patterns related to the primes. > I admit I didn't read this very carefully, but it reminds me of > something I've seen a few years ago and as far as I know is an open > problem. It also looks a bit more natural than the OP's definition. > Take the difference sequence of the primes, and iteratively take the > difference sequence of the previous sequence, but make sure you take > all differences in absolute value. > 2 3 5 7 11 13 15 17 19 . . . . > 1 2 2 4 2 2 2 2 . . . . > 1 0 2 2 0 0 0 . . . . > 1 2 0 2 0 0 . . . . > 1 2 2 2 0 . . . . > The claim is that the leftmost diagnoal is always 1 (except for the > initial 2). This seems to be some sort of delicate statement about the > distribution of primes. Has anyone seen this claim? It's problem A10 in Guy, Unsolved Problems In Number Theory. See Odlyzko, Iterated absolute values of differences of consecutive primes, Math. Comput. 61 (1993) 373-380, MR 93k:11119. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: diff. geom. book opinion wanted I'm looking for a good modern book on differential geometry, as comprehensive as possible, but also good for self-study (I know I'm asking too much!). I've been suggested Lee's Introduction to Smooth Manifolds. Does anybody have it? can you comment on it? Any suggestions, opinions, comments, etc. appreciated. nojb. === Subject: Re: diff. geom. book opinion wanted Have you looked at Micael Spivak's, or Do Carmo's book? They are both very comprehensive. Lurch > I'm looking for a good modern book on differential geometry, as > comprehensive as possible, but also good for self-study (I know I'm > asking too much!). I've been suggested Lee's Introduction to Smooth > Manifolds. > Does anybody have it? can you comment on it? > Any suggestions, opinions, comments, etc. appreciated. > nojb. === Subject: Re: diff. geom. book opinion wanted > I'm looking for a good modern book on differential geometry, as > comprehensive as possible, but also good for self-study (I know I'm > asking too much!). I've been suggested Lee's Introduction to Smooth > Manifolds. Yes, this is exactly the book you want. It's by far the best book on the subject I've seen, significantly better than Spivak volume one or Morita, which are the next best two. It's modern, comprehensive, detailed, and has plenty of examples. At UCLA many of the grad students studying for the quals (including myself) are using this book to learn the smooth manifold material. The class that was supposed to teach that material actually covered something else, so we've had to learn it on our own, and this book has been a lifesaver. -- John Leo http://www.halfaya.org/leo/ === Subject: Re: diff. geom. book opinion wanted nojb. I'm looking for a good modern book on differential geometry, as > comprehensive as possible, but also good for self-study (I know I'm > asking too much!). I've been suggested Lee's Introduction to Smooth > Manifolds. > Yes, this is exactly the book you want. It's by far the best book on the > subject I've seen, significantly better than Spivak volume one or Morita, > which are the next best two. It's modern, comprehensive, detailed, and has > plenty of examples. At UCLA many of the grad students studying for the > quals (including myself) are using this book to learn the smooth manifold > material. The class that was supposed to teach that material actually > covered something else, so we've had to learn it on our own, and this book > has been a lifesaver. === Subject: DNA Profile statistics http://mygate.mailgate.org/mynews/sci/sci.math/a6c0abbcc1da1b974cf349471b97f 2 26.111419%40mygate.mailgate.org What area of statistics am I looking for - multivariate regression analysis ?. Could anyone point me to where am I likely to find similar real-world problems occuring ? I suspect the 1:1 billion to 1:100 billion sorts of figures for a random ,false, match with DNA profiles (ie 2 separate people with the same DNA profile ) are derrived from something like equal chance of occurance. So 10 loci and 2 possibles on each loci. The average number of alleles for each of the 10 SGM loci is 14. So permutating 2 from 14 ,10 times in effect. But for ancestral reasons some alleles are far more common than others,usually something like a normal distribution with definite peaks. eg my profile, none are more rare than 9 percent frequency of occurance in the UK caucasian population and they all occur within plus or minus 2 alleles of the median for UK caucasians. Is there any way of determining how much greater the chance of a false match there is in that subset compared to the general case.? Data for UK caucasians from the UK forensic science service. Table of Median (M) and (M + or -1 ) and (M + or -2) allele frequencies of occurance (% ,per cent, of the population) for the 10 UK FSS NDNAD loci Locus / M / M+-1 / M +- 2 VWA 27 71 88 THO1 30 45 56 D6,D8 33 68 84 FGA 19 49 63 D21 26 51 74 D18 16 43 71 D2 14 28 39 D16 29 74 82 D19 38 78 91 D3 26 65 84 So for VWA (say) there are 27 % of people with pairs (17,anything) or (anything,17) 71% (16 or 17 or 18,anything) plus inverse as above and 88% (15 or 16 or 17 or 18 or 19, anything) plus inverse For THO1 the distribution is seriously skewed so the peak is at 9.3 and very little above 9.3. So it would be valid for THO 1 to be Median +0 and minus 4 for the range instead of +-2. In that case for THO1 the relevent line in above table would be THO1 30 55 99 [M / M+0,-2 / M+0,-4 ] yes 99 percent have THO1 alleles 6 or 7 or 8 or 9 or 9.3 ,the median being 9.3 Or to put it another way anyone having just one locus/allele outside of the M +,-2 range is ,intuitively I would have said, far less likely to find themselves falsly implicated by a false match to a future scene of crime DNA profile. Multiplying each together ,twice for each loci in the above table so .88*.88*.56*.56 ...... gives simplistically a probability of someone falling in the median +-2 (including 99% for THO1) as only about 0.35 per cent. Well as far as I am concerned I'm nothing special as far as pure-bred 'English' . I have a bit of Irish and French ancestry, I cannot see myself being in a rarified 0.35% of the population. So putting the other way around - plain multiplication of frequencies/probabilities is erroneous ,I suspect, but I cannot quantify it. +++++++++ What you didn't know about DNA profiling and what they aren't telling you about DNA profiles http://www.nutteing2.freeservers.com/dnapr.htm or nutteing3 in a search engine e mail nutteing2@quickfindit.....com (just one dot) -- === Subject: Re: DNA Profile statistics > Data for UK caucasians from the UK forensic science service. > Table of Median (M) and (M + or -1 ) and (M + or -2) allele frequencies > of > occurance (% ,per cent, of the population) for > the 10 UK FSS NDNAD loci > Locus / M / M+-1 / M +- 2 Could you amplify? What is this the median of? A locus is a particular place on the chromosome, I presume. If that sequence consists of p bases, 4 possible bases at each position, then in principle there are 4^p possible values of the sequence. score is being assigned, which has a median value. What is that score? What does M +-1 mean? Plus or minus 1 WHAT? One sigma? > VWA 27 71 88 > THO1 30 45 56 > D6,D8 33 68 84 > FGA 19 49 63 > D21 26 51 74 > D18 16 43 71 > D2 14 28 39 > D16 29 74 82 > D19 38 78 91 > D3 26 65 84 > So for VWA (say) there are 27 % of people with pairs (17,anything) or > (anything,17) > 71% (16 or 17 or 18,anything) plus inverse as above > and 88% (15 or 16 or 17 or 18 or 19, anything) plus inverse I can't make sense of this without understanding what these scores mean. What does inverse mean here? OK, got some answers to my questions here: www.sbs.auckland.ac.nz/info/schools/ biotechnology/dnaprofiling.pdf DNA profiling is based on sequences of repeating pairs, i.e. GAGAGAGAGA. The numerical score is the number of repeats. You get two scores, corresponding to the same locus on the two copies of each gene (one from mom, one from dad). I think you're asking about whether an assumption about independence holds for these scores. That is the basis of the one in 10 billion or whatever calculations. The PDF I cited says that they are but I gather from googling that it's at least a little controversial, along with other probability assumptions in the model. Include the keyword independence in your search and you'll turn up a bunch of stuff, like this one: www.aic.gov.au/conferences/medicine/seasteal.pdf DNA profiling independence: 3260 matches - Randy === Subject: Re: DNA Profile statistics http://mygate.mailgate.org/mynews/sci/sci.math/46f8f070954f4ca1a48c7a4c86070 a 7e.111419%40mygate.mailgate.org > Data for UK caucasians from the UK forensic science service. > Table of Median (M) and (M + or -1 ) and (M + or -2) allele frequencies > of > occurance (% ,per cent, of the population) for > the 10 UK FSS NDNAD loci > Locus / M / M+-1 / M +- 2 > Could you amplify? What is this the median of? A locus is > a particular place on the chromosome, I presume. If that > sequence consists of p bases, 4 possible bases at each > position, then in principle there are 4^p possible values > of the sequence. > score is being assigned, which has a median value. What > is that score? What does M +-1 mean? Plus or minus 1 WHAT? > One sigma? > VWA 27 71 88 > THO1 30 45 56 > D6,D8 33 68 84 > FGA 19 49 63 > D21 26 51 74 > D18 16 43 71 > D2 14 28 39 > D16 29 74 82 > D19 38 78 91 > D3 26 65 84 I think you're asking about whether an assumption about > independence holds for these scores. That is the basis > of the one in 10 billion or whatever calculations. > The PDF I cited says that they are but I gather from > googling that it's at least a little controversial, > along with other probability assumptions in the > model. Include the keyword independence in your search > and you'll turn up a bunch of stuff, like this one: > www.aic.gov.au/conferences/medicine/seasteal.pdf > DNA profiling independence: 3260 matches > - Randy My own DNA profile (caucasian) slightly altered for obvious reasons is (17,19)(8,9.3)(13,13)(20,22)(29,29)(13,15)(18,19)(12,12)(12,14)(16,18) the numbers are alleles for,2 for each of 10 loci VWA,THO1,D8S1179,FGA,D21S11,D18S51,D2S1338,D16S539,D19S433,D3S1358 So for locus D3S1358 (shortened to D3) on chromosome 3 the alleles ,one from father and one from mother ,are 16 and 18. That is 16 times the basic DNA repeat and 18 times that repeat for that locus. What I call the medians for each of these 10 loci is the allele which has the most frequent occurance within a racial sub-group - here UK caucasians. So maxima occur at alleles 17,9.3,13,21,30,14,20,11,14,15 for each of the 10 loci the corresponding maximum 'allele frequencies' are in percentages % 27 / 30 / 33 / 19 / 26 / 16 / 14 / 29 / 38 / 26 In forensic science terms the ideal would be 6 percent if there was an even spread across the average of 14 possible alleles per locus. But there are distinct peaks in reality. So median for D3 is 15 ,M +-1 is 14 or 15 or 16 and M +-2 is 13 or 14 or 15 or 16 or 17 Returning to my profile the frequencies in percentages for each allele are 27,9 / 11,30 / 33,33 / 14,16 / 23,23 / 12,14 / 9,11 / 29,29 / 9,38 / 25,14 so the rarest is 9 percent ie 9% of UK caucasians have this allele 18 on locus D2S1338 Right the way through the determination of statistics relating to probability of occurance within the population all these 20 numbers are considered independent (Bayes formula). I am uncomfortable with the juxtoposition of myself being within M +-2 of each median for each locus which I would have thought would include a large section of the population say 30 per cent. In conjunction with multiplying all the M +-2 ( in the table in previous post ) frquencies together and arriving at a combined frequency of only 0.1 to 0.35 percent. I have an uncomfortable feeling that I should be a rarity to have my profile to be within 5 number spread in each of twenty and at the same time the maths 'show' I was only one of 300 to 1000 in the general population to be so 'blessed'. Paul Nutteing,Hampshire,England +++++++++++++++++++++++++ What you didn't know about DNA profiling and what they aren't telling you about DNA profiles http://www.nutteing2.freeservers.com/dnapr.htm or nutteing3 in a search engine e mail nutteing2@quickfindit.....com (just one dot) -- === Subject: Re: dumb (but new) math joke > Q: What's the highest number in group theory > A: a belian Q: What's purple and commutes? A: an abelian grape -- http://hertzlinger.blogspot.com === Subject: Re: dumb (but new) math joke Joseph Hertzlinger pushed briefly to the shed door: ^ ^ > Q: What's the highest number in group theory ^ > ^ > A: a belian ^ ^ Q: What's purple and commutes? ^ ^ A: an abelian grape Oh dear, oh dear ... I thought we were talking about /new/ maths jokes here ... OK, a new one off the top of my head (inspired by an incident during a quantum mechanics lecture many moons ago): Q: Who used to paint Heisenberg's house for him? A: The ladder operator Andy -- No, you claim the magpie is to blame for all the worlds ills, based on your ignorance of magpies. (4a7391c12e538ef306d33d71c9482221@TeraNews) Malcolm@malcsplace.com === Subject: Re: Enumerating all cycles sharing a given edge > I'm looking for an efficient algorithm for enumerating all cycles in > undirected graph that share the same (predetermined) edge. Suppose the edge connects the nodes A and B. If you were to remove the edge from the graph, any cycle containing that edge becomes a path between A and B. Conversely, any A-B path in that graph becomes a cycle when the edge is added to it again. So now you just have to get a path-finding algorithm. No doubt there are many. Jaap === Subject: Re: Enumerating all cycles sharing a given edge > I'm looking for an efficient algorithm for enumerating all cycles in > undirected graph that share the same (predetermined) edge. I would > think efficient in this case still equals exponential, but is better > than enumerating all cycles of a graph? Any references to literature > would be welcome as well. > As pointed out already, a worst case runtime would indeed be > exponential. What you will want to implement is a backtracking algorithm > which prunes off a chunk of the vertex set with each subsequent choice... > not much else you can do than this, unless you know some guaranteed > structure about the graph. > Jim Is it possible to achieve this in linear time per discovered cycle? Also, unfortunately, there are present some additional conditions that make backtracking (BFS) less than ideal. Is there any known approach that is based on breadth first search instead? Jack === Subject: Re: Euler's Proof Martin, result, i.e., the product form, then expanded it to show the relationship.. My question was, how did Euler go about finding that remarkable relationship, IN THE FIRST PLACE, between ordinary numbers and the primes? See the other responses for references that describe his approach. > Where can I find a description of the gist of Euler's proof equating the > two > forms of the zeta function, i.e,, the infinite series of ordinary numbers, > to the product series involving prime numbers? > The proof is very straightforward. > Start from the product representation, expand and multiply: > zeta(s) = prod_p 1/(1-p^-s) > = prod_p (1 + 1/p^s + 1/p^{2s} + 1/p^{3s} + ...) > = sum_{n=1}^oo (1/n^s) > -mb === Subject: Even number of terms in a cyclotomic polynomial How it is proved that if a cyclotomic polynomial p(n) has an even number of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? === Subject: Re: Even number of terms in a cyclotomic polynomial > How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, (ii) that Phi_n(1) = 1 if n has more than one prime factor and Phi_n(1) = p if n is a power of the prime p. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Even number of terms in a cyclotomic polynomial > How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, (ii) that Phi_n(1) = 1 if n has more than one prime factor and Phi_n(1) = p if n is a power of the prime p. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Even number of terms in a cyclotomic polynomial > How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, (ii) that Phi_n(1) = 1 if n has more than one prime factor and Phi_n(1) = p if n is a power of the prime p. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Even number of terms in a cyclotomic polynomial > How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, (ii) that Phi_n(1) = 1 if n has more than one prime factor and Phi_n(1) = p if n is a power of the prime p. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Even number of terms in a cyclotomic polynomial How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that > (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, > (ii) that Phi_n(1) = 1 if n has more than one prime factor > and Phi_n(1) = p if n is a power of the prime p. > Let me ask how it is proved that if a(n)=3 (the number of terms) then > n is of the form n = 2^i * 3^j ? Don Coppersmith now answered my question here : http://mathforum.org/discuss/sci.math/t/526744 but I don't see this in Google . Is there a problem with posting to the Math Forum ? Bill === Subject: Re: Even number of terms in a cyclotomic polynomial >> >> How it is proved that if a cyclotomic polynomial p(n) has an even >> number of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? >> >> Use the facts that >> (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, >> (ii) that Phi_n(1) = 1 if n has more than one prime factor >> and Phi_n(1) = p if n is a power of the prime p. >> Let me ask how it is proved that if a(n)=3 (the number of terms) then >> n is of the form n = 2^i * 3^j ? > Don Coppersmith now answered my question here : > http://mathforum.org/discuss/sci.math/t/526744 That's a very nice argument -- it didn't come up on my newsfeed either. A simpleminded and less general alternative. If Phi_n has three terms, then Phi_n = x^{2m} + a x^m + 1. As the zeros have absolute value 1 then |a| <= 2. But a is an integer, we can't have a = 0 and we can't have |a| = 2 for then Phi_n would be a square of a polynomial. So a = +-1. Then Phi_n has zeroes exp(+- 2pi/(3m)) or zeroes exp(+- pi/(3m)). They include a primitive 3m-th or 6m-th root of unity so n = 3m or 6m and phi(n) = 2m. This is only possible for the stated values of n. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Even number of terms in a cyclotomic polynomial How it is proved that if a cyclotomic polynomial p(n) has an even number > of non-zero terms a(n) then a(n)=2 and n is a power of 2 ? Use the facts that > (i) for n >= 2 the cyclotomic polynomial Phi_n is reciprocal, > (ii) that Phi_n(1) = 1 if n has more than one prime factor > and Phi_n(1) = p if n is a power of the prime p. > Let me ask how it is proved that if a(n)=3 (the number of terms) then > n is of the form n = 2^i * 3^j ? Checking the n's for which a(n) = 5 gives the sequence : 5,10,20,25,40,50,80,100,125,160,200,... and these appear to be of the form 5^n*2^m so it looks that there is some pattern here ,given m the sequence of numbers such that a(n)=m is nice . Dan === Subject: example of Lindenbaum-Tarski algebra Consider a theory for sentential calculus T which formulas are {A,B,C,D,0,1}. In T the following hold |- A v B -> C |- C & D |- A & B -> 0. |- A -> D I have to draw the related L-T algebra lattice. Is the following right? 1 / / / C D / / / A v B / / / A B / / / 0 === Subject: Re: example of Lindenbaum-Tarski algebra > Consider a theory for sentential calculus T which formulas are > {A,B,C,D,0,1}. In T the following hold > |- A v B -> C > |- C & D > |- A & B -> 0. > |- A -> D > I have to draw the related L-T algebra lattice. Is the following right? > 1 > / > / > / > C D > / > / > / > A v B > / > / > / > A B > / > / > / > 0 Not sure, but I think [C&D] = [C] = [D] = [~(A&B)] = [1] | | [AvB] / / [A] [B] / / [A&B] = [0] Where, if ~ isn't in your language, ~X is defined to be X->0. -- G.C. === Subject: Re: Factorial/Exponential Identity, Infinity > Ross has claimed that if f(n) = n!/( (n/2)! 2^n) ) > then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he > reconsider. Ross, consider this, which you may easily verify yourself: f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. Thus f(2n+2) > f(2n) > 1 for every integer n > 2, > and f(2n) gets further away from 1 as n increases, > and, in fact, diverges towards +oo. > f(n) = n! / ( (n/2)! 2^n ) > f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) ) > f(2n) = (2n)! / ( n! 2^(2n) ) > f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2) > = (2n+1)(2n+2) / ( (n+1) 4 ) > = (2n+1)2/4 > = n + 1/2 > Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than > f(2n). That seems to be a counterexample. My point is that f(2n+2) is (n+1/2) times greater than f(2n). There is no integer n for which (n+1/2) is infinite. > (2n+2) / 2n = 1 + 1/n > lim (1 + 1/n) = 1 But this is irelevant. What one concludes from my result is that f(2n) > f(2)*n! for all n > 1. And since n! -> +oo, amd f(1) <> 0. f(2n) -> +oo. > I'm more interested in contradictions to the steps of the proof, yet I > still have to figure out how to respond to the proferred > counterexample, unless you do. > Ross My counterexample proves your argument contains at least one error. An acknowledgement of that fact would be in order. The problem of where those errors occur is all yours. === Subject: Re: Factorial/Exponential Identity, Infinity > Ross has claimed that if f(n) = n!/( (n/2)! 2^n) ) > then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he > reconsider. Ross, consider this, which you may easily verify yourself: f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. Thus f(2n+2) > f(2n) > 1 for every integer n > 2, > and f(2n) gets further away from 1 as n increases, > and, in fact, diverges towards +oo. f(n) = n! / ( (n/2)! 2^n ) f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) ) > f(2n) = (2n)! / ( n! 2^(2n) ) f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2) > = (2n+1)(2n+2) / ( (n+1) 4 ) > = (2n+1)2/4 > = n + 1/2 Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than > f(2n). That seems to be a counterexample. > My point is that f(2n+2) is (n+1/2) times greater than f(2n). There > is no integer n for which (n+1/2) is infinite. (2n+2) / 2n = 1 + 1/n > lim (1 + 1/n) = 1 > But this is irelevant. What one concludes from my result is that > f(2n) > f(2)*n! for all n > 1. > And since n! -> +oo, amd f(1) <> 0. f(2n) -> +oo. I'm more interested in contradictions to the steps of the proof, yet I > still have to figure out how to respond to the proferred > counterexample, unless you do. Ross > My counterexample proves your argument contains at least one error. > An acknowledgement of that fact would be in order. > The problem of where those errors occur is all yours. n+1/2 is the quotient of f(2n+2)/f(2n). What we're looking for in this as a test of possible convergence is that that quotient should be equal to or have a limit evaluation of one. What I want to make clear is the application of that as a test or as a counterexample that the same reasoning would imply the divergence of Stirling's identity, which is proven by different means. So I think I have a proof that doesn't depend on successive terms demonstrating equality, and I look to a well-known proof of a similar identity and see that its successive terms don't approach equality. I want to resolve that as a line of argument against the result, and start by saying it would invalidate another unrelated result, which is known to be true and accurate. I'm saying the successive term quotient argument would be invalid for the same reaons it is invalid against the well-known result, Stirling's identity. Now, I haven't verified Stirling's identity, to be frank it is beyond my current knowledge and would require weeks. Here's something I like to consider, where I assume lim n-> oo n! / ( (n/2)! 2^n) = 1 It's that (n!)^2 / ( ((n/2)!)^2 2^2n ) = 1 (n!)^4 / ( ((n/2)!)^4 2^4n ) = 1 (n!)^8 / ( ((n/2)!)^8 2^8n ) = 1 etcetera, from n! / ( (n/2)! 2^n) = ( (n/2)! 2^n) / n! That also leads me to think sqrt(n!) / ( sqrt((n/2)!) 2^(n/2) ) = 1 that is x E Z | lim n->oo ( n! ^(2^x) ) / ( ((n/2)!)^(2^x) (2^n)^(2^x) ) = 1 It's a good argument, the quotient argument, I don't yet know how to counter it directly, so I counter it indirectly by showing the counterexample against a well-known good result. If the argument affects the F/E identity proof, then it would affect the Stirling identity proof. So my response is I understand that and am considering it, and don't consider it to affect the rationale of the proof, tell it to Stirling. I'm happy because it, the argument, doesn't represent a contradiction to the progression of the proof. About the factorial/exponential identity proof, I think there are gaps in the proof, that is to say, I haven't fully demonstrated all the aspects of the proof. It's only been put forth on this thread on sci.math within the past several days, I imagine it will get further review. So, please attack the proof with logical, rational arguments so I can see if it holds. Ross === Subject: Re: Factorial/Exponential Identity, Infinity I've found more errors in my derivation. The (n/2)! term is to be squared, and, the 2 is exponentiated by n-1. That's pretty simple, I should have caught that. We have: P = ( n! / ( (n-k)! k! ) ) p^k q^(n-k) k = n-k = n/2 -> P = ( n! / ((n/2)!)^2) p^(n/2) q^(n/2) P = p = q = 1/2 -> 1/2 = n! / ( ((n/2)!)^2 (2^(n/2)) (2^(n/2)) ) 1/2 = n! / ( ((n/2)!)^2 2^n ) 1 = n! / ( (n/2)!^2 2^(n-1) ) lim n->oo n! / ( (n/2)!^2 2^(n-1) ) = 1 = lim n->oo ( (n/2)!^2 2^(n-1) ) / n! Those aren't acceptable errors. My error was that I said the incorrect (n/2)! (n/2)! = 2 (n/2)! instead of the correct (n/2)!(n/2)! = (n/2)!^2. In consideration of Stirling's identity, lim n->oo n! / ((n/e)^n sqrt(n2pi)) = 1 there is that: lim n->oo ( (n/2)!^2 2^(n-1) e^n) / ( n^n sqrt(n) sqrt(2pi) ) = 1 lim n->oo ( (n/2)!^2 (2e)^n ) / ( 2 n^n sqrt(n) sqrt(2pi) ) = 1 Ross === Subject: Re: Factorial/Exponential Identity, Infinity > I've found more errors in my derivation. The (n/2)! term is to be > squared, and, the 2 is exponentiated by n-1. > That's pretty simple, I should have caught that. > We have: > P = ( n! / ( (n-k)! k! ) ) p^k q^(n-k) > k = n-k = n/2 - P = ( n! / ((n/2)!)^2) p^(n/2) q^(n/2) > P = p = q = 1/2 -> The above, P = p = q, looks like you are claiming that in 2*n tosses of a fair coin, the probability of getting exactly the same number of heads as tails is 1/2. Infortunately, this only is true for n = 1. The probability of an equal number of heads and tails strictly decreases as n increases. > 1/2 = n! / ( ((n/2)!)^2 (2^(n/2)) (2^(n/2)) ) The above is only true when n = 2, so the rest of the derivation, which depends on it being true for all n, is also wrong. === Subject: Re: Factorial/Exponential Identity, Infinity > I've found more errors in my derivation. The (n/2)! term is to be > squared, and, the 2 is exponentiated by n-1. That's pretty simple, I should have caught that. We have: P = ( n! / ( (n-k)! k! ) ) p^k q^(n-k) k = n-k = n/2 -> P = ( n! / ((n/2)!)^2) p^(n/2) q^(n/2) P = p = q = 1/2 - The above, P = p = q, looks like you are claiming that in 2*n tosses > of a fair coin, the probability of getting exactly the same number > of heads as tails is 1/2. > Infortunately, this only is true for n = 1. The probability of an > equal number of heads and tails strictly decreases as n increases. 1/2 = n! / ( ((n/2)!)^2 (2^(n/2)) (2^(n/2)) ) > The above is only true when n = 2, so the rest of the derivation, > which depends on it being true for all n, is also wrong. I claim that in all the possible infinite sequences of coin tosses that half of them have equal numbers of heads and tails, P=1/2, where there is a one half probability of each coin toss being heads, p=1/2 and an equal probability of being tails, p=1/2. Consider the case with two coin tosses. Here are the possible results: 00 01 10 11 Notice that two of the sequences, one half of those four, have equal numbers of ones and zeros, those being 01 and 10. Consider with four coin tosses: 0000 0001 0010 0011 - 0100 0101 - 0110 - 0111 1000 1001 - 1010 - 1011 1100 - 1101 1110 1111 Only 3/8 of the sequences have equal numbers of ones and zeros, exactly n! / ((n/2)!^2 2^n). Uh oh. I need to prove that for infinite sequences that half of the possible sequences have equal numbers of ones and zeros. Damnit! Let me tell you where I got that idea, it was while we were talking about Cantor & Infinity, I have this notion that it is true. I reread this and see that I did not explain it there. That would be insufficient foundation to presume that then. I would have to actually prove that to use it. I write a program to look at some of the values that it calculates. I notice some patterns in the values. The program calculates P = f(n) = n! / (n/2)!^2 2^n for n being powers of two. I have the program multiply P by n, n f(n). I notice that the value of n f(n) about doubles in even powers of two. I determine a function on n to divide this away. I'm observing something like this: f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537 f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567 f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574 f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576 That seems to imply to me that the limit of that function in the extreme is somewhere slightly greater than 1.595. Yet, over all of the variables as opposed to even powers of two there is quite the range of values, the odd powers of two evaluate to around 1.128... with a different function. Oh well, I'll probably find a bunch more errors. Anyways, I'm interested in any known constants that have a rational functional relation to around 1.595. If you have some information in relation to the above please do note it. I still need to figure out the whole half the sequences have equal numbers of zeros and ones thing, as I still think it's true, but I need to figure it out. Ross === Subject: Re: Factorial/Exponential Identity, Infinity === > :Subject: Re: Factorial/Exponential Identity, Infinity > :> === > :> :Subject: Re: Factorial/Exponential Identity, Infinity > :> : > :> : === > :> ::Subject: Re: Factorial/Exponential Identity, Infinity > :> :: > :> ::> I write a program to look at some of the values that it calculates. I > :> ::> notice some patterns in the values. The program calculates P = f(n) > :> ::> = n! / (n/2)!^2 2^n for n being powers of two. I have the program > :> ::> multiply P by n, n f(n). I notice that the value of n f(n) about > :> ::> doubles in even powers of two. I determine a function on n to divide > :> ::> this away. I'm observing something like this: > :> :: :> ::> f(1024) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59537 > :> ::> f(4096) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59567 > :> ::> f(16304) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59574 > :> ::> f(65536) (n n!) / (n/2)!^2 2^( (log_2 n)/2-1) 2^n ~= 1.59576 > :> :: :> ::> That seems to imply to me that the limit of that function in the > :> ::> extreme is somewhere slightly greater than 1.595. Yet, over all of > :> ::> the variables as opposed to even powers of two there is quite the > :> ::> range of values, the odd powers of two evaluate to around 1.128... > :> ::> with a different function. Oh well, I'll probably find a bunch more > :> ::> errors. Anyways, I'm interested in any known constants that have a > :> ::> rational functional relation to around 1.595. > :> :: :> : > :> :Did you consider sqrt(8/pi)? > :> : > : :> Also, if f(n)=(n*n!)/((n/2)!^2 2^n 1/2sqrt(n)), then f(2n) will tend to > :> sqrt(8/pi) anyway - you don't need to only look at f(2^(2n)). > :> Maybe if you use the gamma function to define (n/2)! when n is odd you can > :> also get that f(n) tends to sqrt(8/pi). > :Hey, great. > :The Inverse Symbolic Calculator at > :http://www.cecm.sfu.ca/projects/ISC/ISCmain.html says that 2sqrt(8/pi) > := 1.5957691216057307117597842397376.... > No! sqrt{8/pi} not 2sqrt{8/pi} since sqrt{8/pi}>1 it is clear that > 2sqrt{8/pi}>2. > :Please explain how you arrived at that value. I had noticed that the > :values for f(2n) approached that value, I just noticed the doubling in > :f(2^n). > The number looked a bit like sqrt{8/pi}. > :f(n) = (n * n!) / ( (n/2)!^2 2^n 2^((log_2 n)/2-1) ) > :Replacing 2^((log_ 2n)/2-1) with sqrt(n)/2 confused me for a second as > :I though you had written 1/2sqrt(n), instead it's 1/2 sqrt(n). > Aren't they the same? > :f(n)=(n*n!)/((n/2)!^2 2^n 1/2 sqrt(n)) > :f(n) = ( sqrt(n) * n! ) / ( (n/2)!^2 2^(n-1) ) > :A writer directed me to Abramowitz and Stegun's Handbook of > :Mathematical Functions I have a copy, 6'th ed., formula 6.1.18 on > :page 256 is a duplication formula for gamma(x) = (x-1)!, y! = > :gamma(y+1): > :gamma(2x) = sqrt(2Pi) 2^(2x - 1/2) gamma(x) gamma(x+1/2) > :I set x=n/2. > :f(x) = ( sqrt(2x) (2x)! ) / ( x!^2 2^(2x) 2^(2x-1) ) > :f(x) = ( sqrt(2x) gamma(2x+1) / ( gamma(x+1)^2 2^(2x) 2^(2x-1) ) > :f(x) = ( sqrt(2x) 2x gamma(2x) / ( (x gamma(x))^2 2^(2x) 2^(2x-1) ) > :f(x) = ( sqrt(2x) 2x sqrt(2Pi) 2^(2x - 1/2) gamma(x) gamma(x+1/2) ) / > :( x^2 gamma(x)^2 2^(2x) 2^(2x-1) ) > :f(x) = 2 sqrt(2x) sqrt(2Pi) gamma(x+1/2) / ( x gamma(x) 2^2x 2^(2x-1) > :) > :f(x) = 4 sqrt(xPi) gamma(x+1/2) / ( x gamma(x) 2^(4x-1) ) > :f(x) = ( sqrt(Pi) gamma(x+1/2) ) / ( sqrt(x) gamma(x) 2^(4x-3) ) > :Formula 6.1.12 shows a form for gamma(x+ 1/2) in terms of gamma(1/2) > :and a production of the odd integers less than 2x over 2^x. It's > :possible to show that production as a function of the factorial and a > :summation of the sequence of the powers of two. Anyways I am looking > :at the other functions for something to reduce gamma(x+1/2)/ gamma(x), > :I think gamma(1/2) = 3/2 gamma(3/2) because gamma(x) = x gamma(x-1), > :6.1.9 gives gamma(3/2) = 1/2 sqrt(Pi), gamma(1/2) = 3/4 sqrt(Pi). As > :gamma(z) = z gamma(z-1), z!/(z-y)! gamma(z-y) = gamma(z). Let's see > :on that: > :gamma(x) = x gamma(x-1) > :gamma(x-1) = (x-1)gamma(x-2) > :gamma(x-2) = (x-2)gamma(x-3) > :gamma(x) = x (x-1) (x-2) gamma(x-3) > :So z = x + 1/2, y = x-1, gamma(x + 1/2 - x + 1) = ((x+1/2)!/x!) > :gamma(3/2) = (gamma(x+3/2)/ gamma(x) ) / gamma(3/2), is that right? > :f(x) = ( sqrt(Pi) gamma(x+1/2) ) / ( sqrt(x) gamma(x) 2^(4x-3) ) > :f(x) = ( sqrt(Pi) gamma(x+3/2) / ( sqrt(x) gamma(x)^2 gamma(3/2) > :2^(4x-3) ) > :f(x) = gamma(x+3/2) / ( sqrt(x) gamma(x)^2 2^(4x-4) ) > :Here I plan to return the variable x to n/2. > :f(n) = gamma( n/2 + 3/2) / ( sqrt(n/2) gamma(n/2)^2 2^(2n-4) ) > :reviewed it many times. > :Please explain your reasoning and thought process about this. > :I'm thinking here that lim n->oo f(n) = sqrt(8/Pi) > Certainly for even n it is. You can get it using inequalities connected > with Stirling's formula (you can also use Gosper's formula although > without having any kind of expression for the error term that wouldn't > quite be good enough. With inequalities on either side though you can use > a sandwich theorem to get the required result. One would expect that it is > likely to hold also for odd n if one took the time to analyze it. > :( sqrt(n) * n! ) / ( (n/2)!^2 2^(n-1) ) = 2 sqrt(2)/sqrt(Pi) > :(sqrt(n) * n! ) / ( (n/2)!^2 2^n ) = sqrt(2)/sqrt(Pi) > :(sqrt(n Pi) n! ) / ( (n/2)!^2 sqrt(2) 2^n) ) = 1 > :I haven't proven that, only observed it. > :Stirling's equation has e in it, that is something for a different > :time. > Ah... > :Do you have a proof of this identity? > I actually used Stirling, I'm afraid to sandwich the limit between > sqrt{8/pi}-g(n) and sqrt{8/pi}+h(n) where g(n),h(n)->0 as n->infinity (at > least or even n). > :Ross I think I can further reduce the equation to get a closed form in relation to gamma. The idea here is to reduce the function to a constant. Euler gives a form for Gamma: this Handbook of Mathematical Functions is 1046 pages long, a math test is one single-sided page of equation and formulae. Chaper Six starting on page 255 is: 6: Gamma Function and Related Functions. Euler's Formula is: Gamma(z) = lim n->oo n!n^z / (n+z)!/(z)! Gamma(z) = lim n->oo n! n^z z! / (n+z)! The function gamma(z) is also given in Euler's infinite product form: Gamma(z) = 1 / z e^(gamma z) II_n=1^oo [ (1+z/n) e^(-z/n)] , |z| oo as Stirling's formula is. Also, I want to prove my formula is another approximation of n! so I can call it the Factorial/Exponential identity, except it has Pi in it, and I haven't proven it. So anyways I should make a program that gives me some values for small values of n for P(n,k) to visually determine relations for when n -> oo. Then, if I can determine a pattern, it might help to determine a finite form, thus deriving a family of identities. Here's one way to start, let's notice that two of the values have no ones or no zeros, zero and one. n! / ( n! 0! 2^n ) = 1 / 2^n n! / n! = 2^n / 2^n 1=1 The idea there is that there are 2^n possible sequences, as is so for any finite number of n, and that two of the sequences have all zeros or all ones, and in consideration of one of them at a time, then 1 / 2^n is the probability of selecting one and the probability of selecting zero from the unit interval. Anyways that's the easy case, we can say with certitude that there is only one sequence with all zeros, and only one sequence with all ones. For having one or more zeros with one or more ones there are infinitely many. Then, when we have P, for example having P = 2 / sqrt( n Pi) for k=n/2, then the number of sequences where 2^n is the count of all the sequences is P * 2^n or 2^(n+1) / sqrt(n Pi). For what value of k will P(n,k) * 2^n = n? There might exist one, as there exist values of k where P(n,k) * 2^n is less than one, for example 1, and values where P(n,k) * 2^n is greater than n, for example our 2^(n+1) / sqrt(n Pi): 1 < n < 2^(n+1) / sqrt(n Pi). Heh heh heh heh! Wait a second: is 2^(n+1) / sqrt(n Pi) greater than n? It is for n = 2, 3, 4, ..., where 1 is less than n for n = 2, 3, 4, .... Anyways if there was a value of k for which there are n or n/2 sequences, that is where P(n,k) = n/ 2^n or n/ 2^(n+1), I wonder what that would be. n! k!(n-k!) = n Here with having one one: n! /(n-1)! 1! 2^n = n/2^n What that implies is that of the 2^n sequences, n many of them have one one and n many of them have one zero. The sequences that have one one are the set of numbers 2^(-n), the sequences that have one zero are the sequences 1-2^(-n). How about having two zeros or two ones? It might seem possible to consider that there would be n^2 of them, but it might also be much larger than that. For the x'th element of 2^(-x), there is a first one in the sequence, then n-x other elements in the sequence to flip the bit. So, the sum of those for each x over n many of the sequences would be the amount that have two ones. The first of these numbers would be .1100..., 3/4. The next would be .10100..., 5/8, then .100100..., 9/16, .1000100..., 17/32, etcera, ((2^(n-1)+1) / 2^n). Then, the next ones would be .01100..., 3/8, .010100..., 5/16, .0100100..., 9/32, etcetera. The numbers with two zeros start with one quarter, .0011..., then .01011..., 3/8, .011011..., 7/16, etcetera. n! / (n-2)! 2! 2^n = x/2^n n^2-n / 2 = x All of these numbers are rationals that have denominators with powers of two, we haven't gotten to any with infinite ones and infinite zeros except as we have been discussing them and about half of them. In base three, n! / k! (n-k)! 1/3^k 2/3^(n-k) = x / 3^n For k = 0, x = 1, for zero and one. (n! / (n-1)! 1!) 1/3 (2/3)^(n-1) = x / 3^n n 2^(n-1) = x For k = 1, x=n 2^(n-1), implying (n 2^(n-1))/3^n of the sequences are having one zero and many ones or twos in the trinary. Ross === Subject: Re: Factorial/Exponential Identity, Infinity Starting with sequences with equal numbers of zeros and ones, I think a new thing to consider are those sequences having 1/4, 1/8, 1/16, etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8, 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros. Rational numbers with one third ones are similar to .100100100(100)... = 1/2 + 1/16 + 1/128 + ... = sum i=0^oo 2^(-3x-1) .010010010(010)... = 1/4 + 1/32 + 1/256 + ... = sum i=0^oo 2^(-3x-2) .001001001(001)... = 1/8 + 1/64 + 1/512 + ... = sum i=o^oo 2^(-3x-3) They start with any finite sequence with one third ones and terminate with any repeating finite sequence of one third ones. There are also irrational numbers of the form. These would be the expressions of the probability of a sequence having 1/3, 1/5, 2/5, etcetera many ones, as n->oo. n! / ( (n/3)! (2n/3)! 2^n ) n! / ( (n/5)! (4n/5)! 2^n ) n! / ( (2n/5)! (3n/5)! 2^n ) n! / ( (n/6)! (5n/6)! 2^n ) n! / ( (n/7)! (6n/7)! 2^n ) n! / ( (2n/7)! (5n/7)! 2^n ) n! / ( (n/8)! (7n/8)! 2^n ) n! / ( (3n/8)! (5n/8)! 2^n ) n! / ( (n/9)! (8n/9)! 2^n ) n! / ( (2n/9)! (7n/9)! 2^n ) n! / ( (4n/9)! (5n/9)! 2^n ) Then, for each of those, consider having integer x more or less than, for example, one third. n! / ( (n/3+x)! (2n/3-x)! 2^n) n! / ( (n/3-x)! (2n/3+x)! 2^n) For example, this would be the kind of sequences with a finite sequence with one more or one less one than one third and a terminating sequence with one third ones. .110(100)... .000(100)... What would these probabilities tell us? Well, it might be possible to easily derive what their limits are as they are evaluated for large n as multiplied by other expressions and then to prepare a whole bunch of forms with various fractions so that if/when the (2 sqrt2) / sqrt(n Pi) case for even numbers of ones and zeros is proven that all the other identities could be proven at once, giving an infinite family of expressions for n! as n->oo in terms of 2^n and ((qn)+-x)! and (((1-q)n)-+x)! for rational q. One: the count of sequences with all zeros, zero. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > Starting with sequences with equal numbers of zeros and ones, I think > a new thing to consider are those sequences having 1/4, 1/8, 1/16, > etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8, > 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros. > Rational numbers with one third ones are similar to > .100100100(100)... = 1/2 + 1/16 + 1/128 + ... = sum i=0^oo 2^(-3x-1) > .010010010(010)... = 1/4 + 1/32 + 1/256 + ... = sum i=0^oo 2^(-3x-2) > .001001001(001)... = 1/8 + 1/64 + 1/512 + ... = sum i=o^oo 2^(-3x-3) > They start with any finite sequence with one third ones and terminate > with any repeating finite sequence of one third ones. > There are also irrational numbers of the form. Any n-ary expansion that eventually becomes periodic (merely repeating some finite sequence of digits) is rational, regardless of the base. === Subject: Re: Factorial/Exponential Identity, Infinity > Starting with sequences with equal numbers of zeros and ones, I think > a new thing to consider are those sequences having 1/4, 1/8, 1/16, > etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8, > 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros. Well, I got into the case where a quarter of the sequence elements are ones or zeros. n! / ( (n/4)!(3n/4)! 2^n ) I problem here with getting an identity that converges to a value is that (n/4)!(3n/4)! >> (n/2)!(n/2)!. I thought I might be able to use Euler's formula for Gamma. Gamma(z) = lim n->oo n! n^z / ((z+1)(z+2)...(z+n)) What I did was set x to n/2 and z to n/4, so that z+x = 3n/4. Gamma(n/4) = (n/2)! (n/2)^(n/4) / ( (3n/4)!/(n/4)! (n/4 -1)! = (n/2)! (n/2)^(n/4) (n/4)! / (3n/4)! (n/2)! (n/2)^(n/4) (n/4) = (3n/4)! Here what I want is a relation between (n/2)!^2 and (3n/4)!(n/4)! (3n/4)!(n/4)! = (n/2)! (n/2)^(n/4) (n/4) (n/4)! (3n/4)!^2 (n/4)!^2 = (n/2)!^2 (n/2)^(n/2) (n/4)^2 (n/4)!^2 (3n/4)! (n/4)! = (n/2)!^2 (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)! (3n/4)! (n/4)! / (n/2)!^2 = (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)! So I think that where I multiplied n!/((n/2)!^2 2^n) by n/(2^(log_2 (n/2-1)) that I could multiply n!/ (n/4)!(3n/4)! 2^n by n/(2^(log_2 (n/2-1))) and also (n/2)^(n/2) (n/4)^2 (n/4)! / (3n/4)!, however, the values I have computed show it going to zero instead of any positive constant, which is what is more or less derived from the case with (n/2), equal numbers of ones and zeros. n= 4, f(n)= 0.66666666666666666666666666666666666666666666666666666666 n= 16, f(n)= 2.98810325476992143658810325476992143658810325476992143658 n= 64, f(n)= 0.26720147600790400813975210460541880927282821458127736128 n= 256, f(n)= 0.00000000407055224032997121708258424900273493808866341188 n= 1024, f(n)= 0.00000000000000000000000000000000000000000005319865455386 n= 4096, f(n)= 0.00000000000000000000000000000000000000000000000000000000 n= 16384, f(n)= 0.00000000000000000000000000000000000000000000000000000000 n= 65536, f(n)= 0.00000000000000000000000000000000000000000000000000000000 I don't know if that's the correct way to repace the variables in the Gamma function. If x = n/2 and y = n/4 =x/2, then: Gamma(x/2) = limit n->oo (2x)! (2x)^(x/2) / ((5x/2)! /(x/2)!) (x/2-1)! = (2x)! (2x)^(x/2) (x/2)! / (5x/2)! (5x/2)! (x/2-1)! = (2x)! (2x)^(x/2) (x/2)! (5x/2)! = (2x)! (2x)^(x/2) (x/2) What I want there is an expresion for (x/2)!, (3x/2)! and x!. (x/2)! = (5x/2)! (x/2-1)! / ( (2x)! (2x)^(x/2) ) On the right side of the equation there I'm looking for factorials of fractions of x, where all the fractions there are higher than one, where the fractions of the factorial terms are to sum to one. (n/4)! = (5n/4)! (n/4-1)! / ( n! n^(n/4) ) That expression doesn't have terms (3n/4)! and (n/2)!, just (n/4)!. I could multiply both sides by (3n/4)! / (n/2)!^2. (n/4)!(3n/4)! / (n/2)!^2 = (5n/4)! (3n/4)! (n/4-1)! / ( n! n^(n/4) (n/2)!^2 ) So that if correct allows me to multiply the thing by (5n/4)! (3n/4)! (n/4-1)! / ( n! n^(n/4) (n/2)!^2) and get the thing to multiply by n/ 2^(log_2(n/2-1)) to get a constant. Let's see, in my program here I have x=n/2, y=n/4, z=3n/4, so 5n/4 is n+y, I have to write a new function to generate n^(n/4) except I can use ((n/4)^(n/4))^4 or (y^y)^4. Calculating (n/4-1)! is a hassle, the other results are pretty much precalculated except for (n+y)!. I could use a better algorithm to calculate intermediate values, I precalculated n!, 2^n, and n^n for 2x, x=1...32, and 2^x for x=1...20. That's about a megabyte. n= 4, f(n)= 0.000014112573173691905592530497270628348207985458404601827860 n= 16, f(n)= 0.000000000000000000000000000000000000000000000007226596805738 n= 64, f(n)= 0.000000000000000000000000000000000000000000000000000000000000 n= 256, f(n)= 0.000000000000000000000000000000000000000000000000000000000000 n= 1024, f(n)= 0.000000000000000000000000000000000000000000000000000000000000 n= 4096, f(n)= 0.000000000000000000000000000000000000000000000000000000000000 n= 16384, f(n)= 0.000000000000000000000000000000000000000000000000000000000000 I guess n^(n/4) does not equal ((n/4)^(n/4))^4, for example 8^2 = 64 while (2^2)^4 = 256. I need to write a new function to generate n^(n/4). In the meantime as y is less than Integer.MAX_VALUE I can use the pow() function without writing a new function for the use of big integers, the value in the denominator would be less. n= 4, f(n)= 1.875000000000000000000000000000000000000000000000000000000000 n= 16, f(n)= 0.696873106062412261962890625000000000000000000000000000000000 n= 64, f(n)= 0.707269486896323441061911497431205955411422732924867762503011 n= 256, f(n)= 45.830017329992275093956651137014803794959323615450153685895410 n= 1024, f(n)= 51109458999.835860517712702106878661662829103279886830226237808560868620 n= 4096, f(n)= 5044451663733146794724285287208666268270792829347.41480644267483340175220440 5 871217611766651003570779252566509 n= 16384, f(n)= 3061445809155957028137997578849768147728793647208631315434302415991260525463 4 5067843599488308481231157303189465650061123640113331981288986506532530952686 1 4023332692667654215384168407952132332888276981790.52300438092694161580877961 1 316129609882583345358716090780820 Another improvement to the program would be for it to generate many more columns in the table with a variety of computed values and have more information to review at a time, and to compute in the background, or to use a large table of precalculated values. Before I thought to try and use the definition of the Gamma function I tried a variety of expressions, here again is where a large variety of expressions calculated for n would be helpful. (n/4)! = (5n/4)! (n/4-1)! / ( n! n^(n/4) ) (n/4) (n/4)! = (5n/4)! (n/4)! / (n! n^(n/4)) (n/4) = (5n/4)! / (n! n^(n/4)) lim n! n^(n/4) (n/4) / (5n/4)! = 1 About the case with k=n/2, here are some calculated values f(n) = (n n!) / ( (n/2)!(n/2)! 2^n 2^(log_2(n/2-1)) ) f(4) = 1.500000000000000000000000000000000000000000000000 f(16) = 1.571044921875000000000000000000000000000000000000 f(64) = 1.589548059967470344452933339596256701042875647544 f(256) = 1.594211517956484839624950947759399703028478483734 f(1024) = 1.595379577150690797583016460816556042629587428727 f(4096) = 1.595671726561313225912210533040091324648758050191 f(16384) = 1.595744772287097827906598649182402930052438187131 f(65536) = 1.595763034241236923518120275255359526357923810327 It appears that lim (n n!) / ( (n/2)!(n/2)! 2^n 2sqrt(n) ) = sqrt(8/Pi). I'll call sqrt(8/Pi) a constant c. This has n! / (n/2)!(n/2)! 2^n = ( 2c ) / ( sqrt(n) ) , besides having sqrt(n) = ( (n/2)!^2 2^n 2c ) / (n!) n = ( (n/2)!^4 2^2n (2c)^2) / (n!)^2 n! = 2c (n/2)!^2 2^n / sqrt(n) n! = (5n/4)! / n^(n/4) (n/4) That gives (5n/4)! / ( n^(n/4) (n/4) ) = 2c (n/2)!^2 2^n / sqrt(n) lim (5n/4)! sqrt(n) / ( n^(n/4) (n/4) (n/2)^2 2^n) = 2c Where sqrt(n) n! / (n/2)!(n/2)! 2^n = 2c (5n/4)! sqrt(n) / ( n^(n/4) (n/4) (n/2)^2 2^n) = sqrt(n) n! / (n/2)!(n/2)! 2^n (5n/4)! = n! (n^(n/4) (n/4)) leading back to n! = (5n/4)! / (n^(n/4) (n/4)) or also (5n/4)! = n! (n/4) n^(n/4) Let's see, n^(n/4) * n^(3n/4) = n^n, n^(n/4) = n^n / n^(3n/4) Stirling has that n! = sqrt(2pi) n^(n+1/2)e^(-n) n! / n^n = sqrt(2Pi) sqrt(n) / e^n n! / n^n = (5n/4)! / ( (n^n)^2 n^(3n/4) n/4) n! / n^n = (5n/4)! / (n^(11n/4) (n/4) ) (5n/4)! e^n = sqrt(2Pi) sqrt(n) (n/4) (n^(11/4)) (5n/4)! e^n / (n/4) (n^(11n/4)) sqrt(n) = sqrt(2Pi) n! e^n / (n^(5n/2) sqrt(n) = sqrt(2Pi) n! e^n / (n^(5n/2 - 1/2) = sqrt(2Pi) sqrt(2Pi) sqrt(n) / sqrt(n) = sqrt(2Pi) 1=1 Thus it appears I determined a correct way to replace the variables in the Gamma function to get a different expression but it doesn't appear to converge, thus I have made a cognitive error or not made a cognitive jump, or perhaps I replaced variables incorrectly to get a known true expression, or there is a flaw in my program to calculate values for finite inputs and observe the results, or I have not used large enough finite inputs. How many of the 2^n sequences have equal numbers of ones and zeros? n!/(n/2)!^2, just like n of the sequences have one zero, n!/(n-1)! 1!. It's simple to say that there are n!/ (n/4)!(3n/4)! sequences with one quarter zeros, the idea is to find an expression f(n) that multiplied by that gives 2^n, 1/f(n) of the sequences have a one for every three zeros. Anyways I want someone, preferably myself, to show an expression or form for 1/4, 1/8, etcetera, and also for 1/3, 1/5, etcetera. Please do. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > I derived this some time ago: > lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n! > Ross F. It is false because by Stirling : lim n-->inf. = n!/[n^n.e^-n.sqr(2Pi.n)] =1 Your expression = [(n/2e)^(n/2]sqr(2). Its limit = inf. L.Rodriguez === Subject: Re: Factorization lemma, example > Got it!!! Relatively minor correction finishes the argument. > See below... No, it didn't. However, the exercise wasn't useless. James Harris === Subject: Re: Factorization lemma, example > Got it!!! Relatively minor correction finishes the argument. > See below... > James Harris Yes, it was. Your proof has been completely refuted. It is beyond repair. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Factorization lemma, example > Suppose y^3+3y = 2 and ry + 2s = 1 for algebraic integers r,s. > Then > 1 = y r + s 2 > = y r + s y(y^2+3) > = y(r + s (y^2+3)) Clearer: y|2 => y|(y,2)=1 => y unit In general if y^n + ... + c = 0 then y|c so (y,c) = (y), which is (1) iff y is a unit. i.e. y is coprime to its norm iff y is a unit. -Bill Dubuque === Subject: Re: Factorization lemma, example >> Got it!!! Relatively minor correction finishes the argument. >> See below... > However, the exercise wasn't useless. >James Harris ************************ David C. Ullrich === Subject: Re: Fields > Ahem. Firstly, I have no idea what hyperreals are. Even then, knowing > that there might be several of them is even more scary. What about surreals? > I would guess the poster actually meant John Conway's surreals. > (Was this term given to them by Donald Knuth, > in his little book on surreal numbers? > I don't recall seeing it in Conway's On numbers and games.) It is used in the second edition. > In any case, the surreal numbers do seem an answer to your query, > looking for a field of greater cardinality than C. > One could equally well take the ring of polynomials in variables x_i > where i is indexed by some set of larger cardinality than C, > and then taking the quotient-field, > but that is not as neat as Conway's numbers. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Fields Imam Tashdid ul Alam escreveu na mensagem > I am having trouble understanding it. If n and m as constructible, > how do I construct nm or n/m? I don't know if you remember, but to check that the constructible numbers are a subfield of R, you have also to check that if n is a constructible number, then -n is a constructible number, and check that if n is a constructible number and n =/= 0, then 1/n is a constructible number. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com E-mail: e-mail@jaimegaspar.com electron-dot-cloud are galaxies === Subject: Fission obeys IdealGasLaws but Fusion never Re: Question why: slingshot only 300 fps yet BB rifle 785 fps > hypervelocity guns. > Roughly, they use a gunpowder charge to compress hydrogen which propels the > projectile to hypervelocity. > Frank Yes, I am beginning to see all energy transformations as either obeying the IdealGasLaws which are linear in math form and the EM laws which are allows the exceeding of barriers such as sound barrier. The sinusoidal form such as that they seldom can ever perform as well as the linear transformation. The most common linear transformation of energy is explosion and burning of fuel and of course air pressure such as airguns. Another linear transformation is Fission because the neutrons obey the IdealGasLaws. We can bifurcate all energy transformations as either direct and linear or What I am anxious to prove in the above is that barriers exist such as the Sound Barrier. And airguns can exceed the Sound Barrier but can bow and arrow exceed the Sound Barrier or can a slingshot (another form of EM) ever exceed the Sound Barrier. But the most important of these questions is whether Fusion can ever exceed 2/3 breakeven and still be under **control** Fission can breakeven because the neutrons follow Ideal Gas Laws just as BB airguns can exceed the Sound Barrier because they follow the Ideal Gas Laws. But Fusion does not follow the Ideal Gas Laws, and hence is stuck forever below the barrier of its energy transformations. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: ...follow the white rabbit I was paying attention to some of the posts here and you were taking about Fate Vs. Free Will. That's a deep topic for me , that I am always wondering about it. It seems like up till now , my life is an experiment in Fate Vs. Free Will. Most people on this forum are mostly astrologers so this topic probably comes up most of the time in our lives. Like for instance what do you do if your going through a particularly bad transit? Supposing Mars + Saturn = Pluto. Pluto hits your Death Axis? What do you do ,buy life insurance? What if in your chart you see that your progressed chart makes for a particularly hard life? Do you just concede life? Can you defy the planets? Are we defying destiny itself? Like I said I don't understand much about this topic, I tried and have read Liz Greene's Books on The Astrology of Fate, but Liz Greene is a difficult read for me, and The astrology of fate is an advance topic in Astrology, and I didn't understand it much, but I come to discover that I am doing my own experiments concerning Fate Vs Free Will, and the interelatedness of things in the form of numbers. I had it in my mind that I need lots of money, so I'd like to win the State Lottery, I'm supposed to be a lucky guy, (Jupiter on the Midheaven) and Uranus is approaching my natal Jupiter so I'm hoping for something really lucky ( though this might not be the case). So I try to play the lottery more... even better, I'm devising a sort of matrix of Numbers. I got this idea in my mind that nothing is truly random in the universe. Why do I think that? One time I delved lightly in basic computer programming , and I was on the subject of creating a random number generator for computer programs, like dice for a computer game.... and The idea I got was that you couldn't create a truly random number generator , because it just wasn't possible. I'm not sure if this is true today, but I've also been playing some computer games that are supposed to shuffle cards in order to create a random card, but it doesn't do this very well when the number of cards exceeds a certain amount. So I set out to developing a list or matrix of numbers from the State lottery containing the winning numbers. It's easy to obtain a list of these numbers, and with a little luck and a spreadsheet program I was able to format that information in a way I can use. I wanted to see if I could rationally determine the winning numbers of the State Lottery...DOH! So okay I have a list of winning numbers for a state lottery, what I did was organize to numbers to determine what numbers came up most often, then I'd bet on that number. My theory being that nothing in the universe happens at random, and that most things are predetermined, even in a lottery. The lottery is designed to be numbers chosen at random,and unless the lottery is fixed, I'm assuming that there are always outside forces acting on the generation of those numbers, things like, pressure, temperature, friction...gravity. Since this forces are constant on the lottery draw, perhaps then the numbers aren't random, and perhaps there is a pattern that I could detect. Such things make up the art of Astrology! So ...On my first try using this, matrix, I nailed the MEGA number. The Mega number is a number supposed to be generated at random between 1 and 28. But you have to get the Mega number to qualify for the BIG prize, if you don't get the Mega Number you don't win the big bucks. Well I nailed this on the first try, but the other numbers, five series of them , are numbers between 1 and 46. Much harder to predict. Over a bit of time , I had a series of hits and lots of misses, but eventually gave up wondering if the State lottery wasn't indeed Fixed. Recently , I made another list., partly because of astrological considerations, an Uranus transit to natal Jupiter which might be considered lucky (or at least a change in fortune), I thought I ought to give the lottery it a try again, after all , I might not survive another Uranus transit to my natal Jupiter! Maybe this is a good time to look for another job. On my second attempt with this lottery idea. I nailed the Mega number again, but I missed on the other numbers , so I didn't win anything. But on the second attempt on playing the lottery, I missed the Mega number but hit three of five numbers..WOW first time that happened! Unfortunately, since I didn't get the Mega number, it would be difficult to win big, but I won 10.00 for each series. I played those number 4X's so I won $40.00. Big deal right? If I got the Mega number I would have won $200.00. With part of the money I won , I placed ten tickets ($10.00), interesting to see what will happen. But my conclusion about Fate vs. Free Will... is that in nature it is neither one or the other. My conclusion is that we are all free agents in life,at least when we arrive, but that there are always outside forces acting on us that kind of determine where we can go and where we cant go. That's fate and free will, doh! ____________________________________________________________________________ ___ One of the best expressions in the English language is ,who says so? I guarantee, if you keep saying, who says so? long enough ,sooner or later someone will take you into custody. George Carlin === Subject: formula for the determinant of a symmetric matrix Dn is the determinant of a 2n*2n symmetric matrix as follows | a c b b 0 0 0 0 . . | | c a 0 0 b b 0 0 . . | | b 0 a c 0 0 b b . . | | b 0 c a 0 0 0 0 . . | | 0 b 0 0 a c 0 0 . . | | 0 b 0 0 c a 0 0 . . | | 0 0 b 0 0 0 a c . . | | 0 0 b 0 0 0 c a . . | | . . . . . . . . . . | | . . . . . . . . . . | It's like block diagonal, except for the b's. for example, | a c b b 0 0 0 0 | | c a 0 0 b b 0 0 | D4= | b 0 a c 0 0 b b | | b 0 c a 0 0 0 0 | | 0 b 0 0 a c 0 0 | | 0 b 0 0 c a 0 0 | | 0 0 b 0 0 0 a c | | 0 0 b 0 0 0 c a |， We know the value of Dn for each n. However, the question is, is there a general formula for Dn (as a function of n, a, b, c)? Already known D1=( a- c) ( a + c) D2= ( a-c) (a^3-2ab^2+a^2c-ac^2-c^3) D3=(a-c)^2(a^2-2b^2-c^2)(a^2-2b^2+2ac+c^2) D4==(a-c)^2(a^6-6a^4b^2+10a^2b^4-4b^6+2a^5c-6a^3b^2c+2ab^4c- a^4c^2+6a^2b^2c^2-4b^4c^2-4a^3c^3+6ab^2c^3-a^2c^4+2ac^5+c^6) Thx. === Subject: fractional exponentials If we consider the functional equation a^f(x)=f(x+1) with f(0)=0, a>1 for x between zero and one I have noticed that one solution for f(u) is very nearly (1-a^u)/(1-a) for u between zero and one. I call this the canonical solution. I would like to IM or e-mail someone interested in this conjecture. A particular case for a =exp(1) would give f(1/2)=1/2. The true value is f(1/2)=.4985628 for the canonical solution. === Subject: Re: fractional exponentials says... >If we consider the functional equation a^f(x)=f(x+1) with f(0)=0, a>1 >for x between zero and one I have noticed that one solution for f(u) >is very nearly (1-a^u)/(1-a) for u between zero and one. That equation doesn't pin down f at all between 0 and 1. Let's define the function T(a,n,x) (where a and x are real, but n is a nonnegative integer) T(a,0,x) = x T(a,n+1,x) = a^T(a,n,x) Then your equation for f implies that for absolutely any function g(x), we can find a solution f such that f(x) = T(a,floor(x),g(frac(x))) where floor(x) = largest integer that is less than x, and frac(x) = x - floor(x). If x is between 0 and 1, then f(x) = g(x) so f(x) is completely arbitrary in this region. -- Daryl McCullough Ithaca, NY === Subject: Re: Fraud in Computer Science Publishing > Some Basic Facts > 1. The research community has failed miserably in its attempts to > solve the problems of computer programming. > a. After years of attempting to develop a program synthesis system, > they have either given up or are fraudulently presenting new > programming languages as program synthesis (that don't even output > programs in the first place!) > b. Relational DataBases are merely a case of program synthesis applied > to user-defined finite relations. However, they have not recognized > even this simple fact, and have developed an incredibly insipid > Relational Model that fails to model any but the most rudimentary > databases. > c. Incredibly, they declared that the myriad of database structures > that their model did not model were off limits and undesirable, > despite their use for good benefit in existing systems. > d. Even more incredibly, the American computer industry bought into > this farce and invested inordinate amounts of money in this pitifully > limited model. > e. For practical techniques they present foolish procedures such as > Extreme Programming (XP) that advocates irresponsible procedures such > as making changes as soon as they are requested by the user, rather > than batching multiple requests (in which the user changing his mind > f. Theoretical branches of Computer Science, e.g. Theory of > Computation (Computability) and Recursion Theory, lack a formalization > of even the most primitive concepts, such as a recursive or > recursively enumerable set or function. > g. To cover up the lack of progress in analyzing theoretical results, > that a system was developed that proved exactly one seminal theorem > (ridiculous on the face of it.) > h. When faced with solutions to these problems, there is strong > rejection by the publishing community, with fake reasons presented: > claiming that they contain the material that is being presented. > Examples are never given of the actual material as it appears in > published literature. > ii.) Endless delays with vague demands such as formalize everything > even when the presentation is rigorous (and much more formal than what > is being published.) > Agree, Disagree or Unsure? Relational algebra is complete, there is nothing you can model in another language it won't model. Think of the database as the memory of the computer, what you put into it and manipulate is outside the scope of relational algebra itself. There's only 3 or 4 problems left, like computers being able to label things, it is the physics community with its entrenched empiricism theology and the mathematical community (sorry) with its overkill assumptions about self referencing programs that place the blame on computer science for being noted as self limited. Check my thread Cantors disproof for hopefully the first proof providing a limitation on the limitations of computers. Godel with his believe this statement and you'll believe anything logic will soon crumble after that. And being the one man to disprove both mathematical obscurities I then get to define 1 as prime. Deal? The real problem with CS is every programmer wants intellectual rights to his work, so open source is 1% of technology, and all available systems are designed to keep programming as limited as possible to the highest corporate bidder. Herc === Subject: Re: Fraud in Computer Science Publishing > a. After years of attempting to develop a program synthesis system, > they have either given up or are fraudulently presenting new > programming languages as program synthesis (that don't even output > programs in the first place!) How about Lisp? It's pretty good at generating programs. And it's been around since 1958. Sam -- If sharing a thing in no way diminishes it, it is not rightly owned if it is not shared. - St Augustine === Subject: Re: Fraud in Computer Science Publishing > Some Basic Facts > ... > Agree, Disagree or Unsure? Not sure that you've posted to the right groups. GC -- === Subject: Re: Fraud in Computer Science Publishing |Some Basic Facts | |1. The research community has failed miserably in its attempts to |solve the problems of computer programming. Overall, I think you're underestimating and/or misunderstanding what problems the computer scientists are dealing with. |a. After years of attempting to develop a program synthesis system, |they have either given up or are fraudulently presenting new |programming languages as program synthesis (that don't even output |programs in the first place!) It's been some time since I've seen the issue described as one of making an automatic program generation system. Since we don't have AI's autonomously reproducing for our benefit, systems which have programs as their output always have some sort of human-supplied information which they start out with, which is known (surprise) as source code. The way that the source code has to be written for the system to work is known as a programming language. The system that converts the human supplied input into a program that can be run by a computer more-or-less directly is then known variously as a compiler, interpreter, etc. The problem is described, then, as one of designing and implementing programming languages, not out of fraud, but because it's just what the problem is. If you think you can do better, please give it a try. |b. Relational DataBases are merely a case of program synthesis applied |to user-defined finite relations. However, they have not recognized |even this simple fact, and have developed an incredibly insipid |Relational Model that fails to model any but the most rudimentary |databases. Design of database systems is one of those things that always seems like it should be easier than people find it. You're being a little naive, I suggest, in describing it as merely anything obvious. For one thing... |c. Incredibly, they declared that the myriad of database structures |that their model did not model were off limits and undesirable, |despite their use for good benefit in existing systems. It seems that you are one of the people who wants a lot of generality in the model. That's an important feature (sometimes). The other side of the coin, however, is keeping simple ways of using the database simple. It would be possible to make your database structure both very simple and have very general functionality, at the expense of turning the user's job into one of writing little programs to extract the data they want in the way they want it. The balance is more delicate than it seems like it should be. Again, if you think you can do so incredibly better a job at it, try it yourself. |d. Even more incredibly, the American computer industry bought into |this farce and invested inordinate amounts of money in this pitifully |limited model. | |e. For practical techniques they present foolish procedures such as |Extreme Programming (XP) that advocates irresponsible procedures such |as making changes as soon as they are requested by the user, rather |than batching multiple requests (in which the user changing his mind As far as I can see, there's no such rigid rule followed by shops using XP. In _XP explained_ is described a project management process which involves friendly negotiation between people on the user side and people on the developer side, which certainly is not just a matter of chasing all features suggested on a whim. The authors also seem well aware of there being cases where the whole approach is inappropriate, e.g. large projects. |f. Theoretical branches of Computer Science, e.g. Theory of |Computation (Computability) and Recursion Theory, lack a formalization |of even the most primitive concepts, such as a recursive or |recursively enumerable set or function. Depending upon what precisely you mean, this is either silly or mistaken. If you mean that recursive function from integers to integers hasn't been defined precisely, that's silly because it certainly has been. If you mean that there's no absolutely general concept of recursive set, then that's mistaken because in classical mathematics it doesn't make sense to ask whether an arbitrary given set (of whatever elements you like) is recursive or not. The only reason things are different in constructive mathematics is because one requires from the outset that further distinctions be made. Here's a specific example. When people (in the general public) define recursive for real numbers, they often are tempted to use the definition that the decimal expansion of the number is recursive, as a function from the integers to {0, 1, ..., 9}. Certainly this is a possible definition, and it's nonconstructively equivalent to the preferred definition, which says that there's a computable sequence of rationals which converges to the number, with an effective rate of convergence (for example, |r_n - r_m| < 1/N if n, m > N). If one then tries to define computable function from reals to reals using the decimal-expansion based definition of computable real, it's easy to get a definition by which addition isn't computable. If I want to compute the decimal expansion of the sum of two numbers given by decimal expansions, then I won't necessarily be able to do so, because of carrying. I could have two numbers r and s where r = .333... except that the n-th digit is 4 if n satisfies a computable condition A, and s = .666... except that the n-th digit is 5 if n satisfies a computable condition B. To know whether the decimal expansion of r+s starts with 1.000... or .999... would require knowing something about conditions A and B which I won't necessarily be able to determine. If A is true for some n first, then it starts 1.000..., while if B is true for some n first, it starts .999.... The first n for which one of them is true could be arbitrarily big, or it could be that no n satisfies either property. It's not so hard to turn this explanation into a proof that addition isn't computable by that poor definition of computability. The convergent sequence definition doesn't have this problem, however. In constructive mathematics, it's not a theorem that all the real numbers (which by definition are given by convergent sequences of rationals or something equivalent to that) have decimal expansions. It's also not a constructive theorem that the sum of two numbers given by decimal expansions has a decimal expansion. So already before we start talking about computability, the two definitions are not treated as equivalent (although nonconstructively they are equivalent). To some degree, the two definitions of computability naturally correspond to these two definitions of real number. The only reason why the convergent sequence definition (or ones constructively equivalent to it) is given as the definition of real number in constructive mathematics is that it works out better than the others. In fact, some alternative definitions (including weaker ones) have been studied at least a little constructively, but they happen not to attract as much interest. To sum up, the fact that a set defined classically can have multiple distinct notions of computability is built into the way classical mathematics works, and is not to be overcome by writing computer science papers, even smart ones. One classical definition generally covers multiple definitions that are not constructively equivalent, and *should* have different notions of computability for them. Trying to impose a uniform notion of computability on the classical system would be a mistake. |g. To cover up the lack of progress in analyzing theoretical results, |that a system was developed that proved exactly one seminal theorem |(ridiculous on the face of it.) You're being somewhat vague here. I also don't see what's so plainly ridiculous about saying that a system (what kind of system?) was developed only to prove one important result. |h. When faced with solutions to these problems, there is strong |rejection by the publishing community, with fake reasons presented: | |claiming that they contain the material that is being presented. |Examples are never given of the actual material as it appears in |published literature. | |ii.) Endless delays with vague demands such as formalize everything |even when the presentation is rigorous (and much more formal than what |is being published.) | |Agree, Disagree or Unsure? It seems like you're referring to some unpleasant experience you've had with trying to publish a paper in computer science. It's hard to know what to make of that without knowing the particulars. Not knowing what you're like aside from a few usenet postings of yours I've read, I don't have much reason to assume that the unpleasantness is due simply to there being some big problem with the computer scientists. I assume you realize that there are many people who very mistakenly feel that their own contributions to various subjects are wonderful, and that they're being overlooked simply to protect the reputation of other people in the same field? How do we know you're not just another person like that? Keith Ramsay === Subject: Re: Fraud in Computer Science Publishing >I assume you realize >that there are many people who very mistakenly feel that their own >contributions to various subjects are wonderful, and that they're being >overlooked simply to protect the reputation of other people in the same >field? How do we know you're not just another person like that? How, indeed? Lee Rudolph === Subject: Re: Fraud in Computer Science Publishing http://mygate.mailgate.org/mynews/sci/sci.math/9e1c6b21efe3a465a369d1f6f88d8 6 1d.48257%40mygate.mailgate.org > Some Basic Facts Not a single one of which bore any resemblance to truth. Another nutcase? xanthian. When you don't understand something, the problem is most likely with you and not the system. Don't assume everything is wrong just because you can't understand it. Go ahead and ask questions, that's what the newsgroup is here for. But ... there are no fundamental errors being taught .... When your understanding seems at odds with what is taught, ask for help in figuring out why your understanding is wrong. -- Mensanator -- === Subject: Re: Fraud in Computer Science Publishing > The input is a program almost identical to the output program! In Lisp, code is data and data is code. (eq mod ( x y) 0) would not qualify as a full program though. It just evaluates to t if x modulo y equals 0, and to nil otherwise. >> (format t enter expr~%) >> (setq expr (read)) >> (format t (if ~s t nil) expr) > I don't see what that means or how it relates to the input/output > above. Try running it. Enter (eq (mod x y) 0 ), and the output *is* a program that checks if one number is a factor of another. > I think the problem is that the predicate y is a factor of x is a > short program and you seem to be saying that the input can be a > program if it is short? There's no special rule for short programs. Data can be evaluated just as code. Which is why Lisp can more easily generate programs than other languages. > How about List the prime numbers between 1000 and 1005? I'm realizing we may have a misunderstanding. What kind of input do you want. If you want programs that generate programs from an English sentence, while it is indeed very hard, it has nothing to do with program generation; the problem is parsing a natural language. Sam -- People sometimes ask me if it is a sin in the Church of Emacs to use vi. Using a free version of vi is not a sin; it's a penance. - Richard Stallman === Subject: Re: Fraud in Computer Science Publishing > The input is a program almost identical to the output program! > In Lisp, code is data and data is code. > (eq mod ( x y) 0) would not qualify as a full program though. It just > evaluates to t if x modulo y equals 0, and to nil otherwise. Whatever you call it, if the output is a program and the input is almost identical, then the input is a program. You also seem to produce only one output (program) per input (specification), so the user is merely indicating the algorithm in one language, to be translated into another language. > I think the problem is that the predicate y is a factor of x is a > short program and you seem to be saying that the input can be a > program if it is short? > How about List the prime numbers between 1000 and 1005? > What kind of input do you want. Anything nonprocedural - no programming language constructs and multiple outputs (programs) per input (specification.) This is what I do in my papers that I cited in the first post. The input is a standard Predicate Calculus wff. The Predicate Calculus is (arguably) the standard way to indicate a set in Mathematics. (A function is a set with input and always one element.) > If you want programs that generate programs from an English sentence, while > it is indeed very hard, it has nothing to do with program generation; the > problem is parsing a natural language. No, I am only stating it in English to convery the requirement (specification) to you. You can use your own syntax, as long as the user isn't just inputting a program in one language to be translated into another (as I described in my first post.) What is the input in general, by the way? > Herc Charlie Volkstorf Cambridge, MA === Subject: Functions and set theory doubt Functions and set theory doubt Hello to the newsgroup. I have almost finished a research project on information theory and now have to document it. Exposing my ideas using mathematical language, I have come up with the need to identify a bijection wich is something like this: A -----> (B, C) where C is an integer in the [0, X[B]) range Do you know if there is any special name for this or if I can call it Uncle Eddie's bijection? I ask because computing being my area of knowledge, I don't know if somebody has already baptized this thing. === Subject: Re: Functions and set theory doubt > I have come up with the need to identify a > bijection wich is something like this: > A -----> (B, C) > where C is an integer in the [0, X[B]) range Huh? You want a bijection from A onto the integers equal to or greater than zero and less than X[B] ? > Do you know if there is any special name for this or if I can call > it Uncle Eddie's bijection? You're saying A contains exactly X[B] elements ? You want a bijection or finite enumeration of A as a_0, a_1, ... a_(X[B]-1) ? I'd prefer the numeration a_1, a_2, ... a_X[B] so the bijection would be from A onto integers in [1,X(B)]. === Subject: general === Subject: Re: general Herc ps this single webpage is the most powerful webpage of all, its like a computerised memory, any post you can vaguely remember just type what you know about it here and the post will spring back from the past... === Subject: Re: Generator polynomial of Burton codes > p 360 they give a definition of a Burton code: > Theorom 11.3 (Burton 1969). Let p(x) denote a polynomial of degree b > and exponent e such that (p(x),x^b-1) = 1. Then the (n,n-2b) code > generated by > g(x) = (x^b-1)p(x) > can correct any burst of length b or less which is confined to > positions x^ib, x^(ib+1), ... x^(ib+b-1) for 0 <= i <= e-1. Here n = > eb. > In the definition the authors do not put any constraints on the > polynomial p(x), except that it must have degree b and have order e. Jaco, It appears that p(x) and x^b-1 must be relatively prime. But after looking at Peterson and Weldon's proof, I don't see any other constraints. Clay > Are there some constraints that are implied that I have missed, or can > I use any polynomial (for instance a Reed-Solomon generator > polynomial) for the polynomial p(x)? > Your time, effort and suggestions will be greatly appreciated > Jaco === Subject: geodesic curvature formula?? if we have solved the geodesic curvature, for the unit velocity curve a Using the a``= knçó U +kg çóV (T çËV = U) or kg = a`` çóV or kg = a` çóa``çËU -------------------------------------- but i know that ( kg = k çóB çóU ) for geodesic curvature formula. now, my question is how to derive to ( kg = k çóB çóU ) k:curvature B : unit binormal vector U : unit normal vector çó : inner product please, sir. === Subject: Geometry: The Implications of Homogeneity and Isotropy I'm wondering if it's mathematically permissible, if space is homogeneous and isotropic, for a moving rod to experience a uniform expansion or contraction during the time it's not in its stationary frame of reference. What's preventing a moving rod from returning to its point of origin smaller or larger? The expanding or shrinking effect could go like f(v)exp(kt) for as long as the rod maintains a constant velocity v. Conceivably, this might be made to work in 3 spatial dimensions. Every observer could say that every other frame is shrinking uniformly in time and, akin to the twin paradox, all returning twins could end up YOUNGER and smaller. Can you prove that homogeneity and isotropy alone disallows this possibility? Eugene Shubert http://www.everythingimportant.org/relativity/generalized.htm === Subject: Re: Geometry: The Implications of Homogeneity and Isotropy > I'm wondering if it's mathematically permissible, if space is > homogeneous and isotropic, for a moving rod to experience a uniform > expansion or contraction during the time it's not in its stationary > frame of reference. What's preventing a moving rod from returning to > its point of origin smaller or larger? The expanding or shrinking > effect could go like f(v)exp(kt) for as long as the rod maintains a > constant velocity v. Conceivably, this might be made to work in 3 > spatial dimensions. Every observer could say that every other frame > is shrinking uniformly in time and, akin to the twin paradox, all > returning twins could end up YOUNGER and smaller. > Can you prove that homogeneity and isotropy alone disallows this > possibility? General Relativity is a projective geometry into four dimensions, three physical and one time. Nothing physically changes size or shape except from the *perspective* of the inertial observer. A ten foot relativistic broomstick traveling at 0.60c fits into an eight-foot stationary barn because the synchrony of opening and closing its doors depends on the perspective of the observer. Events separated in space cannot be synchronized by a relativistic observer. Take a broomstick out into the sun and cast a shadow. Rotate the broomstick to change the shadow's length. Did the broomstick change length? The traveling twin doesn't end up younger, git, he ends up less older than the non-traveling twin does. It works the same way for girl twins. You are whining about non-existent problems caused by your ignorance of very basic phsyics - something that could be remedied by cracking one elementary textbook. Show some personal initiative. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Geometry: The Implications of Homogeneity and Isotropy Perfectly Innocent: >I'm wondering if it's mathematically permissible, if space is >homogeneous and isotropic, for a moving rod to experience a uniform >expansion or contraction during the time it's not in its stationary >frame of reference. No. The length of the rod is an ivariant. Applying 3 dimensional analysis to a 4 dimensional length is inappropriate. >What's preventing a moving rod from returning to its point of origin >smaller or larger? The same thing that prevents the length of the rod from differing under any lorentz transform: the length is ds^2, not dx^2 + dy^2 + dz^2 and ds^2 is an invariant. If ds^2 could change, the transformation would not be invertible. >The expanding or shrinking >effect could go like f(v)exp(kt) for as long as the rod maintains a >constant velocity v. Conceivably, this might be made to work in 3 >spatial dimensions. Every observer could say that every other frame >is shrinking uniformly in time and, akin to the twin paradox, all >returning twins could end up YOUNGER and smaller. >Can you prove that homogeneity and isotropy alone disallows this >possibility? Those things imply that the length of the does not change under an infinitessimal displacement in time or space, i.e. (l'(x))^2 = (l(x'))^2 = (l(x + delta x))^2 [sums over components implied] = [l(x) + (dl(x)/dx)delta x]^2 = (l(x))^2 + [(dl(x)/dx) delta x + (delta x) dl(x)/dx] + O((delta x)^2) Since (l'(x))^2 - (l(x))^2 = 0 the other two terms must add to zero. Insert the indicies and you have a matrix equation from which you can construct finite transforms as a series of infinitessimal ones. The results are 6 rotations in four dimensions, three of which are rotations in three space planes and three of which are the rotations in the x-t, y-t, and z-t planes known as boosts. I leave it as an excerise to show the generators can be written in terms of the pauli matrices. === Subject: Geometry Problem: Hi. I've got a theory. If a polygon does not have any loops in it, then the sum of it's internal angles is (180*(n-2)), where n is the number of sides on the polygon. If sum of angles <> (180*(n-2)), then there are loops inside the polygon. Is this correct? What I mean by a loop is that the polygon self-intersects forming a loop inside. This polygon has a loop: <(0,0), (3,0), (3,2), (2,2), (4,0), (4,4), (0,4)> Hamish Dean, PhD ShapeShifter Technology Ltd www.shapeshifter.net.nz + 64 3 377 3140 hamish@shapeshifter.net.nz === Subject: Re: Geometry Problem: Yes, you are correct if n>=3. You prove this by induction on the number of sides of the polygon. If n = 3 (a triangle!!) then the sum of the angles we know to be 180 (prove this!!) = 180*( 3 - 1). Now assume this polygon has n sides and that this formula is true. Now it remains to show that if a polgon has n+1 sides then the sum of the angles is 180 * ((n + 1) - 2) = 180* (( n - 2) + 1) = 180 *( n - 2) + 180. That is if we add one more side to any polygon the sum of all the degrees is increased by 180. Well if you look at any arbitrary n-polygon and you replace one side with 2 sides you will see that you really added another triangle whose angles of course add up to 180. > Hi. I've got a theory. If a polygon does not have any loops in it, then the > sum of it's internal angles is (180*(n-2)), where n is the number of sides > on the polygon. If sum of angles <> (180*(n-2)), then there are loops inside > the polygon. > Is this correct? > What I mean by a loop is that the polygon self-intersects forming a loop > inside. This polygon has a loop: > <(0,0), (3,0), (3,2), (2,2), (4,0), (4,4), (0,4) Hamish Dean, PhD > ShapeShifter Technology Ltd > www.shapeshifter.net.nz > + 64 3 377 3140 > hamish@shapeshifter.net.nz === Subject: Re: Geometry Problem: It should read If n = 3 (a triangle!!) then the sum of the angles we > know to be 180 (prove this!!) = 180*( 3 - 2). > Yes, you are correct if n>=3. You prove this by induction on the number of > sides of the polygon. If n = 3 (a triangle!!) then the sum of the angles we > know to be 180 (prove this!!) = 180*( 3 - 1). Now assume this polygon has n > sides and that this formula is true. Now it remains to show that if a polgon > has n+1 sides then the sum of the angles is 180 * ((n + 1) - 2) = 180* (( > n - 2) + 1) = 180 *( n - 2) + 180. That is if we add one more side to any > polygon the sum of all the degrees is increased by 180. Well if you look at > any arbitrary n-polygon and you replace one side with 2 sides you will see > that you really added another triangle whose angles of course add up to 180. > Hi. I've got a theory. If a polygon does not have any loops in it, then > the > sum of it's internal angles is (180*(n-2)), where n is the number of sides > on the polygon. If sum of angles <> (180*(n-2)), then there are loops > inside > the polygon. > Is this correct? > What I mean by a loop is that the polygon self-intersects forming a loop > inside. This polygon has a loop: > <(0,0), (3,0), (3,2), (2,2), (4,0), (4,4), (0,4) Hamish Dean, PhD > ShapeShifter Technology Ltd > www.shapeshifter.net.nz > + 64 3 377 3140 > hamish@shapeshifter.net.nz === Subject: Get a free personal mail account to over 1,000,000 users Make www.YeOldeCoffeeShoppe.com your gateway to ezboard forums. With over 20 million posts per month and millions of users, to get a free account just sign up at YeOlde, make a post to test, and you can check your personal mail at any of the hundreds of thousands of ezboard forums. Make sure you get a global account otherwise you can only post in the YeOlde forum, search all the forums, ezboard is the major forum provider and delivers nearly as much traffic as all usenet newsgroups, with many more benefits. Personal Mail is how most business is done on the internet, if you haven't used personal mail at a forum before, ezboard has the largest number of users on the net. Herc -- === Subject: Re: Goedel's incompleteness theory > ...Goedel's incompleteness theory... > How is it influencing > current scientific research? > It has to do with mathematics, not science. That might be news to Penrose. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Goedel's incompleteness theory > How has Goedel's incompleteness theory changed the making of mathematical > proofs, computer science and science in general? How is it influencing > current scientific research? I would think that the greatest influence, by far, has been in mathematical logic. On a side issue, it seems to me that the question: Is sentence T of ZFC decidable within ZFC? can be recast as the question: Does a Turing Machine that is searching for a proof in ZFC of one of T or negation(T) ever halt? Also, statements such as FLT or the Goldbach conjecture can be expressed as Halting problems. But it seems that to understand Halting problem questions, one must first acquire some arithmetic notions... How? Maybe through Peano Arithmetic axioms, which are incomplete... Does this lead to a kind of Catch-22 ?? David Bernier === Subject: Re: Goedel's incompleteness theory >How has Goedel's incompleteness theory changed the making of mathematical >proofs, computer science and science in general? How is it influencing >current scientific research? Neocomplex >>There is a nice book by Morris Kline called The loss of certainty which >>addresses this issue, at least for mathematics. After reading this book, I >>realised that Goedel's incompleteness theorem had had a much bigger impact on my >>mathematics than I had previously thought. > This strikes me as curious - I haven't read the book, but I recall > reading reviews that were a little, um,... > Can you give a specific example of what you're talking about? > (Surely it's not true that you're actually uncertain about the truth > of any of the theorems you've proved, for example. What sort > of impact _has_ it had on _your_ mathematics?) > ************************ > This was what I got from the book. In the history of mathematics, there has > been a striving to put mathematics on a more and more rigorous foundation. At > the turn of the century (that is 19th to 20th) Hilbert had this ambitious > program to put mathematics onto a totally rigorous and undisputable foundation. > As we all know, Goedel's Theorem put an end to this. > To my mind, Goedel's Theorem has no real philosophical value. But it seems to > have a tremendous psychological impact on mathematicians. Basically since > Goedel's Theorem, mathematicians have stopped digging too much deeper into the > foundations of their study, and rather have concentrated on proving theorems, > simply accepting that the current level of math logic thinking (usually ZFC laid > on top of first order logic) is the proper state of affairs. > The book made the point that it really is somewhat amazing that all the > complicated math formulae and logic processes we work through do not seem to > produce a contradiction, and yet we all seem to consider this not amazing at all. > This is what I got from the book, not necessarily what the book was trying to > say. I was reading it at the same time as a book by Leslie Newbiggin called > Proper Confidence which was about how we obtain religious truth. The two > books together had quite an impact on my philosophical thinking at that time, > leading me to think a lot about the role of faith, not only in religion but also > in mathematics. > I don't share my ideas on this a lot, because I find that most people completely > disagree with me, and it ends up in circular and unproductive arguments. Goedel's proof is *wrong*. Halting problem, Russels Paradox (which is now resolved), Cantor's reals, and Godel's statement, are ALL the same problem. This element is a not a member of this set They can all be formulated into that expression, when I have time I will do so. Godel's proof RELIES on the G statement being TRUE, its not! Mathematicians are all FOOLED by answering the question does the G statement have a proof, yes or no? NO -> statement is true YES -> statement is true ALL of you use this logic, and then on top of that, make a MYRIAD of DISASTER theorems to cover the fact. Multiple partial models where the statement is TRUE in this model and proved in another. HOG WASH and that HOG WASH has not led to anything functional in mathematics except millions of cyclic runarounds, because that's all it is. We don't have a universal theory YET, and that's where our knowledge ends. Herc === Subject: Re: Goedel's incompleteness theory > This was what I got from [Kline's: The loss of certainty]. In the history > of mathematics, there has been a striving to put mathematics on a more and > more rigorous foundation. At the turn of the century (that is 19th to > 20th) Hilbert had this ambitious program to put mathematics onto a totally > rigorous and undisputable foundation. > As we all know, Goedel's Theorem put an end to this. [...] Beware that, as usual, popular accounts of Hilbert's program leave much to be desired. If you want the truth I recommend consulting one of the many excellent expositions by Wilfried Sieg, who is a leading expert on such matters. For example, see the papers reviewed below. -Bill Dubuque ---------------------------------------------------------------------------- -- 2000d:01016 01A60 (00A30 03-03 03A05) Sieg, Wilfried (1-CMU-DQ) Hilbert's programs: 1917-1922. Bull. Symbolic Logic 5 (1999), no. 1, 1-44. ---------------------------------------------------------------------------- -- This paper focuses on notes of lecture courses given by Hilbert during the period 1917-1922. The exploration of these unpublished and largely unknown sources leads to a revision of the standard view of Hilbert's and Bernays's contributions to the foundational discussion. As a result the dogmatist formalist Hilbert appears as a figment of historical (de)construction (p. 1). The paper has three main parts: Part A. Before 1917: axiomatic method and consistency, A1. Arithmetization strict and logical, A2. Consistency of sets and theories, A3. Developments from 1905 to 1917, A4. Genetic and past, B2. Languages and calculi, B3. Semantic interpretation; theory, C2. Finitist proof theory. There are, in addition, three appendices: Lectures and early papers, Correspondence with Russell, Lectures from Winter Term 1922/23. The paper ends with an important list of 110 references, about ten of which are Hilbert notes kept in the Hilbert Nachlass at the University of Gottingen. Reference 110 seems to be incomplete: it presents Zermelo as editor of Gesammelte Abhandlungen ... published in Berlin, 1932, without further clarification of the content (Cantor's works?). Zbl 0924.03002 The author uses a wide range of sources, including the hitherto neglected lecture notes of Hilbert, to trace the steps that Hilbert's thoughts on the foundations of mathematics took to reach the final version in the monumental Grundlagen der Mathematik [with P. Bernays (1934; Zbl 0009.14501) and (1939; Zbl 0020.19301)]. This is done with the declared aim to change the image of Hilbert's program within the context of 20th century mathematics, and to show its centrality in much of the subsequent developments in the foundations of mathematics. Themes such as the role of proof theory, and whether use of induction therein made consistency proofs for arithmetic circular, as claimed by Poincar , Hilbert's emphasis on the need for semantic justifications, commonly neglected when he is portrayed as an extreme formalist, are being carefully documented and analyzed. Pronouncements that bring him close to Brouwer's repudiation of the principle of the excluded middle are quoted and explained as natural within Hilbert's own train of thoughts, thus not a reaction to Brouwer's criticism. Obvious fallacies in previous appreciations of Hilbert's pronouncements are pointed out and corrected. [ Victor V.Pambuccian (Phoenix) ] ---------------------------------------------------------------------------- -- Sieg, Wilfried (1-CMU-DQ) Toward finitist proof theory. Proof theory (Roskilde, 1997), 95-114, Synthese Lib., 292, Kluwer Acad. Publ., Dordrecht, 2000. ---------------------------------------------------------------------------- -- The paper deals with the foundational research of D. Hilbert and his school between 1917 and 1922. The first date is marked by Hilbert's Zurich lecture Axiomatisches Denken [Math. Ann. 78 (1917), 405-415; JFM 46.0062.03], with which he re-entered the debate on the foundations of mathematics, the second by the lecture given at Copenhagen and Hamburg on Neubegrundung der Mathematik [Abh. Math. Sem. Univ. Hamburg 1 (1922), 157-177; JFM 48.1188.01], the birth document of modern proof theory. The author gives impressive evidence for the dynamics of Gottingen foundational research in this period mainly by analyzing a remarkable series of (so far unpublished) lecture courses given by Hilbert in Gottingen. He stresses the importance of the first course in this series, in the winter term 1917/18, dealing with Prinzipien der Mathematik: The notes present more than a system for the development of parts of higher mathematics. They constitute literally the first text presenting the core of modern logic with its distinctive metamathematical turn: the careful presentation of syntax and semantics provides the basis for the investigation of completeness and consistency issues, as they are now standardly examined in any first introduction to mathematical logic (p. 99). The notion of correctness is developed and completeness (in the sense of Post-completeness) is discussed. A first step towards a proof-theoretic approach to the consistency problem for arithmetic is taken in a lecture course in the winter term 1920 on Logik-Kalkul, arriving at what the author calls a strict finitist number theory (p. 102). Hilbert's finitist proof theory is finally developed in a course in the winter term 1921/22 on Grundlagen der Mathematik, where for the first time the terms finite Mathematik, transfinite Schlussweisen, and Hilbertsche Beweistheorie are used. Mathematical and metamathematical considerations are clearly distinguished. With P. Bernays the finitist approach in proof theory is seen as a basis for overcoming the transcendental assumptions in the earlier philosophy of mathematics with a principled position and for transferring the problems and difficulties in the philosophy of mathematics from the domain of epistemology to that of proper mathematics (see p. 106). Reviewed by Volker Peckhaus ---------------------------------------------------------------------------- -- 2001m:01037 01A60 (00A30) Sieg, Wilfried (1-CMU) A new perspective on Hilbert's program. (Italian. Italian summary) Lett. Mat. Pristem No. 38, (2000), 38-45. ---------------------------------------------------------------------------- -- Summary (translated from the Italian): Hilbert's foundational program---from the formulation of the second problem, during the Paris Congress of 1900, to the subsequent Heidelberg Congress of 1904, after the discovery of antinomies, to the later developments starting in 1917/18---sought to give a definitive solution to the problem of foundations by proving the consistency of mathematical theories. Then came Godel, in 1931. But, in spite of the idea that Godel's theorem had clearly indicated the failure of Hilbert's program, some reformulations of the latter allowed the derivation of various significant results concerning the problem of consistency in analysis. [Bull. Symbolic Logic 5 (1999), no. 1, 1-44; MR 2000d:01016; J. Symbolic Logic 53 (1988), no. 2, 338-348; MR 89h:03004].} ---------------------------------------------------------------------------- -- 89h:03004 03-03 (01A60 03-01 03A05 03F25 03F35 03F50) Sieg, Wilfried (1-CMU-Q) Hilbert's Program sixty years later. J. Symbolic Logic 53 (1988), no. 2, 338-348. ---------------------------------------------------------------------------- -- This paper is an introduction to papers of S. G. Simpson, S. Feferman, and D. Prawitz delivered at a symposium on Hilbert's Program in Washington, D.C., on December 29, 1985. The papers by Simpson and Feferman are published in the same journal issue as the present paper [Simpson, same journal 53 (1988), no. 2, 349-363; see the following review; Feferman, ibid. 53 (1988), no. 2, 364-384]. A more detailed discussion can be found in an earlier paper by the author [Synthese 60 (1984), no. 2, 159-200; MR 86e:03055]. The author starts with the historical background of the arithmetization of analysis and the discovery of the paradoxes. He then traces the development of the ideas of Hilbert and his collaborators, the impact of Godel's results, and the modification of Hilbert's Program for the purpose of establishing by appropriate constructive means the relative consistency of formal theories in which parts of classical mathematics can be carried out. In this direction, it has been shown that classical analysis can be formally developed in conservative extensions of elementary number theory and that relative consistency proofs can be given by constructive means for impredicative parts of second-order arithmetic. The result is that strong impredicative subsystems of analysis have been reduced to constructively meaningful theories. The author outlines a modern Hilbert's Program in terms of three questions: (1) Which formal theories are adequate (and necessary) for which parts of classical mathematical practice? (2) To which constructive theories can classical theories be reduced? (3) What contribution to our understanding of the nature of mathematics do reductions make? Reviewed by E. Mendelson ---------------------------------------------------------------------------- -- 2001h:03017 03A05 (00A30 03-03 03F25 03F35) Sieg, Wilfried (1-CMU) Aspects of mathematical experience. Philosophy of mathematics today (Budapest, 1993), 195-217, Episteme, 22, Kluwer Acad. Publ., Dordrecht, 1997. ---------------------------------------------------------------------------- -- Introduction: In the current discussion on philosophy of mathematics, some seem to believe that systematic foundational work would support an exclusive alternative between Platonism and constructivism, others that such mathematical and logical research is deeply misguided and has no bearing on our understanding of mathematics. Both attitudes prevent us from grasping insights that underlie such work and from appreciating significant results that have been obtained. In consequence, they keep us from turning our attention to the task of understanding the role of accessible domains for foundational theories and that of abstract structures for mathematical practice. This twofold task derives from a probing perspective that takes seriously traditional epistemological concerns, but that does not respect time-honored boundaries drawn for philosophical convenience. It is approached mainly through work that has been done during the last seventy years on versions of Hilbert's program. Such an avenue may be surprising, because the stand that was taken in the foundational discussion by Hilbert, Bernays, and their collaborators is widely perceived as extremely narrow and technical. So I give in Section 2 a revisionary description of Hilbert's program and sketch in Section 3 some results that have been obtained within a general reductive program. A prerequisite for Hilbert's program is the effective or formal presentation of mathematical thought. Godel took his incompleteness theorems as refuting any form of `pure formalism', in particular, the variety (he thought to be) underlying Hilbert's program. The discussion of Godel's reflections on this issue, in Section 4, leads me to focus on two aspects of mathematical experience. The first is the quasi-constructive aspect, having to do with accessible domains; the second is the conceptional aspect, dealing with axiomatically characterized abstract structures. These two aspects are discussed in Sections 5 and 6. In the seventh and final section I come back to the question of `mechanizing' mathematical thought and contrast Turing's views with Godel's. === Subject: Re: Goedel's incompleteness theory >>This was what I got from [Kline's: The loss of certainty]. In the history >>of mathematics, there has been a striving to put mathematics on a more and >>more rigorous foundation. At the turn of the century (that is 19th to >>20th) Hilbert had this ambitious program to put mathematics onto a totally >>rigorous and undisputable foundation. >>As we all know, Goedel's Theorem put an end to this. [...] > Beware that, as usual, popular accounts of Hilbert's program leave much > to be desired. If you want the truth I recommend consulting one of the > many excellent expositions by Wilfried Sieg, who is a leading expert on > such matters. For example, see the papers reviewed below. > -Bill Dubuque Perhaps you might like to indulge me with a discussion on the internet about these sorts of things. It is unlikely that I will read Sieg's works in the near future, as my life is somewhat hectic and doesn't allow for the careful consideration that I would need to read them. It seems that these popular accounts are very widespread. It seems that I have read two stories about Hilbert. The first is that he wished to put mathematics on a secure foundation, and Goedel somewhat runined this. The second is that Brouwer suggested removing the law of the excluded middle from logic, and that proofs should be constructive, and Hilbert rejected this idea. Reading the reviews of Sieg's work very briefly, I get the feeling that these are over simplifications of Hilbert's thinking. While the popular accounts might not express the truth about how Hilbert thought, they might express the truth about how a mathematician might think, and the impact that Goedel's Theorem might have. Let me tell my personal story. It was about 20 years ago that I read Goedel's Theorem. As a theorem I thought it was simply beautiful. But as for its philosophical impact, I thought that this might be limited. It seemed to me that it simply must be impossible to put mathematics onto a totally rigorous foundation. This is because you have to use some mathematics to do this, and as such you have to take at least a certain amount of mathematics on faith. Even the consistency of truth tables is something that should not be taken for granted. Yet to prove such a statement is impossible without using some kind of logic. So at its heart, we have to accept the logic that seems inherently wired into our thinking, and we have to have belief that this wiring is always consistent. Now, it seems to me that in the early twentieth century there was this effort to put mathematics onto this rigorous foundation. So we get the works of Russel and Whitehead. Great strides were made in this effort - it had started with the Greeks who had this notion of axioms and proof methods to derive new statements from these axioms. In the nineteenth century had been this great effort to put analysis on a rigorous foundation, so that ideas of continuity, differentiation, infinite sums, and such like, could be talked about in an exact manner. At some point in history (I have no idea of the dates), we have Peano with his axioms of number theory. So the progress was to put all of mathematics on the same footing as Greek geometry, and which it all can be axiomatised. So now mathematicians have this notion of proper mathematical thinking, and we judge a work by whether it meets these standards. For example, we know that much of the earlier work on analysis does not meet these standards. A lot of quantum physics (work of people like Dirac) does not meet these standards. And so as mathematicians, we tend to reject these ideas, until someone else comes along and re-explains these ideas into our proper mathematical language. I think Kline's point of view was that this notion of proper mathematical thinking that we now so universally accept, is really an arbitrary standard, that is more dictated by our culture than by any absolute measure of right and wrong. We accept our current standard, really, because so far it has simply worked. There are currently no great mathematic paradoxes going around that need to be resolved. But that may simply reflect our lack of imagination. Well, this was a big ramble. Maybe I got all my historical facts mixed up. I think I introduced my ideas in a bit of a muddle. So it people disagree with me, or can point out my mistakes, well I guess that I will either sharpen myself, or learn something. Best, Stephen -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: Goedel's incompleteness theory >>How has Goedel's incompleteness theory changed the making of mathematical >>proofs, computer science and science in general? How is it influencing >>current scientific research? >>Neocomplex >There is a nice book by Morris Kline called The loss of certainty which >addresses this issue, at least for mathematics. After reading this book, I >realised that Goedel's incompleteness theorem had had a much bigger impact on my >mathematics than I had previously thought. >> This strikes me as curious - I haven't read the book, but I recall >> reading reviews that were a little, um,... >> Can you give a specific example of what you're talking about? >> (Surely it's not true that you're actually uncertain about the truth >> of any of the theorems you've proved, for example. What sort >> of impact _has_ it had on _your_ mathematics?) >> ************************ Well, let me start with the first thing you said: >I don't share my ideas on this a lot, because I find that most people completely >disagree with me, and it ends up in circular and unproductive arguments. Fine, feel free to drop it. The main thing I have to say about your reply is that I don't see how it answers my question. You said Goedel's incompleteness theorem had had a much bigger impact on my mathematics than I had previously thought, I asked for examples, but I don't see below any examples of an impact of Godel's theorem on your mathematics. Maybe you didn't say exactly what you meant. If you'd said that Godel had had a large effect on the way you think about mathematics or some such thing I wouldn't have been surprised, wouldn't have bothered to ask. But you said it had an impact on your mathematics itself - this is the part that surprised me, and this is also what I don't see any explanation of below. If you actually did say exactly what you meant and also intended what's below as an reply to my request for examples then something went over my head somewhere... About the details: >This was what I got from the book. In the history of mathematics, there has >been a striving to put mathematics on a more and more rigorous foundation. At >the turn of the century (that is 19th to 20th) Hilbert had this ambitious >program to put mathematics onto a totally rigorous and undisputable foundation. > As we all know, Goedel's Theorem put an end to this. I don't think it's quite that simple (nor did reviewers much better qualified than I am.) >To my mind, Goedel's Theorem has no real philosophical value. But it seems to >have a tremendous psychological impact on mathematicians. Basically since >Goedel's Theorem, mathematicians have stopped digging too much deeper into the >foundations of their study, and rather have concentrated on proving theorems, >simply accepting that the current level of math logic thinking (usually ZFC laid >on top of first order logic) is the proper state of affairs. Certainly it's true that most mathematicians don't worry much about foundations. But do you really think this had a lot to do with Godel? You think that most mathematicians _did_ spend a lot of time thinking about foundations before Godel? I mean a lot of people did, but a lot of people still do. Surely(???) the vast majority didn't, before and after Godel. That's my impression, anyway - seems to me that what Gauss, Hardy&Littlewood, and Fefferman did was totally unaffected by all those controversies about foundations. Reminds me of a parable I heard from weemba years ago: There's these spiders in the basement - spiderwebs covering all the walls, stretching from wall to wall through all the empty space. Then there's a flood and all the spiderwebs are swept away. After the waters recede you see these spiders working furiously to recreate something like the original profusion of webs. It was urgent, because they thought the spiderwebs were holding up the walls... >The book made the point that it really is somewhat amazing that all the >complicated math formulae and logic processes we work through do not seem to >produce a contradiction, and yet we all seem to consider this not amazing at all. It's not amazing to me at all. >This is what I got from the book, not necessarily what the book was trying to >say. I was reading it at the same time as a book by Leslie Newbiggin called >Proper Confidence which was about how we obtain religious truth. The two >books together had quite an impact on my philosophical thinking at that time, >leading me to think a lot about the role of faith, not only in religion but also >in mathematics. ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory >> ...Goedel's incompleteness theory... >> How is it influencing >> current scientific research? >> It has to do with mathematics, not science. >That might be news to Penrose. Not sure where your tongue is located here... Seems clear to me (and to many people who know better than I do) that a lot of things would be news to Penrose. ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory >>The book made the point that it really is somewhat amazing that all the >>complicated math formulae and logic processes we work through do not seem to >>produce a contradiction, and yet we all seem to consider this not amazing at all. > It's not amazing to me at all. Yes, I think that this is where we disagree (I mean I could argue all the other points, but I think that this is the crux of it all). I happen to think that you have been brainwashed by western culture, and that is why you find it all completely natural, and not amazing at all. Actually, I am also brainwashed by this same culture and I am quite happy to use these tools without a second thought when I am writing my papers. But then when I step back, and think about why or whether it all works, I am simply amazed. Since this is an emotional state, I cannot really tell you that you are wrong not to be amazed. (I might add that I read Kline's book in a vacuum. Maybe his history was all messed up - I simply was not knowledgable enough to know, neither had I ever read any of the reviews. But I did like his conclusions, and to some extent some of his conclusions (or rather what I think are his conclusions, because to be honest I am a fast and not careful reader) survive even if his historical accounts are totally messed up.) Finally, I think that my mathematics is influence by Goedel, because like most everyone else I am happy to accept the current cultural level of acceptable math rigor and use it when I write my papers. And I had thought (perhaps eroneously) that had Goedel's Theorem not been true, that today we would have a different level of culturally acceptable rigor. You don't notice this infuence on my papers, because the same influence is upon your papers also, and you think that it is perfectly natural. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: Goedel's incompleteness theory >The book made the point that it really is somewhat amazing that all the >complicated math formulae and logic processes we work through do not seem to >produce a contradiction, and yet we all seem to consider this not amazing at all. >> It's not amazing to me at all. >Yes, I think that this is where we disagree (I mean I could argue all the other >points, but I think that this is the crux of it all). >I happen to think that you have been brainwashed by western culture, and that is >why you find it all completely natural, and not amazing at all. Actually, I am >also brainwashed by this same culture and I am quite happy to use these tools >without a second thought when I am writing my papers. But then when I step >back, and think about why or whether it all works, I am simply amazed. Since >this is an emotional state, I cannot really tell you that you are wrong not to >be amazed. >(I might add that I read Kline's book in a vacuum. Maybe his history was all >messed up - I simply was not knowledgable enough to know, neither had I ever >read any of the reviews. But I did like his conclusions, and to some extent >some of his conclusions (or rather what I think are his conclusions, because to >be honest I am a fast and not careful reader) survive even if his historical >accounts are totally messed up.) >Finally, I think that my mathematics is influence by Goedel, because like most >everyone else I am happy to accept the current cultural level of acceptable math >rigor and use it when I write my papers. And I had thought (perhaps eroneously) >that had Goedel's Theorem not been true, that today we would have a different >level of culturally acceptable rigor. There's no way to know what things would be like if things were different - wondering what things would be like if the Sun were made of green cheese is problematic enough, and wondering what things would be like if a mathematical fact were not so is much more worse. But I don't see at all what Godel's theorem has to do with what is and what is not an acceptable level of rigor. (Not denying that what's acceptable is a cultural thing, I just don't see how it has anything to do with Godel - I don't see that he had anything much to say about what a proof _is_, he said something about what's provable. Seems to me that the people responsible for modern notions of rigor were much earlier.) >You don't notice this infuence on my >papers, because the same influence is upon your papers also, and you think that >it is perfectly natural. Possibly. Or perhaps I notice the same influence but don't see what it has to do with Godel... oversimplifying things, one might suggest that even though Godel showed the Hilbert program was doomed to failure the Hilbert program still has a great influence. ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory > But I don't see at all what Godel's theorem has to do with what is > and what is not an acceptable level of rigor. (Not denying that > what's acceptable is a cultural thing, I just don't see how it has > anything to do with Godel - I don't see that he had anything > much to say about what a proof _is_, he said something about > what's provable. Seems to me that the people responsible for > modern notions of rigor were much earlier.) I think that my sense was that Goedel's impact was more psychological. What it did is to stop most mathematicians from digging any deeper into the foundations of the subject. >>You don't notice this infuence on my >>papers, because the same influence is upon your papers also, and you think that >>it is perfectly natural. > Possibly. Or perhaps I notice the same influence but don't see what > it has to do with Godel... oversimplifying things, one might suggest > that even though Godel showed the Hilbert program was doomed to > failure the Hilbert program still has a great influence. Yes, essentially I agree with you here. I have somehow argued myself into a corner into saying that it is Goedel as opposed to the other mathematicians of those times that have had a great influence on the way I do mathematics, and so I would like to retract that consequence of what I am saying. (Because I don't think it is essential to my point of view.) -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: Goedel's incompleteness theory >> But I don't see at all what Godel's theorem has to do with what is >> and what is not an acceptable level of rigor. (Not denying that >> what's acceptable is a cultural thing, I just don't see how it has >> anything to do with Godel - I don't see that he had anything >> much to say about what a proof _is_, he said something about >> what's provable. Seems to me that the people responsible for >> modern notions of rigor were much earlier.) >I think that my sense was that Goedel's impact was more psychological. What it >did is to stop most mathematicians from digging any deeper into the foundations >of the subject. >You don't notice this infuence on my >papers, because the same influence is upon your papers also, and you think that >it is perfectly natural. >> Possibly. Or perhaps I notice the same influence but don't see what >> it has to do with Godel... oversimplifying things, one might suggest >> that even though Godel showed the Hilbert program was doomed to >> failure the Hilbert program still has a great influence. >Yes, essentially I agree with you here. I have somehow argued myself into a >corner into saying that it is Goedel as opposed to the other mathematicians of >those times that have had a great influence on the way I do mathematics, and so >I would like to retract that consequence of what I am saying. Okie dokie. That was the part I didn't get. (You should get a retraction form in the mail in a few days - fill it out, send it in and we're set.) >(Because I don't >think it is essential to my point of view.) ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory <57EXa.66691$Ho3.9837@sccrnsc03> A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x >> But I don't see at all what Godel's theorem has to do with what is >> and what is not an acceptable level of rigor. (Not denying that >> what's acceptable is a cultural thing, I just don't see how it has >> anything to do with Godel - I don't see that he had anything much to >> say about what a proof _is_, he said something about what's >> provable. Seems to me that the people responsible for modern notions >> of rigor were much earlier.) > I think that my sense was that Goedel's impact was more psychological. > What it did is to stop most mathematicians from digging any deeper > into the foundations of the subject. Surely, the foundations of mathematics is a live topic. The conversation is carried on by a relatively small number of mathematicians and philosophers. The majority of mathematicians do not investigate foundational issues, but hasn't it always been that way? Do you think that, pre-Goedel, most mathematicians were actively investigating foundations? Of course, interest in the topic must have been at a peak somewhere right before Goedel, due to Russell's paradox and (prior to that) the work to provide a rigorous foundation for analysis. But I don't see any clear indication that interest waned due to Goedel. I also suppose that, even at its peak, mathematical foundations was a topic far removed from Joe Mathematician. too soon. Perhaps one of us has an inaccurate historical perspective. Could be me. -- This page contains information of a type (text/html) that can only be viewed with the appropriate Plug-in. Click OK to download Plugin. --- Netscape 4.7 error message === Subject: Re: Goedel's incompleteness theory > But I don't see at all what Godel's theorem has to do with what is > and what is not an acceptable level of rigor. (Not denying that > what's acceptable is a cultural thing, I just don't see how it has > anything to do with Godel - I don't see that he had anything > much to say about what a proof _is_, he said something about > what's provable. Seems to me that the people responsible for > modern notions of rigor were much earlier.) > I think that my sense was that Goedel's impact was more psychological. What it > did is to stop most mathematicians from digging any deeper into the foundations > of the subject. >>You don't notice this infuence on my >>papers, because the same influence is upon your papers also, and you think that >>it is perfectly natural. > Possibly. Or perhaps I notice the same influence but don't see what > it has to do with Godel... oversimplifying things, one might suggest > that even though Godel showed the Hilbert program was doomed to > failure the Hilbert program still has a great influence. > Yes, essentially I agree with you here. I have somehow argued myself into a NOOOO now you've fallen for his brainwashing, you were right the first time, % ****I happen to think that you have been brainwashed by western culture, and that is ****why you find it all completely natural, and not amazing at all. Actually, I am %%%% that had Goedel's Theorem not been true, that today we would have a different %%%% level of culturally acceptable rigor > corner into saying that it is Goedel as opposed to the other mathematicians of > those times that have had a great influence on the way I do mathematics, and so > I would like to retract that consequence of what I am saying. (Because I don't > think it is essential to my point of view.) This is how he lured you.. >I don't share my ideas on this a lot, because I find that most people completely >disagree with me, and it ends up in circular and unproductive arguments. Fine, feel free to drop it. **** subtle but emotional attack The main thing I have to say about your reply is that I don't see how it answers my question. You said Goedel's incompleteness theorem had had a much bigger impact on my mathematics than I had previously thought, I asked for examples, but I don't see below any examples of an impact of Godel's theorem on your mathematics. Maybe you didn't say exactly what you meant. If you'd said that Godel had had a large effect on the way you think about mathematics or some such thing I wouldn't have been surprised, wouldn't have bothered to ask. But you said it had an impact on your mathematics itself - this is the part that surprised me, and this is also what I don't see any explanation of below. _____________ Then upholding the LIES LIES LIES that Godels proof is intrinsic. Hilbert will be back. Try to identify when you use this faulty logic.... does the G statement have a proof, yes or no? NO -> statement is true YES -> statement is true In this proof : G = this statement has no proof ~G -> ~this statement has no proof -> G has a proof -> G [contradiction] -> G P = this statement is false ~P -> ~this statement is false -> P is true -> P [contradiction] -> P And the only answer for this UNBREAKABLE proof that relies on answering the self_proving_statement does it have a proof? is..... the first one is allowed and the second one isn't. Herc This statement has no proof by Herc Pinocchio says 'my nose is big'. The nose knows, no big nose, so the nose grows. Pinocchio knows the nose grows, so it goes, no more nose, my big nose shows, to show it grows when I say big nose with no nose to show, then my nose goes, when my nose shows I said big nose so then it flows. === Subject: Re: Goedel's incompleteness theory >>This was what I got from [Kline's: The loss of certainty]. In the history >>of mathematics, there has been a striving to put mathematics on a more and >>more rigorous foundation. At the turn of the century (that is 19th to >>20th) Hilbert had this ambitious program to put mathematics onto a totally >>rigorous and undisputable foundation. >>As we all know, Goedel's Theorem put an end to this. [...] > Beware that, as usual, popular accounts of Hilbert's program leave much > to be desired. If you want the truth I recommend consulting one of the > many excellent expositions by Wilfried Sieg, who is a leading expert on > such matters. For example, see the papers reviewed below. > -Bill Dubuque > Perhaps you might like to indulge me with a discussion on the internet about > these sorts of things. It is unlikely that I will read Sieg's works in the near > future, as my life is somewhat hectic and doesn't allow for the careful > consideration that I would need to read them. > It seems that these popular accounts are very widespread. It seems that I have > read two stories about Hilbert. The first is that he wished to put mathematics > on a secure foundation, and Goedel somewhat runined this. The second is that > Brouwer suggested removing the law of the excluded middle from logic, and that > proofs should be constructive, and Hilbert rejected this idea. Reading the > reviews of Sieg's work very briefly, I get the feeling that these are over > simplifications of Hilbert's thinking. Hilbert wanted to use strictly finite reasoning--he was much more demanding than Brouwer. Hilbert's idea was to argue about the _symbols_ of mathematics, of which there are only finitely many, and to prove a consistency theorem. It was Godel's _second_ incompleteness theorem that put paid to that. > While the popular accounts might not express the truth about how Hilbert > thought, they might express the truth about how a mathematician might think, and > the impact that Goedel's Theorem might have. > Let me tell my personal story. It was about 20 years ago that I read Goedel's > Theorem. As a theorem I thought it was simply beautiful. But as for its > philosophical impact, I thought that this might be limited. It seemed to me > that it simply must be impossible to put mathematics onto a totally rigorous > foundation. This is because you have to use some mathematics to do this, and as > such you have to take at least a certain amount of mathematics on faith. Even > the consistency of truth tables is something that should not be taken for > granted. Yet to prove such a statement is impossible without using some kind of > logic. So at its heart, we have to accept the logic that seems inherently wired > into our thinking, and we have to have belief that this wiring is always consistent. > Now, it seems to me that in the early twentieth century there was this effort to > put mathematics onto this rigorous foundation. So we get the works of Russel > and Whitehead. So far as rigour was concerned Russell and Whitehead was (were?) a step backward from Frege. Even though there was a contradiction in Frege's work, I still claim that. A return to Fregean rigour came with the Poles and especially Tarski and Lesniewski > Great strides were made in this effort - it had started with the > Greeks who had this notion of axioms and proof methods to derive new statements > from these axioms. In the nineteenth century had been this great effort to put > analysis on a rigorous foundation, so that ideas of continuity, differentiation, > infinite sums, and such like, could be talked about in an exact manner. At some > point in history (I have no idea of the dates), we have Peano with his axioms of > number theory. So the progress was to put all of mathematics on the same > footing as Greek geometry, and which it all can be axiomatised. > So now mathematicians have this notion of proper mathematical thinking, and we > judge a work by whether it meets these standards. For example, we know that > much of the earlier work on analysis does not meet these standards. A lot of > quantum physics (work of people like Dirac) does not meet these standards. And > so as mathematicians, we tend to reject these ideas, until someone else comes > along and re-explains these ideas into our proper mathematical language. > I think Kline's point of view was that this notion of proper mathematical > thinking that we now so universally accept, is really an arbitrary standard, > that is more dictated by our culture than by any absolute measure of right and > wrong. We accept our current standard, really, because so far it has simply > worked. There are currently no great mathematic paradoxes going around that > need to be resolved. But that may simply reflect our lack of imagination. > Well, this was a big ramble. Maybe I got all my historical facts mixed up. I > think I introduced my ideas in a bit of a muddle. So it people disagree with > me, or can point out my mistakes, well I guess that I will either sharpen > myself, or learn something. > Best, Stephen > -- > Stephen Montgomery-Smith > stephen@math.missouri.edu > http://www.math.missouri.edu/~stephen -- G.C. === Subject: Re: Goedel's incompleteness theory >How has Goedel's incompleteness theory changed the making of mathematical >proofs, computer science and science in general? How is it influencing >current scientific research? Neocomplex >>There is a nice book by Morris Kline called The loss of certainty which >>addresses this issue, at least for mathematics. After reading this book, I >>realised that Goedel's incompleteness theorem had had a much bigger impact on my >>mathematics than I had previously thought. > This strikes me as curious - I haven't read the book, but I recall > reading reviews that were a little, um,... > Can you give a specific example of what you're talking about? > (Surely it's not true that you're actually uncertain about the truth > of any of the theorems you've proved, for example. What sort > of impact _has_ it had on _your_ mathematics?) > ************************ > This was what I got from the book. In the history of mathematics, there has > been a striving to put mathematics on a more and more rigorous foundation. At > the turn of the century (that is 19th to 20th) Hilbert had this ambitious > program to put mathematics onto a totally rigorous and undisputable foundation. > As we all know, Goedel's Theorem put an end to this. Godel's first incompleteness theorem showed that one can't have a _monolithic_ foundation for all of mathematics. It doesn't stop each part of mathematics having a foundation. Godel's second incompleteness theorem showed that one can't have a proof of consistency even of a small part of mathematics (a fragment of number theory). Hilbert was hoping to have a consistency proof, and, what's more, a finitary one. Surely one of the reasons for the pursuit of rigour in the 19th century was the counterexamples in analysis which showed that even a Cauchy could _be_ wrong. The worry was not _maybe_ it's wrong but rather it _really_ was wrong. Once Weierstrass, Cantor, etc, put it right (let's hope) the 20th century mathematicians could relax a little. > To my mind, Goedel's Theorem has no real philosophical value. But it seems to > have a tremendous psychological impact on mathematicians. Basically since > Goedel's Theorem, mathematicians have stopped digging too much deeper into the > foundations of their study, and rather have concentrated on proving theorems, > simply accepting that the current level of math logic thinking (usually ZFC laid > on top of first order logic) is the proper state of affairs. > The book made the point that it really is somewhat amazing that all the > complicated math formulae and logic processes we work through do not seem to > produce a contradiction, and yet we all seem to consider this not amazing at all. > This is what I got from the book, not necessarily what the book was trying to > say. I was reading it at the same time as a book by Leslie Newbiggin called > Proper Confidence which was about how we obtain religious truth. The two > books together had quite an impact on my philosophical thinking at that time, > leading me to think a lot about the role of faith, not only in religion but also > in mathematics. > I don't share my ideas on this a lot, because I find that most people completely > disagree with me, and it ends up in circular and unproductive arguments. > -- > Stephen Montgomery-Smith > stephen@math.missouri.edu > http://www.math.missouri.edu/~stephen -- G.C. === Subject: Re: Goedel's incompleteness theory > Surely, the foundations of mathematics is a live topic. The > conversation is carried on by a relatively small number of > mathematicians and philosophers. The majority of mathematicians do > not investigate foundational issues, but hasn't it always been that > way? > Do you think that, pre-Goedel, most mathematicians were actively > investigating foundations? > Of course, interest in the topic must have been at a peak somewhere > right before Goedel, due to Russell's paradox and (prior to that) the > work to provide a rigorous foundation for analysis. But I don't see > any clear indication that interest waned due to Goedel. I also > suppose that, even at its peak, mathematical foundations was a topic > far removed from Joe Mathematician. > too soon. Perhaps one of us has an inaccurate historical > perspective. Could be me. I really have no idea about how much people worried about foundations in the old days. But it seems to me that they must have worried about it a lot. Here is my reasoning, which admitedly is based completely upon suposition and guesswork. In the 17th and 18th centuries, people didn't have a good grasp of what limits, differentiation and infinite sums were. Their level of understanding was the same as what I had when I learned calculus in high school. It was all intuitive. Now I know that these concepts worried me a lot. Even things like the definition of a function were completely unclear to me. If I was worrying about foundations so much, it would greatly surprize me if people at that time did not have similar worries. I do recall reading somewhere that infinite sums like sum (-1)^n, and paradoxes surrounding them, were of great concern. Now comes along the 19th century, and rigorous notions of continuity and differentiation are invented, using epsilons and deltas. It really is very clever stuff, and now it is taught in senior level analysis classes, so that these days I guess we would no longer consider it as part of foundations of mathematics, it is simply part of a mathematicians vocabulary. Mathematicians moved on to worry about the foundations of set theory, because by now people were beginning to accept set theory as the common language of mathematics. (Now when did that happen?) Again you have paradoxes that need to be resolved like Russels paradox (the set of all sets that are not elements of themselves). Again, people must have talked a lot about this stuff. But these days, I must say that I don't think I have met any mathematician who worries about foundations. I do meet mathematicians who do advanced set theory, like Martin's axiom and unattainable ordinals, but I am not sure that this would count any more as foundations, and anyway the number of mathematicians I meet who do this stuff is very very few - if you major in these subjects it is hard to get a job. Basically we are taught about set theory, and how to express all our math problems in it, and it seems to give answers with such finality. For example, what is a function - it is simply a set of ordered pairs such that (a,b) and (a,c) can only both be in the set if b=c. The answer is so clear and workable, and so much better than the vague and imprecise concepts the 17th century mathematicians must have had. Thus there are no burning issues today that force mathematicians to worry about these issues. There really are no difficult paradoxes in mathematics that are today left outstanding. You could try to dig deeper, and ask questions as to why truth tables always give the same tautologies as those given by some basic axioms along with modus ponens, but the search seems pointless and fruitless. But I think that one day in the future, someone will find a serious problem with our present way of doing mathematics, and perhaps in a century or so the language of math will no longer be set theory, but something completely different and unexpected. Who can say? -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: Goedel's incompleteness theory > How has Goedel's incompleteness theory changed the making of mathematical > proofs, computer science and science in general? How is it influencing > current scientific research? I don't know, but Godel showed that arithmetic A was incomplete by producing a statement S that was not provable from it and by doing that in such a way that it was clear that if A was augmented to B with extra axioms and rules so that S could be proved from B, then some further statement T would not be provable from B, and so on. Now, the statement S, though of interest to logicians, was not, I think, of great interest to other mathematicians. Then along came Paris/Harrington/Kirby who produced a statement R of _mathematics_ (as opposed to logic) that was not provable from A though it was true and stated in the language of A. One wonders if the Riemann Hypothesis, say, that can be stated in the language of A, might be like R--true but not provable from A. Consider absolute geometry (= Euclidean geometry without the parallel axiom), one can complete it to Euclidean geometry or hyperbolic geometry and then answer certain questions (for example what is the sum of angles in a triangle) in two different ways according to which axiom is added. Now, suppose that A _doesn't_ settle the Riemann Hypothesis but one of two extra axioms can be added to settle it one way or the other. Suppose further that those two axioms, though inconsistent with one another, are equally intuitively acceptable just like the Euclidean and hyperbolic parallel axioms. What then? -- G.C. === Subject: Re: Goedel's incompleteness theory <57EXa.66691$Ho3.9837@sccrnsc03> A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x > Godel's first incompleteness theorem showed that one can't have a > _monolithic_ foundation for all of mathematics. It doesn't stop each > part of mathematics having a foundation. How do you get that out of the incompleteness theorem? In what sense does ZF, say, fail to provide a monolithic foundation? Just what do you mean when you say foundation? -- Destiny is a funny thing. Once I thought I was destined to become Emperor of Greenland, sole monarch over its 52,000 inhabitants. Then I thought I was destined to build a Polynesian longship in my garage. I was wrong then, but I've got it now. -- The Tick === Subject: Re: Goedel's incompleteness theory >This was what I got from [Kline's: The loss of certainty]. In the history >of mathematics, there has been a striving to put mathematics on a more and >more rigorous foundation. At the turn of the century (that is 19th to >20th) Hilbert had this ambitious program to put mathematics onto a totally >rigorous and undisputable foundation. >>As we all know, Goedel's Theorem put an end to this. [...] >> Beware that, as usual, popular accounts of Hilbert's program leave much >> to be desired. If you want the truth I recommend consulting one of the >> many excellent expositions by Wilfried Sieg, who is a leading expert on >> such matters. For example, see the papers reviewed below. >> -Bill Dubuque >> Perhaps you might like to indulge me with a discussion on the internet about >> these sorts of things. It is unlikely that I will read Sieg's works in the near >> future, as my life is somewhat hectic and doesn't allow for the careful >> consideration that I would need to read them. >> It seems that these popular accounts are very widespread. It seems that I have >> read two stories about Hilbert. The first is that he wished to put mathematics >> on a secure foundation, and Goedel somewhat runined this. The second is that >> Brouwer suggested removing the law of the excluded middle from logic, and that >> proofs should be constructive, and Hilbert rejected this idea. Reading the >> reviews of Sieg's work very briefly, I get the feeling that these are over >> simplifications of Hilbert's thinking. >Hilbert wanted to use strictly finite reasoning--he was much more >demanding than Brouwer. Hilbert's idea was to argue about the _symbols_ >of mathematics, of which there are only finitely many, And in fact _all_ mathematical reasoning _is_ expressed in terms of finitely many symbols... >and to prove a >consistency theorem. It was Godel's _second_ incompleteness theorem >that put paid to that. I think completeness would be better than consistency here. ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory >[...] >Thus there are no burning issues today that force mathematicians to worry about >these issues. There really are no difficult paradoxes in mathematics that are >today left outstanding. You could try to dig deeper, and ask questions as to >why truth tables always give the same tautologies as those given by some basic >axioms along with modus ponens, but the search seems pointless and fruitless. Not to focus on irrelevant details, but it doesn't involve much digging to determine why truth tables give the same tautologies as suitable sets of axioms and inference rules - this is a basic result in mathematical logic that you can easily find proofs of in books (one half is the (or a) Soundness Theorem, and one half is the Completeness Theorem. The Completeness Theorem is the non-trivial half - if I recall correctly it was first proved for predicate logic by Godel.) Or maybe it's not irrelevant - if one were unaware of the Completeness Theorem that might lead to some nervousness about what might be missing from our deductive systems... >But I think that one day in the future, someone will find a serious problem with >our present way of doing mathematics, and perhaps in a century or so the >language of math will no longer be set theory, but something completely >different and unexpected. Who can say? Who indeed. Some people in current threads on sci.math have been suggesting that the fact that a problem may be found someday indicates that there is a problem at present. Makes more sense to me to wait until the problem is found before worrying about the fact that it hasn't been fixed. ************************ David C. Ullrich === Subject: Re: Goedel's incompleteness theory > How has Goedel's incompleteness theory changed the making of mathematical > proofs, computer science and science in general? How is it influencing > current scientific research? > Neocomplex It opened up the possibility that conjectures in Mathematics that seemed true but that nobody had ever been able to prove (e.g. FLT) could in fact be true but not provable. It expanded the search for Mathematical proofs into a search for a proof or a demonstration (proof) that there is no proof. However, note that questions (conjectures) of the form (there exists X)P(X), where P(X) is a recursive predicate, cannot be shown to be true but unprovable! Can you see why? Charlie Volkstorf Cambridge, MA === Subject: Re: Goedel's incompleteness theory <57EXa.66691$Ho3.9837@sccrnsc03 Godel's first incompleteness theorem showed that one can't have a > _monolithic_ foundation for all of mathematics. It doesn't stop each > part of mathematics having a foundation. > How do you get that out of the incompleteness theorem? > In what sense does ZF, say, fail to provide a monolithic foundation? > Just what do you mean when you say foundation? Since ZF contains number theory one can prove a Godelian incompleteness theorem about it. I.e. one can produce a statement G in the language of ZF that is true, but, granted the consistency of ZF, is unprovable. ZF fails to provide a foundation for any branch of mathematics in which G is a theorem that one would like to prove. -- G.C. === Subject: Re: Goedel's incompleteness theory >Godel's first incompleteness theorem showed that one can't have a >_monolithic_ foundation for all of mathematics. It doesn't stop each >part of mathematics having a foundation. >>How do you get that out of the incompleteness theorem? >>In what sense does ZF, say, fail to provide a monolithic foundation? >>Just what do you mean when you say foundation? > Since ZF contains number theory one can prove a Godelian > incompleteness theorem about it. I.e. one can produce a statement G in > the language of ZF that is true, but, granted the consistency of ZF, is > unprovable. ZF fails to provide a foundation for any branch of > mathematics in which G is a theorem that one would like to prove. My question is: how does going to a non-monolithic foundation save you from this? For example; could a less ambitious foundation, one that is solely concerned with number theory avoid both incompleteness and inconsistency where ZF could not? === Subject: Re: Goedel's incompleteness theory >Godel's first incompleteness theorem showed that one can't have a >_monolithic_ foundation for all of mathematics. It doesn't stop each >part of mathematics having a foundation. >>How do you get that out of the incompleteness theorem? >>In what sense does ZF, say, fail to provide a monolithic foundation? >>Just what do you mean when you say foundation? > Since ZF contains number theory one can prove a Godelian > incompleteness theorem about it. I.e. one can produce a statement G in > the language of ZF that is true, but, granted the consistency of ZF, is > unprovable. ZF fails to provide a foundation for any branch of > mathematics in which G is a theorem that one would like to prove. > My question is: how does going to a non-monolithic foundation > save you from this? For example; could a less ambitious > foundation, one that is solely concerned with number theory > avoid both incompleteness and inconsistency where ZF could not? My point was: the fact that one cannot have one foundation for the whole enterprise does not stop one from having different foundations for different parts of the whole. If the job of a foundation is to give certainty to what is derived from it, then a lot of piecemeal foundations may provide no less certainty than one all-encompassing foundation. -- G.C. === Subject: Re: Goedel's incompleteness theory >[...] >Thus there are no burning issues today that force mathematicians to worry about >these issues. There really are no difficult paradoxes in mathematics that are >today left outstanding. You could try to dig deeper, and ask questions as to >why truth tables always give the same tautologies as those given by some basic >axioms along with modus ponens, but the search seems pointless and fruitless. > Not to focus on irrelevant details, but it doesn't involve much > digging to determine why truth tables give the same tautologies > as suitable sets of axioms and inference rules - this is a basic > result in mathematical logic that you can easily find proofs of > in books (one half is the (or a) Soundness Theorem, and one > half is the Completeness Theorem. The Completeness Theorem > is the non-trivial half - if I recall correctly it was first proved > for predicate logic by Godel.) Mention of truth tables and tautologies suggests that propositional calculus was being discussed. I think that completeness theorem was proved by Bernays and Post indepently. > Or maybe it's not irrelevant - if one were unaware of the Completeness > Theorem that might lead to some nervousness about what might > be missing from our deductive systems... >But I think that one day in the future, someone will find a serious problem with >our present way of doing mathematics, and perhaps in a century or so the >language of math will no longer be set theory, but something completely >different and unexpected. Who can say? > Who indeed. Some people in current threads on sci.math have been > suggesting that the fact that a problem may be found someday > indicates that there is a problem at present. Makes more sense to > me to wait until the problem is found before worrying about the > fact that it hasn't been fixed. > ************************ > David C. Ullrich -- G.C. === Subject: Re: Going from base n to base 10 > Does the map F that goes from the natural numbers base n to the > natural numbers base 10, defined by > F(some k-digit number in base n) = (rightmost digit)(n^0)+(next digit > to left)(n^1) + (next digit after that)(n^2) +... + (leftmost > digit)(n^{k-1}) > a bijection? > Also, is there any special name for this map? The canonical base > conversion map? or something? > The natural numbers are the natural numbers are the natural numbers. > You are confusing a natural number with its representation. > Let N be the naturals and N* all sequences of natural numbers or zeros > which (the sequences) are eventually zero. > You have two maps: > T:N -> N* such that n = T(n)_0 + T(n)_1*10 + T(n)_2*10^2 + ... > where T(n)=(T(n)_0, T(n)_1, T(n)_2, ...) > and, (for base b), > S:N -> N* such that n = S(n)_0 + S(n)_1*b + s(n)_2*b^2 + ... > It is non-trivial that both T and S exist and are bijections but, given > that, TS^(-1) and ST^(-1) are bijections and are, I think, what you are > looking for. In less technical language, I think it would be easy to show, or already has been shown, that any real number can be represented as a power series numeral on any integral base whatever from ( 0, +inf). For instance why not represent something in base 667. It would be cumbersome to find enough different symbols for all of the different numbers up to 667, but in theory, could be done. An interesting question here is Could the base be irrational? if so. what would you use for symbols? RJ Pease === Subject: graph homomorphism Suppose f : G ---> G' is a graph homomorphism, and let C be a connected cycle in G. Is it true that f(C) is a connected cycle in G'? I know this is true (and indeed goes the other way, too) for a graph isomorphism, but I'm not sure in the case where we only have a homomorphism of graphs. Thx. LW === Subject: Re: graph homomorphism > Suppose f : G ---> G' is a graph homomorphism, and let C be a > connected cycle in G. Is it true that f(C) is a connected cycle in G'? > I know this is true (and indeed goes the other way, too) for a graph > isomorphism, but I'm not sure in the case where we only have a > homomorphism of graphs. Go back to the definition of homomorphism, and ask yourself: What does f do to C's vertices? What does f do to C's edges? Is that result then necessarily a connected cycle if C is? > Thx. > LW -- --------------------------- | BBB b barbara minus knox at iname stop com | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: graph homomorphism === Subject: graph homomorphism >Suppose f:G -> G' is a graph homomorphism, and let C be a connected >cycle in G. Is it true that f(C) is a connected cycle in G'? f:V1->V2 is homomorphism of simple graphs (V1,E1), (V2,E2) when for all x,y in V1, ((x,y) in E1 ==> (f(x),f(y)) in E2) Is this the right definition? How is it extended to non-simple graphs? Let x1, x2, x3, x4, x5, x6, x1 be a cycle in G, is f(x1), f(x2), f(x3), f(x4), f(x5), f(x6), f(x1) a cycle in G' when f(x1) = f(x4) ? ---- === Subject: Re: graph homomorphism > Suppose f : G ---> G' is a graph homomorphism, and let C be a > connected cycle in G. Is it true that f(C) is a connected cycle in G'? > I know this is true (and indeed goes the other way, too) for a graph > isomorphism, but I'm not sure in the case where we only have a > homomorphism of graphs. What definition of graphs and graph homorphisms do you use? For any natural number n consider the graph C_n given by vertices {1,...,n}, edges {(1,2),...,(n-1,n),(n,1)} such that (i,j) is the only edge from i to j. At least in the case of Graphs without loops a connected cycle in G is then just a graph homomorphism from C_n to G for a suitable natural number n>1. c: C_n ---> G in G, composing with f yields another cycle fc: C_n ---> G ---> G' in G' The formal argument above also works if your graphs have loops or multiple edges, but you might end up with closed pathes that do not really look like cycles. Marc === Subject: Re: graph homomorphism No. For example, C_6 ---> K_2. In other words, C_6 is 2-colorable. It is not always true for odd length cycles either. Chip > Suppose f : G ---> G' is a graph homomorphism, and let C be a > connected cycle in G. Is it true that f(C) is a connected cycle in G'? > I know this is true (and indeed goes the other way, too) for a graph > isomorphism, but I'm not sure in the case where we only have a > homomorphism of graphs. > Thx. > LW === Subject: Graph theory/Matlab Originator: fab@soda.csua.berkeley.edu (Fabio Rojas) Hey - is there a graph theory package (shareware or otherwise) for Matlab? Basically, I want to program Matlab to do one very simple procedure - take the adjacency matrix of a graph, tell me how many components it has and give me a vector of average shortest path lengths for each connected component. If Matlab can already to this, please tell me. Fabio === Subject: Re: Greatest Unsolved Problems > Not just trolling here... I would really like to hear your opinion: p ln x = x^n p is real and n is an integer. find p=f(n) I hate to be harping about polynomials again, but that is what it's all about. === Subject: Group generators Is there an algorithm to determine the minumum number of generators for a finite group ? Clearly, for a cyclic group it is 1, for a symmetric permutation group or a dihedral group it is 2 but what about any finite group ? === Subject: Re: Group generators >Is there an algorithm to determine the minumum number of generators for a >finite group ? Clearly, for a cyclic group it is 1, for a symmetric >permutation group or a dihedral group it is 2 but what about any finite >group ? I am not exactly sure what you are asking here. Obviously there is an algorithm to solve this problem. For n = 1, 2, 3,... try all subsets of the group of size n, and check whether or not they generate the group. When you find an n-element subset that generates the group, return n. Are you asking whether there is an efficient programmable algorithm? I think the answer to that is no in general, but in practice, for most groups that are not enormously large, it would not be difficult to answer it. For p-groups (i.e. groups of order a power of a prime p), it is easy, because the number of generators is equal to the rank of G/[G,G]G^p. Given an arbitrary group G, I would proceed as follows. 1. Check if G is cyclic. 2. If not, calculate the rank (= number of invariant factors) n of G/[G,G]. Conjecture that the answer is r := Max(2,n). Generally speaking, if G can be generated by r generators, then a random subset of G of size r has a reasonably high probability of generating G, so try some random r-subsets, and see if they generate G. 3. If this does not work, then life gets more interesting, and you would need to investigate the abstract structure of G further. I wonder what the smallest group is for which 1 and 2 will not work? There are examples like (A_5)^20, which need 3 generators, but that is a very large group. Derek Holt. like A_5 === Subject: Re: Halting proof(s) > |-|erc says... >> I don't know what you are talking about, but if HH is a computer >> program, then we can prove that it isn't equal to Halt. >//----------- >//A 01L B >//A 11R B >//B 01R C >//B 10L E >//C 00R D >//C 10L A >//D 01L A >//D 10R D >//E 01L HALT >//E 10L C >//------------- >This is HH, I'm pretty sure its the bona fide Halt algorithm :) >This is actually TM number 9320303603822 where the transition sequence >above is mapped to the number, where state HALT is 0, L is 0, R is 1, and the >later transitions are more significant, all the smaller size TMs with under 5 >states >were added to that number. So 01LB is (1*12+0*6+2) * 24^0 = 14. >i.e there are (2*2*6)^10 5 state TMs. > http://members.ozemail.com.au/~chess3/tm3.html >You can give it any parameter in unary, and the claim is it will output the >value for Halt for that number by terminating to the right with either 01 or >00. > Well, there is an easy proof that the claim is false. Remember, by definition > of Halt, if n is the Godel number of a program, then Halt(n) = 0 if that program > halts on input n, and Halt(n) = 1, if it never halts on input n. > So, given your program HH, we define a new function g(n) by > g(n) = if HH(n) = 0, then loop forever > else output 0 > We compute the Godel number k of g, and consider the computation > g(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, and > otherwise, Halt(k) = 1. > So, consider the two possibilities: > 1. g(k) halts, and Halt(k) = 0. But by definition of g above, > we know that g(k) only halts if HH(k) is nonzero. So in this > case, it is clear that Halt(k) != HH(k), so HH does not compute > Halt. > 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, > we know that the only way for g(k) to fail to halt is if HH(k)=0. > So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly > compute Halt. Which is it 1 or 2? You did a specific proof for pi/4, why can't you disprove HH? not convincing in the least when you have an actual program and use a conceptual proof for all programs. If your Halting proof is valid then you can demonstrate it for the program given. The flaw in your proof is exactly the same I've been pointing out the last 3 posts, try to think ahead where your argument is leading I said the observable options for g(k) are different to the options listed so: Observered Your 'conclusion' g(k) halted : g(k) halts g(k) hasn't halted : g(k) halts g(k) hasn't halted : g(k) never halts You then substantiate this by concluding we know the values of f. ???? This is a different example, think again. Like I said 10 posts ago, Halt not being computable is a different meaning to no TM possibly calculates Halt, as long as the TM is unidentifiable, such as HH which does not exhibit any characteristics to differentiate it from Halt. http://members.ozemail.com.au/~chess3/tm3.html Its been trialled for 10^12 iterations with continually changing pattern, after you prove it does or doesn't halt then you can proceed with your examination of the values of g(k). Good luck! Herc === Subject: Re: Halting proof(s) |-|erc says... >> So, consider the two possibilities: >> 1. g(k) halts, and Halt(k) = 0. But by definition of g above, >> we know that g(k) only halts if HH(k) is nonzero. So in this >> case, it is clear that Halt(k) != HH(k), so HH does not compute >> Halt. >> 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, >> we know that the only way for g(k) to fail to halt is if HH(k)=0. >> So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly >> compute Halt. >Which is it 1 or 2? I don't know. But if you know that either A or B holds, and you know that A implies C, and you also know that B implies C, then you can conclude that C holds. For example: Either it will rain tomorrow, or it won't. If it doesn't rain tomorrow, then I will water my garden. If it does rain tomorrow, then it will rain on my garden. I don't know whether it will rain or not, but I don't need to know which is the case to know that my garden will get water. >You did a specific proof for pi/4, why can't you disprove HH? I did. My proof works for any program whatsoever. >not convincing in the least when you have an actual program and use a >conceptual proof for all programs. If I prove something about all programs, and then you give me a particular program, then it is necessarily true about that particular program. That's the whole point of proving general statements. >If your Halting proof is valid then you >can demonstrate it for the program given. I did. >I said the observable options for g(k) are different to the options listed so: >Observered Your 'conclusion' >g(k) halted : g(k) halts >g(k) hasn't halted : g(k) halts >g(k) hasn't halted : g(k) never halts >You then substantiate this by concluding we know the values of f. ???? I don't need to *know* whether g(k) will ever halt or not. I know that *either* it will halt, or it won't. In either case, I can prove that f does not compute the halting program. >This is a different example, think again. >Like I said 10 posts ago, Halt not being computable is a different meaning to >no TM possibly calculates Halt You were wrong when you said it 10 posts ago, and you are still wrong. Halt not being computable means exactly that no TM can possibly calculate Halt. -- Daryl McCullough Ithaca, NY === Subject: Re: Halting proof(s) > |-|erc says... >> So, consider the two possibilities: >> 1. g(k) halts, and Halt(k) = 0. But by definition of g above, >> we know that g(k) only halts if HH(k) is nonzero. So in this >> case, it is clear that Halt(k) != HH(k), so HH does not compute >> Halt. >> 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, >> we know that the only way for g(k) to fail to halt is if HH(k)=0. >> So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly >> compute Halt. >Which is it 1 or 2? > I don't know. But if you know that either A or B holds, > and you know that A implies C, and you also know that B > implies C, then you can conclude that C holds. > For example: Either it will rain tomorrow, or it won't. > If it doesn't rain tomorrow, then I will water my garden. > If it does rain tomorrow, then it will rain on my garden. > I don't know whether it will rain or not, but I don't need to know > which is the case to know that my garden will get water. Then why did you provide a specific proof for the known computable pi/4? g(k) has halted >g(k) hasn't halted No, there is a third observable: the definition of g. If we look at the definition of g, we can see that if f(k)=0, then g(k) will never halt, and that if f(k)!=0, then g(k) will halt. ...We can look at the definition of g and we can look at the value of f(k). That confirmationn of your Halting proof is completely irrelevant then! A only implies C because we know why A holds, your proof makes assumptions. >You did a specific proof for pi/4, why can't you disprove HH? > I did. My proof works for any program whatsoever. it works for programs that are known to be computable. the conclusion is that we cannot know a function to compute the Halting problem, nothing more. >If your Halting proof is valid then you >can demonstrate it for the program given. > I did. >I said the observable options for g(k) are different to the options listed so: >Observered Your 'conclusion' >g(k) halted : g(k) halts >g(k) hasn't halted : g(k) halts >g(k) hasn't halted : g(k) never halts >You then substantiate this by concluding we know the values of f. ???? > I don't need to *know* whether g(k) will ever halt or not. I know > that *either* it will halt, or it won't. In either case, I can prove > that f does not compute the halting program. Here are the specifics for your proof applied to HH: Well, there is an easy proof that the claim is false. Remember, by definition of Halt, if n is the Godel number of a program, then Halt(n) = 0 if that program halts on input n, and Halt(n) = 1, if it never halts on input n. So, given your program HH, we define a new function g(n) by g(n) = if HH(n) = 0, then loop forever else output 0 ** forall(n) HH(n) has not yet terminated, we DONT KNOW if HH(n) = 0 or not, so regardless_of_the_value of HH(n), g(n) will loop waiting for HH to give a result. We compute the Godel number k of g, and consider the computation g(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, and otherwise, Halt(k) = 1. So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above, we know that g(k) only halts if HH(k) is nonzero. So in this case, it is clear that Halt(k) != HH(k), so HH does not compute Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, we know that the only way for g(k) to fail to halt is if HH(k)=0. So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly compute Halt. Its option 2, g(k) hasn't finished running yet, its waiting for the value of HH(k) in its subprogram. Halt(k) is not known. The only way for g(k) to not to halt is if HH(k) = 0, ***or**** HH(k) hasn't finished computing yet. No contradiction with Halt(k) = HH(k). Herc === Subject: Re: Halting proof(s) |-|erc says... >> I don't know. But if you know that either A or B holds, >> and you know that A implies C, and you also know that B >> implies C, then you can conclude that C holds. >> For example: Either it will rain tomorrow, or it won't. >> If it doesn't rain tomorrow, then I will water my garden. >> If it does rain tomorrow, then it will rain on my garden. >> I don't know whether it will rain or not, but I don't need to know >> which is the case to know that my garden will get water. >Then why did you provide a specific proof for the known computable pi/4? I thought the more specific case would be more illustrative. It certainly isn't needed, logically. >>You did a specific proof for pi/4, why can't you disprove HH? >> I did. My proof works for any program whatsoever. >it works for programs that are known to be computable. By computable, you mean that they always halt? Once again, we can do a case split: Case 1: forall n, f(n) halts. Case 2: for some n, f(n) does not halt. In case 2, we *know* that f(n) cannot equal Halt(n), because f(n) is undefined. So the only case that we need to worry about is case 1. As I said earlier, you don't need to *know* whether case 1 or case 2 holds. You prove that 1. Case 1 -> f does not compute Halt 2. Case 2 -> f does not compute Halt 3. Case 1 or Case 2. 4. Therefore, f does not compute Halt. It's exactly like my proof that my lawn will be watered tomorrow. >the conclusion is that we cannot know a function to compute the >Halting problem, nothing more. We know more than that. We know that there does not exist a computable function that computes the halting problem. >Here are the specifics for your proof applied to HH: >of Halt, if n is the Godel number of a program, then Halt(n) = 0 >if that program halts on input n, and Halt(n) = 1, if it never halts >on input n. >So, given your program HH, we define a new function g(n) by > g(n) = if HH(n) = 0, then loop forever > else output 0 >** forall(n) HH(n) has not yet terminated, we DONT KNOW if HH(n) = 0 > or not, so regardless_of_the_value of HH(n), g(n) will loop > waiting for HH to give a result. You don't understand. We don't have to *run* g(n). The only thing we use g(n) for is that we write the program for g, and then compute the Turing Machine number of g (called k). The proof shows that HH(k) cannot be equal to Halt(k). >We compute the Godel number k of g, and consider the computation >g(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, and >otherwise, Halt(k) = 1. >So, consider the two possibilities: > 1. g(k) halts, and Halt(k) = 0. But by definition of g above, > we know that g(k) only halts if HH(k) is nonzero. So in this > case, it is clear that Halt(k) != HH(k), so HH does not compute > Halt. > 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, > we know that the only way for g(k) to fail to halt is if HH(k)=0. > So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly > compute Halt. >its waiting for the value of HH(k) >in its subprogram. Halt(k) is not known. The only way for g(k) to not >to halt is if HH(k) = 0, ***or**** HH(k) hasn't finished computing yet. Nobody is running g(k). We're not observing g(k), we are looking at its program and making deductions from that. -- Daryl McCullough Ithaca, NY === Subject: Re: Halting proof(s) >So, consider the two possibilities: > 1. g(k) halts, and Halt(k) = 0. But by definition of g above, > we know that g(k) only halts if HH(k) is nonzero. So in this > case, it is clear that Halt(k) != HH(k), so HH does not compute > Halt. > 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, > we know that the only way for g(k) to fail to halt is if HH(k)=0. > So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly > compute Halt. > You don't understand---we don't run g(k) at all. We can look > at the program for g(k) to know that if g(k) halts, then > HH(k) is nonzero and Halt(k)=0, and if g(k) fails to halt, > then HH(k) = 0 and Halt(k)=1. But I demonstrated that if g(k) fails to halt then HH(k) is not necessarily 0. True or false? Herc === Subject: Re: Halting proof(s) |-|erc says... >> You don't understand---we don't run g(k) at all. We can look >> at the program for g(k) to know that if g(k) halts, then >> HH(k) is nonzero and Halt(k)=0, and if g(k) fails to halt, >> then HH(k) = 0 and Halt(k)=1. >But I demonstrated that if g(k) fails to halt then HH(k) is not >necessarily 0. No, you didn't. The definition of g(k) says that if HH(k) is nonzero, then g(k) will halt, (and so Halt(k) will be 0 and thus unequal to HH(k)) and if HH(k) is zero, then g(k) will not halt (and so Halt(k) will be 1 and thus unequal to HH(k)). The third possibility is that HH(k) never halts (in which case HH(k) is again unequal to Halt(k)). -- Daryl McCullough Ithaca, NY === Subject: Re: Halting proof(s) > |-|erc says... >> You don't understand---we don't run g(k) at all. We can look >> at the program for g(k) to know that if g(k) halts, then >> HH(k) is nonzero and Halt(k)=0, and if g(k) fails to halt, >> then HH(k) = 0 and Halt(k)=1. >But I demonstrated that if g(k) fails to halt then HH(k) is not >necessarily 0. > No, you didn't. > The definition of g(k) says that if HH(k) is nonzero, then g(k) > will halt, (and so Halt(k) will be 0 and thus unequal to HH(k)) > and if HH(k) is zero, then g(k) will not halt (and so Halt(k) > will be 1 and thus unequal to HH(k)). The third possibility > is that HH(k) never halts (in which case HH(k) is again unequal > to Halt(k)). I understand your proof, but 10 years ago they didn't teach that it applies to undecidable programs aswell. You really need the base step that you know the values of f, or HH to make the proof work, because g not finished doesn't imply anything. g can hang for 2 reasons! OK you've covered that as a third option now where HH never halts. This is an assumption. HH is computable and hasn't finished calculating yet is indistinguishable from HH never halts. So we can't infer HH != Halt. Its very odd your proof works by examining the values of f, and then it supposedly works doubly when this step is not justified and you use the high level reasoning by assuming knowledge about the program in question. Herc === Subject: Re: Halting proof(s) this is showing up in comp.theory not in sci.math on my reader so posted twice.. > |-|erc says... >> You don't understand---we don't run g(k) at all. We can look >> at the program for g(k) to know that if g(k) halts, then >> HH(k) is nonzero and Halt(k)=0, and if g(k) fails to halt, >> then HH(k) = 0 and Halt(k)=1. >But I demonstrated that if g(k) fails to halt then HH(k) is not >necessarily 0. > No, you didn't. > The definition of g(k) says that if HH(k) is nonzero, then g(k) > will halt, (and so Halt(k) will be 0 and thus unequal to HH(k)) > and if HH(k) is zero, then g(k) will not halt (and so Halt(k) > will be 1 and thus unequal to HH(k)). The third possibility > is that HH(k) never halts (in which case HH(k) is again unequal > to Halt(k)). I understand your proof, but 10 years ago they didn't teach that it applies to undecidable programs aswell. You really need the base step that you know the values of f, or HH to make the proof work, because g not finished doesn't imply anything. g can hang for 2 reasons! OK you've covered that as a third option now where HH never halts. This is an assumption. HH is computable and hasn't finished calculating yet is indistinguishable from HH never halts. So we can't infer HH != Halt. Its very odd your proof works by examining the values of f, and then it supposedly works doubly when this step is not justified and you use the high level reasoning by assuming knowledge about the program in question. Herc === Subject: Has Perelman's proof remained solid? Hey all, I was just curious if anyone has any information regarding the current validity of Perelman's Thurston's geometrization proof. I saw his talk at Penn in April, but have not seen any information since then regarding the current status of his work or if any problems have been found. marc === Subject: Help finding a Carroll Problem I am looking for a problem referenced by Martin Gardner, Chapter 11 in his collection entitled The 2nd Scientific American Book of Mathematical Puzzles & Diversions, Simon and Schuster, 1961. Gardner references Lewis Carroll telling that Carroll was fond of inventing quaint and enormously complicated problems of this sort. Eight are to be found in the appendix of his Symbolic Logic. Gardner continues speaking of a large problem about judges not smoking tobacco which was solved in 60's by Kemeny. Can you give me some help on where tofind the text of the problem? Sergio Martino s.martino@tno.it includes the following problem devised by Raymond Smullyan (who has written several delightful books on logic): === Subject: Re: Help finding a Carroll Problem NNTP-Proxy-Relay: library2.airnews.net If you're a Yahoo member, you could try posting this question to: http://groups.yahoo.com/group/lewiscarroll Some of the dingbats there know all about Lewis Carroll.