mm-332 Subject: Re: Rationals are Uncountable > Let X be the set of all rational number in the interval (-1,1). > Let x_n be a sequence that contains every member of X. > Construct the sequences a_n and b_n as described by Cantor. By now, it should be obvious that the sequences a_n and b_n > must converge to an irrational number. > Let r be this number. Define q to be a rational number with the same sign > as r and |q| < |r|. Create a new sequence x'_n: x'_1 = x_1 - q > x'_2 = x_2 -q > ... Construct new sequences a'_n and b'_n from x'_n. > a'_n and b'_n will converge to a new irrational > number r' where |r'| < |r|. > Repeat this process an infinite number of times. > x'_n will be a sequence of rational numbers > that does not include 0 even though 0 is in > the range of X'. If each of your X' sets is displaced by some rational amount, q, from the previous one, how do you even know that the sum of these displacements will ever converge (in the reals), much less still allow your final X' to contain 0 ? That is, how do you know that the q's will add up to a finite number no greater than 1? Its obvious that the sum of all q is less than r. |r| < 1 |q1| < |r| > |q2| < |r| - |q1| > |q3| < |r| - |q1| - |q2| > ... In that case, what prevents one of your x'_n, assuming them still to be all rational, from being zero? If you have a sequence of rationals covering the rational interval (-1,1), then no (cumulative) translation of that sequence by a rational distance less than 1 will allow it not to contain 0. === Subject: Re: Rationals are Uncountable That is, how do you know that the q's will add up to a finite number no greater than 1? r lies in the range (-1,1). > |q| < |r| < 1 > Let q_1 be the first q. q_1 > q_2 > q_3 ... q_n is a strictly monotone decreasing series. > 1/2 > 1/3 > 1/4 > ... is a strictly decreasing sequence whose series diverges. When you do x'_n = x_n -1/2 x''_n = x'_n - 1/3 x'''_n = x''_n - 1/4 and so on, you run into prroblems. === Subject: Re: Rationals are Uncountable > That is, how do you know that the q's will add up to a finite number no > greater than 1? r lies in the range (-1,1). |q| < |r| < 1 Let q_1 be the first q. q_1 > q_2 > q_3 ... q_n is a strictly monotone decreasing series. > 1/2 > 1/3 > 1/4 > ... is a strictly decreasing sequence whose series > diverges. > When you do > x'_n = x_n -1/2 > x''_n = x'_n - 1/3 > x'''_n = x''_n - 1/4 > and so on, you run into prroblems. I realized this right after I hit the send button. That is why I pos a second proof: |r| < 1 |q_n| < |r| - |q_1| - |q_2| - ... - |q_n-1| Let x_n be a sequence that contains every rational in X = (-1,1). Define x'_n to be isomorphic to x_n if there exists a linear tranformation that converts x_n into x'_n. For example: x'_n = 2 * x_n + 1 x'_n is isomorphic to x_n and contains every rational in the interval (-1,3). Assume the intersection of all intervals a_n and b_n (construc from x_n) consists of the irrational number, r. There is no isomorphic sequence of x_n that contains every rational in the interval (-1-r,1-r). Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable That is, how do you know that the q's will add up to a finite number > no greater than 1? r lies in the range (-1,1). > |q| < |r| < 1 > Let q_1 be the first q. q_1 > q_2 > q_3 ... q_n is a strictly monotone decreasing series. > 1/2 > 1/3 > 1/4 > ... is a strictly decreasing sequence whose series diverges. When you do x'_n = x_n -1/2 x''_n = x'_n - 1/3 x'''_n = x''_n - 1/4 and so on, you run into prroblems. I realized this right after I hit the send button. > That is why I pos a second proof: |r| < 1 > |q_n| < |r| - |q_1| - |q_2| - ... - |q_n-1| Let x_n be a sequence that contains every > rational in X = (-1,1). Define x'_n to be isomorphic to x_n if > there exists a linear tranformation that > converts x_n into x'_n. I suspect you mean a linear mapping, rather than a linear transformation, as the latter must send 0 to 0. You can say that X' is order-isomorphic to X without new definitions, provided your linear mappiing does not reverse the order. For example: > x'_n = 2 * x_n + 1 x'_n is isomorphic to x_n and contains > every rational in the interval (-1,3). Assume the intersection of all intervals > a_n and b_n (construc from x_n) > consists of the irrational number, r. There is no isomorphic sequence of x_n > that contains every rational in the interval (-1-r,1-r). But if your linear mapping translates rationals into rationals, it will transform rational endpoints to rational endpoints, which -1-r and 1-r are not. And if, as must be the case here, your mapping translates every rational to an irrational, then there are NO rationals at all in the new interval, (-1-r,1-r), but it will still have gaps. === Subject: Re: Rationals are Uncountable Let x_n be a sequence that contains every rational in X = (-1,1). Define x'_n to be isomorphic to x_n if there exists a linear tranformation that converts x_n into x'_n. > I suspect you mean a linear mapping, rather than a linear > transformation, as the latter must send 0 to 0. > You can say that X' is order-isomorphic to X without new definitions, > provided your linear mappiing does not reverse the order. For example: x'_n = 2 * x_n + 1 x'_n is isomorphic to x_n and contains every rational in the interval (-1,3). Assume the intersection of all intervals a_n and b_n (construc from x_n) consists of the irrational number, r. There is no isomorphic sequence of x_n that contains every rational in the interval (-1-r,1-r). > But if your linear mapping translates rationals into rationals, it will > transform rational endpoints to rational endpoints, which -1-r and 1-r > are not. And if, as must be the case here, your mapping translates every > rational to an irrational, then there are NO rationals at all in the > new interval, (-1-r,1-r), but it will still have gaps. No sequence order isomorphic to x_n can represent ANY rational in the interval (-1-r, 1-r). Why is this? x_n is supposed to contain every rational. Why would I need a different sequence to get all of the rationals in (-1-r, 1-r)? This other sequence must contain rational numbers not in x_n. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable Let x_n be a sequence that contains every > rational in X = (-1,1). Define x'_n to be isomorphic to x_n if > there exists a linear tranformation that > converts x_n into x'_n. I suspect you mean a linear mapping, rather than a linear transformation, as the latter must send 0 to 0. You can say that X' is order-isomorphic to X without new definitions, provided your linear mappiing does not reverse the order. For example: > x'_n = 2 * x_n + 1 x'_n is isomorphic to x_n and contains > every rational in the interval (-1,3). Assume the intersection of all intervals > a_n and b_n (construc from x_n) > consists of the irrational number, r. There is no isomorphic sequence of x_n > that contains every rational in the interval (-1-r,1-r). But if your linear mapping translates rationals into rationals, it will transform rational endpoints to rational endpoints, which -1-r and 1-r are not. And if, as must be the case here, your mapping translates every rational to an irrational, then there are NO rationals at all in the new interval, (-1-r,1-r), but it will still have gaps. No sequence order isomorphic to x_n can represent ANY rational > in the interval (-1-r, 1-r). Which order on x_n are you talking about, the order of n in N or the order of x_n in Q? They must be different. > Why is this? x_n is supposed to contain > every rational. Why would I need a different sequence to get all > of the rationals in (-1-r, 1-r)? If x_n is rational and r is irrational then x'_n = x_n - r is irrational. This other sequence must contain rational numbers not in x_n. Not if of the form I described. === Subject: Re: Rationals are Uncountable No sequence order isomorphic to x_n can represent ANY rational in the interval (-1-r, 1-r). > Which order on x_n are you talking about, the order of n in N or the > order of x_n in Q? They must be different. I don't understand your question. I think n in N. The order of x_n in Q can't be of order type 1. (There is no countable function that contains every member of Q in linear order.) Why is this? x_n is supposed to contain every rational. Why would I need a different sequence to get all of the rationals in (-1-r, 1-r)? > If x_n is rational and r is irrational then x'_n = x_n - r is irrational. This other sequence must contain rational numbers not in x_n. > Not if of the form I described. Let me explain my proof in more detail. Let X0 be all rationals in the interval (-1,1). Let x0_n be a countable sequence that contains every member of X0. Construct the sequences a0_n and b0_n as described by Cantor. The intersection of all (a0_n, b0_n) contains an irrational number, r0. Choose a rational number, q1, that has the same sign as r0 and |q1| < |r0|. Construct a new sequence, x1_n, as follows: x1_n = x0_n - q1 x1_n is a sequence that contains every rational number in the set X1 = (-1-q1, 1-q1). Construct a1_n and b1_n. a1_n and b1_n have some interesting properties. For example, if a0_n = x0_i and b0_n = x0_j then a1_n = x1_i and b1_n = x1_j. IOW, the same indices of x0_n that make up a0_n and b0_n must make up a1_n and b1_n. Since r0 was not 0, there are intervals (a0_i, b0_i) that do not contain 0. Let j be the index of the first such interval. Similarly, there is a first interval, (a1_k, b1_k), that does not contain 0. j must be less than k. I am pretty sure this allows me to prove the following: If x0_m = 0 and x1_n = 0 then m < n. By subtracting q1 from every member of x0_n I am forcing 0 to appear later in x1_n than it did in x0_n. The intersection of a1_n and b1_n contains an irrational number, r1. Choose rational q2 to have the same sign as r1 and |q2| < |r1| Repeat the process I describe. 0 must appear later in the sequence x2_n than it does in x1_n. By repeating this process, I can force 0 to appear at infinity in some xk_n. Define q = q1 + q2 + q3 + ... + qk By construction, |q| < |r0|. q is a rational number. q is in the range of X0. The sequence x0_n does not contain q. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable No sequence order isomorphic to x_n can represent ANY rational > in the interval (-1-r, 1-r). [...] * According to Cantor, the rationals are countable: Georg Cantor, Ueber eine Eigenschaft des Inbegriffs aller reellen algebraischen Zahlen, J. f. Math. 77 (1874), 258-262. earle * === Subject: Re: Rationals are Uncountable No sequence order isomorphic to x_n can represent ANY rational > in the interval (-1-r, 1-r). [...] * The rationals are countable. Here's a proof: http://planetmath.org/encyclopedia/ ProofThatTheRationalsAreCountable.ht ml earle * === Subject: Re: Rationals are Uncountable No sequence order isomorphic to x_n can represent ANY rational in the interval (-1-r, 1-r). > The rationals are countable. Here's a proof: > http://planetmath.org/encyclopedia/ ProofThatTheRationalsAreCountable.html I have given surjections from N to Q is this thread. (Standard proof the rationals are countable.) You can find a proof of just about anything if you look hard enough. I found these in mathematics arXiv: There are less transcendental numbers than rational numbers http://front.math.ucdavis.edu/math.GM/0305326 Countable structure does not have a free uncountable auorphism group http://front.math.ucdavis.edu/math.LO/0010305 Intersection properties of open sets, II http://front.math.ucdavis.edu/math.LO/9605209 Peano Arithmetic may not be interpretable in the monadic theory of orders http://front.math.ucdavis.edu/math.LO/9308219 The complexity of the reals in inner models of set theory http://front.math.ucdavis.edu/math.LO/9501203 You might think these authors are crackpots, but many of them are highly respec in their field. You should read the abstracts of some of these papers. The last paper lis states: In fact we prove that if $M$ is an inner model of set theory and the set of reals in $M$ is analytic then either all reals are in $M$ or else $aleph _1^M$ is countable All of the proofs that the rationals are countable rely on showing that the rationals have the same cardinality as the set of all natural numbers. I star this thread based on a proof I discovered showing that no set can contain every computable natural number. This proof questions the very definition of countable. The proof you refer to states: Note that the set of numbers with a given height is finite. This proof relies on the existence of the set of all natural numbers. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable <3Pmdndh6FqkP_Ird4p2dnA@comcast.com> <-oKdnfK9mv7KBIXdRVn-hA@comcast.com> Discussion, linux) Cancel-Lock: sha1:zgj6f0U8fJVUl75L+rz3a1TLgdk= > You should read the abstracts of some of these papers. > The last paper lis states: > In fact we prove that if $M$ is an inner model of set theory > and the set of reals in $M$ is analytic then either > all reals are in $M$ or else $aleph _1^M$ is countable Aha! That's what you thought was funny. Russell, you twit, they're talking about models of set theory within set theory, not the model of reals. The author is *not* contradicting Cantor. -- Now I'm informing all of you that the people arguing against me are EVIL, yes they are real, live EVIL people as mathematics is that important, so it's important enough for Evil itself to send minions like them. -- on Evil's interest in Algebraic Number Theory === Subject: Re: Rationals are Uncountable You should read the abstracts of some of these papers. The last paper lis states: In fact we prove that if $M$ is an inner model of set theory and the set of reals in $M$ is analytic then either all reals are in $M$ or else $aleph _1^M$ is countable > Aha! That's what you thought was funny. > Russell, you twit, they're talking about models of set theory within set > theory, not the model of reals. The author is *not* contradicting > Cantor. I'm not contradicting Cantor, either. Obviously, the reals are uncountable if the rationals are. There are models of ZFC where the reals are countable. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable > You should read the abstracts of some of these papers. > The last paper lis states: In fact we prove that if $M$ is an inner model of set theory > and the set of reals in $M$ is analytic then either > all reals are in $M$ or else $aleph _1^M$ is countable Aha! That's what you thought was funny. Russell, you twit, they're talking about models of set theory within set theory, not the model of reals. The author is *not* contradicting Cantor. I'm not contradicting Cantor, either. > Obviously, the reals are uncountable if the rationals are. There are models of ZFC where the reals are countable. Then why do you keep arguing that the rationals are uncountable, as in the subject of this thread? > Russell > - 2 many 2 count > === Subject: Re: Rationals are Uncountable > You should read the abstracts of some of these papers. > The last paper lis states: In fact we prove that if $M$ is an inner model of set theory > and the set of reals in $M$ is analytic then either > all reals are in $M$ or else $aleph _1^M$ is countable >> Aha! That's what you thought was funny. >> Russell, you twit, they're talking about models of set theory within set >> theory, not the model of reals. The author is *not* contradicting >> Cantor. >I'm not contradicting Cantor, either. >Obviously, the reals are uncountable if the rationals are. >There are models of ZFC where the reals are countable. Uh, no. There is a true fact that someone might express using mean by it is simply wrong. (Assuming that ZFC is consistent) there do indeed exist countable models of ZFC. In such a model every set appears countable if we look at it from the outside. But the sentence the reals are countable is _false_ in those models. Probably you'd be better off getting the trivialities that people are explaining to you straight first, before trying to move on the actual mathematical logic... >Russell >- 2 many 2 count ************************ === Subject: Re: Rationals are Uncountable <3Pmdndh6FqkP_Ird4p2dnA@comcast.com> <-oKdnfK9mv7KBIXdRVn-hA@comcast.com> <87k73ahssr.fsf@phiwumbda.org> Discussion, linux) Cancel-Lock: sha1:s11Pqaph14xtjaE+DMB8cKUaKUQ= > You should read the abstracts of some of these papers. > The last paper lis states: In fact we prove that if $M$ is an inner model of set theory > and the set of reals in $M$ is analytic then either > all reals are in $M$ or else $aleph _1^M$ is countable >> Aha! That's what you thought was funny. >> Russell, you twit, they're talking about models of set theory within set >> theory, not the model of reals. The author is *not* contradicting >> Cantor. > I'm not contradicting Cantor, either. > Obviously, the reals are uncountable if the rationals are. Uh huh. And so are the natural numbers. If you prove that Q is uncountable, then you've proved that ZFC is inconsistent. > There are models of ZFC where the reals are countable. I won't even begin to point out why this doesn't mean that R is really countable. I hesitate for two reasons: (1) I'm not well acquain with the model theory that explains this result and (2) you surely haven't the background necessary to understand it. I'd probably bungle the explanation, but even if (by some stroke of luck) I got it just right, somehow I doubt you'd understand the result. In any case, those models also satisfy the sentence that R is uncountable, since they are models of ZFC. -- At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom that evening, Robert Dees urges us Ôto network on behalf of the people of Iraq.' -- Naomi Klein reports on Microsoft's efforts to further democracy. === Subject: Re: Rationals are Uncountable <3Pmdndh6FqkP_Ird4p2dnA@comcast.com> <-oKdnfK9mv7KBIXdRVn-hA@comcast.com> Discussion, linux) Cancel-Lock: sha1:hHDJaBy4QRWCwt4rSHzJziJVLds= > You can find a proof of just about anything if you look hard enough. > I found these in mathematics arXiv: > There are less transcendental numbers than rational numbers > http://front.math.ucdavis.edu/math.GM/0305326 > Countable structure does not have a free uncountable auorphism group > http://front.math.ucdavis.edu/math.LO/0010305 > Intersection properties of open sets, II > http://front.math.ucdavis.edu/math.LO/9605209 > Peano Arithmetic may not be interpretable in the monadic theory of orders > http://front.math.ucdavis.edu/math.LO/9308219 > The complexity of the reals in inner models of set theory > http://front.math.ucdavis.edu/math.LO/9501203 > You might think these authors are crackpots, > but many of them are highly respec in their field. Why would we think they're crackpots, aside from the first one? I don't know what makes one think that, say, PA *ought* to be interpretable in the monadic theory of orders. I don't know diddly about the monadic theory of orders. I don't know why that complexity On the other hand, you are incapable of taking countable intersections don't have the same problems as you, you know). How is it that you can now understand enough mathematics to determine whether, say, Velickovic and Woodin got it right when they analyzed the complexity of reals in inner models of set theory? also proofs that Cantor is wrong. arXiv.org (gosh, what are the odds?). How does that prove the point that one can prove anything if he tries hard enough? Crackpots don't have *proofs*. They are deluded. Footnotes: [1] It is for this reason that I haven't bothered to figure out what this nonsense with sequences is supposed to mean. You can't understand basic mathematical arguments, but now you're presenting some complica construction involving iterating sequences. I simply can't imagine that you've gained any insight worth my time in figuring out what the heck you're on about. -- There are people [...] who think it's socially acceptable to level accusations of mental illness in insulting exchanges to make points[...] [They] are rather sick [them]selves, and in reality, are sociopathic. --- , evidently a self-described sociopath === Subject: Re: Rationals are Uncountable > You can find a proof of just about anything if you look hard enough. You can find claims of proof of just about anything, but not all such claims stand up to careful scrutiny, viz, your claim that the rationals are uncountable and your claim thet the reals are countable. === Subject: Re: Rationals are Uncountable No sequence order isomorphic to x_n can represent ANY rational > in the interval (-1-r, 1-r). Which order on x_n are you talking about, the order of n in N or the order of x_n in Q? They must be different. I don't understand your question. When you say sequence order, why are you not referring to the position of the x_n's in the sequence x_1, x_2, x_3, ... instead of their position in the rationals? > I think n in N. > The order of x_n in Q can't be of order type 1. > (There is no countable function that contains every > member of Q in linear order.) If your linear order refers to the order as members of the field of rationals, correct, but it is the isue of what ordering you are ref erring to that is at issue. Why is this? x_n is supposed to contain > every rational. Why would I need a different sequence to get all > of the rationals in (-1-r, 1-r)? If x_n is rational and r is irrational then x'_n = x_n - r is irrational. This other sequence must contain rational numbers not in x_n. Not if of the form I described. Let me explain my proof in more detail. Let X0 be all rationals in the interval (-1,1). > Let x0_n be a countable sequence that contains every member > of X0. Construct the sequences a0_n and b0_n as described > by Cantor. The intersection of all (a0_n, b0_n) contains an > irrational number, r0. Choose a rational number, q1, that has the same sign as r0 and > |q1| < |r0|. Construct a new sequence, x1_n, as follows: > x1_n = x0_n - q1 x1_n is a sequence that contains every rational number in > the set X1 = (-1-q1, 1-q1). Construct a1_n and b1_n. a1_n and b1_n have some interesting properties. > For example, if a0_n = x0_i and b0_n = x0_j then > a1_n = x1_i and b1_n = x1_j. IOW, the same indices of x0_n that make up a0_n and b0_n > must make up a1_n and b1_n. Since r0 was not 0, there are intervals (a0_i, b0_i) that do not > contain 0. Let j be the index of the first such interval. Similarly, > there is a first interval, (a1_k, b1_k), that does not contain 0. > j must be less than k. I am pretty sure this allows me to prove the > following: If x0_m = 0 and x1_n = 0 then m < n. By subtracting q1 from every member of x0_n I am forcing > 0 to appear later in x1_n than it did in x0_n. The intersection of a1_n and b1_n contains an irrational > number, r1. Choose rational q2 to have the same sign as r1 and > |q2| < |r1| Repeat the process I describe. 0 must appear later in the > sequence x2_n than it does in x1_n. By repeating this process, I can force 0 to appear at > infinity in some xk_n. Define q = q1 + q2 + q3 + ... + qk > By construction, |q| < |r0|. > q is a rational number. > q is in the range of X0. The sequence x0_n does not contain q. A dispacement transformation os any subset of the reals meas a mapping of form f(x) = x + r for some real number r. The final result of all your infinite resequencing is a displacement transformation of this type for some rreal number r. If (-1,1) is transformed to (-1+r, 1+r) then the set of rationals in (-1,1) is either transformed into a set of all rationals with rational LUB of 1 + r and rational GLB of -1 + r when r is rational, or a set of all irrationals with irrational LUB of 1 + r and irrational GLB of -1 + r when r is irrational. If r is rational, the the image of (-1,1) contains all rationals strictly between -1+r and 1+r, but if r is irrational, then the image contains no rationals at all. > Russell > - 2 many 2 count > === Subject: Re: Rationals are Uncountable Define q = q1 + q2 + q3 + ... + qk By construction, |q| < |r0|. q is a rational number. q is in the range of X0. The sequence x0_n does not contain q. > A dispacement transformation os any subset of the reals meas a mapping > of form f(x) = x + r for some real number r. > The final result of all your infinite resequencing is a displacement > transformation of this type for some rreal number r. Some rational number, q. > If (-1,1) is transformed to (-1+r, 1+r) then the set of rationals in > (-1,1) is either transformed into a set of all rationals with rational > LUB of 1 + r and rational GLB of -1 + r when r is rational, or a set of > all irrationals with irrational LUB of 1 + r and irrational GLB of > -1 + r when r is irrational. > If r is rational, the the image of (-1,1) contains all rationals > strictly between -1+r and 1+r, but if r is irrational, then the image > contains no rationals at all. q is a rational number. |q| < 1. Resequencing (-1,1) to (-1-q, 1-q) shows that the xk_n does not contain 0. The original sequence, x0_n, does not contain q. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable > q is a rational number. |q| < 1. > Resequencing (-1,1) to (-1-q, 1-q) shows that the xk_n does > not contain 0. The original sequence, x0_n, does not contain q. If the Cantor theorem is to apply, we need only consider sequences, x_n, which cover the set X, in this case the rationals in (-1,1). If q is rational and the original sequencing, x_n, contains every rational in (-1,1), then any translation of (-1,1) by a rational, x -> x + q, with |q| < 1, produces a new x'_n which still contains every rational in (-1+q,1+q), including 0. It is only if you are not trying to apply the Cantor theorem that you can allow the x_n to omit values of X. === Subject: Re: Rationals are Uncountable q is a rational number. |q| < 1. Resequencing (-1,1) to (-1-q, 1-q) shows that the xk_n does not contain 0. The original sequence, x0_n, does not contain q. > If the Cantor theorem is to apply, we need only consider sequences, x_n, > which cover the set X, in this case the rationals in (-1,1). > If q is rational and the original sequencing, x_n, contains every > rational in (-1,1), then any translation of (-1,1) by a rational, x -> x > + q, with |q| < 1, produces a new x'_n which still contains every > rational in (-1+q,1+q), including 0. 0 is provably not in x'_n. Russell - 2 many 2 count === Subject: Re: Rationals are Uncountable > q is a rational number. |q| < 1. > Resequencing (-1,1) to (-1-q, 1-q) shows that the xk_n does > not contain 0. The original sequence, x0_n, does not contain q. If the Cantor theorem is to apply, we need only consider sequences, x_n, which cover the set X, in this case the rationals in (-1,1). If q is rational and the original sequencing, x_n, contains every rational in (-1,1), then any translation of (-1,1) by a rational, x -> x + q, with |q| < 1, produces a new x'_n which still contains every rational in (-1+q,1+q), including 0. 0 is provably not in x'_n. > Russell > - 2 many 2 count Then there was some rational p in (-1,0) noy in x_n. === Subject: Infinite interpolation. Many folks here and elsewhere have considered plausible ways of constructing a continuous, even smooth, function to go through infinitely many given points. Typically these would be (1, y_1) (2, y_2) (3, y_3) etc, where {y_n} is some given sequence. Power-towers are a popular choice for this! The above task is to smoothly extend a given function f:N-->R to f:R-->R . The original domain, (N above), needn't be restric to N, though. It could be any countable domain; even a converging sequence, like the numbers 1/n (rather than n); or where the members are more spread out, like n^2. And indeed this latter type seem to be easier to handle! My first thought was to try to extend Lagrange's polynomial interpolation formula for a finite domain, to an infinite one. Surprisingly to me, this method seems to work when the domain is sparse, e.g. {n^2 | n in N}, but not for N itself! Specifically:- given a sequence {y_n | n in N} , find a smooth function f such that f(n^2) = y_n for all n. An answer seems to be... oo i=/=K (x^2 - i^2) f(x) = SUM y_I * PROD ----------- ...with obvious mofification for K=1 i>=1 K^2 - i^2 any other countable domain. This is correct for arguments K^2, and seems to converge for all arguments. The formula seems to work for the given original domain, the squares, or for any similar one such that the sum of the reciprocals converges. But it does not converge for the simpler domain N, and the like. Is this already known? Is it significant? Does the formula give reasonable answers for other arguments? ------------------------------------------------------------- --------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ------------------------------------------------------------- --------------- -- Q: What is a tachyon? A: A gluon that isn't completely dry. ------------------------------------------------------------- --------------- -- === Subject: Re: Infinite interpolation. > Specifically:- given a sequence {y_n | n in N} , find a smooth function f > such that f(n^2) = y_n for all n. An answer seems to be... > oo i=/=K (x^2 - i^2) > f(x) = SUM y_I * PROD ----------- ...with obvious mofification for > K=1 i>=1 K^2 - i^2 any other countable domain. This can be simplified using the product formula for sin( pi t ) == pi t product (1 - t^2/n^2), n = 1..oo, and then you get the formula Daryl just pos. The formula is valid for well-behaved sequences y_n. If you multiply each term by exp(-(x-K)^2), it becomes valid for a much larger class of sequences. === Subject: Re: Infinite interpolation. > Many folks here and elsewhere have considered plausible ways of constructing > a continuous, even smooth, function to go through infinitely many given > points. > Typically these would be (1, y_1) (2, y_2) (3, y_3) etc, where {y_n} is > some given sequence. Study of this problem, where the interpolant is to be an analytic function on a given domain G in the complex plane (and the set where it is defined to start with is any set without a limit point in G), can be found in many textbooks on complex variables. The condition for existence is usually sta as the sequence of values does not grow in modulus too fast. -- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Infinite interpolation. >My first thought was to try to extend Lagrange's polynomial interpolation >formula for a finite domain, to an infinite one. Surprisingly to me, >this method seems to work when the domain is sparse, e.g. {n^2 | n in N}, >but not for N itself! >Specifically:- given a sequence {y_n | n in N} , find a smooth function f > such that f(n^2) = y_n for all n. >An answer seems to be... > oo i=/=K (x^2 - i^2) >f(x) = SUM y_I * PROD ----------- ...with obvious mofification for > K=1 i>=1 K^2 - i^2 any other countable domain. I guess that should be y_K in this formula. >This is correct for arguments K^2, and seems to converge for all arguments. Huh? If the sum converges you'd get f(n) = y_n, not f(n^2) = y_n. But I don't know why you think the sum would converge. Write it as sum_{k=1}^infinity y_k g_k(x). Clearly, for any fixed non-integer x, the sum will diverge if y_k grows rapidly enough. 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: Infinite interpolation. Bill Taylor says... >Many folks here and elsewhere have considered plausible ways of constructing >a continuous, even smooth, function to go through infinitely many given points. >Typically these would be (1, y_1) (2, y_2) (3, y_3) etc, where {y_n} is >some given sequence. Power-towers are a popular choice for this! >The above task is to smoothly extend a given function f:N-->R to f:R-->R Here's an idea. Note that sin(pi x)/(pi x) is equal to 0 whenever x is an integer, except for the exception x=0, in which case we can stipulate its value to be 1. So the function f(x) = sum n=1 to infinity of y_n sin(pi (x-n))/(pi (x-n)) would seem to give the value f(n) = y_n (assuming that the sum converges). -- Daryl McCullough Ithaca, NY === Subject: Engelking's General Topology I'm looking for a copy of ryszard engelking's general topology. I've searched every book-dealer/book-search-database on the internet, but haven't been able to turn up a single copy. The publisher in Germany insist on payment in check or cash, plus they take hella long to send it, so this would be my last resort. Can anyone tell me where I can purchase a copy of this book...? or perhaps a magnanimous someone could sell a copy to me (provided it's reasonably priced, since I'm a very poor student)??? Please help... Thanks! === Subject: Converse of Coprimality Game (GCDs>1) In this one-person game, the rules are similar is in my coprimality game, except here, one places (instead of 1, 2, 3, 4,...) the integein order, from the EKG sequence into the grid. Sequence here: http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A064413 And we want *NO* adjacent integer to be coprime with each other. I have not attemp this at all. Is it possible/easy for any untrivial n's? (The EKG sequence is defined by a(m+1) is lowest positive integer not among {a(1),a(2),...a(m)} where a(m+1) has at least one common prime divisor with a(m).) 1,2,4,6,3,9,12,8,10,... My coprimality game's rules as written in-part in my original post: (thread: http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8& threadm=266426e1.0401 280041.5b6ea997%40posting.google.com&prev=) >I was playing around with this today, and found it somewhat addicting. A 1-person game/ puzzle (with a score), which involves a little math. >First, start with a n-by-n grid drawn on paper. >I suggest a grid of at least n=5, and (for serious fun) perhaps an n >of at least 10. >So, you start by writing 1 in any of the grid's n^2 squares. >You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the 4 >adjacent to the 3, etc, in a chain of increasing integein such >away that (ideally...) all n^2 integers are placed into the grid, one >integer (and ONLY one integer) per square. >(By adjacent, I mean immediately next to in the direction of up, >down, left, or right, but *not* diagonally.) >And, oh by the way, the integers are to be placed so that EVERY pair >of adjacent squares contains two integers which are COPRIME with each >other... >(Again, adjacent is as defined above.) >So, you place the integers into the grid as far as you can until your >path cannot be continued for whatever reason. >And your score is the last integer you were able to write into a >square. . >I had intended long ago to post this game idea, which is based upon >the puzzle/math problem at: >http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2f >df.0211011659.79913415%40posting.google.com&rnum=22&prev= >(I may have actually already mentioned this exact idea, but more >likely a 2-player variation. And, anyway, I have a question regarding >it, so as to justify a new thread.) Leroy Quet === Subject: Re: Converse of Coprimality Game (GCDs>1) > In this one-person game, the rules are similar is in my coprimality > game, except here, one places (instead of 1, 2, 3, 4,...) the > integein order, from the EKG sequence into the grid. > Sequence here: > http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anu m=A064413 And we want *NO* adjacent integer to be coprime with each other. I have not attemp this at all. Is it possible/easy for any untrivial n's? (The EKG sequence is defined by a(m+1) is lowest positive integer not > among {a(1),a(2),...a(m)} where a(m+1) has at least one common prime > divisor with a(m).) 1,2,4,6,3,9,12,8,10,... > Obviously, players should start at the 2, not the 1. (I just copy/pas straight from the EIS, which includes the 1 {I believe} so the sequence is then a permutation of the positive integers.) And there is no solution for 2 <= n <= 4 , at least. Leroy Quet My coprimality game's rules as written in-part in my original post: > (thread: > http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8& threadm=266426e1.04012 80041.5b6ea997%40posting.google.com&prev=) > I was playing around with this today, and found it somewhat > addicting. A 1-person game/ puzzle (with a score), which involves a little math. First, start with a n-by-n grid drawn on paper. I suggest a grid of at least n=5, and (for serious fun) perhaps an n of at least 10. So, you start by writing 1 in any of the grid's n^2 squares. You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the > 4 adjacent to the 3, etc, in a chain of increasing integein such away that (ideally...) all n^2 integers are placed into the grid, one integer (and ONLY one integer) per square. (By adjacent, I mean immediately next to in the direction of up, down, left, or right, but *not* diagonally.) And, oh by the way, the integers are to be placed so that EVERY pair of adjacent squares contains two integers which are COPRIME with each other... (Again, adjacent is as defined above.) So, you place the integers into the grid as far as you can until your path cannot be continued for whatever reason. And your score is the last integer you were able to write into a square. > .... I had intended long ago to post this game idea, which is based upon the puzzle/math problem at: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2f df.0211011659.79913415%40posting.google.com&rnum=22&prev= (I may have actually already mentioned this exact idea, but more likely a 2-player variation. And, anyway, I have a question regarding it, so as to justify a new thread.) > === Subject: Re: Simple Question ... > That is true, but it is also true that, for example, the function > f(x) = ßoor(x) is increasing. The difference is that ßoor(x) is not > strictly increasing. Contrary to what the previous poster said, when > mathematicians say that ßoor(x) is increasing, they are not using casual > speech or being careless about it, they are correctly using technical > language. The standard and universally accep mathematical definition of > f is increasing is that f(y) >= f(x) for y >= x. Unfortunately there is no such standard and universally accep mathematical > definition of natural number or even of positive integer, so ambiguities > can arise! What I find fascinating is that with increasing Anglo-Saxon mathematics follow the Bourbaki-ist view, while they do not with positive. -- === Subject: Re: proposition/logical question Thanks lots again, Dear Smith. -- John Woo > I'm not sure this's the correct group to post for proposition question. anyway, the problem (similar to smullyan's puzzle) is two islands x and y, people there are either T (True,always tells truth), F (false, always lie) and N (neutral, sometimes T othertimes F). other facts are, gold is in at least one of these two islands and, if any people N is in island x, then gold is in both X and Y. question if we can ask one question to one of these three kinds of people in order to infer which island contains gold, what effective question can we ask? Consider: Let the proposition S be you always say Ôyes' when asked if 2 + 2 = 4. > Let P be any (decideable) proposition. Then consider the question ÔIs S is equivalent[1] to P?'. P is either true or false and there are three types of respondent: Case 1: respondent is T Answer Ôyes': P is true (It's claimed to be equivalent to a true statement > and the statement of equivalency is true). Answer Ôno' : P is false (It's not equivalent to a true statement.) Case 2: respondent is F Answer Ôyes': S is falsely claimed to be equivalent to P; since S is > false, it follows that P must be true. Answer Ôno': S is falsely claimed to be inequivalent to P; since S is > false, it follows that P is false. Case 3: respondent is N > Nothing can be inferred about P from the answer. So, if you ask the question Ôis the statement there is gold on island X > equivalent to the statement you always answer Ôyes' when asked if > 2 + 2 = 4?' and the answer is Ôyes', you dig on island X; if the answer > is no, you dig on island Y. (If you have chosen an N, then there's gold on > both islands; the choice is forced by what happens if you have chosen > T or F.) [1] Ôequivalent' here meaning that either both are true or both > are false. > Thanks lots, P.A.C. Smith. but I still have no idea on how to formally prove it (without using true/false table). e.x. to prove F answer No idicate Gold is on other island, namely on Y, to formalize statements, there are X: gold on x f : always lie S: 2 + 2 = 4 > Wrong S. > You have asked an F whether X is equivalent to a universally true > statement, which is basically just asking ÔIs X true?' and he says ÔNo.' > Were you to ask a T, you would get the opposite response, so you don't get > any information. > You have to use a proposition whose truth depends on the nature of the > respondent (ÔYou claim that it is the case that 2 + 2 = 4' rather than ÔIt > is the case that 2 + 2 = 4') in order to ensure that answering ÔYes' means > ÔX is true' rather than ÔI claim that X is true'. F: say no so, f say, meaning f = , no equivalent meaning != , so > I can't follow this. f = ( X != S) since S is true (by maths), thus X != S is X != true (by substitute) , so X = false, hence gold no in x island Did I prove it correctly and formally ? > Not sure. Here's my attempt: > X: gold on x > S: speaker claims that 2 + 2 = 4 > (Notation: !X = not X, (X <=> Y) = X equivalent to Y, 1 = true, 0 = false) > Assume speaker is F. Therefore S is false. > F says no: > asser: (!(X <=> S)) = 1 > so in fact: (X <=> S) = (X <=> 0) = 1 > ie, X is false. > F says yes: > asser: (X <=> S) = 1 > so in fact: !(X <=> S) = !(X <=> 0) = (X <=> 1) = 1 > ie, X is true. > OTOH, assume speaker is T. Therefore S is true. > T says no: > asser: !(X <=> S) = 1 > so in fact: !(X <=> S) = !(X <=> 1) = 1 > ie, X is false - as in F case. > T says yes: > asser: (X <=> S) = 1 > so in fact: (X <=> S) = (X <=> 1) = 1 > ie, X is true - as in F case. > If S were still false here, we'd be in trouble. > -- > P.A.C. Smith > The vast majority of Iraqis want to live in a peaceful, free world. > And we will find these people and we will bring them to justice. === Subject: Re: word reduction > cannoneer - 1 = canner > heighten - 8 = hen > alanines - 9 = alas > advertent - 10 = advert I don't have any for 2 to 7. You wouldn't accept fourteen - 4 = teen, sixteen - 6 = teen, seventeen - 7 = teen, would you? -- === Subject: Re: word reduction >> cannoneer - 1 = canner >> heighten - 8 = hen >> alanines - 9 = alas >> advertent - 10 = advert >> I don't have any for 2 to 7. >You wouldn't accept >fourteen - 4 = teen, >sixteen - 6 = teen, >seventeen - 7 = teen, >would you? No, I wouldn't. Nobody has formalized the rules, but I'd say that it's cheating to have the number as one of the elements of a compound word. Oh, I found: catwoman - 2 = caman (which, according to the OED, is a stick or club used in the game of shinty) cartwoman - 2 = carman (I don't know what cartwoman means, but it does get 613 hits in Google). 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: word reduction Robert Israel You wouldn't accept ... would you? > No, I wouldn't. But wilt thou kindly accept the submission of 1000 + reckoner = Archimedes (See for example:) Asks Rainer Rosenthal r.rosenthal@web.de === Subject: Re: word reduction > Oh, I found: > catwoman - 2 = caman (which, according to the OED, is a stick or > club used in the game of shinty) Thinking of obscure games, let me suggest, in honor of Leroy, piquet - 3.14159... David === Subject: Inequality Poser Injector-Info: news.mailgate.org; posting-host=210.21.106.12; posting-account=61944; posting-date=1075348346 X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 5e43aad58cce294381fa81222f38ee 00.61944%40mygate.mailgate.org Show that, for real a,b,c > 0, 27[(a + b)(b + c)(c + a)]^2 >= 64abc(a + b + c)^3 -- Pos via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Re: Choosing a Math Grad school > Hi guys, I'd like some info about math PhD programs, if anyone can please help. A > concern I have are rankings, in particular: how much do they matter when you > make *final* decisions? I've my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago, > Texas-Austin, Princeton, Illinois-Urbana. My area of interest is operator > algebras (thus the first three) or Harmonic analysis and PDE (thus the > latter four). Depending on place-offers (and assuming that one of them can > be excluded), do any of you guys know reasons for preferring one to the > other? (modulo that these are all great places for analysis research). For instance, suppose you get an offer from UCLA and Illinois-Urbana or Penn > State, do you obviously go for UCLA because it is higher ranked? My > intention is to follow an academic career (professorship), so future > placement prospects are crucial to me. Thanks for any comments, particularly from mathematicians that have been or > have visi the above mentioned universities and may be able to give me > some hands-on impressions! > Don't forget to factor in social and cultural environment. Princeton is a lot different than Berkeley. Austin is not Los Angeles. Where do you want to spend the next few years of your life? === Subject: Re: Harris and Zelazny > I just got a copy of Roger Zelazny's Creatures of Light and Darkness > for a friend, and noticed the following quote at the start: Generations pass away and others go on, > since the time of the ancestors. > They that build buildings, > their places are no more. > What has been done with them? ((( four more stanzas ))) --Harris 500, 6:2-9 > Is this our Harris? I this refers to the Harris Papyrus in the British Museum. - William Hughes === Subject: Re: Harris and Zelazny I just got a copy of Roger Zelazny's Creatures of Light and Darkness for a friend, and noticed the following quote at the start: Generations pass away and others go on, since the time of the ancestors. They that build buildings, their places are no more. What has been done with them? ((( four more stanzas ))) --Harris 500, 6:2-9 Is this our Harris? I this refers to the Harris Papyrus in the British Museum. - William Hughes My God!!! Do you mean he's been at this not just for over a decade, but for thousands of years!!!??!!!! Cheers - Chas === Subject: Re: Harris and Zelazny I just got a copy of Roger Zelazny's Creatures of Light and Darkness >for a friend, and noticed the following quote at the start: >Generations pass away and others go on, >since the time of the ancestors. >They that build buildings, >their places are no more. >What has been done with them? >((( four more stanzas ))) > --Harris 500, 6:2-9 >Is this our Harris? >>I this refers to the Harris Papyrus in the British Museum. >> - William Hughes > My God!!! Do you mean he's been at this not just for over a decade, > but for thousands of years!!!??!!!! Cheers - Chas Maybe it just seems that way. === Subject: stereographic projection and circles Im trying to prove that circles are mapped to circles by stereographic projection, but i just had an idea that seemed to make the problem trivial. of course, that leads me to believe there is something wrong with it. This is it: Let f: S^2 --> C_infty be the stereographic projection of the unit 2-sphere onto the extended plane. This map is bijective (right? I have explicit formulas for f and its inverse), and therefore has an inverse, say g=f^{-1}. Because of this, showing that circles in S^2 map to circles in C_infty is the same as showing circles in C_infty map to circles in S^2. With that in mind, take three points in C_infty that define a circle, say z1,z2,z3, then g(z1),g(z2),g(z3) are all distinct and therefore define a plane in R^3. This plane cuts through S^2, and it is already well known that the intersection of a plane and a sphere is a circle (or empty set or point, but that doesnt really matter). So does this show everything that needs to be shown? What went wrong (if anything)? I am also aware of the fact that a circle on S^2 containing (0,0,1) maps to a line in C_infty, but I am not sure how that fits into this argument (which may be the source of the problem in the first place). Thanks for any help === Subject: Re: stereographic projection and circles Pos-And-Mailed: yes === > Im trying to prove that circles are mapped to circles by stereographic > projection, but i just had an idea that seemed to make the problem > trivial. of course, that leads me to believe there is something wrong > with it. This is it: Let f: S^2 --> C_infty be the stereographic projection of the unit > 2-sphere onto the extended plane. This map is bijective (right? I have > explicit formulas for f and its inverse), and therefore has an > inverse, say g=f^{-1}. Because of this, showing that circles in S^2 > map to circles in C_infty is the same as showing circles in C_infty > map to circles in S^2. With that in mind, take three points in > C_infty that define a circle, say z1,z2,z3, then g(z1),g(z2),g(z3) > are all distinct and therefore define a plane in R^3.... Yes, but if z4 is any other point on your circle in C_infty, how do you know that g(z4) lies on the same plane? You could have star from three points on any curve whatever in C_infty, so would you then conclude that any such curve is mapped to a circle? Ken Pledger. === Subject: radius of convergence for series here are two series sum_{n=0}^infty a^n z^n , where ain C and sum_{n=0}^infty k^n z^n , where kin Z{0} It seems to me these two series would have the same radius of convergence. I mean, Z _is_ contained in C, so how is the second any different? These series came up as two different problems (in Conway's complex variables book). by the way, I got 1/|a| for the first, so I assume 1/|k| would work for the second. Is there anything different between the two? thanks you very much.... === Subject: Re: radius of convergence for series >here are two series >sum_{n=0}^infty a^n z^n , where ain C >and >sum_{n=0}^infty k^n z^n , where kin Z{0} >It seems to me these two series would have the same radius of >convergence. I mean, Z _is_ contained in C, so how is the second any >different? These series came up as two different problems (in Conway's >complex variables book). by the way, I got 1/|a| for the first, so I >assume 1/|k| would work for the second. Well then the two series do _not_ have the same radius of convergence! 1/|a| is not the same as 1|k|. But what you have in mind here, as oppposed to what >Is there anything different between the two? >thanks you very much.... ************************ === Subject: Men Like Gods Sarfatti Lecture 2 on Metric Engineering the Zero Point Dark Energy for UFO Super Technology of Men Like Gods (H.G. Wells) http://books.fantasticfiction.co.uk/n0/n1391.htm?authorid=3082 http://www.troynovant.com/Franson/Wells/Men-Like-Gods.html RovelliÍs formalism in http://www.cpt.univ-mrs.fr/~rovelli/book.pdf e^I(x) = eu^I(x)dx^u = ñgravitational fieldî Cartan 1-form i.e. his eq. 2.1 p.41 I in Minkowski space, but Rovelli says u in Tangent vector space TxM not in base space M. This is odd since the local tangent space has the Minkowski metric and the curved base space has guv metric. The globally ßat solution is eu^I(x) = (Kronecker Delta)u^I i.e. eq 2.44 p.45 My elastic-plastic Kleinert ñworld crystalî distortion field must be L^u(x) = [e^uI(x) - (Kronecker Delta)^uI]dx^I = (Loop Gravity Quantum of Area)arg(Vacuum Coherence),u (S1) ,u is ordinary partial derivative Next introduce the antisymmetric ñspin connectionî that will apply to spinning tops for example. The spin connection is also a Cartan 1-form W^IJ(x) = Wu^Ijdx^u i.e. eq. (2.2) p. 41 There are 6 of these generators of the Lie algebra so(3,1)of the 6 parameter Lorentz group of 3 space-time rotations (boosts) and 3 space-space rotations (total angular momentum) WIJ = - WJI Note that L^u(x) is the local gauge force compensating field restoring the broken 4-parameter translation group symmetry T4. However, while the T4 and so(3,1) are intertwined as a ñsemi-direct productî and as seen in the Thomas precession and the Sagnac effect, there is no local gauge force compensating field for Lie algebra so(3,1) when we use a torsion-free spin connection to define the tidal force tensor curvature that intimately uses so(3,1). Thus, the generally covariant partial derivative of General Relativity in the Cartan formalism starts from DuV^I = V^I,u + Wu^IJV^J i.e. (2.3) p. 41 More generally the generally covariant Cartan exterior derivative on a 1-form v^I is the 2-form DV^I = dV^I + W^IJ/V^J i.e. (2.4) p. 42 Where d is the Cartan exterior derivative and / is the wedge product. What is the relation of / to Clifford algebra? Note that the Lorentz group Lie algebra so(3,1) is the fundamental connection field from the breaking of the global symmetry of T4 whose Lie algebra is total energy-momentum. Total ñmomenergyî (Wheeler) is not a natural concept in General Relativity because locally variable curvature breaks global translational symmetry even though curvatureÍs definition involves the rotations of so(3,1) since local curvature is the anholonomy disclination ñBerry phaseî in the orientation of a vector parallel transpor around a closed space-time loop as the area of the loop shrinks to zero. Torsion T^I means that the loop in tangent fiber space is broken or disloca for a closed loop in base space or vice versa. That is, with torsion, closed loops do not map to closed loops in parallel transport. The condition of zero torsion gauge force field is the vanishing Cartan torsion field 2-form T^I = De^I = de^I + W^Ije^J = 0 i.e. (2.6) p. 42 Again note that it is the spin connection from so(3,1) that plays the vital role! Finally the tidal relative acceleration or timelike geodesic deviation is also a Cartan 2-form from the zero torsion so(3,1) spin connection (which allows fermions as well as bosons) R^IJ = dW^IJ + W^IK/W^KJ i.e. (2.8) p. 4.2 i.e. tidal curvature is the nonlinear self-organizing spin connection covariant exterior derivative on itself! This is why there are curved vacuums independent of real mass-energy sources! Gravity acts on itself in a non-Abelian way. The Einstein local field equation with the zero point energy cosmological /zpf term is in Cartan formalism is the Euler-Lagrange equation [IJKL]e^I/R^J^K + /zpf e^I/e^J/e^K = 0 i.e. (2.9) p. 42 where [IJKL] is the antisymmetric 4-symbol Note the cubic nonlinearity in the zero point energy / ñmetric engineeringî term. The dynamical action density is ~ e^I/e^J/R^K^L + /zpf e^I/e^J/e^K/e^L contrac with [IJKL] to make an invariant scalar i.e. eq. (2.10) p. 42 Sarfatti Lecture 1 (typo correc) on the Zero Point Dark Energy Metric Engineering of UFOs http://www.livingreviews.org/Articles/Volume4/2001-1carroll/ . Carroll, S.M., ñThe Cosmological Constant, Living Rev. Relativity, 4, (2001), ñGeneral relativity is a paradigmatic example of a scientific theory of impressive power and simplicity. The cosmological constant, meanwhile, is a paradigmatic example of a modification, originally introduced [81] to help fit the data, which appears at least on the surface to be superßuous and unattractive. Its original role, to allow static homogeneous solutions to Einstein's equations in the presence of matter, turned out to be unnecessary when the expansion of the universe was discovered [131], and there have been a number of subsequent episodes in which a nonzero cosmological constant was put forward as an explanation for a set of observations and later withdrawn when the observational cosmological constant can be interpre as a measure of the energy density of the vacuum. This energy density is the sum of a number of apparently unrela contributions, each of magnitude much larger than the upper limits on the cosmological constant today; the question of why the observed vacuum energy is so small in comparison to the scales of thought to be easier to imagine an unknown mechanism which would set it precisely to zero than one which would suppress it by just the right amount to yield an observationally accessible cosmological constant. This checkered history has led to a certain reluctance to consider further invocations of a nonzero cosmological constant; however, recent years have provided the best evidence yet that this elusive quantity does play an important dynamical role in the universe.î Note in the above excerpt: interpre as a measure of the energy density of the vacuum. This energy density is the sum of a number of apparently unrela contributions, each of magnitude much larger than the upper limits on the cosmological constant today; the question of why the observed vacuum become a celebra puzzleî I allege I have solved this problem using the idea of ñvacuum coherenceî missing from the orthodox theory in the precise way I use it. The idea of ñvacuum condensateî is in orthodox theory. It is a rela idea, but not exactly the way I mean it. EinsteinÍs GR local geometrodynamical field equation with the cosmological term /guv is Ruv [CapitalEth] (1/2)Rguv + /guv = 8pi(G/c^4)Tuv (1) Impose the large-scale coarse-grained isotropic homogeneous Friedman-Robert-Walker solution ds^2 = -(cdt)^2 + a^2(t)Ro^2[(dr)^2/(1 [CapitalEth] kr^2) + r^2dO^2] (2) r is dimensionless and the Her MajestyÍs Royal NavyÍs navigational ñcelestial sphereî spherical angular line element is dO^2 = (dtheta)^2 + (sintheta)^2(dphi)^2 (3) in usual polar coordinates for latitude theta and longitude phi on the 2D sphere of unit radius. a(t) is dimensionless = R(t)/Ro. Subscript o means ñnowî. k is +1 (closed universe in 3D like a sphere) or 0 (spatially ßat like an infinite plane in EuclidÍs geometry) or [CapitalEth] 1 like a hyperboloid. Both k = 0 and k = -1 are open universes of infinite spatial extent. The cosmological redshift z of retarded radiation from a co-moving source in the ñHubble ßowî where the Cosmic Black Body Radiation (CBR) is maximally isotropic from the past till now obeys the equation a(past) = [1 + z(now)]^-1 (4) The symmetric stress-energy density tensor Tuv for ñstuffî on the RHS of eq (1) is of the form Tuv = (energy density + pressure)UuUv + pressure guv = (energy density)[(1 + w)UuUv + wguv] (5) Uv is the dimensionless 4-velocity dx^u/ds of this ñcosmic ßuidî stuff. w = 0 for cold matter (6a) w = 1/3 for electromagnetic radiation (far field) (6b) w = -1 for any kind of zero point vacuum ßuctuation (ZPF) of any quantum field including string and brane fields (6c) An ñexotic vacuumî is any kind of virtual stuff with w = -1 and a non-vanishing pressure. has ñdarkî gravity and anti-gravity properties bending EM radiation signals for example. Dark matter exotic vacuum in a clump will act like a convex converging gravity lens. Dark energy in a clump will act like a diverging gravity lens. Dark matter makes a gravity red shift and dark energy makes a gravity blue shift. The global cosmic time t (~ 1/(Absolute temperature of CBR) as a convenient measure) dependent Hubble parameter is H(t) = a^-1(da/dt) (7) EinsteinÍs tensor field equation (1) forced into the highly symmetric ñKilling vector fieldî mold of FRW eq (2) simplifies to TWO ordinary differential equations: H^2 = 8pi(G/c^2)(energy density) + c^2//3 [CapitalEth]kc^2/a^2Ro^2 (8a) Note that H^2 is analogous to NewtonÍs Laplacian of the Gravity Potential Energy of ANY Source Stuff (real or virtual in sense of quantum field theoryÍs ñonî or ñoffî ñmass shellî) per unit test analogy to NewtonÍs Poisson equation in this spatially homogeneous strictly large-scale approximation is the second equation a^-1d^2a/dt^2 = -(4pi/3)(G/c^2)[(energy density) + 3(pressure)] + c^2//3 = -(4pi/3)(G/c^2)(energy density)[(1 + 3w) + c^2//3 to be continued PROBLEM SET #1 1.1 Term Paper Project # 1 What does Hal Puthoff write about zero point energy and gravity and its relation to ñinterstellar ßightî and UFOs? Go to http://www.earthtech.org/publications/ Compare what Puthoff and Davis profess with the physics in these lectures. === Subject: Re: Fresnel integral Alexander Schmidt > I want to prove the convergence of the integral from 0 to +oo over the > function sin(x^2) dx. Others have pos answebut you might like to draw part of the graph to see why it should converge, much as an alternating series may converge by the Liebniz test. > 3.I would also be interes in resources concerning the Drichlet integral. The basics are in the Schaum's Outline series, Advanced Calculus, by Spiegel, first edition. I haven't seen the newer second edition. LH === Subject: Re: Flying Saucer Physics a lot of supposition, monsieur Sarfatti, but I've always thought that almost all of the remote viewing folks -- the ones that libera teh technology form the military, after they pretended to become embarrassed & hid it and so on -- were mostly spooks. (the supposition comes-in, because almost my sole source is Art Bell's Ghost-to-Ghost, BC, which is brought to you buy Hicks, Muse, the largest merger & acquisition company on Earth. and that was a couple of years ago .-) *your* supposition is taht security clearances are ipso- facto proof of veracity, or even comprehension. Section II: People Who Believe in Zero-point Energy n.b. for NewBies: Jack *never* responds to newsgroups > technology. Hal has been working on this problem for many decades and > has held high USG security clearances and has been privy to reliable > information that the saucers are real and are alien. Otherwise he > would not be working on the problem. However, IMHO Hal's theories, both > of the zero point energy and of the gravity field will not solve the > corresponds to what Paul Hill called the acceleration field in > Unconventional Flying Objects and to negative matter propulsion in > the sense of Robert Forward's paper on the idea of Herman Bondi also > studied by Joseph Stalin's WWII Los Alamos Manhattan Project physics > spy-master Yacov Terletskii. > The zero point exotic vacuum Josephson tunnel current density is ~ > sin(phase control parameter). The Daily [Yale] Voice from the Mausoleum, thus quoth: Perhaps my judgment is impaired by the fact that I am Jewish, Zionist, and generally sympathetic to Great Britain, but to me it is clear that the guy is a nut. Most people within the current establishment seem to think so as well and dismiss him as a quirky and entertaining oddity. The reality is unfortunately much more sinister. According to the Center for Responsible Politics in his current bid for the White House alone LaRouche raised over $5.5 million in small donations. According to the Federal Election Commission as late as last April his fundraising figures surpassed those of Lieberman, Dean and Graham. According to the last report from three months ago he still has more money than Kucinich, Mosley-Braun and Sharpton combined, all three of them politicians with nationwide standing. The implications of his fundraising success are astounding. He is ignored by the media, locked out of debates, and has as much chance of winning as a proverbial snowball in hell and yet he raised this truly massive amount of campaign cash. The only possible explanation is that he has a powerful grassroots organization. Just think about it: there is an ultraradical organization out there but the only reason we ever hear about it is because its leader compulsively runs for president and because a member at some point probably shoved a ßier in your hand. The LaRouche experience shows how a radical organization can use the electoral system to fund its activities and spread its message. thus quoth: as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler (Don Fowler was the DNC Chair in Ô96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in Ô07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: JSH: Logical position is clear I'd already had a bad thought, before I saw this origination of youNBB. if all of HSJ's corresponbdents star one HSJ item, each day ... he'd just cut & paste, faster. nevermind! it must be true that in general for algebraic integer x, f(x) is NOT an algebraic integer. > Right. Except for x = 1, x = 2, x = 5, x = 8, x = 10, and, > oh, infinitely many other values of x. > Right. You say that (5 b_2(x) + 2)/sqrt(7) cannot be an > algebraic integer because 2/sqrt(7) is not an algebraic > integer. Never mind the fact that if x = 1, b_2(x) = -sqrt(-14), > so that (5 b_2(1) + 2)/sqrt(7) = -5 sqrt(-2) + sqrt(7), which IS an algebraic integer. Gee, how could that happen ??? > Haven't you said it was impossible ? And never mind the fact that in the factorization of the > Decker polynomial, for *any* x > 0, you can find a value w(x) > which is a nonunit divisor of 7, such that (5 b_2(x) + 2)/w(x) IS an algebraic integer, even though 2/w(x) never is. > Or, look at it this way. Clearly (7 + 7) is divisible by > 7, and of course 7/7 + 7/7 = 1 + 1 is the sum of two algebraic > integers. On the other hand, (5 + 9) is ALSO divisible by > 7, even though neither 5/7 nor 9/7 are algebraic integers. The Daily [Yale] Voice from the Mausoleum, thus quoth: Perhaps my judgment is impaired by the fact that I am Jewish, Zionist, and generally sympathetic to Great Britain, but to me it is clear that the guy is a nut. Most people within the current establishment seem to think so as well and dismiss him as a quirky and entertaining oddity. The reality is unfortunately much more sinister. According to the Center for Responsible Politics in his current bid for the White House alone LaRouche raised over $5.5 million in small donations. According to the Federal Election Commission as late as last April his fundraising figures surpassed those of Lieberman, Dean and Graham. According to the last report from three months ago he still has more money than Kucinich, Mosley-Braun and Sharpton combined, all three of them politicians with nationwide standing. The implications of his fundraising success are astounding. He is ignored by the media, locked out of debates, and has as much chance of winning as a proverbial snowball in hell and yet he raised this truly massive amount of campaign cash. The only possible explanation is that he has a powerful grassroots organization. Just think about it: there is an ultraradical organization out there but the only reason we ever hear about it is because its leader compulsively runs for president and because a member at some point probably shoved a ßier in your hand. The LaRouche experience shows how a radical organization can use the electoral system to fund its activities and spread its message. thus quoth: as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler (Don Fowler was the DNC Chair in Ô96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in Ô07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: 7-space I have not solved the 7th degree polynomial roots. However I have skir around it on this link, http://mypeoplepc.com/members/jon8338/polynomial/id2.html Jon Giffen === Subject: Deconvolution of peaks (help needed) hi all i want some help in the topic of deconvolution i m new to this field and i have to make project in it the problem is i have to identify and deconvolute the peaks given by the data of x-ray defraceter. help me out in finding the algorithms for deconvolution of peaks i came ac one algorithm fourier self deconvolution but i have not material on this topic give me some guidence and some liks where i can get real understanding plz reply as soon as possible. Regards Dhaval === Subject: needed matlab code for FFT and IFFT hi all i need matlab code for FFT and IFFT plz send me as soon as possible to my email id regards Dhaval === Subject: Re: needed matlab code for FFT and IFFT > hi all > i need matlab code for > FFT and IFFT Y = fft(X). There! > plz send me as soon as possible to my email id Nope. You post here, you read here. > regards > Dhaval Same. Steven === Subject: Re: needed matlab code for FFT and IFFT > hi all > i need matlab code for > FFT and IFFT > plz send me as soon as possible to my email id I like to assign homework problems to my numerical analysis class, too. Which text are you using? === Subject: Re: Linear Context Free Grammar > Another, maybe more probable, source for your question is the fact that > I use(d) --as many other people do-- ``linear language'' for > ``linear context-free language''. Of course, one can put the ``linear > restriction Ô' [i.e. at most one nonterminal symbol in the right-hand > sides of the productions] upon context-sensitive (or type 1) grammars > or upon unrestric (type 0) grammars. It is an interesting exercise > to determine/characterize the corresponding language families. Yes, that's the reason! I understood it! Thank you for your reply. Behrang. === Subject: Reimann hypothesis and decimal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbS229756; can someone please tell me what the reimann hypothesis is and possibly give me an example i would appreciate it a lot you can send me an e-mail thank you. === Subject: Re: Reimann hypothesis and decimal X-ID: JTAb5aZD8e5zC1hEWmn7pq4xqC1zZYEWv0vFfcsBvg-+xqvttbGSE5 xococacolaqt88@aol.com (kiki) schrieb: >can someone please tell me what the reimann hypothesis is and possibly give me an example i would appreciate it a lot you can send me an e-mail thank you. The guy's name is Riemann -- Thomas === Subject: Re: Reimann hypothesis and decimal >can someone please tell me what the reimann hypothesis is If zeta(s) = 0 and 0 < Re(s) < 1 then Re(s) = 1/2. >and possibly give me an example I have no idea what an example of this would be. >i would appreciate it a lot you can send me an e-mail Have you decided yet what to do with the million bucks? ************************ === Subject: Re: Reimann hypothesis and decimal |>can someone please tell me what the reimann hypothesis is | |If zeta(s) = 0 and 0 < Re(s) < 1 then Re(s) = 1/2. | |>and possibly give me an example | |I have no idea what an example of this would be. zeta(0.5+14.1347251417347...*i) = 0, zeta(0.5+20.0220396387715...*i) = 0, zeta(0.5+25.0108575801457...*i) = 0, etc. === Subject: Re: Reimann hypothesis and decimal >|>can someone please tell me what the reimann hypothesis is >| >|If zeta(s) = 0 and 0 < Re(s) < 1 then Re(s) = 1/2. >| >|>and possibly give me an example >| >|I have no idea what an example of this would be. >zeta(0.5+14.1347251417347...*i) = 0, >zeta(0.5+20.0220396387715...*i) = 0, >zeta(0.5+25.0108575801457...*i) = 0, >etc. Uh, right. Not sure exactly what 14.1347251417347... denotes - this is more stating that an example exists than actually giving it. (Ok, you were just joking...) > ************************ === Subject: Re: Reimann hypothesis and decimal >>zeta(0.5+14.1347251417347...*i) = 0, >>zeta(0.5+20.0220396387715...*i) = 0, >>zeta(0.5+25.0108575801457...*i) = 0, >>etc. >Uh, right. Not sure exactly what 14.1347251417347... >denotes - this is more stating that an example exists >than actually giving it. >(Ok, you were just joking...) Well, continuing the typical mathematician tendency to degrade humor into dry mathematics, let me point out that it's sort of a pity the Riemann hypothesis is couched in terms of complex analysis. I don't have the formulas to hand right now but I recall there are natural scalings of zeta which move the critical line to the real axis, so that RH becomes the statement, All the zeros of the following function are real. The revised function(s) I have seen are clearly real on the real line and oscillate -- that is, DU's objection can be answered by noting points where the function alternates in sign, which doesn't quite give an example of a zero but certainly _proves_ that an answer exists (instead of simply _asserting_ existence). of real-valued inequalities which are known to be equivalent to RH (some of which are more number-theoretic, making the connection between RH and primes more plausible). This would avoid the confusion on the part of interest amateurs who can't understand how sum(1/n^s) could ever be zero in the first place. We now return you to your local humor newsgroup. dave === Subject: Re: Reimann hypothesis and decimal ... > Well, continuing the typical mathematician tendency to degrade humor > into dry mathematics, let me point out that it's sort of a pity the > Riemann hypothesis is couched in terms of complex analysis. There is a reformulation in real analysis. See . (That reminded me that it is almost 20 years ago that I was involved in this...) -- === Subject: Re: hmm by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbTn29777; >> so where does x^y = y^x when x > 0 and y > 0? >> joe >x = y = 1 ? **********************************************8 Well, actually for all x = y > 0 ... Saludos Tonio === Subject: Re: hmm > so where does x^y = y^x when x > 0 and y > 0? > x = y = 1 ? > **********************************************8 > Well, actually for all x = y > 0 ... Of course. Such trivial solutions aside, the only positive integer solutions for x and y are given by (x,y)=(2,4) and (x,y)=(4,2). For _all_ real solutions of the equation x^y = y^x, see my sci.math . David Cantrell === Subject: Re: Coprimality: One-Person Game (& question) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbUB29807; >I want to mention this example of a 7-by-7 game-result I got today. >It is not remarkable for its score, which is unimpressive (41). (And I >DID erase...) >But it is notable(somewhat) because its path has >(purposefully-crea) rotational-symmetry. My guess is, if you`re looking for more symmetry (perhaps such as this) you should introduce more symmetry into the rule's of the game. This is because not every integer in the square has an adjacent integeger to the top, bot, left right; in my opinion it is, in a way, an asymmetry. But it is easy to change the situation to get rid of it- just redefine top, bot, left, right adjacent to include going from one end and out the other. For example, the top adjacent integer for 2 in the square below would be 28; the right adjacent integer to 16 would be 4, etc. Of course, you will always get lower scores for a given n by n square. >3 2 1 12 13 14 15 >4 9 10 11 * 17 16 >5 8 * 20 19 18 * >6 7 * 21 * 35 36 >* 24 23 22 * 34 37 >26 25 * 31 32 33 38 >27 28 29 30 41 40 39 >Since the center is 21, the multiples of 3 and 7 are symmetrically >placed, as are the multiples of 2 (which would have been the case >whatever the center integer was). [snip] === Subject: Re: sum of series? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbTx29771; >I was wondering if there's a way to evaluate exactly what this series >converges to: >infinity k^k >SUM ------------ >k = 1 (k+1)^(k+1) -Robert Hi Robert, k^k / (k+1)^(k+1) = (1-1/(k+1))^(k+1) * 1/k >= 0.25/k. So, by comparing with the harmonic series, your series diverges towards infinity. (Note that (1-1/(k+1))^(k+1) is monotonically increasing and converges to 1/e). Best wishes Torsten. === Subject: Re: non-explicit differential equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbUM29824; >Are there some standard existence and/or uniqueness results for non >explict differential equations? >Actually I'm interes in the linear case, i.e. equations of the form >Ex' = Ax + b >with or without constant coefficients. >Can anyone give me a reference? >best markus Hi Markus, such systems are called ÔDifferential algebraic systems'. You'll have a lot of hits if you make an internet search. Best wishes Torsten. === Subject: Re: Ideals of Z[x] by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbTp29784; >> anjan208@rediffmail.com >> _ What are all the ideals of the polynomial ring Z[x]? >> Z[x] is an Euclidean domain and >> Euclidean domains are principal ideal domains. >Is the ideal genera by 2 and x a principal ideal? no, if we assume it is genera by some p(x) then {h(x)*p(x)|h in Z[x]} = ==+<2>={x*h(x)+2*k(x)|h,k in Z[x]} since 2 is in , we have 2=h(x)*p(x) and there for we have to have deg h=0 and deg p=0 this will give p(x)=1 or p(x)=2 if p(x)=2 we canot have x in if p(x)=1 we will get 1=x*h(x)+2*k(x) which canot happen since the constant term in right is even we will end up with a contradiction there for is not a principle ideal actuall it is the ideal of all polynomials with even constant term === Subject: Re: Rationals are Uncountable by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbhj29872; ...................... >unlimi means bigger than infinite. ******************************************************** Bigger than infinite? Could you define PRECISELY this? ********************************************************** >> (2) there actually *is* such an unlimi set. >Sets really exist? >I'm not interes in mathematics that might >have anything to do with reality. ************************************************************* ***** Oh, dear! Then you're not interes at all in mathematics at all! Maths has NOT, and I'd add it MUST NOT have, any relation whatsoever with no reality at all. Maybe you meant to say that you're exclusively interes in what nowadays is known as applied mathematics, but this too is based in unreal mathematics, so you better make up your mind quick and maybe look for some other field of interest... Saludos Tonio >Russell >- 2 many 2 count === Subject: Group Theory by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbUY29817; Dear all: Let's say G is an h-group if for any natural number n there there exists at most a finite number of subgroups(=sbgps.) of G of index n (for example finitely genera groups are h-groups). It's easy to show that G is an h-group iff for any normal sbgp. K of G of finite index (=f.i.) and for any finite simple group S there exists only a finite number of normal sbgps. L of K s.t. K/L is isomorphic to S.[1] So let be given an h-group G, a f.i. normal sbgp. K of G and a finite simple group S, and let L1,..., Lr be all the normal sbgps. of K s.t. K/Li isomorphic to S. We can see at once that for all pair of different indexes 1 <= i,j <= r we have that LiLj = K. My question is: can we bound somehow, probably by imposing some additional conditions on G, or K, the number r in terms of n? For example, we could demmand that K/Li be a chief factor of G (thus Li is a normal sbgp. OF G which is maximal with respect to being contained in K), and if G is say solvable, or even locally solvable, then S would be abelian...or we could begin by supposing that n is the least natural number for which there is a normal sbgp. K with some quotient isomorphic to S; we could even begin at first considering sbgps. of index AT MOST n, for some natural n, or we could consider only sbgps. Li which are normal IN G, etc. If, for example, we take K = G and we ask how many different normal sbgps. Li of G are there s.t. G/Li is isom. to S, then auatically G/Li is a chief factor of G and thus we should, probably, ask about these factors under so and so conditions. The above question popped up pretty surprisingly during my PhD research and I didn't pay much attention to it back then. Any insight, hint, recommendation, quote, etc. will be much apprecia. Saludos Tonio [1] Wilson, John S., Groups Satisfying the Maximal Condition for Normal Subgroups, Math. Z. 118, 107-114 (1970), Lemma 1. === Subject: Re: Quaternions and the determinant. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QFk3J07414; Sorry, I think I should generally be a little more specific when asking questions, 2nd try: Is (iii), below, already known? (the original (iv), from the first posting was seen to be incorrect). >My relations are: >Let Q(A) be the quaternion algebra over the reals. >For x = (x_1)i + (x_2)j + (x_3)k + (x_4)1, define >res: Q(A) -> R^3, res(x) = (x_1, x_2, x_3) and >tes: Q(A) -> R , tes(x) = x_4 >Then for all x, y, z in Q(A) with x_4 = y_4 = z_4 = 0 the >following holds: >(i) res(x) scalar-product res(y) = - tes(x*y) >(ia) ||res(x)|| = sqrt( - tes(x*x) ), euclidean norm >(ii) res(x) c-product res(y) = res(x*y) >(iii) det( res(x), res(y), res(z) ) = -tes(x*y*z) For example, a proof that res(x) scalar-product (res(y) c-product res(y)) = = det( res(x), res(y), res(z) ), using the above: res(x) scalar-product (res(y) c-product res(z)) = = res(x) scalar-product res(y*z), from (ii) = - tes( x*(y*z) ), from (i) = - tes( x*y*z ) = det( res(x), res(y), res(z) ), from (iii) q.e.d. Just a check: Let I be the (3 by 3) identity matrix, then det(I) = det( res(i), res(j), res(k) ) = = - tes( i*j*k ) = - tes( -1 ) = 1 C. Dement === Subject: scholarship information by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QHlKd17198; A somewhat unpleasant situation: I'm an American student studying physics in Germany (which in it of itself, is not unpleasant ). I originally came here 92-93 as a high school exchange student with a scholarship from YFU, Youth For Understanding - as part of a Congress/Bundestag initiative) I have a German Vordiplom in physics and, 3 years ago, had all the credits needed to take my final exams and begin my Diplom thesis (thesis usually lasts a year or so). Then I became a father and had to put my studies on hold for two years. When I'd finally gotten back round to my studies, I already had 2-3 private math works written up. rethinking some of my plans. First, I reinven the quaternions- just playing around with some scratch paper one day at home (before that, they'd exis to me only as a buzzword...) Secondly, I scored the highest grade possible on my final exam in functional analysis. I now have a work 60-70% written up and a math professor at my university who sais that it may even have chances of getting published on its own (a little luck would also be nice). In any case, I am told that I could use my own, original work (which has been a study project of mine for the past 6 months) as either (a) a diplom thesis in math or (b) a doctorial thesis (in which case I should say my work is only 30-40% complete). However, under the German system, both (a) and (b) are impossible since I am stictly a physics major. If I go back and study math to get the 3-4 credits I need, it would put me at least a year back (not factoring in the time I would gain by my thesis head-start). If begin a thesis in physics, I would have to dive completely back into the subject and, worse, would be forced to forget completing my math project(s) for, to say the least, a long time. The head prof. of mathematics at my university said last week I'll fight for you if I, indeed, wan to change my major over to math. I thought this was all fine and dandy and walked around quite pleased throught most of last week. However, I talked with my mother-in-law yesterday... ouch!: You are a father and aren't getting any younger (I'm 28). You should be more worried about where you are going to work ...etc.... than about changing your major...etc. Therefore, this morning I went and talked with my professor about scholarships, knowing a small scholarship would probably work wonders (both financially and with finding work, afterwards). Problem: there really aren't that many. It's not like in the States- since German tuition is basically free to those who qualify- however, supporting a family is *not* free (to offset the lack of scholarships, there do seem to be a lot of prize competitions, however. Indeed, my professor said I could technically try this- but that a one time cash reward of 500 Euros or so probably would not help much in the long run.) Do you have any suggestions? C. Dement === Subject: Re: Pure abstract demonstration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QIt9q22492; > I think, You'll be surprised with some elementary >developments: Some exact relation can be closed as f(t). > However some advanced developments in that same relation >would show some F(t), where once f(t) is of n-degree so >F(t) is of (n-1)n degree. Somebody should notice, that >the best idea for F(t) to be coherent or homogeneous to f(t) is >just F(t) = [f(t)]^(n-1). To some small and prime values of >n once beginning with n=3 such properity can be checked >using results of division F(t)/f(t). There could be also shown, > that for f(t) once the oldest coefficient is 1 so Abs.th. = -(A+B) > but for F(t) once the oldest coefficient is 1 so >Abs.th. = -(A+B)^(n-1) instead of [-(A+B)]^(n-1) = (A+B)^(n-1) > What generally and sufficiently shows impossibility of F(t) >to be coherent to f(t). > See also, that from other side value A+B shows possibility >to be divided by t as some potential root in f(t) and in F(t). > Some other combination of parameters and changing values for >this very relation will show some possible factorisation of >any of Absolute therms.( I used to address and achieve in main >part such developments at 22-th May of 2001 ) > Parameters A and B represents here >some of parametric values taken from Abel formulae: > A = a^n once B = b^n or B = n^(nu-1) b^n or inverse . > Once in general used input is X=T+B; Y=T+A; Z=T+A+B >where T=n^u abtp always but for n=3 p=1 > and the question was to solve X^n +Y^n = Z^n > for n prime numbers bigger than 2 and for X;Y;Z integers > Main part of FLT ? Hint: compare [X^n + Y^n]/(X+Y) to some s^n or n*s^n > and express it in a;b;t;p parameters exclusively... > f(t)= t^n - 2 n^u abtp - a^n - b^n once including > so called first fall of FLT: X;Y;Z not divided by n Sir, The biggest possibility is in some kind of Your mistake. However lets check Your revelations for n=3: You claim Your parameters: T=3^u *abt once A=a^3 and B=b^3 or 3^(3u-1)*b^3 Already it is known for n=3, that 3 should divide Z;X or Y so lets take 3/X ; X=T+B ; B=3^(3u-1) b^3 also does Your parameters fits ? Ls=(T+A+B)^3=(T+A)^3 +(T+B)^3=Rs ? Ls= T^3 +3T^2(A+B) +3T(A+B)^2 +(A+B)^3 Rs= 2T^3 +3T^2(A+B) +3T(A^2 +B^2) +A^3 +B^3 then Rs-Ls= T^3 -6TAB -3AB(A+B) = 0 ; T^3 -3AB(2T+A+B) = 0 3^3u a^3 b^3 t^3 -3*a^3 3^(3u-1)*b^3 [2*3^u abt +a^3 +3^(3u-1) b^3]=0 and really axtracting 3^3u a^3 b^3 we'll have: t^3 - (2*3^u abt +a^3 +3^(3u-1) b^3] = 0 What is just Your f(t) for n=3 and for the fall X/3 Now You suggest Y^2 -YX +X^2 = s^3, what is also correct, but how to express everything in a;b;t;p parameters? I can see, that X+Y in this case should be some cube too: X+Y = T+B+T+A = 2T+A+B = 2*3^u abt +a^3 +3^(3u-1) b^3 also Your t^3 = X+Y and once Y^2 -YX +X^2 = s^3 so Z = t*s now t*s = T+A+B; t^3 = 2T +A+B: T = t^3 -ts = t(t^2 -s); but also once T = 3^u abt = t(t^2 -s): t^2 -s = 3^u ab and finally s = t^2 -3^u ab *****(how smart are those numbers) What You sugess then before: Ls= s^3 = X^2 -XY +Y^2 =Rs Ls=(t^2 -3^u ab)^3 = = t^6 -3*t^4 3^u ab +3*t^2 3^2u a^2 b^2 - 3^3u a^3 b^3 Rs={3^u abt +3^(3u-1) b^3}^2 -[3^u abt +3^(3u-1) b^3][3^u abt +a^3] + +{3^u abt + a^3}^2 ...... I can feel, that there is really some more accounts as one web page could hold so I'll do it on some paper and come back very soon Juan C.C. As he looked, the brilliance spread and spread... from the Divine Comedy of Dante Alighieri And so on my current entertainment is going to be finished: > There was some significant chance of likely correct > proof of P. Fermat origine > Roman B. Binder > Yours Ro === Subject: Re: Building myself a robotic girlfriend by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QJtLG27333; A psychiatrist would be a nice choice. === Subject: Re: problem with a calculation/Gleichungsaufgabe by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QLTgm02821; Wie das andere gesagte Plakat multiplizieren Kreuz und vereinfachen das Problem, welches die genaue reale Antwort (20/17)(8-sqrt(13)) ist Wo sqrt Quadratwurzel ist Traurig f.9fr den Deutschen. Like the other poster said C multiply and simplify the problem The exact real answer is (20/17)(8-sqrt(13)) Where sqrt is square root Sorry for the German. === Subject: Basic algebra problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QM1bl05871; I can't solve the following problem (though it seems to be easy): Prove that in a commutative infinite ring with a multiplicative identity there can't exist a finite number of non-invertible elements !=0. === Subject: Re: Basic algebra problem >I can't solve the following problem (though it seems to be easy): >Prove that in a commutative infinite ring with a multiplicative identity there can't exist a finite number of non-invertible elements !=0. Well, you have had one solution - mine is similar. Assume finitely many nonivertibles and hence infinitely many invertibles. Let a be noninvertible. Then so is ax, for any invertible x, so there must exist infinitely many distinct invertible x_1,x_2,x_3,... with ax_1 = ax_2 = ax_3 = .... Now a(x_i x_1^-1 - 1) = 0 for all i, and we have infinitely many distinct nonivertibles x_i x_1^-1 - 1. Derek Holt. === Subject: compactness vs sequential compactness by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QMj7X09346; The Eberlein-Smulian theorem says that compactness and sequential compactness are equivalent in the weak topology of some normed space. The closed unit ball S in l^1 is not compact in the weak topology. The Banach-Alaoglu theorem tells us that the closed unit ball S in l^1 is compact in the weak* topology. My question is: Is S sequentially compact in the weak* topology? If yes, that would be a great counterexample of the equivalence between compactness and sequential compactness. At first sight, it doesn't seem impossible since a weakly* convergent subsequence could imply a weakly convergent subsequence resulting in a contradiction, for S is not weakly compact. Thanks === Subject: Re: compactness vs sequential compactness >The Eberlein-Smulian theorem says that compactness and sequential compactness are equivalent in the weak topology of some normed space. >The closed unit ball S in l^1 is not compact in the weak topology. >The Banach-Alaoglu theorem tells us that the closed unit ball S in l^1 is compact in the weak* topology. >My question is: >Is S sequentially compact in the weak* topology? Presumably the weak* topology obtained from regarding l^1 as the dual of c_0. >If yes, that would be a great counterexample of the equivalence between compactness and sequential compactness. Surely you meant If not, that would be...? Anyway, it's easy to see that the unit ball of l^1 is weak* sequentially compact: If you have a sequence, a diagonal argument gives a subsequence that converges pointwise, and it follows that the subsequence is weak* convergent. Similarly, if X is separable then it's easy to see that the unit ball of X* is weak* sequentially compact. But I really doubt that this is true in general, in fact I wouldn't be surprised if it were equivalent to X being separable. In particular, I bet that if S is a large enough set then the unit ball of l^1(S), ie L^1(counting measure on S), is not weak* sequentially compact. All you need is a sequence in the unit ball of l^1(S) which does not have a pointwise convergent subsequence - I bet you can give an example of that, say, by some sort of diagonal construction with S = power set of N. >At first sight, it doesn't seem impossible since a weakly* convergent subsequence could imply a weakly convergent >subsequence resulting in a contradiction, for S is not weakly compact. >Thanks ************************ === Subject: Re: compactness vs sequential compactness >>The Eberlein-Smulian theorem says that compactness and sequential compactness are equivalent in the weak topology of some normed space. >>The closed unit ball S in l^1 is not compact in the weak topology. >>The Banach-Alaoglu theorem tells us that the closed unit ball S in l^1 is compact in the weak* topology. >>My question is: >>Is S sequentially compact in the weak* topology? >Presumably the weak* topology obtained from regarding l^1 as the dual >of c_0. >>If yes, that would be a great counterexample of the equivalence between compactness and sequential compactness. >Surely you meant If not, that would be...? >Anyway, it's easy to see that the unit ball of l^1 is weak* >sequentially compact: If you have a sequence, a >diagonal argument gives a subsequence that converges >pointwise, and it follows that the subsequence is >weak* convergent. >Similarly, if X is separable then it's easy to see that the unit >ball of X* is weak* sequentially compact. But I really doubt >that this is true in general, in fact I wouldn't be surprised if >it were equivalent to X being separable. The final conjecture there is wrong, the unit ball of X* can be weak* sequentially compact for non-separable X. (See below.) >In particular, I bet that if S is a large enough set then the >unit ball of l^1(S), ie L^1(counting measure on S), is not >weak* sequentially compact. All you need is a sequence >in the unit ball of l^1(S) which does not have a pointwise >convergent subsequence - I bet you can give an example >of that, say, by some sort of diagonal construction with >S = power set of N. No, that's wrong, the unit ball of l^1(S) is always weak* sequentially compact. (Given a sequence f_n, the union of the supports of the f_n is countable; now use the proof for l^1(N) above.) But here's a counterexample to weak* Eberlein-Smulian: Let X = L^infinity([0,1]). For every x in [0,1], choose L_x in the unit ball of X* such that L_x(f) = f(x) for all f continuous at x. Choose a convergent sequence of distinct points x_n. Then it's easy to see that L_x_n has no weak* convergent subsequence. >>At first sight, it doesn't seem impossible since a weakly* convergent subsequence could imply a weakly convergent >>subsequence resulting in a contradiction, for S is not weakly compact. >>Thanks >************************ > ************************ === Subject: Re: compactness vs sequential compactness >[...] >But here's a counterexample to weak* Eberlein-Smulian: >Let X = L^infinity([0,1]). For every x in [0,1], choose L_x >in the unit ball of X* such that > L_x(f) = f(x) >for all f continuous at x. Choose a convergent sequence >of distinct points x_n. Then it's easy to see that L_x_n >has no weak* convergent subsequence. Realized a few minutes later that one can give a much simpler counterexample, based on the same idea. Since you don't seem interes I'll just explain to myself: Let X = l^infinity. Define L_n x = x_n. Then it's easy to see that the sequence L_n has no convergent subsequence. >At first sight, it doesn't seem impossible since a weakly* convergent subsequence could imply a weakly convergent >subsequence resulting in a contradiction, for S is not weakly compact. >Thanks >>************************ >************************ > ************************ === by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH6M21322; > History question: Did linear algebra originate together with quantum mechanics in 1st half >> of 20th century? No. >>Matrix theory was big in the mid 19th (Cayley and Hamilton and all >>that. Linear associative algebras (note the s) were studied under >>the name hypercomplex systems by Wedderburn and others in the late >>19th. Abstract fields, in particular finite fields, were probably >>origina by Galois around 1829-31 (he died in the latter year). >>Finite fields are often called Galois fields and GF(q) is the field of >>q elements. But despite all that pre-history, I really think that >>abstract vector spaces, the proper study of linear algebra is early >>20th, although I could be wrong. === Subject: Re: Please read my preprint for a proof of Goldbach Conjecture by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH6s21292; >> The point of this repetitive postings is that I can not see >> why I am wrong. I would not like to stop defending my thesis >> until I am totally convinced that I was wrong. >It is not me who has to convince you that you are wrong. It is you who >has to convince me that you are right. That's how proofs work: The >person writing the proof has to convince the people reading it that it >is correct. I am not convinced, therefore it is not a proof. >By the way, this was just the first problem in the proof, and that is >when I stopped reading. Fix it, and I may read on to the next problem. I am sorry that my last almost-auatic response might be rude to you. Recently, my life is going on in fake. By fake, I mean the kind of fake pride that has been caused by posting a possible proof. It is indeed a fake since there does exist a mathematician who seriously reads the paper and makes an objection. So, in order to make my life authentic, could you give me other ßaws assuming that Definition 2.2 makes sense? In your book, assuming something completely nonsense might be just upsetting, but if you would be inspired, please give me a reply. Your reply would be the most exciting event in my present life. I will try to find a good explanation of Definition 2.2; I could not yet regard it as a ßaw. P.S. By the way, the one on the equichordal point problem was rejec by the Bulletin of London Mathematical Society; I just received a letter of denial from the Bulletin of London Mathematical Society. Again, I was so sure about the proof (sure as 1+1=2)that I am in a great predicament of what my standard is worth of...... Hisanobu Shinya === Subject: Re: Please read my preprint for a proof of Goldbach Conjecture P.S. By the way, the one on the equichordal point problem was rejec > by the Bulletin of London Mathematical Society; I just > received a letter of denial from the Bulletin of London Mathematical > Society. Again, I was so sure about the proof (sure as 1+1=2)that I am in > a great predicament of what my standard is worth of...... Rejection by a journal doesn't mean that a paper is wrong. It is more likely to mean that the referees/editors think it is (relatively) unimportant and/or uninteresting. -- === Subject: Re: Basic algebra problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH7w21364; >I can't solve the following problem (though it seems to be easy): >Prove that in a commutative infinite ring with a multiplicative identity there can't exist a finite number of non-invertible elements !=0. ************************************************************* **** Hi: Well, it's easy but slightly tricky: Let's suppose R is our ring and first assume there's some non zero non-invertible element a which also is a zero divisor ==> for any unit u in R we get that au is a non-zero (because u cannot be a zero divisor) non-invertible element (since otherwise ALSO a would be invertible), and thus we get at once a contradiction. So we can assume R is an infinite commutative integer domain with a finite number of non-zero non-invertible elements, say a1,...,an. We then know that R* = {invertible elements of R} is infinite, but then again for any u in R* we get that a1*u is non-invertible and non-zero, so a1*u = aj for some 1<= j <= n, and since R* is infinite there are two DIFFERENT units u,w in R* such that: a1u = a1w ==> a1(u-w) = 0 ; but we assumed R is an integral domain, and a1 is no zero, so...can you complete the argument? Good luck! Tonio === Subject: more readable Re: Any news on odd perfect numbers? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH9221444; I published my own naive geometrical argument that there are no odd perfect numbers on Jan 14th here http://mathquest.com/discuss/sci.math/m/569936/570941 and my non-standard notation irrita a reader or two - apologies. My outline argument is that since Euler showed an odd perfect number must be a multiple of a square, then if we consider how summing the factors must fill in an array of squares equalling the odd prime, a contradiction emerges when we look at the square holding some or all of the subfactors of the factor which is a square [which I helpfully name Ôsq']. If anyone friendly has the patience, I'd be grateful for an opinion. It's not long, and of course will gladly answer any questions. Mark G. === Subject: C product of two orientable smooth manifolds by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH8T21406; I have a question about the c product of two orientable smooth manifolds. Proof that M and N are two orientable smooth manifolds if and only if MxN is an orientable smooth manifold. Thanks a lot for your answeJorge. === Subject: Re: C product of two orientable smooth manifolds > I have a question about the c product of two orientable smooth manifolds. Proof that M and N are two orientable smooth manifolds if and only if MxN is an orientable smooth manifold. Thanks a lot for your answeJorge. If you're dealing with the case of _closed_, connec manifolds M and N, you can give a relatively simple proof of the fact that M, N orientable implies MxN orientable -- the result follows from the fact that H_n(M^n;Z) = Z if M is orientable and it's trivial otherwise. Combine that with the fact that the Kuenneth Formula implies that H_m(M^m) tensor H_n(N^n) = H_{m+n}(MxN) if M, N are orientable (this also needs the fact that H_{m-1}(M^m) is free abelian if M is orientable). I don't see any easy way to get the proof of the other implication using these ideas -- there's a result (presen as a Lemma in Hirsch, Differential Topology) which would actually let you take care of both directions: If 0 --> xi' --> xi --> xi'' --> 0 is a short exact sequence in the category of vector bundles (over manifolds, say), then 4.1 Lemma. Two of xi', xi, xi'' are orientable if and only if the third is. -- since the pullback of an orientable bundle is itself orientable, you can exploit that lemma in various ways to get the result. (For instance, the tangent bundle to MxN is the Whitney sum of the pullbacks of the tangent budles of M, N under the projections -- the lemma then says that MxN is orientable if M, N are. On the other hand, if MxN is orientable, then the pullbacks of the tangent bundles of M, N are orientable according to the lemma -- but, then, pulling _those_ back by injecting [for instance] M into MxN by m |--> (m,n0) for some n0 in N shows that M's tangent bundle is orientable, proving that M is an orientable manifold ... ) === mm-332 === Subject: Re: JSH: Mathematical clarity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RJQ4520935; > [.snip.] >> Let's reduce this to a simpler situation: >> Theorem. If A, B, and C are algebraic integers and p is a prime >> integer, > AB = Cp, > Oh yes, the key hypothesis. I have written this too many times and I guessed I just assumed it. ^ >> and C is coprime to p, then there exist >> algebraic integers r and s such that: >> 1. r*s = p. >> 2. A/r and B/s are algebraic integers, >> 3. (A/r)*(B/s) = C. >> Proof: Define r = GCD(A, 7) and s = GCD(B, 7). QED. >> {A stronger theorem can be proved ... this one is sufficient for >>the present purposes.} > [.snip.] === Subject: Re: JSH: Mathematical clarity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RJPn720872; [snip] >I'm the misunderstood and persecu genius. >You're the . *Incapable*. They are *incapable*. Maurice === Subject: Re: S-curve equation? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RKi8T30786; Can anyone please write down an equation for y=f(x) where x goes from >any positive value to another? >Thanks alot, >Mike Google Gompertz equation for a nice s-shape that is often used to model the adoption of a new technology vs. time === Subject: Re: S-curve equation? >Can anyone please write down an equation for y=f(x) where x goes from >>any positive value to another? >>Thanks alot, >>Mike > Google Gompertz equation for a nice s-shape that is > often used to model the adoption of a new technology vs. > time > Also the simple Gaussian curve, integra along the normalized axis (parameter z). In essence it is the product of two competing exponential processes and approach 0 and 1 asymptotically (to your level of confidence if viewed statistically, or level of precision if looked at diffusively). Each unit of progression along the parameter corresponds identically to the Ôstandard deviation' or the Gausian diffusion constant. It does not account for age structure very well. The gompertz one seems to be the more useul one in that respect for evolution in time of aged populations. MK === Subject: Re: Past posteconsider facts by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RKi8830790; >> What if you were faced with arguments over the existence of sqrt(2), > or sqrt(-1)? Imagine if as a group mathematicians had risen to protect the status > quo, and rejec sqrt(-1)? > sqrt(-1), sqrt(2), and a host of other things we now use today at one time >> were not accep by the mathematicians of the day. BUT, in math, the truth >> always comes out eventually, so you have nothing to worry about. YOU might >> not be able to convince anybody of anything (as indeed appears to be the >> case), BUT if you are right, eventually someoe else will be able to. (oh >> BTW, history shows that railing against the establishment is not a very >> effective method, so your probably accomplishing even less with your rants >> and fits). Comforting thought, eh? >I'm expressing my frustration with liars and cheats. >It seems to me that it'd be worse to just shut-up and hope, when I can >express my opinions about these contemptible people. >Besides, it's a way to blow off steam. >After all, eventually, my work will *be* the establishment. > Hey I have a new idea on this. Why don't you start a new thread in which you discuss what is wrong with the RickDecker example? A. === Subject: Re: The most linear sigmoid by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RLtFt06383; Nothing is wrong with numerical approach, but I would like to have the analytical solution if it exist. I want to linearize the sigmoid function on the interval [-1,1] which it could be reduce to [0,1] (symmetry). My linearization criterion is the integral of quadratic error between the sigmoid function and the linear function (x+1)/2. I don't mind extremities mappings. My recents calculations reduced the problem to integrate ln(1+exp(-ax))dx but this problem seems to be impossible to solve analytically ! Thierry. >What is wrong with using a numerical approach? You want a exact answer? >I'm not sure I understand your problem. You have a linear function from [0, >1] to [1/2, 1] and you want to approximate this with a sigmoid function >with some unspecified slope. Is that right? >Do you require that the approximating function maps 0 to 1/2 and 1 to 1. In >that case your sigmoid does not work. >-Michael. >> I'm looking for solving analytically this equation : >> f1(x) = (x+1) / 2 linear >> f2(x) = (1/(1+exp(-ax)) sigmoid >> d ( int( (f1(x)-f2(x))^2, x=0..1) ) / da = 0 >> I find a numerical solution : a = 2.46 >> The problem is to adjusting the sigmoid slope (a) for having the most >closed sigmoid to the linear function. >> Thanks in advance for your help, >> Thierry Hoinville. === Subject: Re: NOT JSH:The Axiom of Determinacy by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0S16CQ24553; >What is a fully explicit, fully formal statement of the Axiom of >Determinacy? I am not sure what a game is. Also, I am not completely >sure about what a forced win would be in the setting of this >generalized notion of game. >---- David Actually by labelling this as NOT JSH, your post will not be read by the people who put anything with JSH in the subject line into their killfiles. === Subject: Re: Galois Group textbook by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SDMkR23356; > I've read Hungerford and Herstein, but they both seem a bit choppy in that > there does not seem to be a development of Galois theory, just a collection > of various theorems. >> Personally I think Ian Stewart's Galois Theory (2nd) is very good >> especially for a beginner to the subject. It develops Galois theory so >> that you get an understanding for its place in mathematics and proves >> the main theorems. As a bonus it includes transcendence proofs for e and >> pi, an explicit constuction of the regular 17-gon and an algorithm for >> calculating Galois groups. It doesn't require too much prerequisites and starts with an overview of >> basic abstract algebra. Maybe a little more on symmetric polynomials and >> group theory would be desirable since it's used in later proofs. (I >> comple group theory by means of van der Waerden's Algebra on these >> points.) Notice: There is a serious error in 1st edition, but it's correc in >> the 2nd one, where also the algorithm for calculating Galois groups is >> presen. So be sure to get the 2nd edition! >You might also look at a Carus monograph by Hadlock on field theory, >which seems very readable and motiva and has lots of solved >problems in it. Tata Institute of Fundamental Research has a good Mathematical Pamphlet on Galois Theory. Hi Mark. Remember me from Rutgers? - Rekha === Subject: Generalized mean. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SDMix23297; Dear Phorum Participants! What it can be said about the following limit: g(E[f(x^(1/n))]^n, when n -> +oo, where x - positive random variable; E[.] - mathematical expectation operator (as supposed it exists). f - continuous strictly monotone function, g - its inverse. Thanks in any advance, Mikhail. === Subject: Re: Generalized mean. >What it can be said about the following limit: >g(E[f(x^(1/n))]^n, when n -> +oo, >where >x - positive random variable; >E[.] - mathematical expectation operator (as supposed it exists). >f - continuous strictly monotone function, g - its inverse. Your parentheses are unbalanced, but I suppose you mean g(E[f(x^(1/n))])^n (it would make less sense to do g(E[f(x^(1/n))]^n) Well, what would you want to say about it? I guess you want the limit to be exp(E[ln(x)]). This will be true if everything is sufficiently nice, but I'm not sure exactly what conditions are necessary. Note that if f is differentiable at 1, n (f(x^(1/n)) - f(1)) -> f'(1) ln(x) as n -> infinity (pointwise in x). I'll assume this limit survives the taking of expec value (which would be true e.g. if 0 < a < x < b a.s. for some constants a and b). So E[f(x^(1/n))] = f(1) + f'(1) E[ln(x)]/n + o(1/n) as n -> infinity. Then if f'(1) <> 0, g is also differentiable at 1 and g(E[f(x^(1/n))]) = g(f(1)) + g'(f(1)) (E[f(x^(1/n))] - f(1)) + o(1/n) = 1 + E[ln(x)]/n + o(1/n) so g(E[f(x^(1/n))])^n -> exp(E[ln(x)]). Note that the result can be false if f'(1) = 0 or if f is non-differentiable at 1. For example, if f(t) = (t-1)^p we should have g(E[f(x^(1/n))])^n -> exp(E[(ln x)^p]^(1/p)) (again, assuming say 0 < a < x < b a.e.) 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: what is a vertice? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SDMjd23352; >
worked on this problem off and on for several years while in
high
school.
>> Picked it up later and worked on it with still no success.
Supposedly
it
>> is a famous problem.
If the diagonals to the opposite sides of a triangle are
equal in
length,
>> then the triangle is isosceles.
Proof???
--
>>
-------------------------------------------------------------
--------
>> Angels do come - just leave a soft place in your heart for
them to land
:)
>>
-------------------------------------------------------------
--------Do
the diagonals connect a vertice with the midpoint of the
opposite
>side?
===
Subject: Re: what is a vertice?

worked on this problem off and on for several years while in
high
school.
> Picked it up later and worked on it with still no success.
Supposedly
it
> is a famous problem.
If the diagonals to the opposite sides of a triangle are
equal in
length,
> then the triangle is isosceles.
Proof???
--
>
-------------------------------------------------------------
--------
> Angels do come - just leave a soft place in your heart for
them to
land :)
>
-------------------------------------------------------------
--------Do
the diagonals connect a vertice with the midpoint of the
opposite
side?


The 3 points that get connec to form a triangle are called its
vertices. Each one of these vertices is a vertex. However, if
you
under the impression that all plurals in English are formed
by adding
s to the singular, then one of the vertices would be called a
vertice,
pronounced vertisee. I have heard software engineers who are
native
speakers of English use the term vertice in this way. This is
similar
to the term serie that shows up periodically in this forum
from people
(never native English speakers as far as I can tell) who
think that
series must be a plural form because it ends with the letter
s, and
that the singular form of series is serie.
Achava
===
Subject: Re: what is a vertice?
[snip]
>> -Do the diagonals connect a vertice with the midpoint of
the
>> opposite side?
> The 3 points that get connec to form a triangle are called
its
> vertices. Each one of these vertices is a vertex. However,
if you
> under the impression that all plurals in English are formed
by adding
> s to the singular, then one of the vertices would be called
a vertice,
> pronounced vertisee. I have heard software engineers who
are native
> speakers of English use the term vertice in this way. This
is similar
> to the term serie that shows up periodically in this forum
from people
> (never native English speakers as far as I can tell) who
think that
> series must be a plural form because it ends with the
letter s, and
> that the singular form of series is serie.
You deserve a kudo ;-) for your explanation of vertice. But
I'm not
sure
that your explanation of serie is quite correct. It may well
be that
some
native speakers of French and German, for whom serie _is_
correct in
their native language (well, except that there needs to be an
accent in the
French word and capitalization for the German), assume
incorrectly that the
word is the same in English. Thus, they may not be thinking
incorrectly
that series is plural and the final s can be dropped to make 
it
singular. Rather, they might merely be thinking, incorrectly,
that the word
in English is the same as in their native language, being
completely
unaware that the English word is actually series.
David
Subject: Re: what is a vertice?
===
[snip]
[...]
This is similar
to the term serie that shows up periodically in this forum
from people
(never native English speakers as far as I can tell) who
think that
series must be a plural form because it ends with the letter
s, and
that the singular form of series is serie.
You deserve a kudo ;-) for your explanation of vertice. But
I'm not
sure
> that your explanation of serie is quite correct. It may
well be that
some
> native speakers of French and German, for whom serie _is_
correct in
> their native language (well, except that there needs to be
an accent in
the
> French word and capitalization for the German), assume
incorrectly that
the
> word is the same in English. Thus, they may not be thinking
incorrectly
> that series is plural and the final s can be dropped to make
it
> singular. Rather, they might merely be thinking,
incorrectly, that the
word
> in English is the same as in their native language, being
completely
> unaware that the English word is actually series.
David
The meaning of Serie in German is slightly different from
that of
series
in English. The nearest equivalent in mathematical
terminology is
Reihe,
e.g. series expansion = Reihenentwicklung in German.
It is understood that Reihe implies summation, whereas Folge
(sequence) is something that can be mapped on the natural
numbers.
There are many different uses of Serie in German having a
meaning
close to the meaning of series in English, e.g.
Fernsehserie (TV series), Unfallserie (series of accidents),
Serienproduktion (mass production), in Serienschaltung (serial
connexion),
Gesetz der Serie (~law of averages).
Hugo
===
Subject: Re: what is a vertice?
> Gesetz der Serie (~law of averages).
??? I know that Gesetz der Serie as a bombastic
description for if something happens many times
then it will happen again.
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: what is a vertice?
Adjunct Assistant Professor at the University of Montana.
[.snip.]
>You deserve a kudo ;-) for your explanation of vertice. But
I'm not
sure
>that your explanation of serie is quite correct. It may well
be that
some
>native speakers of French and German, for whom serie _is_
correct in
>their native language (well, except that there needs to be
an accent in
the
>French word and capitalization for the German)
Spanish works. A single series is una serie, while several of
them
would be series. No accents in either one.
--
===
Subject: Re: what is a vertice?
>But I'm not sure
> that your explanation of serie is quite correct. It may
well be that
some
> native speakers of French and German, for whom serie _is_
correct in
> their native language (well, except that there needs to be
an accent in
the
> French word and capitalization for the German), assume
incorrectly that
the
> word is the same in English. Thus, they may not be thinking
incorrectly
> that series is plural and the final s can be dropped to make
it
> singular. Rather, they might merely be thinking,
incorrectly, that the
word
> in English is the same as in their native language, being
completely
> unaware that the English word is actually series.
David
Very interesting. I was unaware of the French and German
words for
series, in spite of many years (very long ago) of studying
French
(somehow the poets and novelists don't seem to use it very
often) and
a very small amount of time studying German (also many years
ago). I
like your explanation better than I like mine, although mine
is
plausible for some people.
Achava
===
Subject: matrix algebra
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SDMiA23306;
To your kind attention:
I have been trying to solve the following problem (without
much success):
I have n matrices Gi: G1,G2,...,Gn, each of which is not
necessarily
symmetric, square (say the matrices dimension is k) and with
full rank, and
whise eigenvalues can be positive and/or negative. I would
like to find it
out if there is a vector a, whose n elements are
a1,a2,...,an, such that the
new square k-dimensional matrix
P=a1*G1+a2*G2+...+an*Gn
has reduced rank, equal to say rI have been trying to solve the following problem (without
much success):
>I have n matrices Gi: G1,G2,...,Gn, each of which is not
necessarily
>symmetric, square (say the matrices dimension is k) and with
full rank,
>and whise eigenvalues can be positive and/or negative. I
would like to
>find it out if there is a vector a, whose n elements are
a1,a2,...,an,
>such that the new square k-dimensional matrix
>P=a1*G1+a2*G2+...+an*Gn
>has reduced rank, equal to say rclosed form solution for this problem, i.e. something that
expresses the
>a1,a2,...,an as combinations of the elements of G1,...,Gn.
A start might be to consider the characteristic polynomial
p(a1,...,an,t) = det(P - tI) of P. If the lowest k-r
coefficients of
this are 0 and the next is nonzero, i.e. 0 is an eigenvalue
of algebraic
multiplicity k-r, then r <= rank(P) <= k-1, and there's a
good chance
that the rank is actually r (although since algebraic and
geometric
rank are not the same, it might not be). So if you can solve
a system
of k-r polynomials in n variables, you might have a solution.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: matrix algebra
> I have n matrices Gi: G1,G2,...,Gn, each of which is not
necessarily
> symmetric, square (say the matrices dimension is k) and
with full
> rank, and whise eigenvalues can be positive and/or
negative. I would
> like to find it out if there is a vector a, whose n elements
are
> a1,a2,...,an, such that the new square k-dimensional matrix
> P=a1*G1+a2*G2+...+an*Gn
> has reduced rank, equal to say r a closed form solution for this problem, i.e. something that
> expresses the a1,a2,...,an as combinations of the elements
of
> G1,...,Gn.
(You say the matrices are not necessarily symmetric and 
don't
give
an alternative reason to think the eigenvalues will be real.
So either
you forgot that eigenvalues can be complex, or you're
restricting to
some other case where matrix eigenvalues are provably real.)
If I understand correctly, your question is whether there is
a linear
combination of n given k x k matrices which is singular (or,
more
generally, whether there is a linear combination of them with
rank r).
The short answer is: this can be hard. In the special case of
n=k,
if you had k matrices with the property that all linear
combinations
of them (except the trivial zero combination) were
non-singular, then
you could use these matrices to make R^k into a division
algebra
(where (e_i)*(e_j) = sum (G_i)_{l,j} e_l ). It is true but not
at all easy to prove that this is possible for k = 1, 2, 4,
and 8 only.
I believe it's true that for n>k there is always a 
nontrivial
linear
combination of the matrices which is singular. When n[...] a larger context: How to find the
>maximal dimension of a linear subspace W of mxn matrices
such that
>every element of W - (0) has rank k (or rank leq k; or rank
geq k).
>Various topological tools can be applied over the reals to
get results
>which generalize the 1, 2, 4, 8 theorem. This is discussed
in the
>classic paper Adams-Lax-lips, On matrices whose real linear
>combinations are nonsingular, Proc. Amer. Math. Soc. 16
(1965)
>318-322; 17 (1966) 945-947.
>Also see K.Y.Lam and P.Yiu, Linear spaces of real matrices of
>constant rank, Linear. Alg. Appl. (to appear).
(That last paper has appeared, in Linear Algebra Appl. 195
(1993), 69--79.
The authors find the maximal dimensions of linear subspaces of
M_k(R)
having rank r=k-1 or k-2.)
I believe all these results consider the problem of _finding_
linearly
independent matrices whose span is nonsingular. In your case,
the
matrices are known and so you just need to decide whether
det(P)
vanishes for some (real) choice of the a_i; but the general
results
I quo will save you the trouble if n is too large.
dave
===
Subject: Re: Squaring A Circle...(almost)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SDMlA23386;
>We all (should) know it is impossible to square a circle
with only a
>straight-edge and compass and pencil.
>But, of course, we can approximately square a circle
However
There is an orthogonal triangle following the natural
equation:
[T^4]-[T^2]-1=0
T=SQRT(GOLDEN RATIO)=SQRT{[SQRT(5)+1]/2}
This triangle has a base (small horizontal)equal to T^1.
Height (bigger vertical)equal to T^2 .
Hypotenuse equal to T^3.
The sides are in geometric ratio=T.
http://www.stefanides.gr/platostriangle.htm
We note that:
If we multiply all sides by pi ,and divide by T we get:
base =pi , vertical=pi*T,and hypotenuse=pi*T^2.
A circle is circumscribed over this triangle
with diameter its hypotenuse=D=pi*T^2
The Circumference of this circle is equal to :
[(pi)^2]*T^2
This is equal to the area of the square on the
vartical side of the orthogonal
triangle=[pi*T]^2=[(pi)^2]*T^2,
which is the product of the base by the
(diameter)hypotenuse=pi*D=
=[pi]*[pi*T^2] , which is a paralelogram,
and the bigger angle= ArcTan(T).
We see ,now that we have in NATURE a THEORETICAL AREA OF A
SQUARE
EQUAL TO A CIRCLE CIRCUMFERENCE.
[Note : the triangle with sides 1,R(V),V
V=GOLDEN NUMBER , R(V)=SQRT(GOLDEN NUMBER)
IS THE MAGIRUS/KEPLER TORTHOGONAL TRIANGLE
Ref:Roger Herz-Fischler
A Mathematical History of Division in Extreme
and Mean Ratio-Wilfrid Laurier University Press-1987.
Panagiotis Stefanides
http://www.stefanides.gr
===
Subject: 0 is no ordinary number
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SEsXQ31002;
what all of you are doing is treating 0 like a variable, or
any normal
number, such as 1,2,3,4,5..... but what you must understand
is that 0 is no
ordinary number. In fact, 0 did not come into existance until
1000 AD. 0 is
a concept, not a number, that is what you must understand.
0 is the opposite of infinity. Infinity is not a 
number,
neither is 0.
there is no -0, just as there is no -infinity. You can not
treat these too
concepts just as you treat an ordinary number, simply because
they are in
essence not numbers. This is a difficult concept to fathom.
0 = Nothing I am fairly confident that most people will agree
with that
statement. Nothing, by definition is:
Something that has no existence.
Something that has no quantitative value; zero
One that has no substance or importance; a nonentity
how can a nonentity possibly fit into another nonentity?
Neither one
exists. how can Nothing / Nothing possibly equal something?
nothing can
never fit into nothing.
Forget about algebra, saying 0 * X = 0 so 0/0 = X is simply
ignorance.
0 is not a normal number, and can not be trea as one! 0 and
infinity are
exceptions to many mathematical rules. Both of them should be
trea as
words or concepts rather than numbers. Instead of computing
formulas, and algebra, that you think makes sence, take a
moment and think
about what 0 actually is. Then this will all make sence to
you.
===
Subject: Re: 0 is no ordinary number
>In fact, 0 did not come into existance until 1000 AD.
Presumably in AD 10, AD 20, and even AD 100 they just put up
with it,
but when they got to AD 1000 it was just too annoying not to
be able
to write three quarters of the date.
And I suppose it explains why there wasn't a year 0.
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname
in the
headers.
FreeBSD rules!
===
Subject: Re: 0 is no ordinary number
> In fact, 0 did not come into existance until 1000 AD.
Right. And the world turned from black-and-white into colour
around 1950.
Pretty grainy colour for awhile, too.
This is obvious because old photographs are in
black-and-white. Can't you
see that?
===
Subject: Re: 0 is no ordinary number
> 0 = Nothing
You're correct that if you define 
Ô0' that way, then
Ô0' is no ordinary number. Here's 
a proof:
A small scoop of ice cream is better than nothing.
A big scoop is better than a small scoop.
Nothing is better than a big scoop of ice cream.
Therefore, nothing is better than nothing, a property
which no ordinary number has.
> nothing can never fit into nothing.
Didn't I just disprove that.
> Forget about algebra,
Done.
===
Subject: Re: 0 is no ordinary number
In sci.math, Steven
 or any normal number, such as 1,2,3,4,5..... but what you
> must understand is that 0 is no ordinary number. In fact,
> 0 did not come into existance until 1000 AD. 0 is a concept,
> not a number, that is what you must understand.
[1] Numbers are concepts.
[2] Numbers don't exist, as such.
0 is the opposite of infinity. Infinity is not a 
number,
> neither is 0. there is no -0, just as there is no -infinity.
> You can not treat these too concepts just as you treat an
> ordinary number, simply because they are in essence not
numbers.
> This is a difficult concept to fathom.
Not really. Very old computer systems had -0, and
mathematically
-0 makes some sense if one uses limits. For example
lim(x->0+) (-x)
might be construed to be -0. Of course 0 is its own arithmetic
inverse in normal math so there's not a lot here.
0 = Nothing I am fairly confident that most people will agree
> with that statement. Nothing, by definition is:
> Something that has no existence.
> Something that has no quantitative value; zero
> One that has no substance or importance; a nonentity
how can a nonentity possibly fit into another nonentity?
> Neither one exists. how can Nothing / Nothing possibly equal
> something? nothing can never fit into nothing.
Forget about algebra, saying 0 * X = 0 so 0/0 = X is simply
> ignorance. 0 is not a normal number, and can not be trea
> as one!
Well, your logic's slightly off, but it's 
clear that, since 0
is
the arithmetic identity, one cannot divide by it.
It's easy enough to prove that, if A is the arithmetic
identity for a general field, then for any B, A*B = A.
Therefore C/A (C != A) and A/A are undefined; the former
because no number B in the field can possibly have A*B =
C; the latter because *every* number B in the field has
A*B = A.
> 0 and infinity are exceptions to many mathematical rules.
> Both of them should be trea as words or concepts
> rather than numbers. Instead of computing formulas, and
> algebra, that you think makes sence, take a moment and think
> about what 0 actually is. Then this will all make sence to
you.
>
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: 0 is no ordinary number
> In sci.math, Steven
> 
[snip]
> Very old computer systems had -0,
Yes, as a distinct entity from +0. But you make it sound as
if this
situation is no longer the case! All systems which conform to
the _current_
internationally accep standard for ßoating-point arithmetic
must have
both +0 and -0.
> Of course 0 is its own arithmetic
> inverse in normal math so there's not a lot here.
It would be clearer to say additive inverse, rather than
arithmetic
inverse. After all, multiplication is also part of
arithmetic...
And below, similarly, rather than arithmetic identity,
additive
identity would be clearer.
[snip]
> it's clear that, since 0 is the arithmetic identity, one
cannot divide
> by it.
But it's not clear that you can't divide by 
it. Indeed --
since you
mentioned computer systems earlier -- note that in
ßoating-point
arithmetic you _can_ divide by it. And of course there are
also systems
in pure mathematics where this can be done. But certainly 0
does not
have a multiplicative inverse. And so if one defines division
in terms of
multiplicative inversion, then one cannot divide by zero,
which is the
point you were making.
David
> It's easy enough to prove that, if A is the arithmetic
> identity for a general field, then for any B, A*B = A.
> Therefore C/A (C != A) and A/A are undefined; the former
> because no number B in the field can possibly have A*B =
> C; the latter because *every* number B in the field has
> A*B = A.
===
Subject: Re: 0 is no ordinary number
In sci.math, David W Cantrell
 In sci.math, Steven
>> > Very old computer systems had -0,
Yes, as a distinct entity from +0. But you make it sound as
if this
> situation is no longer the case! All systems which conform
to the
_current_
> internationally accep standard for ßoating-point arithmetic
must have
> both +0 and -0.
Hmm...I think you're right though I've not 
studied the issue
extensively. There are also issues with undernormalized
numbers,
basically bitpatterns with exponent -0x400 (represen in
excess-0x400 as all zeroes).
I was thinking integers at the time.
> Of course 0 is its own arithmetic
>> inverse in normal math so there's not a lot here.
It would be clearer to say additive inverse, rather than
arithmetic
> inverse.
I claim a brain-o.
> After all, multiplication is also part of arithmetic...
> And below, similarly, rather than arithmetic identity,
additive
> identity would be clearer.
[snip]
>> it's clear that, since 0 is the arithmetic identity, one
cannot divide
>> by it.
But it's not clear that you can't divide by 
it. Indeed --
since you
> mentioned computer systems earlier -- note that in
ßoating-point
> arithmetic you _can_ divide by it.
Yes, one can -- and get Inf or NaN. Neither are real although
in
the case of computers they are a compromise, much like
airplanes.
Not everyone will be happy.
> And of course there are also systems
> in pure mathematics where this can be done.
Not within a true field -- although one can get pretty far by
extending the reals.
> But certainly 0 does not
> have a multiplicative inverse. And so if one defines
division in terms of
> multiplicative inversion, then one cannot divide by zero,
which is the
> point you were making.
Precisely.
David
> It's easy enough to prove that, if A is the arithmetic
>> identity for a general field, then for any B, A*B = A.
>> Therefore C/A (C != A) and A/A are undefined; the former
>> because no number B in the field can possibly have A*B =
>> C; the latter because *every* number B in the field has
>> A*B = A.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: 0 is no ordinary number
[bla bla bla]
> Then this will all make sence to you.
I don't want anything to make sence to me. I 
don't know what
sence
means, but it doesn't sound nice.
By the way, are you a friend of that uniteralist guy? or are
you just
dumb all by yourself?
A giant of a vegetable
Robert Louis Stevenson
Treasure Island
===
Subject: Re: 0 is no ordinary number
>Subject: 0 is no ordinary number
>what all of you are doing is treating 0 like a variable, or
any normal
>number, such as 1,2,3,4,5..... but what you must understand
is that 0 is
no
>ordinary number. In fact, 0 did not come into existance
until 1000 AD. 0
is
>a concept, not a number, that is what you must understand.
>0 is the opposite of infinity. Infinity is not a 
number,
neither is 0.
>there is no -0, just as there is no -infinity. You can not
treat these
too
>concepts just as you treat an ordinary number, simply
because they are
in
>essence not numbers. This is a difficult concept to fathom.
>0 = Nothing I am fairly confident that most people will agree
>with that statement.
Then most people would be wrong. 0 = Nothing is false.
>Nothing, by definition is:
>Something that has no existence.
Something with no existence is a Null.
>Something that has no quantitative value; zero
Wrong. A Null is NOT 0. Nulls and 0s are distinct. A 0 is an
identity
element in addition:
A + 0 = A
A Null is NOT an identity element. Any arithmetic operation
involving
a Null returns a Null:
A + Null = Null
When comparing numbeone of the following is true:
A>0 A=0 A<0
If A is a Null, then all the comparisons are false. Nothing
does not
equal 0.
>One that has no substance or importance; a nonentity
>how can a nonentity possibly fit into another nonentity?
Neither
one
>exists. how can Nothing / Nothing possibly equal something?
>nothing can never fit into nothing.
This confusion is caused by your mistaken belief that
Nothing=0.
Once you get over that, all will be clear.
>Forget about algebra, saying 0 * X = 0 so 0/0 = X is simply
>ignorance.
On whose part?
>0 is not a normal number, and can not be trea as one!
Wrong. 0 is a normal number. It's Nulls that are not trea
normally.
>0 and infinity are exceptions to many mathematical rules.
So?
>Both of them should be trea as words or concepts rather than
>numbers.
I won't argue about infinity, but you are wrong 
about 0.
>Instead of computing formulas, and algebra, that you think
>makes sence, take a moment and think about what 0
>actually is.
0 is not the problem. The only thing one has to think about
are
Nulls, since as I showed above, special care must be used when
dealing with them. Since all Null comparisons are false, how
does
one determine whether
A = B
if both are Null? The comparison returns false even though
they
are the same. You have to make a special case comparison
using the IsNull() property:
IF (IsNull(A) AND IsNull(B)) OR (A=B) THEN TRUE ELSE FALSE
>Then this will all make sence to you.
If the above doesn't make sense to you, you will have no
future as
a database administrator.
--
===
Subject: Re: 0 is no ordinary number
Discussion, linux)
Cancel-Lock: sha1:GgeNloKHnvzw65wJ8KQWB9MtH4A=
> what all of you are doing is treating 0 like a variable, or
any normal
> number, such as 1,2,3,4,5..... but what you must understand
is that 0
> is no ordinary number. In fact, 0 did not come into
existance until
> 1000 AD.
Goodness, that *is* unusual! When did the other numbers come
into
existance?
> 0 is a concept, not a number, that is what you must
understand.
> 0 is the opposite of infinity.
Man, you *do* know lots of stuff. I had no idea.
> Infinity is not a number, neither is 0. there is no -0, just
as
> there is no -infinity.
Again, I'm fascina. No -infinity. Had no idea.
> You can not treat these too concepts just as you treat an
ordinary
> number, simply because they are in essence not numbers.
This is a
> difficult concept to fathom.
> 0 = Nothing I am fairly confident that most people will
agree with
> that statement.
Sure, who could disagree with such an obvious fact?
> Nothing, by definition is: Something that has no
> existence.
Zero is something that has no existence! Of course! It's all
clear
now. (Except, I thought that it got its existance in 1000 AD.
Has it
since been downgraded, or is existance not the same as
existence?
Well, it'll all come clear soon enough, I'm 
sure...)
> Something that has no quantitative value; zero One that has
no
> substance or importance; a nonentity
> how can a nonentity possibly fit into another nonentity?
Neither
> one exists. how can Nothing / Nothing possibly equal
something?
> nothing can never fit into nothing.
> Forget about algebra, saying 0 * X = 0 so 0/0 = X is simply
> ignorance.
Golly, now *that* I knew.
> 0 is not a normal number, and can not be trea as one! 0 and
> infinity are exceptions to many mathematical rules. Both of
them
> should be trea as words or concepts rather than numbers.
> Instead of computing formulas, and algebra, that you think
makes
> sence, take a moment and think about what 0 actually is.
Then this
> will all make sence to you.
It already makes much more sence than before, thanks!
--
When I am grown to man's estate
I shall be very proud and great,
and tell the other girls and boys
not to meddle with my toys. --Robert Louis Stevenson
===
Subject: Re: 0 is no ordinary number
>[...] Then this will all make sence to you.
Well _that's_ a scary notion...
************************
===
Subject: Re: 0 is no ordinary number
> what all of you are doing is treating 0 like a variable, or
any normal
> number, such as 1,2,3,4,5..... but what you must understand
is that 0 is
no
> ordinary number. In fact, 0 did not come into existance
until 1000 AD. 0
is
> a concept, not a number, that is what you must understand.
>
The *NUMERAL* zero did not come into existence until 1000 AD.
What is
important about zero is that it allowed positional notation
for numbers.
This allowed us to begin to deal with very large numbeand
also to
devise simple algorithms for adding and multiplying big
numbers. However
the numbers themselves, were unaffec by any of this. That is,
if you
believe that numbers have an existence separate from any
particular
notation for talking about particular numbers. That is a
belief shared
by most modern mathematicians, who believe in the real
numbers yet can
prove that we could never denote any more than a relatively
small
handful of them. But I've heard it said that others feel
differently
(constructionists? I'm not really up on this) and you may be
one of them.
> 0 is the opposite of infinity. Infinity is not a 
number,
neither is 0.
> there is no -0, just as there is no -infinity. You can not
treat these
too
> concepts just as you treat an ordinary number, simply
because they are
in
> essence not numbers. This is a difficult concept to fathom.
>
It's interesting that this concept is concretely realized in
visualizing
the set of extended complex numbers as a sphere, with infinity
at the
north pole and 0 at the south pole. In that respect 0 and
infinity truly
are mirror images of each other, under the map z -> 1/z.
> 0 = Nothing I am fairly confident that most people will
agree with
that
> statement. Nothing, by definition is:
> Something that has no existence.
> Something that has no quantitative value; zero
> One that has no substance or importance; a nonentity
>
Absolutely wrong. Zero is not nothing is an important concept
to get.
*Nothing* is complete absence of anything. *Zero* is
something very
particular, namely . . . zero. It's a point in the complex
plane; it's a
statement that I don't owe you money and you 
don't owe me
money; it's
the quantity of purple giraffes in my back pocket at this
very moment.
0 is SOMETHING. That is a concept you should meditate on.
> how can a nonentity possibly fit into another nonentity?
You're right. That's why zero is not nothing.
Neither one
> exists. how can Nothing / Nothing possibly equal something?
nothing
can
> never fit into nothing.
Can you conceptualize the difference between saying, I am
nowhere, and
saying, I am standing at the point (0,0) in the Euclidean
plane. It's
like the difference between meeting someone at the corner of
1st St and
1st Avenue, versus not showing up at all. Once you are
anywhere at all,
you might as well be at zero, by a simple translation of
coordinate
systems.
===
Subject: Re: Basic algebra problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SFpba03429;
Antonio
>>Prove that in a commutative infinite ring with a
multiplicative
>>identity there can't exist a finite number of 
non-invertible
>>elements !=0.
>************************************************************
*****
>Hi:
>Well, it's easy but slightly tricky: Let's 
suppose R is our
ring and
>first assume there's some non zero 
non-invertible element a
which
>also is a zero divisor ==> for any unit u in R we get that
au is a
>non-zero (because u cannot be a zero divisor) non-invertible
element
>(since otherwise ALSO a would be invertible), and thus we
get at once
>a contradiction.
I can't understand the argument that makes you conclude that
the ring
is integral. Do you mean that u1!=u2 => a*u1!=a*u2, thus the
assumption of having a zero divisor leads to the existence of
infinitely many non-invertible elements? But a*u1 = a*u2 <=>
a*(u1-u2)=0
that can hold even if u1!=u2 since a is a zero divisor.
>So we can assume R is an infinite commutative integer domain
with a
>finite number of non-zero non-invertible elements, say
a1,...,an.
>We then know that R* = {invertible elements of R} is
infinite, but
>then again for any u in R* we get that a1*u is
non-invertible and
>non-zero, so a1*u = aj for some 1<= j <= n, and since R* is
>infinite there are two DIFFERENT units u,w in R* such that:
>a1u = a1w ==> a1(u-w) = 0 ; but we assumed R is an integral
domain,
>and a1 is no zero, so...can you complete the argument?
>Good luck!
>Tonio
I've thought of another way of solving this problem basing 
on
your
solution.
Assuming that the number of non-invertible elements is finite,
we
show that there exists an infinite number of invertible
elements u
such that a1*u=a1. Having a1*(u-1)=0 implies u-1 to be
non-invertible,
and since all such u-1 are pairwise different, we've got an
infinite
number of non-invertible elements. This argument does not
require the
ring to be integral. Is it correct?
Thanks for your help,
Igor
===
Subject: Re: Basic algebra problem
> Antonio
Prove that in a commutative infinite ring with a 
multiplicative
>identity there can't exist a finite number of 
non-invertible
>elements !=0.
*************************************************************
****
Hi:
Well, it's easy but slightly tricky: Let's 
suppose R is our
ring and
first assume there's some non zero 
non-invertible element a
which
also is a zero divisor ==> for any unit u in R we get that au
is a
non-zero (because u cannot be a zero divisor) non-invertible
element
(since otherwise ALSO a would be invertible), and thus we get
at once
a contradiction.
I can't understand the argument that makes you conclude that
the ring
> is integral. Do you mean that u1!=u2 => a*u1!=a*u2, thus the
> assumption of having a zero divisor leads to the existence
of
> infinitely many non-invertible elements? But a*u1 = a*u2 <=>
a*(u1-u2)=0
> that can hold even if u1!=u2 since a is a zero divisor.
>
So we can assume R is an infinite commutative integer domain
with a
finite number of non-zero non-invertible elements, say
a1,...,an.
We then know that R* = {invertible elements of R} is infinite,
but
then again for any u in R* we get that a1*u is non-invertible
and
non-zero, so a1*u = aj for some 1<= j <= n, and since R* is
infinite there are two DIFFERENT units u,w in R* such that:
a1u = a1w ==> a1(u-w) = 0 ; but we assumed R is an integral
domain,
and a1 is no zero, so...can you complete the argument?
Good luck!
Tonio
I've thought of another way of solving this problem basing 
on
your
> solution.
> Assuming that the number of non-invertible elements is
finite, we
> show that there exists an infinite number of invertible
elements u
> such that a1*u=a1. Having a1*(u-1)=0 implies u-1 to be
non-invertible,
> and since all such u-1 are pairwise different, we've got 
an
infinite
> number of non-invertible elements. This argument does not
require the
> ring to be integral. Is it correct?
Thanks for your help,
> Igor
******************************
Of course you're right, Igor. That was a stupid mistake I
made up
there.
But I'm happy to see that my argument (the correct part of
it...)
helped you to solve the problem. I'd only say, to be 
pedantly
accurate
now, that if a1,a2,...,an are the non-zero non invertible
elements of
R, then for AT LEAST one index 1 <= i <= n there exist
infinite units
u s.t. ai*u = ai (it can happen that a1*u = a2 for infinite
units u,
a2*u = a3 for inf. units, etc. where we've re-indexed the
non-units to
get this, but then at some moment we'll come back to some ai
already
taken, and then we're done).
Good luck!
Tonio
===
Subject: Re: Quaternions and the determinant.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SGJP206041;
Question:
Is (iii) or the first corollary, below, already known?
My relations are:
Let Q(A) be the quaternion algebra over the reals.
For x = (x_1)i + (x_2)j + (x_3)k + (x_4)1, define
res: Q(A) -> R^3, res(x) = (x_1, x_2, x_3) and
tes: Q(A) -> R , tes(x) = x_4
Then for all x, y, z in Q(A) with x_4 = y_4 = z_4 = 0 the
following holds:
(i) res(x) scalar-product res(y) = - tes(x*y)
(ia) ||res(x)|| = sqrt( - tes(x*x) ), euclidean norm
(ii) res(x) c-product res(y) = res(x*y)
(iii) det( res(x), res(y), res(z) ) = -tes(x*y*z)
Corollaries:
(1) If tes(x*y) = tes(y*z) = 0, then
res(x) c-product res(y) c-product res(z) = res(x*y*z)
Proof of (1):
(res(x) c-product res(y)) c-product res(z) =
= res(x*y) c-product res(z), by (ii)
= res((x*y)*z), again by (ii) ; since tes(x*y) = 0
= res(x*(y*z))
= res(x) c-product (res(y) c-product res(z)),
by (ii), since tes(y*z) = 0, q.e.d.
(2)
res(x) scalar-product (res(y) c-product res(y)) =
= det( res(x), res(y), res(z) )
Proof of (2):
res(x) scalar-product (res(y) c-product res(z)) =
= res(x) scalar-product res(y*z), from (ii)
= - tes( x*(y*z) ), from (i)
= - tes( x*y*z )
= det( res(x), res(y), res(z) ), from (iii)
q.e.d.
Just a check: Let I be the (3 by 3) identity matrix, then
det(I) = det( res(i), res(j), res(k) ) =
= - tes( i*j*k ) = - tes( -1 ) = 1
C. Dement
===
Subject: Is there a Boolean Algebra for Sagan Smog in 1800s
appalachian
mountain Jungles in Africa and Ebola Hemorrhagic Fever
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SIRDc18081;
Is there a Boolean Algebra for Sagan Smog in 1800s
appalachian mountain
Jungles in Africa and Ebola Hemorrhagic Fever. It is
interesting that it is
there and then it is gone.
My Space is where I want to be Sites: 
Lookingfora.0catch

or 
Angelfire
Bob L Petersen
===
Subject: a Diff. Geo. problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SIRED18096;
can anyone help me on this one:
Let $H$ be a smooth manifold and $Q$ its submanifold
such that $Q$ is the set of all critical points of $G: C to H$
and a path transverse to $Q$, $gamma :Ito H$
one have $G^{-1}(gamma)$ a smooth submanifold
thanks
===
Subject: Re: Pure abstract demonstration
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SJl5Y25130;
>> I think, You'll be surprised with some elementary
>>developments: Some exact relation can be closed as f(t).
>> However some advanced developments in that same relation
>>would show some F(t), where once f(t) is of n-degree so
>>F(t) is of (n-1)n degree. Somebody should notice, that
>>the best idea for F(t) to be coherent or homogeneous to
f(t) is
>>just F(t) = [f(t)]^(n-1). To some small and prime values of
>>n once beginning with n=3 such properity can be checked
>>using results of division F(t)/f(t). There could be also
shown,
>> that for f(t) once the oldest coefficient is 1 so Abs.th. =
-(A+B)
>> but for F(t) once the oldest coefficient is 1 so
>>Abs.th. = -(A+B)^(n-1) instead of [-(A+B)]^(n-1) =
(A+B)^(n-1)
>> What generally and sufficiently shows impossibility of F(t)
>>to be coherent to f(t).
>> See also, that from other side value A+B shows possibility
>>to be divided by t as some potential root in f(t) and in
F(t).
>> Some other combination of parameters and changing values
for
>>this very relation will show some possible factorisation of
>>any of Absolute therms.( I used to address and achieve in
main
>>part such developments at 22-th May of 2001 )
>> Parameters A and B represents here
>>some of parametric values taken from Abel formulae:
>> A = a^n once B = b^n or B = n^(nu-1) b^n or inverse .
>> Once in general used input is X=T+B; Y=T+A; Z=T+A+B
>>where T=n^u abtp always but for n=3 p=1
>> and the question was to solve X^n +Y^n = Z^n
>> for n prime numbers bigger than 2 and for X;Y;Z integers
>> Main part of FLT ?
>> Hint: compare [X^n + Y^n]/(X+Y) to some s^n or n*s^n
>> and express it in a;b;t;p parameters exclusively...
>> f(t)= t^n - 2 n^u abtp - a^n - b^n once including
>> so called first fall of FLT: X;Y;Z not divided by n
> Sir,
> The biggest possibility is in some kind of Your mistake.
> However lets check Your revelations for n=3:
>You claim Your parameters:
> T=3^u *abt once A=a^3 and B=b^3 or 3^(3u-1)*b^3
>Already it is known for n=3, that 3 should divide Z;X or Y
>so lets take 3/X ; X=T+B ; B=3^(3u-1) b^3
>also does Your parameters fits ? Ls=(T+A+B)^3=(T+A)^3
+(T+B)^3=Rs ?
> Ls= T^3 +3T^2(A+B) +3T(A+B)^2 +(A+B)^3
> Rs= 2T^3 +3T^2(A+B) +3T(A^2 +B^2) +A^3 +B^3
>then Rs-Ls= T^3 -6TAB -3AB(A+B) = 0 ; T^3 -3AB(2T+A+B) = 0
>3^3u a^3 b^3 t^3 -3*a^3 3^(3u-1)*b^3 [2*3^u abt +a^3
+3^(3u-1) b^3]=0
>and really axtracting 3^3u a^3 b^3 we'll have:
> t^3 - (2*3^u abt +a^3 +3^(3u-1) b^3] = 0
> What is just Your f(t) for n=3 and for the fall X/3
>Now You suggest Y^2 -YX +X^2 = s^3, what is also correct,
>but how to express everything in a;b;t;p parameters?
> I can see, that X+Y in this case should be some cube too:
> X+Y = T+B+T+A = 2T+A+B = 2*3^u abt +a^3 +3^(3u-1) b^3
>also Your t^3 = X+Y and once Y^2 -YX +X^2 = s^3 so Z = t*s
> now t*s = T+A+B; t^3 = 2T +A+B: T = t^3 -ts = t(t^2 -s);
>but also once T = 3^u abt = t(t^2 -s): t^2 -s = 3^u ab
>and finally s = t^2 -3^u ab *****(how smart are those 
numbers)
> What You sugess then before:
> Ls= s^3 = X^2 -XY +Y^2 =Rs
>Ls=(t^2 -3^u ab)^3 =
> = t^6 -3*t^4 3^u ab +3*t^2 3^2u a^2 b^2 - 3^3u a^3 b^3
>Rs={3^u abt +3^(3u-1) b^3}^2 -[3^u abt +3^(3u-1) b^3][3^u
abt +a^3] +
> +{3^u abt + a^3}^2 ......
> I can feel, that there is really some more accounts as one
web
>page could hold so I'll do it on some paper and come back
> very soon
> Juan C.C.
Dear Mr. Juan
Thanks for Your attention, and I'll help You
a little with the first results for this case for n=3 .
Continuing You should get Ls - Rs as down:
t^6 -3^(u-1) abt^4 +2*3^(2u) a^2 b^2 t^2 -3^u ab[a^3
+3^(3u-1) b^3]t +
-a^6 -2*3^(3u-1) a^3 b^3 -3^(6u-2) a^6 = 0
What can be expressed as product of:
[t^3 -2*3^u abt -a^3 -3^(3u-1)b^3][t^3 -3^u abt +a^3
+3^(3u-1)b^3]= 0
where the first is just equation t^3 = X+Y but the secound 
eq.:
t^3 -3^u abt +a^3 +3^(3u-1) b^3 = 0 represents some split of
possible natural values of t for to solve s^3 = X^2 +XY +Y^2
...
For other estimations as for Z^2 - ZX - X^2 = (3^u bt + a^2)^3
the primary equation is coming back without functional
alterations.
But for this very estimation our primary equation is not
multiplied
with some constant value of some our parameters a or b but
with
certainly different function f'(t) .
Nature of natural numbers can not cover all FLT functions
so ideally but opens such performance - camoußage:
We can't find some defect because of some 
absoluthe therm will
be not divided by changing value or to realise Eisenstein
criterion.
Instead, we should not tolerate any changes of realisation of
s^3.
For n=3 instead of acceptable [f(t)]^2 we have f(t)*f'(t)
what is
very sufficient for to dismiss, that some natural values will
solve
f(t). For bigger prime numbers it will be very clear, that
even if
we get from s^n estimation some f(t)^(n-1) * f?(t) = 0 so
f?(t)
never will be f?(t) = f(t) because the absoluthe therm of f(t)
is -(A+B) but the f(t)^(n-1) *f?(t) = F(t) absoluthe therm is
possibly to be induc as -(A+B)^(n-1) ...
Q.E.D
with Best Regards
Roman B. Binder
> As he looked, the brilliance spread and spread...
> from the Divine Comedy of Dante Alighieri

>> And so on my current entertainment is going to be finished:
>> There was some significant chance of likely correct
>> proof of P. Fermat origine
>> Roman B. Binder
>> Yours Ro
===
Subject: short note on odd perfect numbers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0SLWAa02828;
.
Grateful for any comments or questions on the short argument
at
http://mathquest.com/discuss/sci.math/m/569936/570941
which hinges on what proportion of an odd square its own
factors must sum
to.
Claiming that if an odd perfect number is a product of a
square and an odd
power of a prime, then the factors of the square sq must sum
to sq-x where x
divides into sq, causing a contradiction.
Ciao, Mark G.
.
===
Subject: Re: JSH: Mathematical clarity
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0T4WHf05774;
>> Another comment on this. Again assume as above that A, B,
and
>>C are algebraic integeand that A * B = p * C. Don't assume
>>that p and C are coprime.
>> Let r = GCD(A, p). Then A/r is an algebraic integer which
is
>>relatively prime to p, and p/r is an algebraic integer
which is
>>relatively prime to A (and hence to A/r).
>No on both counts. Say p = rs for some coprime algebraic
integers r
>and s. Let A= r^2, B = s^2, C = p. Then gcd(A,p) = r, A/r =
r which is
>not coprime to p.
Oh, ouch. That was dumb. Well, I did think it might be too
good to be true. I will just have to fall back to the original
claims: p and C are coprime, and r = gcd(A, p) and s = gcd(B,
p),
and (A/r)*(B/s) = C.
>For the other case, say p = r^2 for coprime algebraic
integers r and
>s,
What is s? [A misprint I think. You just mean p = r^2.]
A = r, B = r, C = 1. Then gcd(A,p) = r, but p/r = r which is
not
>coprime to A.
Right. Thanks.
>--
>============================================================
==========
>It's not denial. I'm just very selective 
about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
>============================================================
==========
magidin@math.berkeley.edu
===
Subject: Re: JSH: Mathematical clarity
...
> Oh, ouch. That was dumb. Well, I did think it might be too
> good to be true. I will just have to fall back to the
original
> claims: p and C are coprime, and r = gcd(A, p) and s =
gcd(B, p),
> and (A/r)*(B/s) = C.
There are still those pesky units lying around.
--
===
Subject: Re: JSH: Mathematical clarity
 Another comment on this. Again assume as above that A, B,
and
>C are algebraic integeand that A * B = p * C. Don't assume
>that p and C are coprime.
Let r = GCD(A, p). Then A/r is an algebraic integer which is
>relatively prime to p, and p/r is an algebraic integer which
is
>relatively prime to A (and hence to A/r).
No on both counts. Say p = rs for some coprime algebraic
integers r
and s. Let A= r^2, B = s^2, C = p. Then gcd(A,p) = r, A/r = r
which is
not coprime to p.
Oh, ouch. That was dumb. Well, I did think it might be too
> good to be true. I will just have to fall back to the
original
> claims: p and C are coprime, and r = gcd(A, p) and s =
gcd(B, p),
> and (A/r)*(B/s) = C.
For the other case, say p = r^2 for coprime algebraic
integers r and
s,
What is s? [A misprint I think. You just mean p = r^2.]
Yeah, I kept trying to come up with an example where both
statements
were false, but for some reason I could not; but here it is.
Say A = r^2s, B = s,p = rs^2, C=r, where r and s are
relatively prime
algebraic integers. Then gcd(A,p) = rs; A/rs = r, which is
not coprime
to p, and p/rs = s, which is not coprime to A.
, sans .sig
===
Subject: Re: little help
f:R^n-->R vanishes of order p in 0 ==> Taylor's expansion of
f has
zero
all terms of degree less than p.
May you suggest me how to prove the above prop.?
First of all, thank you for your answer.
> You can't prove it as sta - you need to assume that f is
> infinitely differentiable (or at least p or maybe p + 1 
times
> differentiable).
Exscuse me, I should have said differentiable=C^{infty}.
> With the right hypotheses, assume that
> those terms in the Taylor expansion are _not_ all 0, and
> show that it follows that f does not vanish to order p.
> (You do this by showing that the lowest-order
> non-zero term dominates for small x.
For example, suppose that f vanishes of order p=3 at 0
with n = 1, then, by definition, f/x^2-->0 (x->0) .
On the other hand, suppose f(0) = f Ô(0) = 0, while f 
Ô'(0)
<> 0.
After multiplying by a constant Taylor's theorem
shows that
f(x) = x^2 + c x^3 + o(x^3), where o(x^3)/x^3 -> 0.
This implies that f(x)/x^2 -> 1, contradiction.
May you give an hint for n>1, please ?
(i suppose we need use multi-index notation, what a truble!)
Sirio
===
Subject: Re: little help
> f:R^n-->R vanishes of order p in 0 ==> Taylor's expansion
of f has
zero
> all terms of degree less than p.
> May you suggest me how to prove the above prop.?
>First of all, thank you for your answer.
>> You can't prove it as sta - you need to assume that f is
>> infinitely differentiable (or at least p or maybe p + 1
times
>> differentiable).
>Exscuse me, I should have said differentiable=C^{infty}.
>> With the right hypotheses, assume that
>> those terms in the Taylor expansion are _not_ all 0, and
>> show that it follows that f does not vanish to order p.
>> (You do this by showing that the lowest-order
>> non-zero term dominates for small x.
>For example, suppose that f vanishes of order p=3 at 0
>with n = 1, then, by definition, f/x^2-->0 (x->0) .
>On the other hand, suppose f(0) = f Ô(0) = 0, while f 
Ô'(0)
<> 0.
>After multiplying by a constant Taylor's theorem
>shows that
> f(x) = x^2 + c x^3 + o(x^3), where o(x^3)/x^3 -> 0.
>This implies that f(x)/x^2 -> 1, contradiction.
>May you give an hint for n>1, please ?
Certainly: use the same argument, with trivial
modifications.
>(i suppose we need use multi-index notation, what a truble!)
>Sirio
************************
===
Subject: Re: little help
> Certainly: use the same argument, with trivial
> modifications.
Ok, this afternoon, afterhaving solved another problem,
concerning with
differential of higher order i'll try, thank you for your
help.
I ask for help in sci.math for the following reasons:
Ptoblem's kernel is :let U denote an open set in M= C^infty
manyfold;
let I_p denote the ideal in C^{infty}(U) of real valued
functions of
class C^infty which vanish at p. (Precisely, p in Usubseteq
M, f :
U --> |R is in I_p if and only if f(p) = 0 and f is C^infty) .
Now consider ideal I^k_p in C^infty (U) defined in this 
manner:
I^k_p : = I_p times .... times I_p (k-times),
in other words,
I^k_p is the the product of I_p k-times with itself.
By definition, f in C^infty (U) has a k-order zero in p if
exists lim_{ xto p} frac { f(x) } { || x-p ||^(k-1) } = 0.
It is clear that f in I^k_p implies f in C^infty (U) has a
k-order zero
in
p .
Now, d^k f_p is intrinsically defined if and only if f vanish
at p of order
k (or k+1 ?!?), then How does Taylor's polynomial of f looks
like?
In other words, is it f= p + x-p df_p (x-x_0) + 1/2! d^2 f_p
(x-x_o)^2 +etc
or
f= p + (x-p) nabla f_p (x-x_0) + 1/2!( nabla^2 f)_p (x-x_o)^2
+etc
where nabla denotes a fixed linear connession on M.?
Regards Sirio
===
Subject: BCH Codes and parity-check matrix?
For a normal binary BCH code of length n = 2^m - 1, n - k <=
m*t we
construct a generator polynomial such that alpha,
alpha^2,...,alpha^{2*t) are roots of the generator polynomial
(where
alpha is a primitive element in GF(2^m). This can be construc
by
multiplying the minimal polynomials phi_i(x) (where phi_i is
the
minimal polynomial of alpha^i).
The parity-check matrix is construc by using the roots:
- -
H = | 1 alpha alpha^2 ... alpha^{n-1} |
| 1 alpha^2 (alpha^2)^2 ... (alpha^2)^{n-1} |
| . . . ... . |
| 1 alpha^t (alpha^t)^2 ... (alpha^t)^{n-1} |
- -
Thus, H*transpose(v) = 0, where v is any valid codeword of
the BCH
code. This is the same as substituting each root in v(x), and
getting
the result as zero.
I want to extend these principles to codes over GF(2^m). I
tried an
example using codes over GF(2^4). I facorized x^13 + 1 in
GF(2^m),
and obtained the following factors (using MAGMA) (these will
each
generate a (13,10) cyclic code with d_{min} = 4):
f1 = x^3 + z^7*x^2 + z^13*x + 1;
f2 = x^3 + z^11*x^2 + z^14*x + 1;
f3 = x^3 + z^13*x^2 + z^7*x + 1;
f4 = x^3 + z^14*x^2 + z^11*x + 1;
(z = alpha)
When I factored f1 over GF(2^12), I got the following:
x + z^568
x + z^3625
x + z^3997
However, when I multiply a polynomial p(x) with f1 in GF(2^4)
to
obtain c(x) and then substitute the roots (z^568, z^3625,
z^3997) into
c(x) in GF(2^12), the results are not zero. Why doesn't this
work?
How can I change this? (I know that the fields are genera 
using
Conway polynomials -> Should I try to use another polynomial
to
generate the GF(2^12) field?) Also, multiplying f1*p in 
GF(2^4)
results in a different polynomial than when I multiply f1*p in
GF(2^12). Why does this work for the binary case, but 
doesn't
work
for the non-binary case?
Your time. effort and suggestions will be greatly apprecia.
Jaco Versfeld
===
Subject: Re: BCH Codes and parity-check matrix?
I want to extend these principles to codes over GF(2^m). I
tried an
> example using codes over GF(2^4). I facorized x^13 + 1 in
GF(2^m),
> and obtained the following factors (using MAGMA) (these
will each
> generate a (13,10) cyclic code with d_{min} = 4):
Why should these codes necessarily have d_min=4? May be they
do, I'm
just curious, because I didn't see how this would follow 
from
the BCH-bound
or any bound that I know of?
> f1 = x^3 + z^7*x^2 + z^13*x + 1;
> f2 = x^3 + z^11*x^2 + z^14*x + 1;
> f3 = x^3 + z^13*x^2 + z^7*x + 1;
> f4 = x^3 + z^14*x^2 + z^11*x + 1;
(z = alpha)
What exactly is alpha here? My guess is that it is a root of
x^4+x+1=0, but unless you know that you are fumbling in the
dark,
and it is hard to guess/reproduce what might have gone wrong.
> When I factored f1 over GF(2^12), I got the following:
> x + z^568
> x + z^3625
> x + z^3997
What is z this time? Is it again something called 
Ôalpha'.
Surely
you are aware of the fact that this Ôalpha' is 
different from
the
Ôalpha' in the construction of GF(16)? Your 
notation suggests
that
you always want alpha to be a primitive element of the field.
So,
if beta=z is a primitive element of GF(2^12), then
alpha=z^273 is
a primitive element of GF(2^4) (caveat: this alpha may have a
different
minimal polynomial, i.e. it may now be that z^273 is a zero of
x^4+x^3+1=0 instead of x^4+x+1=0. (273 here = 4095/15).
Something very fishy about these zeros anyway. If f1 is a
factor
of x^13+1, then all these roots should have order 13, i.e.
the exponents
should be multiples of 4095/13=315, but they aren't.
However, when I multiply a polynomial p(x) with f1 in GF(2^4)
to
> obtain c(x) and then substitute the roots (z^568, z^3625,
z^3997) into
> c(x) in GF(2^12), the results are not zero. Why doesn't
this work?
No wonder! I guess that what happened is that you have named
two different
things Ôalpha' (or z), and this has lead to a 
confusion.
> How can I change this? (I know that the fields are genera
using
> Conway polynomials -> Should I try to use another
polynomial to
> generate the GF(2^12) field?) Also, multiplying f1*p in
GF(2^4)
> results in a different polynomial than when I multiply f1*p
in
> GF(2^12). Why does this work for the binary case, but
doesn't work
> for the non-binary case?
In binary case no powers of alpha appear as coefficients, so
you didn't
have a chance to make this mistake.
What you should do, is to
1) generate the field GF(2^12), call a primitive element
alpha=z
2) identify the subfield GF(2^4) within this 
field: let
beta=z^273,
then GF(2^4)=GF(2)(beta)
3) you may want to find out the minimal polynomial of beta, so
that
you can properly map elements of this subfield into a copy of
GF(2^4)
genera by some other means.
If any of this is not crystal clear, then follow the earlier
advice
and study a book on abstract algebra.
Good luck!
Jyrki Lahtonen, Turku, Finland
===
Subject: Re: Structure of Finite topological spaces
> As one progress up the hierarchy of topological spaces
> T_0, T_1, Hausdorff, Regular, Normal
> a strong bias toward infinite spaces becomes apparent.
> Once a finite space has been refined to T_1, it
> is discrete and no further refinements are possible.
> Also once a space has been refined to slightly less than
normal
> it's basically a subspace of a Tychonov cube, that is
> a subspace of some product of closed intervals of the real
line.
> Thus for topology to innovate spaces beyond variants of
real number
> spaces, it needs look to the lessor spaces. My opinion is
the lessor
> spaces, especially below Hausdorff, are descriptive of
black holes and
> their surrounds. Suggestions for topological spaces that
aren't so real
> number like?
> -- structure of finite topological spaces
> U minimal when for all open V proper subset U, V = nulset
> open U ==> U minimal
> iff for all open V, (U subset V or U,V disjoint)
> iff for all open V, (U / V = nulset or U / V = U)
> Let S be a finite space and let
> M = M_S = { U | open nonnul U minimal }
> M is finite pairwise disjoint collection of open sets.
> For all open nonnul U, some V in M with V subset U
> Let A = { U1,.. Uj } be the open nonnul subsets of U
> chose some V in A with minimal size; assume W open
> |W / V| = 0 or |V| by minimal cardinality of V
> V / W = nulset or V / W = V as S is finite
> thus V in B
> In a finite space
> every nonnul open set contains a minimal nonnul open set
> cl /M = S. Assume x not in cl /M.
> Some open U nhood x with U / /M = nulset
> Some minimal open nonnul V subset U; V in M
> V = V / /M subset U / /M = nulset, which cannot be
> If S is T0, M is a collection of singletons.
> If S is T1, M is the collection of all singletons.
> Conversely, if M is the collection of all singletons, S is
T1
> -- application
> Let n = min { |U| : U in M }
> S is n-resolvable and each of the
> n pairwise disjoint dense sets has cardinality |M|
> In addition, S is not n+1 resolvable, and
> no dense set has cardinally < |M|
> The n pairwise disjoint dense subsets of S are
> D_1 = { chose one from U | U in M }
> D_2 = { chose another one from U | U in M }
> = { chose one from UD | U in M }
> ...
> D_(j+1) = { chose yet another one from U | U in M }
> = { chose one from U - (D_1 /../ D_j) | U in M }
> ...
> D_n
> ----
Interesting idea. However, it would help if you would go into
some of the
motivations for your definitions. The topology of 
infinite sets
has the
benefit of being conceptually simplified due to 
its geometric
interpretation. ie: it would be interesting if one could give
a geometric
interpretation based on finite graphs.
l8r, Mike N. Christoff
===
Subject: Re: Structure of Finite topological spaces

Subject: Re: Structure of Finite topological spaces
>Interesting idea. However, it would help if you would go
into some
>of the motivations for your definitions. The topology of
infinite
>sets has the benefit of being conceptually 
simplified due to
its
>geometric interpretation. ie: it would be interesting if one
could
>give a geometric interpretation based on finite graphs.
Unlike you, I'm not a visual mathematician. 
I'm a symbolic
mathematician
that works best with formulas instead of pictures. I was
motiva during
discussion of Ôresolvable spaces' between Fred 
and I. It's a
small area
and easy to structure. Below are my quick notes easily
extending this
notion to spaces with finite topologies. If you have 
finite
graph
interpretation, it would be agreeable addition to this
leisurely ramble.
-- structure of spaces with finite topology
U minimal when for all open V subset U, V = nulset or V = U
open U ==> U (topologically) minimal
iff for all open V, (U / V = nulset or U / V = U)
iff for all open V, (U / V = nulset or U subset V)
M = M_S = { U | open nonnul U minimal }
M pairwise disjoint collection of open sets
-- finite topology S. finite space ==> 
finite topology
finite subbase ==> finite base ==> 
finite topology ==> finite
subbase
for all open nonnul U, some V in M with V subset U. Let
B = { V | nonnul V open subset U } = { U1,.. Uk }; V1 = U1
V_(j+1) = Vj / U_(j+1) if nonnul, = Vj otherwise; V = Vk
nonnul V in M. If open W with nonnul V/W: V/W in B
some j with V/W = Uj; if j = 1: V subset V1 = U1 subset W
if 1 < j: V subset V_(j-1) = Vj / Uj subset Uj subset W
M nonnul finite
cl /M = S. If x not in cl /M:
some open U nhood x with disjoint U,/M; some V in M with V
subset U
V = V / /M subset U / /M = nulset which cannot be
T0 space S ==> for all U in M, |U| = 1. If distinct a,b in U
in M:
some open V with (a in V, b not in V) or (a not in V, b in V)
open U/V proper subset U; U/V = nulset which cannot be
S is T1 iff S finite discrete
if S infinite: some distinct x1,x2,... in S
{ S - {x1,.. xj} | j in N } infinite set open sets, not so!
-- application
let n = n_S = min { |U| : U in M }; n may be infinite
S is n-resolvable with each of the n pairwise disjoint
dense sets having cardinality |M|
In addition, S is not m resolvable for m > n
no dense set has cardinally < |M|.
The n pairwise disjoint dense subsets of S are for j < n
D_j = { chose one from U - /{ D_i | i < j } | U in M }
----
===
Subject: Fresnel integral again
sorry for asking this question again, but I still cannot
prove the
convergence of the integral from 0 to +oo over the function
sin(x^2).
In my yesterdays posting somebody told me to do int. by parts
(see below),
but I cannot figure out how this should work.
Can someone perhaps show the approach step by step ?
Thank you all sooo much....
Regards
> To show the integral from 1 to oo converges, let x =
sqrt(t) and
integrate
> by parts.
Can anyone explain how the integration by part works out here
in detail ?
I don't get it...
I tried with:
u(t) = 1 -> u'(t) = 0
v'(t) = sin(t) -> v(t) = -cos(t)
leads to 1 * (-cos(t)) - Integral of 0 * sin(t)
This does not lead to anything.
Also tried
u(t) = sin(t) -> u'(t) = cos(t)
v'(t) = 1 -> v(t) = t
Does not lead to anything either...
Something I do might be wrong :-(
===
Subject: Re: Fresnel integral again
X-ID: EqlmjiZBreMMNx14gKMnwmBlEyqIB1N-fMZEoARMdcuo+R81lO54UK
Alexander Schmidt  schrieb:
>sorry for asking this question again, but I still cannot
prove the
>convergence of the integral from 0 to +oo over the function
sin(x^2).
>In my yesterdays posting somebody told me to do int. by
parts (see below),
>but I cannot figure out how this should work.
>Can someone perhaps show the approach step by step ?
>Thank you all sooo much....
...
>Can anyone explain how the integration by part works out
here in detail ?
>I don't get it...
> I tried with:
>u(t) = 1 -> u'(t) = 0
>v'(t) = sin(t) -> v(t) = -cos(t)
>leads to 1 * (-cos(t)) - Integral of 0 * sin(t)
>This does not lead to anything.
>Also tried
>u(t) = sin(t) -> u'(t) = cos(t)
>v'(t) = 1 -> v(t) = t
>Does not lead to anything either...
Ah, you did the substition wrong. Int sin(x^2)dx does not
transform to
Int sin(u) du with u = x^2. If u=x^2 then du = 2x*dx =
2sqrt(u)dx.
Ergo dx = du/(2sqrt(u)) and you have to evaluate the integral
Int [1..infty] sin(u) du / (2sqrt(u).
So, have another try!
Thomas
>Something I do might be wrong :-(
--
Thomas
===
Subject: Re: Fresnel integral again
> Alexander Schmidt  schrieb:
> [snip]
> Ah, you did the substition wrong. Int sin(x^2)dx does not
transform to
> Int sin(u) du with u = x^2. If u=x^2 then du = 2x*dx =
2sqrt(u)dx.
> Ergo dx = du/(2sqrt(u)) and you have to evaluate the
integral
> Int [1..infty] sin(u) du / (2sqrt(u).
> So, have another try!
I still don't understand what you guys are doing here ;-(((
Can't anyone explain this stuff in a bit more detail - step
by step ?
I have totally lost the context ....
It's late and I have tried much... so I need your help now !
;-(
===
Subject: Re: Decidable problem ?
> Giving a number m which is the code of a Turing machine, is
it
> possible to
> decide if m is such as its associa Turing machine will
always
go
> to
> the
> right whatever the entry is ?
I would say yes. Think of the TM as a direc graph, where each
node
represents a state, and each arc is the result of the TM
reading a
particular symbol from the tape while in that state. Ignore
nodes
(states) unreachable from the starting node.
 You can solve the halting problem if you can find all
> of the unreachable states.
 Russell
> - 2 many 2 count
No, I am talking about nodes that are not part of the direc
graph
that begins with the starting node, not that can be reached by
executing the TM. In other words, if you have a state 99 and
nothing
transitions to 99, then we ignore that node.
Now reread my proof and see if you agree, ok?
It is important to realize there is nothing special
> about the halt instruction. We can replace halt with
> move left and the proof still holds. Turing used the
> write 1 instruction is his original paper.
Nope. In that case (replace move left with halt in my proof)
you
would have left and right transitions along the way, instead
of just
right transitions, and you wouldn't be assured of having a
direct (no
node visi more than once) path to the halt node. A left
transition
logic could prevent you from ever following the complete
path. (You
could get into an infinite loop, oscillating between the left
and
right transitions.) As I sta in my proof, All previous arcs
along
this path transition to the right. Thus every such node is
reading a
symbol from the initial tape, not one that was written along
the way.
Let's say your direc graph has a path from the start node
consisting of all right transitions except ending with a left
transition. Is there necessarily an initial input tape value
that
follows that path?
PS I do agree that you can't decide if any particular node
(state) is
reached by an arbitrary TM plus input, and consequently by
all inputs
to an arbitrary TM (due to Kleene's s-m-n theorem, actually.
The
formal reduction to the Halting Problem in my own
formalization of
Computer Science is (aA)HALT(I,A) => (aA)HALT(s11(I,J),A) =>
(aA)HALT(I,J) => HALT(I,J) where s11(I,J) = TM # I with J
substitu
for its initial input tape. See my paper at
http://www.arxiv.org/html/cs.lo/0003071 .) However, we can
tell if
any of the left transitions are reached, because of the
special
behavior of a program that only transitions right.
If a proof seems convertible to an erroneous proof, it is
instructive
to examine the erroneous proof to see where the problem
really lies.
The Invariance Theorem, on which Gregory Chaitin's 
Algorithmic
Information Theory (AIT) is based (e.g. An Introduction to
Kolmogorov
Complexity and Its Applications, Ming Li and Paul Vitanyi,
page 96),
can be shown to be ßawed in this manner.
Consider two programming language (bases of computing), the
programs
in one always being twice the size of the programs in the
other. Then
there is in fact no bound to the difference in the size of the
smallest program that outputs a given number, contrary to the
Invariance Theorem. If you carry out the proof ci above with
these
two languages, you will discover the ßaw: the assumption that
it
takes the same size program to initialize the input to a given
constant. One example would be Turing Machines themselves!
While one
language can take 1 character per digit, Turing Machines take
many
instructions per digit (symbol) to initialize the input tape.
So your reasoning is correct - except for the critical
difference
between transitioning right vs. transitioning either left or
right.
> Russell
> - 2 many 2 count
===
Subject: Re: Decidable problem ?
It is important to realize there is nothing special
about the halt instruction. We can replace halt with
move left and the proof still holds. Turing used the
write 1 instruction is his original paper.
> Nope. In that case (replace move left with halt in my
proof) you
> would have left and right transitions along the way,
instead of just
> right transitions, and you wouldn't be assured of having a
direct (no
> node visi more than once) path to the halt node. A left
transition
> logic could prevent you from ever following the complete
path. (You
> could get into an infinite loop, oscillating between the
left and
> right transitions.) As I sta in my proof, All previous arcs
along
> this path transition to the right. Thus every such node is
reading a
> symbol from the initial tape, not one that was written
along the way.
> Let's say your direc graph has a path from the start node
> consisting of all right transitions except ending with a
left
> transition. Is there necessarily an initial input tape
value that
> follows that path?
Maybe I misunderstood the original question.
If you can find such a path then there is an input tape that
will cause the TM to move left. But, if the question is:
Given a number m which is the code of a Turing machine,
is it possible to decide if TM(m) will always go to the right?
Then the answer is that you can identify some such TM(m),
but not every such TM(m).
I assume the input tape is blank. I also assume that the TM
instruction set allows the TM to write to the tape without
moving the tape head. This allows state n to read a
symbol written by a previous state without moving the head.
If we assume the tape head always moves after a write
then your proof may work.
I'm not convinced the halting problem is unsolvable, so I
am just quoting the authorities here. Turing gives a proof
that it is impossible to determine, for every member of the
Russell
- Learning is easy. Unlearning is harder.
===
Subject: Re: Decidable problem ?
Maybe I misunderstood the original question.
> If you can find such a path then there is an input tape that
> will cause the TM to move left. But, if the question is:
Given a number m which is the code of a Turing machine,
> is it possible to decide if TM(m) will always go to the
right?
> Then the answer is that you can identify some such TM(m),
> but not every such TM(m).
If there is a left transition (connec to the start node) then
it
won't always go to the right for all inputs, because there 
is
an input
that will reach this left transition. If there is no left
transition,
then it will always go to the right for all inputs. Thus we
can
decide if any given TM will always go to the right for all
inputs,
which is what we're looking for.
> I assume the input tape is blank.
No, the problem was sta as whatever the entry is (meaning for
all
inputs.)
> I also assume that the TM
> instruction set allows the TM to write to the tape without
> moving the tape head.
I assumed the original model. In the above case, does the
problem
consider this to be a transition not to the right, or not a
transition? If we don't move the head, then after going to
the next
state we will necessarily follow the arc associa with the
symbol
just written (or halt), and can replace the first arc with an
arc that
goes to the state that this arc goes to, thus removing the
motionless
transition, if it's not considered to be a transition that
violates
the always right check. If it is considered to be a
transition not
to the right, then we simply treat it as a left transition in
the
proof.
> Russell
> - Learning is easy. Unlearning is harder.
===
Subject: Re: Decidable problem ?
> If there is a left transition (connec to the start node)
then it
> won't always go to the right for all inputs, because there
is an input
> that will reach this left transition.
This is the part I question.
Just because there there is a left transition connec to the
initial state doesn't mean that node will ever be execu.
There can be some intervening set of instructions, including
Russell
- 2 many 2 count
===
Subject: Re: Decidable problem ?
>
If there is a left transition (connec to the start node) then
it
won't always go to the right for all inputs, because there 
is
an input
that will reach this left transition.
This is the part I question.
> Just because there there is a left transition connec to the
> initial state doesn't mean that node will ever be execu.
If all arcs on the path leading to it are right transitions,
would the
tape input consisting of their symbols follow that path?
In the model where stationary transitions are included, we
can ignore
them or not, depending on whether we want to count that as not
transitioning right.
> There can be some intervening set of instructions, including
Russell
> - 2 many 2 count
===
Subject: Re: Decidable problem ?
> If there is a left transition (connec to the start node)
then
it
> won't always go to the right for all inputs, because there
is an
input
> that will reach this left transition.
This is the part I question.
Just because there there is a left transition connec to the
initial state doesn't mean that node will ever be execu.
> If all arcs on the path leading to it are right
transitions, would the
> tape input consisting of their symbols follow that path?
> In the model where stationary transitions are included, we
can ignore
> them or not, depending on whether we want to count that as
not
> transitioning right.
>
There can be some intervening set of instructions, including
The TM would have to move left to read them.
I think your method works.
It looks like it can be modified to determine if a TM
Red ßags went up in my mind when you said
you could find every unreachable node.
With a little work you might have a solution to the
halting problem.
Russell
- 2 many 2 count
===
Subject: Re: Decidable problem ?
> The TM would have to move left to read them.
> I think your method works.
> It looks like it can be modified to determine if a TM
Red ßags went up in my mind when you said
> you could find every unreachable node.
> With a little work you might have a solution to the
> halting problem.
> Russell
> - 2 many 2 count
Great! I agree with everything you're saying - except your
last
point. After you learned about Turing Machines, wasn't the
next thing
that you learned Turing's 1937 proof of the unsolvability of
the
Halting Problem? Or do you disagree with Turing?
Not that I mind people disagreeing. I love it when people
have enough
sense to Question Authority and not be fooled by the BSers.
But
Turing is for real. Maybe because he was (arguably) the
world's first
computer programmer? After all, those programmers can't BS
around.
Their stuff has to run!
===
Subject: gaussian filter and pascal's triangle
Let f be the Gaussian density of some random variable X
with mean zero and variance one. Thus
f(x) = 1/sqrt(2*pi) * exp(-x^2/2)
Let g be a scaling of f,
g(x) = a*f(x)
Let j be odd, with j=2*n+1. Choose the constant a such that
g(-n) + g(-n+1) + ... + g(0) + ... + g(n) = 1
or for j even, with j=2*n, such that
g(-n/2) + ... + g(-1/2) + g(1/2) + ... + g(n/2) = 1
For example, for j=5 one obtains,
a = 1.009218543
and
g(-2) = .8718189528/2^4
g(-1) = 3.907221470/2^4
g(0) = 6.441919150/2^4
g(1) = g(-1)
g(2) = g(-2)
finally
round(2^4*g(-2)) = 1
round(2^4*g(-1)) = 4
round(2^4*g(0)) = 6
round(2^4*g(1)) = 4
round(2^4*g(2)) = 1
Some other values:
j=1: [1]
j=2: [1 1]/2
j=3: [1 2 1]/2^2
j=4: [1 3 3 1]/2^3
j=5: [1 4 6 4 1]/2^4
Iow, precisely Pascal's triangle!
Does someone know why this is so?
Wilbert
===
Subject: Re: Non-standard analysis
When you post the same message to two groups you should
_c-post_ instead of posting the messages separately!
That means that you put sci.math,sci.math.research on the
Newsgroups line when you make the post. The reason this
is better is that then replies pos in one group will appear
in the other group. If the question is of interest to both
groups then the replies should be too.
(Here for example, if you'd c-pos the message then
when I was reading sci.math.research I would have seen that
you'd already got a correct answer in sci.math, so I would
not have bothered giving essentially the same answer in
sci.math.research.)
************************
===
Subject: Re: Non-standard analysis
> Consider the collection of sets I1, I2, I3, ... , In, ...
> Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In =
{1,2,3,...,n},...
> for natural numbers n. That is, Ij is the set of natural
numbers less
> than or equal to j.
> Define a set S to be finite if there is a 1-1 
mapping of S
onto In for
> some natural number n.
> Define a finite set S to be larger than a 
natural number n,
if there
> is a natural number m and a 1-1 map of S onto Im with m > n.
> Consider further the following countable collection of
sentences:
> 1. There exists a finite set S1 larger than 1.
> 2. There exists a finite set S2 larger than 2.
> ...
> n. There exists a finite set Sn larger than n.
> ...
> Clearly any finite subset of this collection is consistent.
[let j be
> the largest index of a sentence in the finite subset, then,
for
> example, Ij+1 provides the model]
> By compactness, the whole collection is consistent and
therefore there
> is a model with the property that there exists a finite set
S which is
> larger than n for any n. This is clearly a contradiction.
Well, yes. Hence one of your assumptions must be erroneous.
How do you
express finite set in first-order logic? 
It's well known that in
second-order logic (finite set is your clue) 
there's only on
model of
the
reals, and hence of the naturals.
If you change your sentences to Ex (x>n), (clearly
expressible in
first-order logic, since it's expressed in 
first-order logic),
then the
contradiction goes away.
> [S finite implies by definition that there 
exists some
natural number
> n, such that S can be mapped 1-1 onto In. But S larger than
n implies,
> again by definition, that S can be mapped 1-1 onto Im for
some natural
> number m > n]
But how do you state this in the language you're 
considering?
===
Subject: Re: Non-standard analysis
...
Well, yes. Hence one of your assumptions must be erroneous.
How do you
> express finite set in first-order logic? 
It's well known that
in
> second-order logic (finite set is your clue) 
there's only on
model of
the
> reals, and hence of the naturals.
If you change your sentences to Ex (x>n), (clearly
expressible in
> first-order logic, since it's expressed in 
first-order
logic), then the
> contradiction goes away.
>
[S finite implies by definition that there exists 
some natural
number
n, such that S can be mapped 1-1 onto In. But S larger than n
implies,
again by definition, that S can be mapped 1-1 onto Im for some
natural
number m > n]
But how do you state this in the language you're 
considering?
Jon. Some questions:
1. Is it fair to say that the nub of the problem is that the
definitions of finite and larger that I have 
sugges cannot be
formula in first-order logic?
2. Robinson also shows compactness for higher order
structures and
correspondiing languages. So my definitions of 
finite and larger
cannot be formula in a higher order language?
3. Presumably similar problems occur with the following set of
sentences S?:
S1. There exists a set, T, of cardinality greater than 1 and
strictly
less than aleph0.
S2. There exists a set, T, of cardinality greater than 2 and
strictly less than aleph0.
...
Sn. There exists a set, T, of cardinality greater than n and
strictly
less than aleph0.
...
Any finite subset of S is consistent, therefore there exists a
model
of S with set T of cardinality greater than n for any n but
strictly
less than aleph0.
===
Subject: Re: Action Device: Destination Alpha Centauri A
charset=iso-8859-1
> It seems that you have missed my earlier post. I have this
device almost
> ready on ßoor of my room. I can test it right now. Not
necessary to hang
> the things in air at this moment I can test it on ßoor. If
this device
> pushes itself and body only in one direction, Houston has a
problem. If
not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret of
friction...
Volker
===
Subject: Re: Action Device: Destination Alpha Centauri A
It seems that you have missed my earlier post. I have this
device
almost
ready on ßoor of my room. I can test it right now. Not
necessary to
hang
the things in air at this moment I can test it on ßoor. If
this device
pushes itself and body only in one direction, Houston has a
problem. If
not,
I have a problem.
> If it pushes itself in one direction you have discovered
the secret of
friction...
The secret of gravity, more likely.
===
Subject: Re: Action Device: Destination Alpha Centauri A
Abhi  schrieb im Newsbeitrag
> It seems that you have missed my earlier post. I have this
device
almost
> ready on ßoor of my room. I can test it right now. Not
necessary to
hang
> the things in air at this moment I can test it on ßoor. If
this
device
> pushes itself and body only in one direction, Houston has a
problem.
If not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret of
friction...
> The secret of gravity, more likely.
========================
Oh Gravity....
Aaa Sikander-E-Hunar Aaazmayain
Tu Teer Aaazmaaa, Hum Jigar Aaazmayain
========================
-Abhi.
===
Subject: Re: Action Device: Destination Alpha Centauri A
> Abhi  schrieb im Newsbeitrag
It seems that you have missed my earlier post. I have this
device
almost
ready on ßoor of my room. I can test it right now. Not
necessary to
hang
the things in air at this moment I can test it on ßoor. If
this device
pushes itself and body only in one direction, Houston has a
problem. If
not,
I have a problem.
> If it pushes itself in one direction you have discovered
the secret of
friction..
Can you explain this? I mean if device pushes body and itself
in space only
in one direction, what it has to do with friction? Tha same
thing can
happen
on ßoor of my room.
-Abhi.
==============================
kuchh tabiyat hii milii thii aisii chain se jiine kii suurat
na hu_ii
jis ko chaahaa use apanaa na sake jo milaa us se muhabbat na
hu_ii
jis se jab tak mile dil hii se mile dil jo badalaa to
fasaanaa badalaa
rasme.n duniyaa kii nibhaane ke liye hamase rishto.n kii
tijaarat na hu_ii
duur se thaa vo ka_ii cheharo.n me.n paas se ko_ii bhii
vaisaa na lagaa
bevafaa_ii bhii usii kaa thaa chalan phir kisii se hii
shikaayat na hu_ii
vaqt ruuThaa rahaa bachche kii tarah raah me.n ko_ii
khilaunaa na milaa
dostii bhii to nibhaa_ii na ga_ii dushmanii me.n bhii adaavat
na hu_ii
================================
===
Subject: Re: Action Device: Destination Alpha Centauri A
Abhi  schrieb im Newsbeitrag
> It seems that you have missed my earlier post. I have this
device
almost
> ready on ßoor of my room. I can test it right now. Not
necessary to
> hang
> the things in air at this moment I can test it on ßoor. If
this
device
> pushes itself and body only in one direction, Houston has a
problem.
If
> not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret of
> friction..
My grandson has a little mechanical rabbit that pushes itself
in one
direction. Should I be calling the nobel prize comittee?
Simply pushing
itself ac the ßoor proves nothing.
You were told to suspend the device on a string. If it moves
to the side
causing a 45degree or better angle with the ceiling (no, not
swinging back
and forth, just pushed to the side and stable) then Houston
indeed has a
problem. If not then... well... everyone has their fifteen
minutes of fame
and you've just exhaus 100 nanoseconds of yours for nothing.
O'
========================================
Dust in the wind, all we are is dust in the wind.
Same old song, just a drop of water in the open sea
all my dreams - fade before my eyes a curiousity,
dust in the wind, all we are is dust in the wind
=========================================
===
Subject: Re: Action Device: Destination Alpha Centauri A
[snip]
> O'
> ========================================
> Dust in the wind, all we are is dust in the wind.
> Same old song, just a drop of water in the open sea
> all my dreams - fade before my eyes a curiousity,
> dust in the wind, all we are is dust in the wind
> =========================================
=================================
baazeechaa-e-atfaal hai duniya mere aage
hota hai shab-o-roz tamaasha mere aage
ik khel hai auraNg-e-sulemaaN mere nazdeek
ik baat hai Ôeijaz-e-maseeha mere aage
juz naam naheeN soorat-e-aalam mujhe manzoor
juz waham naheeN hastee-eiya mere aage
hota hai nihaaN gard meiN sehara mere hote
ghisata hai jabeeN KHaak pe dariya mere aage
mat pooch ke kya haal hai mera tere peeche ?
too dekh ke kya rang tera mere aage
sach kahte ho, KHudbeen-o-KHud_aaraa na kyoN hooN ?
baiTha hai but-e-aainaa_seemaa mere aage
fir dekhiye andaaz-e-gul_afshaani-e-guftaar
rakh de koee paimaanaa-o-sahba mere aage
nafrat ka gumaaN guzare hai, maiN rashk se guzaraa
kyoN kar kahooN, lo naam na uska mere aage
imaaN mujhe roke hai jo khiNche hai mujhe kufr
ka'aba mere peeche hai kaleesa mere aage
aashiq hooN, pe maashooq_farebee hai mera kaam
majnooN ko bura kehti hai laila mere aage
KHush hote haiN par wasl meiN yoN mar naheeN jaate
aayee shab-e-hijaraaN ki tamanna mere aage
hai mauj_zan ik qulzum-e-KHooN, kaash, yahee ho
aata hai abhee dekhiye kya-kya mere aage
go haath ko jumbish naheeN aaNhoN meiN to dam hai
rehne do abhee saaGHar-o-meena mere aage
ham_pesha-o-ham_masharb-o-ham_raaz hai mera
ÔGHalib' ko bura kyoN kaho achchaa mere aage !
=================================
-Abhi.
===
Subject: Re: Action Device: Destination Alpha Centauri A
Message-Id: <1075406894.32179.0@lotis.uk.clara.net>
Abhi  schrieb im Newsbeitrag
> It seems that you have missed my earlier post. I have this
device
almost
> ready on ßoor of my room. I can test it right now. Not
necessary to
hang
> the things in air at this moment I can test it on ßoor. If
this
device
> pushes itself and body only in one direction, Houston has a
problem.
If not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret of
> friction..
> Can you explain this?
Friction will cause the part in contact with the ßoor
to move less than parts not in contact. This test would
therefore prove nothing.
> I mean if device pushes body and itself in space only
> in one direction, what it has to do with friction? Tha same
> thing can happen on ßoor of my room.
The device will move on your ßoor but not in space.
Release a compressed spring on the ßoor and it will ßy
up but fall back. To show it will work in space, you must
get it to hover unsuppor or ßy up to the ceiling and
stay there as you claimed it would.
George
===
Subject: Re: Action Device: Destination Alpha Centauri A
charset=iso-8859-1
>
Abhi  schrieb im Newsbeitrag
> It seems that you have missed my earlier post. I have this
device
almost
> ready on ßoor of my room. I can test it right now. Not
necessary to
> hang
> the things in air at this moment I can test it on ßoor. If
this
device
> pushes itself and body only in one direction, Houston has a
problem.
If
> not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret of
> friction..
Can you explain this? I mean if device pushes body and itself
in space
only
> in one direction, what it has to do with friction? Tha same
thing can
happen
> on ßoor of my room.
They are not the same thing. Anyone can crawl on the ßoor.
Take a picture as soon as your device hangs in the air.
Lots of Greetings!
Volker
===
Subject: Re: Action Device: Destination Alpha Centauri A
> Abhi  schrieb im Newsbeitrag
 Abhi  schrieb im Newsbeitrag
It seems that you have missed my earlier post. I have this
device
almost
ready on ßoor of my room. I can test it right now. Not
necessary
to
hang
the things in air at this moment I can test it on ßoor. If
this
device
pushes itself and body only in one direction, Houston has a
problem.
If
not,
I have a problem.
> If it pushes itself in one direction you have discovered
the secret
of
friction..
Can you explain this? I mean if device pushes body and itself
in space
only
in one direction, what it has to do with friction? Tha same
thing can
happen
on ßoor of my room.
> They are not the same thing. Anyone can crawl on the ßoor.
> Take a picture as soon as your device hangs in the air.
God is very dramatic. Years of efforts depend only on those
few magical
moments.
-Abhi.
============================
ham ko junuu.N kyaa sikhalaate ho ham the pareshaa.N tumase
ziyaadaa
chaak kiye hai.n hamane aziizo.n chaar garebaa.N tumase
ziyaadaa
chaak-e-jigar muhataaj-e-rafuu hai aaj to daaman sirf lahuu
hai
ek mausam thaa ham ko rahaa hai shauq-e-bahaaraa.N tumase
ziyaadaa
jaao tum apanii baam kii Khaatir saarii lave.n shamo.n kii
katar lo
zaKhmo.n ke mahar-o-maah salaamat jashn-e-chiraaGaa.N tumase
ziyaadaa
ham bhii hameshaa qatl hue our tum ne bhii dekhaa duur se
lekin
ye na samajhe hamako huaa hai jaan kaa nukasaa.N tumase
ziyaadaa
za.njiir-o-diivaar hii dekhii tumane to Majrooh magar ham
kuuchaa-kuuchaa dekh rahe hai.n aalanm-e-zi.ndaa.N tumase
ziyaadaa
===============================
===
Subject: Re: Action Device: Destination Alpha Centauri A
charset=iso-8859-1
>
Abhi  schrieb im Newsbeitrag
>
Abhi  schrieb im Newsbeitrag
> It seems that you have missed my earlier post. I have this
device
> almost
> ready on ßoor of my room. I can test it right now. Not
necessary
to
> hang
> the things in air at this moment I can test it on ßoor. If
this
> device
> pushes itself and body only in one direction, Houston has a
problem.
> If
> not,
> I have a problem.
If it pushes itself in one direction you have discovered the
secret
of
> friction..
 Can you explain this? I mean if device pushes body and
itself in
space
> only
> in one direction, what it has to do with friction? Tha same
thing can
> happen
> on ßoor of my room.
They are not the same thing. Anyone can crawl on the ßoor.
Take a picture as soon as your device hangs in the air.
God is very dramatic. Years of efforts depend only on those
few magical
> moments.
There is no god. There was no effort. You are the only one
trying to create
drama here.
Greetings!
Volker
===
Subject: Re: Action Device: Destination Alpha Centauri A
 [snippety Ôgain]
===================================================
manzil na de charaaG na de hausalaa to de
tinke kaa hii sahii tu magar aasaraa to de
mai.n ne ye kab kahaa ke mere haq me.n ho javaab
lekin Khaamosh kyuu.N hai tuu ko_ii faisalaa to de
baraso.n mai.n tere naam pe khaataa rahaa fareb
mere Khudaa kahaa.N hai tuu apanaa pataa to de
beshak mere nasiib pe rakh apanaa iKhtiyaar
lekin mere nasiib me.n kyaa hai bataa to de
============================================
> Hmm, seems kind of pointless posting in a language the other
> correspondent
> doesn't understand, doesn't it. Guess that 
defeats the
purpose of
> communications, but at least there is no expectation that
the other
> party
> understands.
 On the other hand, for those that won't listen, I guess
talking in
> another
> language is about as useful as not talking at all.
 You should build your action device. If you can't, then
(for you at
least)
> it is an unobtainable goal and all of your rhetoric is just
a lot of
> wind.
It seems that you have missed my earlier post. I have this
device
almost
ready on ßoor of my room. I can test it right now. Not
necessary to
hang
the things in air at this moment I can test it on ßoor. If
this device
pushes itself and body only in one direction, Houston has a
problem. If
> not,
I have a problem.
Till that moment, I am just talking to God ......
=========================
Ye Khudaa....
Aaa Sikander-E-Hunar Aaazmayain
Tu Teer Aaazmaaa, Hum Jigar Aaazmayain
=========================
What I am pasting is ghazals, shers of Urdu Poetry. Perhaps
you know,
> there
is lot of similarity in Hindi and Urdu. So we Hindi speaking
Indian
people
are fond of Urdu poetry.
-Abhi.
> Abhi,
> Considering the simplicity of your device, and your
incredible zeal (at
> least to talk about it), doesn't it seem like 
you're
stretching this out
a
> little long? I mean, at 4pm you could decide to do it, at
4:10pm you
could
> finish building it and at 4:15pm you could take it out for a
spin!
OK.
(1) Stand up and walk ten meters on ground. You can do it in
few seconds.
(2) Now everything is burning before you over ten meter path.
Can you walk
on this fire path for few seconds?
> I can see how you took seventeen years to get this far, I
suspect it will
> take you at least a year to drink a glass of water at the
pace you're
> moving.
This Pilot dream star 17 years ago. This Big Bang episode
star 15
years ago. But actual Physics Theory development star only in
November
1999. And working over this idea of Action Device star in
last week of
September 2002. Day and nights, I have spent non-stop 16
months of my life
on this idea.
It is just like landing of rovers on some planet. Slight
mistake at the
time
of landing can blow up efforts of hundreds of people
alongwith millions of
dollars. It just takes few seconds for this rover to actually
descend on
some planet.
If this descend turn out to be like Opportunity, very good.
If it turn
out
to be like Spirit, still OK.
But if it turn out to be like Beagle-2, then what?
> I'm afraid that Urdu poetry doesn't do it 
for me, which is
why I sent you
a
> few japanese rhymes.
=====================
Galat nahi ke zamane main lajawaab hain hum
Samajh sake na koi, woh kitaab hai hum
=====================
> O'
> PS: Should I be looking out my window sometime this decade
for you to
ßy
> by?
There are some things in our life or physics where efforts,
logic, math,
physics is not enough. Pythagoras, Newton can not help you.
When scales are
evenly balanced, at that time, you feel His awesome power.
I have read in stories about peoples who have gone through
such situations
and talking about such things. Now, it is my turn. I am
feeling His
awesome
power.
-Abhi.
======================
Tere Aazad Bande Ki, Na Yeh Duniya Na Woh Duniya
Yahan Jeene Ki Pabandi, Wahan Marne Ki Pabandi
======================
===
Subject: Special doubleperiodic function in R^2
X-ID: EGeyLiZEge6hHLtkTw7G0YKeZOg6KQlsT6fcgh7z5Hl2Bw7qBn+dUz
I'm looking for a real analytic function f: |R^2 -> |R such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
(b) f(x,y) <> 0 everywhere else
(c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
Is there a function like this? I have my doubts. If one
imposes weaker
conditions on f like continuity, C^r or C^infty: Is it
possible to
construct a function like this?
I tried - of cource - functions like f(x) =sin(Pi*x)sin(Pi*y)
or
sin(Pi*x)+sin(Pi*y), but they don't satisfy condition (b)
Thomas
--
Thomas
===
Subject: Re: Special doubleperiodic function in R^2
> I'm looking for a real analytic function f: |R^2 -> |R 
such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
> (b) f(x,y) <> 0 everywhere else
> (c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
Is there a function like this? I have my doubts. If one
imposes weaker
> conditions on f like continuity, C^r or C^infty: Is it
possible to
> construct a function like this?
I tried - of cource - functions like f(x) =sin(Pi*x)sin(Pi*y)
or
> sin(Pi*x)+sin(Pi*y), but they don't satisfy condition (b)
Thomas
f(x) =(sin(Pi*x))^2 + (sin(Pi*y))^2
===
Subject: Re: Special doubleperiodic function in R^2
> I'm looking for a real analytic function f: |R^2 -> |R 
such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
> (b) f(x,y) <> 0 everywhere else
> (c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
Is there a function like this? I have my doubts. If one
imposes weaker
> conditions on f like continuity, C^r or C^infty: Is it
possible to
> construct a function like this?
I tried - of cource - functions like f(x) =sin(Pi*x)sin(Pi*y)
or
> sin(Pi*x)+sin(Pi*y), but they don't satisfy condition (b)
f(x, y) = |sin(x/pi)| + |sin(y/pi)| satisfies (a), (b) and
(c), I
guess, but it is only continuous. Maybe combining it with a
small
exponential function near the lattice points give a C^1
function.
Just my 2 cents,
Duran Castore (duran_castore@yahoo.com)
===
Subject: Re: Special doubleperiodic function in R^2
>I'm looking for a real analytic function f: |R^2 -> |R such
that:
>(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
>(b) f(x,y) <> 0 everywhere else
>(c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
>Is there a function like this?
Certainly. It is equivalent to construct a real-analytic
function g:T -> R on the torus T = R^2/(2Z)^2 such that
g is 0 at exactly one point of T (without loss of generality,
the point of T corresponding to (0,0)); then the composition
f = g o p, where p is the covering projection R^2 -> T,
will have properties (a), (b), and (c).
One such g is the square of the distance from (0,0) mod (2Z)^2
to (x,y) mod (2Z)^2, where distance on T is to be defined
in terms of a real-analytic Riemannian metric (of which there
is a large supply).
There are many alternative constructions. I am not up enough
on elliptic functions to give any explicit formulas, but I'm
sure they're easy enough to find.
Lee Rudolph
===
Subject: Re: Special doubleperiodic function in R^2
I'm looking for a real analytic function f: |R^2 -> |R such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
> (b) f(x,y) <> 0 everywhere else
> (c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
>
Elliptic functions might be helpful in this !?!?
By design they are complex valued functions of a complex
variable,
but you can identify R^2 with C as the domain, and do some
tricks
with the range (apply squared complex absolute value or
someting
like that). Elliptic functions will have poles, but you can
surely
level those off.
You probably can't do better than C^infinity 
this way, I'm
afraid.
Jyrki Lahtonen, Turku, Finland
===
Subject: Re: Special doubleperiodic function in R^2
I'm looking for a real analytic function f: |R^2 -> |R such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
>> (b) f(x,y) <> 0 everywhere else
>> (c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
>Elliptic functions might be helpful in this !?!?
> By design they are complex valued functions of a complex
variable,
> but you can identify R^2 with C as the domain, and do some
tricks
> with the range (apply squared complex absolute value or
someting
> like that). Elliptic functions will have poles, but you can
surely
> level those off.
Blimey, why bother with elliptic functions?
Won't something like sin^2(pi x) + sin^2(pi y) work much
easier?
--
===
Subject: Re: Special doubleperiodic function in R^2
> Blimey, why bother with elliptic functions?
> Won't something like sin^2(pi x) + sin^2(pi y) work much
easier?
Good question! May be it was the term Ôdoubly 
periodic' that
sent
me into a wrong direction.
Jyrki
===
Subject: Re: Special doubleperiodic function in R^2
X-ID: GlkPniZVYeyVG+tdv4IJXFvj8okgTdEnYH4NNXBRPB9x+C4jlsCrwZ
Robin Chapman  schrieb:
>Blimey, why bother with elliptic functions?
>Won't something like sin^2(pi x) + sin^2(pi y) work much
easier?
Sorry. It seems I produced a lot of hot wind... I figured that
out
myself almost the same moment I pos my question.
Thomas
--
Thomas
===
Subject: Re: Special doubleperiodic function in R^2
X-ID: ThJzXOZOZeaEyMybNKfrqCG3ci9+i4TLSOOqossF2PcmnwD5uHYWZ1
Jyrki Lahtonen  schrieb:
I'm looking for a real analytic function f: |R^2 -> |R such
that:
(a) f(x,y) = 0 if (x,y) in ZxZ (integer lattice)
>> (b) f(x,y) <> 0 everywhere else
>> (c) f(x+2k,y+2m) = f(x,y) for all (k,m) in ZxZ
>Elliptic functions might be helpful in this !?!?
>By design they are complex valued functions of a complex
variable,
>but you can identify R^2 with C as the domain, and do some
tricks
>with the range (apply squared complex absolute value or
someting
>like that). Elliptic functions will have poles, but you can
surely
>level those off.
>You probably can't do better than C^infinity 
this way, I'm
afraid.
I was thinking like that too. But what about f(x) =
sin^2(Pi*x) +
sin^2(Pi*y) ? Can't believe, that I didn't see 
this at once...
Jyrki Lahtonen, Turku, Finland
--
Thomas
===
Subject: Catalan Generator, anyone?
I need a string generator program which produces
all valid bracket clustei.e.
a
a*a
(a*a)*a a*(a*a)
(a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
you get the idea.
One could drop the a and ) and insert (*
recursively
(*
((** (*(* etc.
which would need weeding out dupes. Or generate
the above string by voters lead.
Anyone have a more elegant solution so I don't have
to reinvent the wheel?
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
Subject: Re: Catalan Generator, anyone?
===
> I need a string generator program which produces
> all valid bracket clustei.e.
> a
> a*a
> (a*a)*a a*(a*a)
> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
> you get the idea.
> One could drop the a and ) and insert (*
> recursively
> (*
> ((** (*(* etc.
> which would need weeding out dupes. Or generate
> the above string by voters lead.
Anyone have a more elegant solution so I don't have
> to reinvent the wheel?
>
Why not think of it as a permutation on {1,2,3,...,n-1} where
n = the
number of a's? This would correspond to which * you wish to
perform
first, second, third, etc. Then analyze the permutation and
determine
how to insert the parenthesis to correspond with that order.
Of course, I'm making it sound easier than it might be to
implement.
--
===
Subject: Re: Catalan Generator, anyone?
>> I need a string generator program which produces
>> all valid bracket clustei.e.
>> a
>> a*a
>> (a*a)*a a*(a*a)
>> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
>> you get the idea.
>> One could drop the a and ) and insert (*
>> recursively
>> (*
>> ((** (*(* etc.
>> which would need weeding out dupes. Or generate
>> the above string by voters lead.
>> Anyone have a more elegant solution so I don't have
>> to reinvent the wheel?
> Why not think of it as a permutation on {1,2,3,...,n-1}
where n = the
> number of a's? This would correspond to which * you wish 
to
perform
> first, second, third, etc. Then analyze the permutation and
determine
> how to insert the parenthesis to correspond with that order.
I don't understand how you could easily turn a permutation
into the kind
of bracketing scheme that is needed; how would you represent
(a*((a*a)*a))?
--
===
Subject: Re: Catalan Generator, anyone?
> I need a string generator program which produces
> all valid bracket clustei.e.
> a
> a*a
> (a*a)*a a*(a*a)
> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
> you get the idea.
> One could drop the a and ) and insert (*
> recursively
> (*
> ((** (*(* etc.
> which would need weeding out dupes. Or generate
> the above string by voters lead.
>
You have answered the question in your title. The number of
ways of
bracketing a string on n terms is the n-1st Catalan number,
which is
(2n-2 choose n-1)/n. So for 2 things, you get (2 choose 1)/2
= 1, for
3 you get (4 choose 2)/3 = 2 and for four, (6 choose 3)/4 = 5
and so
on. You asked about generating functions. Well, let a(n) be
the
number of bracketings and
f(x) the sum of a(n)x^n. Since every bracketing starts by
choosing a
number i strictly between 0 and n and bracketing the first i
and the
last n-i. This leads to the equation a(n) is the sum from i
to n-1 of
a(i)a(n-1) and this gets you an equation satisfied by f,
something
like
f(x)^2 + 1 = f(x) (or something similar, it is easy to work
out).
This is a quadratic in f, which can be solved for f in closed
form,
involving a square root. Using the Taylor expansion for the
square
root finally gets you the solution. It is a beautiful example
of
generating functions.
===
Subject: Re: Catalan Generator, anyone?
> I need a string generator program which produces
> all valid bracket clustei.e.
> a
> a*a
> (a*a)*a a*(a*a)
> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
> you get the idea.
Actually, I don't quite get the idea from your examples.
Just noting that in any initial segment, the number of
left brackets must be greater than or equal to the number
of right brackets quickly leads to a solution of the
problem I think you're asking about, so maybe I 
don't get
what you're asking.
===
Subject: Re: Catalan Generator, anyone?
Hauke Reddmann a .8ecrit:
> I need a string generator program which produces
> all valid bracket clustei.e.
> a
> a*a
> (a*a)*a a*(a*a)
> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
> you get the idea.
> One could drop the a and ) and insert (*
> recursively
> (*
> ((** (*(* etc.
> which would need weeding out dupes. Or generate
> the above string by voters lead.
Anyone have a more elegant solution so I don't have
> to reinvent the wheel?
>
Dichoy algorithm should work
Initialisation list_of_1word = a
nword = (pword)*(qword) forall choices of p+q = n and 1 I need a string generator program which produces
> all valid bracket clustei.e.
> a
> a*a
> (a*a)*a a*(a*a)
> (a*(a*(a*a))) (((a*a)*a)*a) (a*a)*(a*a)...
> you get the idea.
> One could drop the a and ) and insert (*
> recursively
> (*
> ((** (*(* etc.
> which would need weeding out dupes. Or generate
> the above string by voters lead.
I don't know what voters lead is, but here's 
how I implemen
it in
scheme:
Given a number n, generate all the pairs (p,q) so p+q=n, then
for each
pair,
find the set of bracketings (using recursion) that divides the
string into
p a's on one side, and q a's on the other. 
Merge all these
bracketings
in SICP style scheme in about 20 minutes (10 to get the
algorithm), if
you'd like to see it let me know. The C version would
probably be just as
easy.
===
Subject: When was zero inven?
In textbooks one finds a suggestion that in Europe zero was
unknown until
around AD 1000. According to this story zero was inven in
India some
time earlier.
This sounds a bit improbable, a half truth, at best. I mean,
naturally
humans must have known the idea of nothing since ages and
even conduc
calculatuions using it (We got three rabbits and ate three
and have now
none, null.
I'd be surprised if zero had not exis in the same way on
tablets or
paprys of, say, a Greek olive merchant. He must have made
inventories to
know when he'd run out of this or that. The simpliest way
would be to have
a column indicating the number of items in stock on the top
and then
reductions below until the item was sold out. The last number
would be
zero, of course, or some symbol indicating nothing.
Anyway, my question is: what is exactly meant by zero was not
known until
it was inven in India?
i
===
Subject: Re: When was zero inven?
There has been lots of groundless speculation in this thread.
How about some actual references to authoritative sources?
Was zero inven or discovered?
===
Subject: Re: When was zero inven?
> There has been lots of groundless speculation in this
thread.
> How about some actual references to authoritative sources?
Bibhutibhusan Datta, Early history of the arithmetic of zero
and infinity
in India, _Bulletin of the Calcutta Mathematical Society_
18:4 (1927)
165-176.
> Was zero inven or discovered?
Yes.
David
===
Subject: Re: When was zero inven?
> This sounds a bit improbable, a half truth, at best....
Applies to all hindsights, much so for all historical ones. A
lot of
research went into this already.
> I'd be surprised if zero had not exis in the same way on
tablets or
> paprys of, say, a Greek olive merchant. He must have made
inventories to
> know when he'd run out of this or that. The simpliest way
would be to have
> a column indicating the number of items in stock on the top
and then
> reductions below until the item was sold out. The last
number would be
> zero, of course, or some symbol indicating nothing.
Thinking in forward and backward directions are two different
things,
the latter is integrative and synthetic ordering, while the
former is
just only analytic.
However to be very honest and to be agreement with you, I was
so
surprised to find many examples, e.g., Apollonius of Perga
(262 to
~190 BC) make such advanced discoveries in Geometry as the
Circle
named after him using bi-polar coordinates. Mind you,
analytical
geometry came way much later. (Ren.8e Descartes (1596-1650)].
http://www-gap.dcs.st-and.ac.uk/~history/Curves/Circle.html
Has this happened without a proper numbering system( which
also means
a measuring system in land measurement) put in place and into
daily
use ?
The Indians and/or Mayans may have used their fingers of the
hands
upto 10 ( Digital has thus derived its name ) to assign place
value
and so on.
But, the ancient Greeks have thought in an Analog way, not
dealing
with precisely quantified/quantized lumps but using lumps that
are
perceived relatively (differentially) distinguished from each
other in
magnitude. So it must have been certainly a more creative
endeavour.
Viewed this way, the impact of the modern digital computer
could be
viewed as negative.
Anyone care to comment?
Narasimham
===
Subject: Re: When was zero inven?
>Subject: Re: When was zero inven?
>Message-id: <676dc11a.0401291451.193d3c85@posting.google.com> This sounds a bit improbable, a half truth, at best....
>Applies to all hindsights, much so for all historical ones.
A lot of
>research went into this already.
>> I'd be surprised if zero had not exis in the same way on
tablets or
>> paprys of, say, a Greek olive merchant. He must have made
inventories to
>> know when he'd run out of this or that. The simpliest way
would be to
have
>> a column indicating the number of items in stock on the
top and then
>> reductions below until the item was sold out. The last
number would be
>> zero, of course, or some symbol indicating nothing.
>Thinking in forward and backward directions are two
different things,
>the latter is integrative and synthetic ordering, while the
former is
>just only analytic.
>However to be very honest and to be agreement with you, I
was so
>surprised to find many examples, e.g., Apollonius of Perga
(262 to
>~190 BC) make such advanced discoveries in Geometry as the
Circle
>named after him using bi-polar coordinates. Mind you,
analytical
>geometry came way much later. (Ren.8e Descartes (1596-1650)].
>http://www-gap.dcs.st-and.ac.uk/~history/Curves/Circle.html
>Has this happened without a proper numbering system( which
also means
>a measuring system in land measurement) put in place and
into daily
>use ?
>The Indians and/or Mayans may have used their fingers of the
hands
>upto 10 ( Digital has thus derived its name ) to assign
place value
>and so on.
>But, the ancient Greeks have thought in an Analog way, not
dealing
>with precisely quantified/quantized lumps but using lumps
that are
>perceived relatively (differentially) distinguished from
each other in
>magnitude. So it must have been certainly a more creative
endeavour.
>Viewed this way, the impact of the modern digital computer
could be
>viewed as negative.
>Anyone care to comment?
What about the abacus? That's not analog.
>Narasimham
--
===
Subject: Re: When was zero inven?
charset=iso-8859-7
===
Ì i A 
ó.8d.98.87.8b.8c
.97.99.95
.92.86.94.9d.92.87
> In textbooks one finds a suggestion that in Europe zero was
unknown until
> around AD 1000. According to this story zero was inven in
India some
> time earlier.
> This sounds a bit improbable, a half truth, at best. I
mean, naturally
> humans must have known the idea of nothing since ages and
even conduc
> calculatuions using it (We got three rabbits and ate three
and have
now
> none, null.
Zero was inven by primitive man, after a successful
intercourse with his
first female companion.
[snip]
> i
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: When was zero inven?
> In textbooks one finds a suggestion that in Europe zero was
unknown until
> around AD 1000. According to this story zero was inven in
India some
> time earlier.
And independently in Central America, the Maya or their
predecessors.
--
http://www.math.ohio-state.edu/~edgar/
===
Subject: One calculus problem .. help needed
Please help me solving the following calculus problem,
Let Omega be an open, bounded subset of R^4 , let u= (u_1,
u_2, u_3,
u_4) : Omega .9a R^4 be a function vanishing on the boundary
such that
1) partial derivative of u_1 with respect to x_2 = partial
derivative
of u_2 with respect to x_1
2) partial derivative of u_1 with respect to x_3 = partial
derivative
of u_3 with respect to x_1
3) partial derivative of u_2 with respect to x_3 = partial
derivative
of u_3 with respect to x_2
The question is, if the Curl of u= (u_1, u_2, u_3, u_4) is
zero?
I don't what is the answer. I guess Curl of u= (u_1, u_2,
u_3, u_4)
need not be zero. But I am unable to construct an example.
Any help in
welcome.
Thank you in advanceá
===
Subject: Types of proof
I was going to provide a list of the various types of proof
that can
be usefully and successfully applied, especially when neither
you (the
all encompassing you) nor the audience knows what's going 
on.
Unfortunately, the University of Western Australia beat me to
it:
http://www.maths.uwa.edu.au/~berwin/teach/M101/
invalid.proofs.html
This provides a good basis for building up proofs of unsolved
problems, or simple proofs for problems solved with ugly
things like
elliptical curves. I'd recommend this site to anyone 
seriously
considering going into (pseudo-)mathematics.
But the university missed one of the most important USENET
methods of
proof:
Proof by Repetition. In the months that I've been lurking in
this
group I've seen some really sophistica uses of this method.
And the
results are fantastic! 0=1, 1 /= 1, reals are countable, the
rationals
are uncountable, the algebraic integers are too small (I was
always
supicious about that!). And this is just the tip of the
cupcake!
It also has a very long history. The great mathematician
Charles
Dodgeson (he was great; he was a mathematician, so ...)
summarised the
method in his epic poem The Hunting of the Snark
JUST the place for a Snark! the Bellman cried,
As he landed his crew with care;
Supporting each man on the top of the tide
By a finger entwined in his hair.
Just the place for a Snark! I have said it twice:
That alone should encourage the crew.
Just the place for a Snark! I have said it thrice:
What I tell you three times is true.
It's too bad all the crew found was a boojum, despite this
very
convincing proof. I think a number of people (4 come to mind)
have
been hunting for snarks, and all they ever find are boojums.
(But now
I wane osophical)
RickO
===
Subject: Re: Types of proof
>I was going to provide a list of the various types of proof
that can
>be usefully and successfully applied, especially when
neither you (the
>all encompassing you) nor the audience knows what's going 
on.
>Unfortunately, the University of Western Australia beat me
to it:
>http://www.maths.uwa.edu.au/~berwin/teach/M101/
invalid.proofs.html
>This provides a good basis for building up proofs of unsolved
>problems, or simple proofs for problems solved with ugly
things like
>elliptical curves. I'd recommend this site to anyone
seriously
>considering going into (pseudo-)mathematics.
I think they missed a few:
Proof by Powerful Theorems
Write down the proposition, then jot down a list of highly
powerful
and/or complica theorems in sequential order and claim that
the end
result follows due to all of these. No one will dare question
the
theorems you ci. Escape before someone inquires what the
theorems
have to do with each other or the actual problem.
Proof by Leap-of-logic
Make general statements that have little to do with the
proposition,
then make a wild jump to completely unrela conclusions.
Favorite of
JSH's.
Proof by Infinite Induction
Show that since something holds for all cases where n=1,2,...
then it
must also hold for infinite cases.
Proof by Hidden Special Case
Midway through the proof introduce some supposed identity out
of
nowhere and pretend like it holds for all values instead of
just the
values you are trying to prove hold.
Proof by Indexing
Use as many different indices as possible. No one will notice
when you
swap a few around at a critical moment.
Proof by Conspiracy Theory
You are right but the establishment is trying to silence you
from
publishing your result and disturbing the status quo.
>Proof by Repetition. In the months that I've been lurking 
in
this
>group I've seen some really sophistica uses of this method.
And the
>results are fantastic! 0=1, 1 /= 1, reals are countable, the
rationals
>are uncountable, the algebraic integers are too small (I was
always
>supicious about that!). And this is just the tip of the
cupcake!
I think it's closely rela to Proof by Conspiracy Theory.
--
I'm not interes in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Square root extraction
X-Complaints-Info: Please send any complaints about misuse to:
abuse@newcastle.ac.uk
I'm trying to compare two versions of Fermat's 
integer
factoring
method. (Still both O(N) time, I'm afraid.)It all basically
comes down
to how fast the root of an integer can be extrac, (so the
conventional version can check if that root is an integer).
For
arbitrary precision numbers (eg too big for a processor to
handle in
hardware in constant time):
1.) I've heard Newton's method takes O(log N) 
time, where N
is the
number we want the root of. Is this correct?
2.) Is there a method that can do it in less than O(log N)
time for
any integer? If yes, what is it called, and how much time
does it
need/where can I find out how much time it needs.
3.) A root-extractor known as the Karatsuba Square Root
algorithm
referred to numbers with n limbs - what are limbs in this
context?
Thanks.
===
Subject: Re: Square root extraction
I'm trying to compare two versions of Fermat's 
integer
factoring
> method. (Still both O(N) time, I'm afraid.)It all 
basically
comes down
> to how fast the root of an integer can be extrac, (so the
> conventional version can check if that root is an integer).
Bad bad bad.
What you want to do is check if the number is an exact square.
If it is then you are interes in its square root, but you
aren't otherwise.
Can 8n+5 ever be an exact square, for integer n?
Generalise.
--
Unpatched IE vulnerability: dragDrop invocation
Description: Arbitrary local file reading through native
Windows dragDrop invocation.
Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0302/12.html
Exploit: http://kuperus.xs4all.nl/security/ie/xfiles.htm
===
Subject: Re: Square root extraction
X-Complaints-Info: Please send any complaints about misuse to:
abuse@newcastle.ac.uk
 wrotE:
I'm trying to compare two versions of Fermat's 
integer
factoring
>> method. (Still both O(N) time, I'm afraid.)It all
basically comes down
>> to how fast the root of an integer can be extrac, (so the
>> conventional version can check if that root is an integer).
>Bad bad bad.
>What you want to do is check if the number is an exact
square.
>If it is then you are interes in its square root, but you
>aren't otherwise.
>Can 8n+5 ever be an exact square, for integer n?
>Generalise.
>
Bearing in mind this is the x^2-y^2=N version of Fermat, and
relying
on modular arithmetic:
(8n+1)^2=8m+1
(8n+2,4,6,0)^2 would be an even number.
(8n+3)^2 = 8m+9 = 8(m+1) +1
(8n+5)^2 = 8m+25 = 8m+24 +1 =8(m+3)+1
(8n+7)^2 = 8m+49 = 8m+48+1 = 8(m+6)+1
So we see that for integer n 8n+5 cannot be an exact square.
However,
I am already taking this into account - it's just that there
is a
limit to how many impossibles like 8n+5 a program for
Fermat's method
can filter out. For one method, I still have to test whether
every
number I haven't been able to skip over like this is a
perfect square,
and that means a root extractor. I must also point out that
you are
incorrect about If it is an exact square you are interes in
it's
square root but not otherwise UNLESS you are going to tell me
a
method for telling if a number is an exact square that's
faster than
root extraction. I know that the correct value of (Q-P)/2 is
the first
perfect square I'll encounter through this method(I 
don't
have the
proof to hand, but it's not hard.) and I'm 
trying to find out,
using
root extraction, if each possible value of that is a perfect
square.
eg (pseudocode)
z=sqrt(y);
if z=int(z) then (y was the perfect square sought) else (it
wasn't -
set up next y and test again.)
Furthermore, both variations are using the same skip and so
it all
comes down to how fast each of the Z root extractions vs. the
(greater
than Z) subtractions and extra comparisons will be so I can
work out
which is the faster method. I'm not seriously considering
using
Fermat's method in an application, but I would like to be
able to make
an accurate analysis. Current analysis suggests that the
second
version mentioned is, if a root extractor takes O(log N)
time, faster
if the factors are not both very near the root,(accentua by
the
presence of more operations like division in Newton's.) but
Newton may
not be the fastest root extractor known.
To read a description of the second version (albeit
unoptimised by
methods like the 8n+5 check - there is mention of a further
improvement to this by Gauss on the page too but I'm eager 
to
get this
compared to the first version before I examine that) go to
http://www.frenchfries.net/paul/factoring/theory/fermat.html
The other, again before I apply the 8n+5 style checks, which
barely
affect comparison between the methods, basically uses the
pseudocode
above - we increment y by a certain x and x by 2 each time.
(x_init=2*ceil(sqrt(N))+1).
===
Subject: Re: Square root extraction
> I'm trying to compare two versions of 
Fermat's integer
factoring
> method. (Still both O(N) time, I'm afraid.)It all 
basically
comes down
> to how fast the root of an integer can be extrac, (so the
> conventional version can check if that root is an integer).
[quoth me]
Can 8n+5 ever be an exact square, for integer n?
Generalise.
[quoth you again]
> So we see that for integer n 8n+5 cannot be an exact
square. However,
> I am already taking this into account - it's just that
there is a
> limit to how many impossibles like 8n+5 a program for
Fermat's method
> can filter out. For one method, I still have to test whether
every
> number I haven't been able to skip over like this is a
perfect square,
> and that means a root extractor.
In that case you didn't give us the full details in advance.
If you'd have said I quickly reject quadratic non-residues
modulo
small primes, then you'd not have received the same reply.
> I must also point out that you are
> incorrect about If it is an exact square you are interes in
it's
> square root but not otherwise UNLESS you are going to tell
me a
> method for telling if a number is an exact square that's
faster than
> root extraction.
Must you? I'm not incorrect, if you look at mean execution
time.
Rejecting quadratic non-residues modulo small primes has a
mean
execution time that is both faster for small numbers and also
asymptotically faster than not doing it. Its worst-case time
is
of course marginally higher as it falls back on finding the
square root, but the worst-case scenario occurs for a very
small
proportion of numbers. It's possible to reject about 99% of
candidates in a time no greater than add-time on most
processors
Put it another way - if it were not faster, why are you doing
it?
--
Unpatched IE vulnerability: WMP local file bounce
Description: Switching security zone, arbitrary command
execution, auatic
email-borne command execution
Reference:
http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307&
L=ntbugtraq&F=P&
S=&P=6783
Exploit: http://www.malware.com/once.again!.html
Subject: Re: Square root extraction
X-Complaints-Info: Please send any complaints about misuse to:
abuse@newcastle.ac.uk
===
Actually, having given the matter more thought, regardless of
factoring it would be nice to know if any known method for
root
extraction is faster than Newton's, so I could write a few
good
arbitrary precision root extractor functions for some of the
older
compilers I use(some of which don't have arbitrary precision
built in)
(and perhaps a better Ônormal' square root 
extractor than
their
current ones.)
===
Subject: Re: Square root extraction
> Actually, having given the matter more thought, regardless
of
> factoring it would be nice to know if any known method for
root
> extraction is faster than Newton's, so I could write a few
good
> arbitrary precision root extractor functions for some of
the older
> compilers I use(some of which don't have arbitrary
precision built in)
> (and perhaps a better Ônormal' square root 
extractor than
their
> current ones.)
Use Newton, but do not use more precision than is absolutely
necessary at any particular stage. That way you only ever do
one full-size operation.
See Bernstein's papers on detecting perfect powers in
essentially
linear time.
--
Unpatched IE vulnerability: Extended HTML Form Attack
Description: C Site Scripting through non-HTTP ports,
stealing cookies, etc.
Published: February 6th 2002
Reference:
http://eyeonsecurity.org/advisories/
multple-web-browsers-vulnerable-to-extend
ed-form-attack.htm
===
Subject: Re: Factoring Gaussian integers
> I'm developing an algorithm for factoring Gaussian
integers. In the case
p = 1 mod 4, p = a^2 + b^2, p | norm(z), p a prime,
then either a + b*i or b + a*i divides z. I'm wondering if
there's a test
as
> to which is the factor. Doing some calculations, it star to
look like
it
> only depended on which octant z was in, but I guess it's
not that simple.
If p = 3 + 2i, say, and Z = m+ni
then m + ni == 0 mod 3 +2i
so i == -mn^(-1) = Q mod 3+2i
if 3 +2Q = 0 mod 13 then 3+2i
is the prime factor otherwise
3-2i
===
Subject: Re: Factoring Gaussian integers
:> I'm developing an algorithm for factoring Gaussian
integers. In the case
:>
:> p = 1 mod 4, p = a^2 + b^2, p | norm(z), p a prime,
:>
:> then either a + b*i or b + a*i divides z. I'm wondering 
if
there's a test
as
:> to which is the factor. Doing some calculations, it star
to look like
it
:> only depended on which octant z was in, but I guess it's
not that
simple.
: Observe that p = a^2 + b^2 = (a+bi)(a-bi) = (b+ai)(b-ai),
so both are
: factors. Zounds! Have we viola unique factorization? No we
: haven't. The ring of Gaussian integers has 4 units, 1, -1,
i, and -i.
: Factors that differ by a unit factor should be considered
the same
: factor. In this case, note that (a-bi)*i = b + ai, so we
see that
: a+bi andn b+ai are in fact the distinct prime factors here.
Sure, but this doesn't work when z is not a real number. In
that case,
only
one of a+bi or b+ai may be a factor; they are not associates.
===
Subject: Re: Factoring Gaussian integers
> :> I'm developing an algorithm for factoring Gaussian
integers. In the
case
> ::> p = 1 mod 4, p = a^2 + b^2, p | norm(z), p a prime,
> ::> then either a + b*i or b + a*i divides z. I'm 
wondering
if there's a
test as
> :> to which is the factor. Doing some calculations, it star
to look
like it
> :> only depended on which octant z was in, but I guess 
it's
not that
simple.
 : Observe that p = a^2 + b^2 = (a+bi)(a-bi) = (b+ai)(b-ai),
so both are
> : factors. Zounds! Have we viola unique factorization? No we
> : haven't. The ring of Gaussian integers has 4 units, 1,
-1, i, and -i.
> : Factors that differ by a unit factor should be considered
the same
> : factor. In this case, note that (a-bi)*i = b + ai, so we
see that
> : a+bi andn b+ai are in fact the distinct prime factors
here.
Sure, but this doesn't work when z is not a real number. In
that case,
only
> one of a+bi or b+ai may be a factor; they are not
associates.
I was very sleepy when I replied to this post (as I am now)
and didn't
quite get the point of it. My claim was only that for p =
4n+1, a+bi
and b+ai were the two non-associa prime factors of p since
b+ai is
an associate of a-bi. I ignored z completely. Now to the
actual
question asked. Why not simply calculate a and b (which can
be done
by formula, although i suspect it is quicker to search) and
then
divide a+bi into z. If the quotient is a Gaussian integer,
then we
are done. If not, then a-bi is the factor. If p|z, then both
are
factors. Letting z = u+vi, then z/(a+bi) =
[(au+bv)/(a^2+b^2)] +
[(av-bu)/(a^2+b^2)]i =
[(au+bv)/p] + [(av-bu)/p]i, so if p|(au+bv) and p|(av-bu),
then a+bi
is a factor, and a similar easy calculation will show whether
or not
a-bi is a factor.
Achava
Subject: Sorting permutations by single element insertions
===
I'm looking for information about the permutation order you
get when
the only allowed operation is to move a single element to
another
place. I don't know the name of this order, which make
searching for
it a bit complica. If anyone knows it, I'd like to be told,
so that
I can go on looking. Below, I add what I am looking for, in
case
someone knows of a source that contains the information.
One way to sort a permutation using this set of operations is
using
the algorithm straight insertion, or insertion sort.
Unfortunately,
this algorithm doesn't always find the minimal 
number of
moves. (The
algorithm will use three operations to sort 4123, but you can
do it
using just one - move the 4 to the other end.) I want to know
which
permutation needs the maximum number of moves, what number
that is,
how you find the number of moves needed in a general case, and
how you
find the moves themselves. Or whether these things are 
unknown,
because the problem is NP-hard or for some other reason.
(I'm writing a paper dealing with permutations, and want to
include
about two sentences and a pointer to more information about
this
permutation order.)
Hillevi Gavel
Department of Mathematics and Physics
M.8alardalens H.9agskola
Sweden
===
Subject: Re: Sorting permutations by single element insertions
I'm looking for information about the permutation order you
get when
> the only allowed operation is to move a single element to
another
> place. I don't know the name of this order, which make
searching for
> it a bit complica. If anyone knows it, I'd like to be 
told,
so that
> I can go on looking. Below, I add what I am looking for, in
case
> someone knows of a source that contains the information.
One way to sort a permutation using this set of operations is
using
> the algorithm straight insertion, or insertion sort.
Unfortunately,
> this algorithm doesn't always find the minimal 
number of
moves. (The
> algorithm will use three operations to sort 4123, but you
can do it
> using just one - move the 4 to the other end.) I want to
know which
> permutation needs the maximum number of moves, what number
that is,
> how you find the number of moves needed in a general case,
and how you
> find the moves themselves. Or whether these things are
unknown,
> because the problem is NP-hard or for some other reason.
(I'm writing a paper dealing with permutations, and want to
include
> about two sentences and a pointer to more information about
this
> permutation order.)
Hillevi Gavel
> Department of Mathematics and Physics
> M.8alardalens H.9agskola
> Sweden
Donald Knuth's planned Volume 4 of TAOCP
http://www-cs-faculty.stanford.edu/~knuth/taocp.html
will have a long chapter on permutation generation. A draft
of this
chapter is available online:
Pre-fascicle 2b (Generating all permutations); at
http://www-cs-faculty.stanford.edu/~knuth/fasc2b.ps.gz
It might be worthwhile to read this before writing a paper on
the
subject.
Hugo Pfoertner
===
Subject: connecness in random graphs
,
I am stuck on problem below.Please suggest a solution for it.
Given a random connec undirec graph G(N,p) where N is total
number of vertices and p is the probability that there exists
an edge
between any two vertices, what is the probability that the
graph will
remain connec if any k vertices (chosen randomly with
equiprobability) and their corresponding edges are removed
from the
graph?
Thanks in advance,
sharad
===
Subject: Re: connecness in random graphs
> I am stuck on problem below.Please suggest a solution for
it.
Given a random connec undirec graph G(N,p) where N is total
> number of vertices and p is the probability that there
exists an edge
> between any two vertices, what is the probability that the
graph will
> remain connec if any k vertices (chosen randomly with
> equiprobability) and their corresponding edges are removed
from the
> graph?
The distribution G(N,p) is fine if you are 
defining random
graphs, but I
don't know what probability distribution you are implying by
generating
connec G(N,p) graphs.
The most plausible interpretation is that you are generating
a graph
from G(N,p) and then if it is not connec, you throw it out.
But this is
not obvious.
A common useage of Ôconnec random graph' 
fixes the number of
edges
and considers all connec graphs with n vertices and k edges
and the
distribution C(n,k) takes one of these graphs uniformly at
random.
(See Asymptotic Properties of Labeled Connec Graphs for
example)
http://citeseer.nj.nec.com/rd/87058740%2C323267%2C1%2C0.25%
2CDownload/http://
citeseer.nj.nec.com/compress/0/papers/cs/15734/http:
zSzzSzmath.ucsd.eduzSz~eb
enderzSzreprintszSz82.ps.gz/bender99asymptotic.ps
Another common generation technique for connec random graphs
is to
start with a random spanning tree (e.g. Prufer sequence) and
then add
random edges.
J
===
Subject: Re: Resolvable Space
<0401241647330.16637-100000@gandalf.math.ukans.edu>
<0401271159350.1036-100000@gandalf.math.ukans.edu>
Subject: Re: Resolvable Space
> An irresovable Hausdorff space (X,t) with no isola points:
> Let X be an infinite set.
> Let t_0 be a Hausdorff topology on X with no isola points.
> Let T be the set of all topologies on X that extend t_0 and
> contain no singletons. By Zorn's lemma, T has maximal
elements.
> Let t be a maximal element of T.
> That's clearly true if D = X, so assume that the 
complement
of D,
> call it C, is nonempty. Since D is t-dense, C is not t-open.
> Maximality of t implies that there is a t-open set U whose
> intersection with C is a singleton {x}. Since {x} is closed,
> U{x} is a nonempty t-open subset of D.
> Inasmuch as t is a maximal element of T, and C is not in t,
> the topology genera by tu{C} must contain a singleton.
>> Still don't get it.
>I don't get what it is that you don't get. 
You don't get why
the
>topology genera by tu{C} contains a singleton, or you don't
get
>why the presence of an open singleton in the topology genera
by
>tu{C} implies the existence of a t-open set U whose
intersection
>with C is a singleton? Let's take it step by step.
Thanks for your patience. Mostly I missed Ôu' in 
Ôtu' as
union.
>Since C is not t-open, the topology genera
>by tu{C} is a proper extension of t.
>Since t is a maximal element of T, and the topology genera by
>tu{C} is a proper extension of t, it follows that the
topology
>genera by tu{C} does not belong to T.
So it has open singleton {x} and as t u {C} is the topology
with
subbase t u {C}, for {x} to be in t u {C}, there's some open
U with
{x} = U / C
Thus Ux is open nonnul and
Ux / C = nulset
So C isn't dense, nor dense D codense.
>T is the set of all topologies on X that extend t_0 and
contain
>no singletons. The topology genera by tu{C} is a topology on
X
>that extends t, which extends t_0. Since the topology genera
by
>tu{C} does not belong to T, we conclude that it contains a
>singleton, which we can call {x}.
Written in the style of a skillful teacher. Much like the
style I was
cultivating as a tutor to explain and explain again in ever
simpler terms
or when attuned to the student in terms and with words
they're familiar
with.
Was I ever taken aback once when I tried to explained how
polynomial
fractions can be handled in much the same way as fractions,
to find
student didn't know fractions.
>The topology genera by tu{C} consists of arbitrary unions of
>finite intersections of elements of tu{C}. Since {x} belongs
to
>that topology, {x} is a union of finite intersections of
elements
>of tu{C}. Since {x} is a singleton, it follows that {x} is a
finite
>intersection of elements of tu{C}. Since t is closed under
finite
>intersections, it follows that {x} is either a t-open set or
else the
>intersection of a t-open set with C. But {x} is not t-open,
because
>the topology t, being an element of T, has no open
singletong.
>Therefore, {x} is equal to the intersection of C with some
t-open
>set, which I call U.
----
===
Subject: Re: Resolvable Space
<0401241647330.16637-100000@gandalf.math.ukans.edu>
Subject: Re: Resolvable Space
> Does every resolvable space X contain two disjoint dense
subsets A
> and B with |A| = |B| = density(X), where density(X) is
defined as
> the minimum cardinality of a dense subset of X? I guess
it's false,
> but I don't know. It's trivially true for 
finite spaces.
>There is an easy T_1 counterexample. Let |X| = m > aleph_0.
Let Z be
>a subset of X, |Z| = aleph_0. Let u be a free ultrafilter on
Z.
>Topologize X as follows: The empty set is open. A nonempty
set A is
>open if and only if it satisfies both of the conditions: |YA|
< m,
>and the intersection of A with Z belongs to u. X is a
T_1-space; X
>is resolvable, since every subset of cardinality m is dense;
Z is a
>countable dense subset of X, but X does not contain two
disjoint
>countable dense sets.
Easy? It's a multi-part exercise.
How did Y pop up? Let it be X ?
tau closed under intersection. If A,B open:
Z / A/B = Z/A / Z/B in u
|X - A/B| = |XA / XB| < m+m = m
tau closed under unions. If Aj's open:
Z/Aj subset Z / /{ Aj } in u
|X - /{ Aj }| = |/{ XAj }| <= |XAj| < m
if x /= y: let U = Xy; x in U; y not in U
|X - Uy| = 1 < m; U/Z = Zy in u; U open
let u be topology for Z that's maximal Hausdorff with no
singletons
tau still topology but now Hausdorff cardinal offense
if x /= y, some open A,B separates x,y:
A/B = nulset; A subset XB; |A| <= |XB| < m; |XA| = m
|D| = m ==> D dense. If A open, A/D = nulset:
D subset XA; |D| <= |XA| < m
countable Z dense; uncountable XZ dense
U subset Z not in u ==> XU open
|X - XU| = |U| < m; Z / XU = ZU in u
dense D, |D| < m ==> Z/D in u. |X - XD| = |D| < m
XD not open; Z / XD = ZD not in u; Z - ZD = Z/D in u
dense D,E, |D|,|E| < m ==> D/E nonnul
Z/D, Z/E in u; Z/D/E in u
finite D ==> D not dense. cl D = D /= X
----
===
Subject: Re: Resolvable Space
<0401241647330.16637-100000@gandalf.math.ukans.eduSubject:
Re: Resolvable Space
Does every resolvable space X contain two disjoint dense
subsets A
and B with |A| = |B| = density(X), where density(X) is defined
as
the minimum cardinality of a dense subset of X? I guess it's
false,
but I don't know. It's trivially true for 
finite spaces.
There is an easy T_1 counterexample. Let |X| = m > aleph_0.
Let Z be
>a subset of X, |Z| = aleph_0. Let u be a free ultrafilter on
Z.
>Topologize X as follows: The empty set is open. A nonempty
set A is
>open if and only if it satisfies both of the conditions: |YA|
< m,
>and the intersection of A with Z belongs to u. X is a
T_1-space; X
>is resolvable, since every subset of cardinality m is dense;
Z is a
>countable dense subset of X, but X does not contain two
disjoint
>countable dense sets.
> Easy? It's a multi-part exercise.
> How did Y pop up? Let it be X ?
Yes. Sorry about that.
===
Subject: Re: Resolvable Space
<0401241647330.16637-100000@gandalf.math.ukans.eduThanks for
your patience. Mostly I missed Ôu' in 
Ôtu' as union.
Oops, I didn't notice that I was using 
Ôu' two ways. My bad.
Sorry
about that!
===
Subject: Re: Resolvable Space
<0401241647330.16637-100000@gandalf.math.ukans.edu>
<0401251015510.18502-100000@gandalf.math.ukans.eduDoes every
resolvable space X contain two disjoint dense subsets A
> and B with |A| = |B| = density(X), where density(X) is
defined as
> the minimum cardinality of a dense subset of X? I guess
it's false,
> but I don't know. It's trivially true for 
finite spaces.
There is an easy T_1 counterexample. Let |X| = m > aleph_0.
Let Z be a
subset of X, |Z| = aleph_0. Let u be a free ultrafilter on Z.
Topologize X as follows: The empty set is open. A nonempty
set A is
open if and only if it satisfies both of the conditions: |YA|
< m,
and the intersection of A with Z belongs to u. X is a
T_1-space; X is
resolvable, since every subset of cardinality m is dense; Z
is a
countable dense subset of X, but X does not contain two
disjoint
countable dense sets.
===
Subject: question about sequences.
If there is a sequence {x_n} such that:
0 <= x_(n+m) <= x_n + x_m
then does the sequences {(x_n)/n} converge ? i have a pretty
strong
feeling that tells me that its convergent. I'm not sure but
maybe the
harmonic series is an exemple of such a sequence. Does anyone
know
anything about such a sequence ? thanks in advance.
===
Subject: Re: question about sequences.
> If there is a sequence {x_n} such that:
0 <= x_(n+m) <= x_n + x_m
then does the sequences {(x_n)/n} converge ? i have a pretty
strong
> feeling that tells me that its convergent. I'm not sure 
but
maybe the
> harmonic series is an exemple of such a sequence. Does
anyone know
> anything about such a sequence ? thanks in advance.
converges. (Don't quote me, though)
Ôcid Ôooh
===
Subject: Re: question about sequences.
> If there is a sequence {x_n} such that:
0 <= x_(n+m) <= x_n + x_m
then does the sequences {(x_n)/n} converge ? i have a pretty
strong
> feeling that tells me that its convergent. I'm not sure 
but
maybe the
> harmonic series is an exemple of such a sequence. Does
anyone know
> anything about such a sequence ? thanks in advance.
*************************************************************
***************
***
Hi:
x_n <= x_(n-1) + x_1 <= x_(n-2) + 2*x_1 <=...<= n*x_1 ==> 0
<= x_n/n <= x_1
So {x_n/n} is bounded, but I can't see whether much more can
be said.
Good luck!
Tonio
===
Subject: Re: question about sequences.
> If there is a sequence {x_n} such that:
0 <= x_(n+m) <= x_n + x_m
then does the sequences {(x_n)/n} converge ? i have a pretty
strong
> feeling that tells me that its convergent. I'm not sure 
but
maybe the
> harmonic series is an exemple of such a sequence. Does
anyone know
> anything about such a sequence ? thanks in advance.
Isn't the sequence x_n=n a counterexample ?
===
Subject: Re: question about sequences.
>
If there is a sequence {x_n} such that:
0 <= x_(n+m) <= x_n + x_m
then does the sequences {(x_n)/n} converge ? i have a pretty
strong
feeling that tells me that its convergent. I'm not sure but
maybe the
harmonic series is an exemple of such a sequence. Does anyone
know
anything about such a sequence ? thanks in advance.
> Isn't the sequence x_n=n a counterexample ?
>
x_n = n satisfies your condition but does not converge.
===
Subject: Re: question about sequences.
>> If there is a sequence {x_n} such that:
>> 0 <= x_(n+m) <= x_n + x_m
>> then does the sequences {(x_n)/n} converge ? i have a
pretty strong
>> feeling that tells me that its convergent. I'm not sure
but maybe the
>> harmonic series is an exemple of such a sequence. Does
anyone know
>> anything about such a sequence ? thanks in advance.
>Isn't the sequence x_n=n a counterexample ?
No. Constant sequences converge.
All such sequences converge.
Let m be any positive integer. The x_km <= k*x_m, and
x_{km+j} <= k*x_m + C for some C. So lim sup x_n/n <= x_m/m.
As m is arbitrary, lim sup x_n/n <= lim inf x_m/m, qed.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558