> All any mathematician, or *anyone* who wishes to challenge the paper > need do, is break that chain. > None can. Many have. ==== > How would a mathematician (or, indeed, any rational person) >> interpret the result that 2 = 1? Here's how: >> Certainly 2 does *not* equal 1. Therefore the argument must contain >> an error. >> Apparently, I am not a rational person. >> I don't reason as you claim above. Either the theory in which the >> proof occurs is inconsistent or there is an error in the proof >> (or both). I do not have a priori knowledge of whether the theory is >> inconsistent or not[2], so I can't see how the mere fact that your >> proof concludes 2=1 indicates your proof is erroneous. If you do not believe the theory in which the proof occurs is > consistent, which is certainly your prerogative, then you will summarily > reject the proof. Either way the conclusion fails. Reject the proof? In what sense would I reject the proof? A proof in an inconsistent theory[1] is still a proof. I wouldn't reject the proof as erroneous simply because the theory is inconsistent. Maybe I would say that proofs in an inconsistent theory are uninteresting, but that's hardly a rejection. I guess you want to claim that a proof that 1=2 must not be a proof about the honest-to-God numbers 1 and 2, since the honest-to-God numbers are distinct. But, I won't follow that path, since I'm not so confident that claims about honest-to-God numbers are all that meaningful. In any case, your simple claim about how rational persons would reason about a proof of 1=2 is looking implausible to me. Footnotes: [1] I am, of course, assuming that in whatever theory the proof of 1=2 is given, we also have |- ~(1=2). -- Jesse Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPR's Morning Edition ==== [snip] The mathematial argument in that paper begins with a truth, and > proceeds by logical steps to a conclusion which then must be true. Only if the steps are error-free. Consider this: That's what a logical step is. > a = b > a^2 = a*b > a^2 - b^2 = a*b - b^2 > (a+b)(a-b) = b(a-b) Everything is fine at this point as a-b=0, so you have 0=0. > a+b = b Then there's a divide by 0 error. In this classic example, the symbols are here used to obscure the reality. Those confused may be taken in by illusion--people can see something as valid if they get confused by the symbols, but at the start you're told a=b, so you have (a+a)(a-a) = a(a-a) and it's not a valid step to divide off a-a, as that's 0, and you can't divide by 0. The trick is that a=b is stated but then b is still shown, so your mind may tell you that a does not equal b, because a doesn't LOOK like b. But how they look doesn't matter to the math, as a=b, tells the tale. > 2a = a > 2 = 1 I noticed a poster DID reply to this classic example, but didn't point out the obvious, possibly because the assumption is that everyone knows why the argument is flawed. Now someone might call it a proof, but it's clearly not, as a proof is correct. That's easy to get confused because you may hear people talking about proof, when they have a *claim* of proof. Most importantly, notice that a short, flawed argument can be handled by showing a break in the logical chain at a single point. That's mathematics. James Harris <3c65f87.0307160558.25b5c60a@posting.google.com> <3c65f87.0307200519.6a5e941b@posting.google.com> <3F1AA941.A19BF836@ix.netcom.com> <87vftxi4ax.fsf@phiwumbda.localnet> ==== [snip] > If you do not believe the theory in which the proof occurs is > consistent, which is certainly your prerogative, then you will summarily > reject the proof. Either way the conclusion fails. Reject the proof? In what sense would I reject the proof? Suit yourself. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== >[...] I guess you want to claim that a proof that 1=2 must not be a proof >about the honest-to-God numbers 1 and 2, since the honest-to-God >numbers are distinct. But, I won't follow that path, since I'm not so >confident that claims about honest-to-God numbers are all that >meaningful. Did you say _exactly_ what you meant here? If so you go even farther in a certain direction than I do, which is impressive. (Or did you just maybe mean that you're not certain that one can know that the listener will hear what you meant to say when you assert something about the htG numbers? Or some other weaker version: Surely you _do_ actually believe that _you_ know what you mean when you say 2+2 = 4 and _you_ are certain that what you have in mind when you say that is in fact true? Again, if not all I can say is wow, a _real_ skeptic.) >In any case, your simple claim about how rational persons would reason >about a proof of 1=2 is looking implausible to me. Footnotes: >[1] I am, of course, assuming that in whatever theory the proof of >1=2 is given, we also have |- ~(1=2). ************************ David C. Ullrich ==== >[snip] >The mathematial argument in that paper begins with a truth, and >proceeds by logical steps to a conclusion which then must be true. >>Only if the steps are error-free. Consider this: > That's what a logical step is. > >a = b >>a^2 = a*b >>a^2 - b^2 = a*b - b^2 >>(a+b)(a-b) = b(a-b) > Everything is fine at this point as a-b=0, so you have 0=0. >>a+b = b > Then there's a divide by 0 error. In this classic example, the > symbols are here used to obscure the reality. If you were defending your proof, you would begin by promptly either ingoring this or pointing our attention to something unrelated. Those confused may be taken in by illusion--people can see something > as valid if they get confused by the symbols, but at the start you're > told a=b, so you have (a+a)(a-a) = a(a-a) and it's not a valid step to divide off a-a, as that's 0, and you > can't divide by 0. The trick is that a=b is stated but then b is > still shown, so your mind may tell you that a does not equal b, > because a doesn't LOOK like b. But how they look doesn't matter to > the math, as a=b, tells the tale. >>2a = a >>2 = 1 > I noticed a poster DID reply to this classic example, but didn't point > out the obvious, possibly because the assumption is that everyone > knows why the argument is flawed. Now someone might call it a proof, > but it's clearly not, as a proof is correct. That's easy to get > confused because you may hear people talking about proof, when they > have a *claim* of proof. Most importantly, notice that a short, flawed argument can be handled > by showing a break in the logical chain at a single point. Then why do you fail to address breaks in your logical chain that others have pointed out? Why do you fail to deal with the counter-examples and counter-proofs? That's mathematics. > James Harris -- Will Twentyman <3F152198.BC118AB@nospam.idiom.com> <3c65f87.0307160558.25b5c60a@posting.google.com> <3c65f87.0307200519.6a5e941b@posting.google.com> <3F1AA941.A19BF836@ix.netcom.com> <87vftxi4ax.fsf@phiwumbda.localnet> <3F1AECAA.80F373D4@ix.netcom.com> <87he5ggv61.fsf@phiwumbda.localnet> ==== >[...] >>I guess you want to claim that a proof that 1=2 must not be a proof >>about the honest-to-God numbers 1 and 2, since the honest-to-God >>numbers are distinct. But, I won't follow that path, since I'm not so >>confident that claims about honest-to-God numbers are all that >>meaningful. Did you say _exactly_ what you meant here? If so you go even > farther in a certain direction than I do, which is impressive. I think I said what I meant. But, we'll see. > (Or did you just maybe mean that you're not certain that one can > know that the listener will hear what you meant to say when you > assert something about the htG numbers? Or some other weaker > version: Surely you _do_ actually believe that _you_ know what you > mean when you say 2+2 = 4 and _you_ are certain that what you have > in mind when you say that is in fact true? Again, if not all I can > say is wow, a _real_ skeptic.) My htG-numbers are meant to stand for a naive, Platonistic view of numbers. That is, one may believe that (1) there really are things denoted by one, two, etc., (2) these things are distinct from any purely formal theory, (3) a particular theory (say, PA) may accurately formalize the properties of htG-numbers or not and (4) any inconsistent theory fails to formalize the properties of the htG-numbers. Now, I'm not claiming that these four claims are just obviously false, but they are not obviously true as far as I can reckon. Since objects like this are what I meant by htG-numbers, if it is not obviously true that there are such things, then claims about what properties hold of htG-numbers are not obviously meaningful. So, I don't think I'm being very skeptical. I've only tried to express the claim: naive Platonism is not obviously true. -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China ==== > Sure, there are mathematicians who will win the Nobel equivalent prize >> anyway, and walk away with the nearly million dollars, even when the >> public doesn't really know if it's b.s., but most mathematicians won't >> ever get the chance. It is quite impossible for a mathematician, as a mathematician, to > win a Nobel prize, since ther is no Nobel prize in mathematics. would be better if he had hyphenated it, but his wording is reasonably > coherent in this short excerpt. On the other hand, how did he come up with nearly a million dollars? As >far as I know the Field's medal comes with 15,000 Canadian dollars. (And >Wiles good not be nominated because he was too old.) The newest entry into the Nobel-equivalent prize for Mathematics is the Abel Prize, which has a monetary component of 6 million NOK (about 750,000 euros, according to their webpage at http://www.abelprisen.no/index_english.html). Using the Currency Coverter at http://www.xe.com/ucc/ I get that 750,000 euros is about $849,400 US dollars, which might reasonably be termed nearly a million dollars. I think it is reasonable to guess that James was refering to the Abel Prize in the throes of his delusions of adequacy. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >I don't know what you mean by counting the primes Ohmigod. There are uncountably many primes! Lee Rudolph ==== : I don't know what you mean by counting the primes : but the Prime Number Theorem was proved : (by Hadamard and de la Vallee Poussin) : without assuming the Riemann Hypothesis, : by showing that there are no zeros in a region : which approaches the critical line asymptotically. But a similar formula, with *tighter bounds*, is equivalent to the Riemann Hypothesis. John Savard ==== So the use of the Riemann Hypothesis in in counting primes is just about the rate of convergence, not about the validity of the formula? What is the ratio of the rate without RH and the rate with RH? : I don't know what you mean by counting the primes > : but the Prime Number Theorem was proved > : (by Hadamard and de la Vallee Poussin) > : without assuming the Riemann Hypothesis, > : by showing that there are no zeros in a region > : which approaches the critical line asymptotically. But a similar formula, with *tighter bounds*, is equivalent to the Riemann > Hypothesis. John Savard ==== If R is a ring with, > for all x in R, x^3 = x > show R is commutative. More can be said: Jacobson proved that if R is a ring in which for every a in R there >is an integer n(a) > 1 depending on a, such that a^n(a) = a, then R >is commutative. > Wow. (See, for example, Theorem 3.1.2 in Herstein's little book >Noncommutative Rings). > No do have. Any hints how he proceeded? Outline: Well, note that it implies Wedderburn's famous result that a finite division ring is commutative, so it's not easy to prove from scratch. In Herstein's proof he first proves Wedderburn's theorem, then he prove the main theorem mentioned above when R is a division ring. (Presumably this proof is actually due to Jacobson). To reduce to the division ring case, first prove that R must be semi-simple, i.e., the Jacobson radical is 0. This implies by a well-know structure theorem that it is a subdirect sum of primitive rings R_i and since each is a homomorphic image of R must also satisfy the hypothesis. Then using the known structure of primitive rings one can concludes that each R_i is a division ring and the result follows. >In Chapter 3 of the above mentioned book Herstein >gives a number of generalizations of this result. > Anything expecially exciting? Not to me. :-) On the other hand, aside from Wedderburn's theorem mentioned above, I never got very excited about commutativity theorems. If you are going to study noncommutative rings, getting Herstein's Noncommutative rings is probably well worthwhile. Edwin ==== > If you are going to study noncommutative rings, getting Herstein's > Noncommutative rings is probably well worthwhile. > Edwin Out of print; the used copies through Amazon are outrageous, but there are some used copies through Barnes and Noble that are reasonable, less than $30. ==== >you may conclude that abb = bbabb = bba, i.e. squares are in the >center of the ring. The rest is plain sailing: >xy = (xy)^3 = x*(yx)^2*y = (yx)^2*xy = >yxy*x^2*y=yxy^2x^2=y^3x^3=yx -- detailing your sketch (abb-bbabb)^2 = abbabb - abbbbabb - bbabbabb + bbabbbbabb = abbabb - abbabb - bbabbabb + bbabbabb = 0 abb-bbabb = (abb-bbabb)^3 = (abb-bbabb)^2 (abb-bbabb) = 0 (bba-bbabb)^2 = bbabba - bbabbabb - bbabbbba + bbabbbbabb = bbabba - bbabbabb - bbabba + bbabbabb = 0 bba-bbabb = (bba-bbabb)^3 = (bba-bbabb)^2 (bba-bbabb) = 0 abb = bbabb = bba xy = xyxyxy = x yxyx y = x yyxyx = yyxxyx = yyyxxx = yx -- extending your method If n > 0 and ring R with for all x in R, x^(n+1) = x show R is commutative. Note x^2n = x^n as x^2n = x^(n+1) x^(n-1) = x^n (ab^n - b^n ab^n)^2 = ab^n ab^n - ab^2n ab^n - b^n ab^n ab^n + b^n ab^2n ab^n = 0 ab^n - b^n ab^n = (ab^n - b^n ab^n)^2 (ab^n - b^n ab^n)^(n-1) = 0 (b^n a - b^n ab^n)^2 = b^n ab^n a - b^n ab^n ab^n - b^n ab^2n a + b^n ab^2n ab^n = 0 b^n a - b^n ab^n = (b^n a - b^n ab^n)^2 (b^n a - b^n ab^n)^(n-1) = 0 ab^n = b^n ab^n = b^n a Is this similar condition sufficent, that all nth powers are in the center, to show R is commutative? Trivial when n = 1 and you've shown when n = 2, but in general I've doubt. ---- ==== >(See, for example, Theorem 3.1.2 in Herstein's little book >Noncommutative Rings). >Well, note that it implies Wedderburn's famous result that a finite >division ring is commutative, so it's not easy to prove from scratch. Hm, even more, viz finite cancellation rings are commutative. >In Herstein's proof he first proves Wedderburn's theorem, then he >prove the main theorem mentioned above when R is a division ring. >(Presumably this proof is actually due to Jacobson). >To reduce to the division ring case, first prove that R must be semi- >simple, i.e., the Jacobson radical is 0. This implies by a well-know >structure theorem that it is a subdirect sum of primitive rings R_i >and since each is a homomorphic image of R must also satisfy the >hypothesis. Then using the known structure of primitive rings one can >concludes that each R_i is a division ring and the result follows. Oh oh, looks like Herstein's beyond my ken as I don't know semi-simple, Jacobson, primitive ring nor it's structure. >The proof also shows that, in fact, such an R is a subdirect sum >of fields. In your case each field must satisfy x^3 - x = 0. >Thus x(x-1)(x+1) = 0. So such a field can only contain 0, 1 and -1 >and must be Z_2 or Z_3. >In Chapter 3 of the above mentioned book Herstein >gives a number of generalizations of this result. ---- ==== Below is a generalization of the ring of algebraic integers. Likely it has been consider, but what the outcome? Is it used or has it been discarded as leading nowhere interesting? Let D be an integral domain with D^* = { 1,-1 } that is with only two units, 1 & -1. K the field of quotients of D F the extension field of algebraic numbers over K An element u of F is an algebraic integer over K when the minimal polynomial of u is in D[x] equivalently some monic polynomial p in D[x] with p(u) = 0 ---- ==== I have made some revisions to this site to complete the method for solving any polyomial to any degree. The polyomial is expressed as a dot product of a curve intersecting a plane. The problem is reduced to a matter of finding where the curve intersects a line. With a translation and rotation of the line into one of the coordinate axes, the roots to the polynomial precipitate. http://mypeoplepc.com/members/jon8338/polynomialroots/ Jon ==== > I have made some revisions to this site to complete > the method for solving any polyomial to any degree. The polyomial is expressed as a dot product of a > curve intersecting a plane. The problem is reduced > to a matter of finding where the curve intersects a > line. With a translation and rotation of the line > into one of the coordinate axes, the roots to the > polynomial precipitate. http://mypeoplepc.com/members/jon8338/polynomialroots/ Jon Here's a test for your method. Solve x^5-x-1 = 0. ==== You *think*? Don't you know if you have completed the method or not? Hmmm. Why not solve a set of test polynomials and compare the results with the results obtained from the Jenkins-Traub method? Post your comparisons when you are ready to share them with others. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I'm a data analyst currently between jobs, not a student; I'm not trying to get over on an assignment, I'm just brushing up. I'm given this problem: If 'c' is a root of F(x)=5x^4-4x^3+3x^2-4x+5, show that '1/c' is also a root. Repeat with G(x)=2x^6+3x^5+4x^4-5x^3+4x^2+3x+2. Given a standard polynomial of degree 12 A_nX^n+A_n-1X^n-1...A_1X...A_0, What conditions must the coefficients Ai satisfy in order that this statement be true: If 'c' is a root of F(x), then '1/c' is also a root. Some of my primal difficulties with the specific nature of language usage in mathematics make this problem weirder for me than it should be. The first two polynomials have no real roots, so I must assume that in order to have any roots at all they have 'potential rational roots'- which because the leading coefficient and the constant are equal happen to be of the form c, 1/c. Since the problem uses no adjectives to describe roots, this seems like the detail I'm supposed to focus on when setting conditions for the coefficients. I experimented with a polynomial like this with a couple of real roots- the second example above with all signs negative except those specified positive in the given conditions. It behaved such that root 'c' implied root '1/c'; on the other hand, I don't currently know how to rule out irrational roots with a proof. To be on the safe side, I would include the characteristics of the example polynomials which prevent any real roots. The conditions are therefore-all coefficients integral; A_12 = A_0 ; A_n, A_n-2, A_n-4...A_2, A_0 > 0, right? ==== > I'm a data analyst currently between jobs, not a student; I'm not > trying to get over on an assignment, I'm just brushing up. > I'm given this problem: If 'c' is a root of > F(x)=5x^4-4x^3+3x^2-4x+5, show that '1/c' is also a root. Repeat with > G(x)=2x^6+3x^5+4x^4-5x^3+4x^2+3x+2. Given a standard polynomial of > degree 12 A_nX^n+A_n-1X^n-1...A_1X...A_0, What conditions must the > coefficients Ai satisfy in order that this statement be true: If 'c' > is a root of F(x), then '1/c' is also a root. Some of my primal difficulties with the specific nature of language > usage in mathematics make this problem weirder for me than it should > be. The first two polynomials have no real roots, so I must assume > that in order to have any roots at all they have 'potential rational > roots'- which because the leading coefficient and the constant are > equal happen to be of the form c, 1/c. Since the problem uses no > adjectives to describe roots, this seems like the detail I'm supposed > to focus on when setting conditions for the coefficients. > I experimented with a polynomial like this with a couple of real > roots- the second example above with all signs negative except those > specified positive in the given conditions. It behaved such that root > 'c' implied root '1/c'; on the other hand, I don't currently know how > to rule out irrational roots with a proof. To be on the safe side, I > would include the characteristics of the example polynomials which > prevent any real roots. > The conditions are therefore-all coefficients integral; A_12 = A_0 > ; A_n, A_n-2, A_n-4...A_2, A_0 > 0, right? > I think you are over-analyzing. Note that the two given polynomials are palindromic. That is, for the first the coefficients are 5, -4, 3, -4, 5 which read the same left to right as right to left. Take that polynomial, plug in 1/c and add up the fractions. You'll get (5 - 4*c + 3*c^2 - 4*c^3 +5*c^4)/c^4 which is zero if c is a root. Whether c is complex or real or rational is not important. -- Paul Sperry Columbia, SC (USA) ==== I think you are over-analyzing. Note that the two given polynomials are > palindromic. That is, for the first the coefficients are 5, -4, 3, > -4, 5 which read the same left to right as right to left. Take that > polynomial, plug in 1/c and add up the fractions. You'll get > (5 - 4*c + 3*c^2 - 4*c^3 +5*c^4)/c^4 which is zero if c is a root. Whether c is complex or real or rational is not important. over-wrought. The quest for chops goes ever onward. ==== > I'm a data analyst currently between jobs, not a student; I'm not > trying to get over on an assignment, I'm just brushing up. > I'm given this problem: If 'c' is a root of > F(x)=5x^4-4x^3+3x^2-4x+5, show that '1/c' is also a root. Let c be a root of F(x)=5x^4-4x^3+3x^2-4x+5. What does that mean? Show that c is not equal to 0. Thus, 1/c makes sense. You want to show that 1/c is a root of F(x). That is, you want to show that F(1/c) is equal to zero by definition of what it means to be a root of F(x). Plug in 1/c into the expression for F(x) and see if you can show that it is equal to zero. Since all you know about c is that it is a root of F(x), you will have to use that information to calculate the value of F(1/c). -- Bill Hale ==== let L be an arc of the trigonometric circle with length alpha degrees (in my exemple alpha = 40¡ and L is the set of the x s.t. 90¡ < x < 130¡). 1) Can we give an integer N s.t. for each n >= N, there is at least a primitive n-th root of unity in L (it seems to me that N=13 works)? 2) More generally can we give N as a function of alpha (not necessary the best N...))? ==== let L be an arc of the trigonometric circle with length alpha degrees > (in my exemple alpha = 40¡ and L is the set of the x s.t. 90¡ < x < 130¡). 1) Can we give an integer N s.t. for each n >= N, there is at least a > primitive n-th root of unity in L (it seems to me that N=13 works)? 2) More generally can we give N as a function of alpha (not necessary > the best N...))? We could simply let N be the first prime larger than 2*360/alpha. Then the Nth roots of unity will then be spaced less than alpha/2 apart on the unit circle, so your interval is bound to get hit twice. And with N chosen prime, only one of the two (or more) roots in the interval could be non-primitive, so you win. (If you know that your interval doesn't pass through unity, then we cannot reduce N to be the first prime larger than 360/alpha, because then the trivial, non-primitive root is no longer an issue). Another solution is to let N be the first power of two larger than 2*360/alpha, which works for similar reasons (every other root of N will be primitive, so one of the two or more hits in the interval will be primitive). Neither of these will be best N's of course. Another possible approach would be to express the endpoint angles (divided by 360) as continued fractions. I'm reminded of an idea by Gosper in HAKMEM (http://www.inwap.com/pdp10/hbaker/hakmem/cf.html): Problem: Given an interval, find in it the rational number with the smallest numerator and denominator. Solution: Express the endpoints as continued fractions. Find the first term where they differ and add 1 to the lesser term, unless it's last. Discard the terms to the right. What's left is the continued fraction for the smallest rational in the interval. (If one fraction terminates but matches the other as far as it goes, append an infinity and proceed as above.) Perhaps this would give the best N; can someone say? Applying this the original example, from 90 to 130 degrees ... The continued fractions for 90/360 and 130/360 are [0;4] and [0;2,1,3,3], I believe. So, [0;3] would be the smallest continued fraction between them, which of course works since there is a 3rd root of unity at 120 degrees. (See http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html for a friendly introduction to continued fractions ...) - Brent ==== > let L be an arc of the trigonometric circle with length alpha degrees > (in my exemple alpha = 40¡ and L is the set of the x s.t. 90¡ < x < 130¡). 1) Can we give an integer N s.t. for each n >= N, there is at least a > primitive n-th root of unity in L (it seems to me that N=13 works)? If I understand this aright then yes. You are looking at an arc of the unit circle, say the set of exp(2 pi i t) for t between a and b. If n > 1/(b-a), i.e., (b-a) > 1/n then there is an integer k with k/n in the interval [a,b]. Then exp(2 pi i k/n) is a k-th root of unity in the given arc. Hence we can take N to be the ceiling of 1/(b-a). > 2) More generally can we give N as a function of alpha (not necessary > the best N...))? See above. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > We could simply let N be the first prime larger than 2*360/alpha. Then the > Nth roots of unity will then be spaced less than alpha/2 apart on the unit > circle, so your interval is bound to get hit twice. And with N chosen prime, > only one of the two (or more) roots in the interval could be non-primitive, > so you win. (If you know that your interval doesn't pass through unity, then > we cannot reduce N to be the first prime larger than 360/alpha, because then that cannot should have said can. oops :-) Actually, we can always reduce to 360/alpha regardless, because if the interval contains unity, then of course it contains a primitive root of unity, namely unity itself (then N becomes 1). - Brent ==== let L be an arc of the trigonometric circle with length alpha degrees > (in my exemple alpha = 40¡ and L is the set of the x s.t. 90¡ < x < 130¡). 1) Can we give an integer N s.t. for each n >= N, there is at least a > primitive n-th root of unity in L (it seems to me that N=13 works)? If I understand this aright then yes. You are looking at an arc of the unit circle, say the set > of exp(2 pi i t) for t between a and b. If n > 1/(b-a), i.e., (b-a) > 1/n then there is an integer > k with k/n in the interval [a,b]. Then exp(2 pi i k/n) is a k-th > root of unity in the given arc. Hence we can take N to be the ceiling > of 1/(b-a). 2) More generally can we give N as a function of alpha (not necessary > the best N...))? See above. I read this differently. He asks for a primitive n-th root of unity, and so we must find an integer k such that k/n is in [a,b] AND (k,n) = 1. It is not obvious to me that we either can or can't do this. We can clearly guarantee any fixed number F of k's over n with the k in sequence by making N larger than the ceiling of 1/F(b-a), but is that good enough? Achava ==== let L be an arc of the trigonometric circle with length alpha degrees > (in my exemple alpha = 40¡ and L is the set of the x s.t. 90¡ < x < 130¡). 1) Can we give an integer N s.t. for each n >= N, there is at least a > primitive n-th root of unity in L (it seems to me that N=13 works)? If I understand this aright then yes. You are looking at an arc of the unit circle, say the set > of exp(2 pi i t) for t between a and b. If n > 1/(b-a), i.e., (b-a) > 1/n then there is an integer > k with k/n in the interval [a,b]. Then exp(2 pi i k/n) is a k-th > root of unity in the given arc. Hence we can take N to be the ceiling > of 1/(b-a). 2) More generally can we give N as a function of alpha (not necessary > the best N...))? See above. I read this differently. He asks for a primitive n-th root of unity, > and so we must find an integer k such that k/n is in [a,b] AND (k,n) = > 1. It is not obvious to me that we either can or can't do this. We > can clearly guarantee any fixed number F of k's over n with the k in > sequence by making N larger than the ceiling of 1/F(b-a), but is that > good enough? > I took the approach of looking at all of the nth roots of one the points are spaced equally along the circle. starting with the first one counterclockwise from (1,0) lable them a1, a2, ..., aN now aJ is a primitive root of 1 iff (J,n)=1 do this reduces to looking at the numbers 1, 2, 3, ..., n that lie between (90n/360) and (140n/360) inclusive [the 90 and 140 are based on the example] and determine if one of numbers is relatively prime to n. For example n=24 would mean that we have to examine 6, 7, 8, 9. (7, 24) = 1 passes [but nne of the others do] It seems to me that the only problem would be in numbers that have small totient values realtive to the number (12, 24, 30, etc). unless you could establish something like: N such that for all n>N and given int [(140-90)n]/360 = [5n/36] consecutive numbers, there exist one (i) such that (i,n) = 1. Notice that this would then generalize to ANY 50 degree wedge. In fact if you could find a constant k such that there exists an N such that for all n>N and given int [kn] consecutive numbers, there exist one (i) such that (i,n) = 1 - you could truly generalize this problem. -Tralfaz ==== > I read this differently. He asks for a primitive n-th root of unity, Doh! Didn't notice primitivity hypothesis :-( Not so easy then .... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > We could simply let N be the first prime larger than 2*360/alpha. thank you for your answer but I think there is a misunderstanding. I want an N s.t. for each n >= N, there is a n-th primitive root of unity in L. Suppose L are the x s.t. 90¡ < x < 270¡, N = 5 but L do not contains any 6-th primitive roots of unity. Isn't it? Nicola ==== It seems to me that the only problem would be in numbers that have small > totient values realtive to the number (12, 24, 30, etc). unless you could > establish something like: N such that for all n>N and given int > [(140-90)n]/360 = [5n/36] consecutive numbers, there exist one (i) such that > (i,n) = 1. but do these remarks give us a method to compute N in the given exemple (in the exemple the arc was 90¡ < x < 130¡)? It doesn't seem trivial to me... ==== First question: > If t is an indeterminate, how can you write the vector . > this is the parametric form of the position vector of the curve in n-space. > Second question: > How did you figure Q1 is a point on the plane? Your picture seems to > indicate this. in vectorform, Q1 = <-a0(a1)/A, -a0(a2)/A, ..., -a0(aN)/A where A is a1^2 + a2^2 + ... + aN^2 but this does not seem to be an obvious > solution to the equation for the hyperplane. > If Q3 = any point on the plane, then (Q3*N/|N|)(N/|N|) = vector from origin to plane that is perpendicular to the plane=Q1. But Q3*N= -a0, so Q1= -a0N/|N|^2 1. t is the intersection of the curve f(x)=a0+a1x+a2x^2+...+anx^n with f(x)=0 or the x-axis. 2. t is the intersection of the curve P(t,t^2,t^3,..,t^n) with the line (x1-q1)/p1=(x2-q2)/p2=(x3-q3)/p3=...=(xn-qn)/pn see site http://mypeoplepc.com/members/jon8338/polynomialroots/ ==== This is probably a beginners questuin, but I have to ask it. When we round 0.5 to the nearest integer, the result is 1. What do we round (negative) -0.5 to? 0 or -1? The int (integer) function rounds up, that means to 0 for any number between -1 (excl.) and 0. But what about the round function? TIA! -- Todor Todorov ==== > This is probably a beginners questuin, but I have to ask it. When we round 0.5 to the nearest integer, the result is 1. Not necessarily. The result could be 0 instead. It depends on the convention being used. There is no single standard. Perhaps you'd find my table at to be helpful. David > What do we round (negative) -0.5 to? 0 or -1? The int (integer) function rounds up, that means to 0 for any number > between -1 (excl.) and 0. But what about the round function? ==== We round off the convential way. When we have a decimal number between two integers a and b we decide if the decimal number is closer to a or b and then you have your answer. If the decimal number is precisely in the middle ((a+b)/2), as in your case, then we pick the larger integer of a and b. The number to the right is always the larger number! When we round off -0.5 you are correct when you say this number is between -1 and 0. Since -0.5 is right in the middle of -1 and 0 we pick the larger of -1 and 0 as the answer. Therefore the answer to your question is 0. > This is probably a beginners questuin, but I have to ask it. When we round 0.5 to the nearest integer, the result is 1. > What do we round (negative) -0.5 to? 0 or -1? The int (integer) function rounds up, that means to 0 for any number > between -1 (excl.) and 0. But what about the round function? TIA! -- Todor Todorov ==== > We round off the convential way. When we have a decimal number between two > integers a and b we decide if the decimal number is closer to a or b and > then you have your answer. If the decimal number is precisely in the middle > ((a+b)/2), as in your case, then we pick the larger integer of a and b. The > number to the right is always the larger number! When we round off -0.5 you > are correct when you say this number is between -1 and 0. Since -0.5 is > right in the middle of -1 and 0 we pick the larger of -1 and 0 as the > answer. Therefore the answer to your question is 0. This method of rounding invites accumulative error. 12.5 = .5 + 1.5 + 2.5 + 3.5 + 4.5 = 1 + 2 + 3 + 4 + 5 = 15 rounded by your method The standard method is to round to the nearest even digit. Thus the above is rounded to 0 + 2 + 2 + 4 + 4 = 12 which is nicer fit. -2 = -1.5 + -.5 = -1 - 0 = -1 by your method = -2 + 0 = -2 by standard method, again a nicer fit. So you see, the standard method is better, but only marginally as -3 = -2.5 + -.5 = -2 + - 0 = -2 by both methods ==== For natural numbers a,b , a>=2 ,b>=2 , let S(a,b):= Sum_{k=1 to k=infty}1/(a^{k^b}) . For which (a,b) the number S(a,b) is irrational ? ============ ==== > For natural numbers a,b , a>=2 ,b>=2 , let > S(a,b):= Sum_{k=1 to k=infty}1/(a^{k^b}) . > For which (a,b) the number S(a,b) is irrational ? A number is rational if and only if its expansion in base a is eventually periodic. Your formula is the expansion in base a. For which a,b will it be eventually periodic? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== For natural numbers a,b , a>=2 ,b>=2 , let > S(a,b):= Sum_{k=1 to k=infty}1/(a^{k^b}) . > For which (a,b) the number S(a,b) is irrational ? A number is rational if and only if its expansion in base a is > eventually periodic. > Your formula is the expansion in base a. For which a,b will it be > eventually periodic? O.K ! My question is to find all pairs (a,b) of positive integers (>=2) for which S(a,b) is irational/ or, another variant of the problem is to find (a,b) for which S(a,b) is rational (as you have observed). Alex/Proposer =============== ==== For natural numbers a,b , a>=2 ,b>=2 , let > S(a,b):= Sum_{k=1 to k=infty}1/(a^{k^b}) . > For which (a,b) the number S(a,b) is irrational ? A number is rational if and only if its expansion in base a is > eventually periodic. > Your formula is the expansion in base a. For which a,b will it be > eventually periodic? HINT: Try to prove that there is a natural number n_0=n_0(a,b) such that a^{(k+1)^b}= a^{2*k^b} - a^{k^b} + 1 , fuer k >= n_0 . == LEMMA ( (Sylvester) , see [4], Seite 119 -German in original:) ,,Jede Zahl x laesst sich auf eine und nur eine Weise in eine Sylvestersche Reihe x = c + Sum{k=1 to k=infty} 1/q_k entwickeln, d.h. eine Reihe, bei der c und q_k ganze Zahlen sind, die den Ungleichungen q_1 >= 2 q_{k+1} >= (q_k - 1)q_k + 1 , (k=1,2,3,...) genuegen. Umgekehrt ist jede Reihe dieser Form konvergent. Ihr Wert ist dann und nur dann rational, wenn in der vorstehend Ungleichung fuer hinreichend grosse Werte von k dauernd das Gleichheitszeichen gilt . REFERENCES : [1-a] G. Cantor , ,,Ueber die einfache Zahlensysteme , Zeitschrift fuer Mathem.und Physik 14 (1869)= Gesammelte Abhandlungen , Seite 35. [1-b] G. Cantor , ,,Zwei Saetze ueber eine gewisse Zerlegung der Zahlen in unendliche Produkte , Zeitschrift fuer Mathem.und Physik 14 (1869)=Gesammelte Abhandlungen , Seite 43. [2] J. Lueroth , ,, Ueber eine eindeutige Entwicklungen von Zahlen in eine unendliche Reihe , Math. Annalen 21 (1883) . [3] O.Perron , ,,Die Lehre von der Kettenbruechen, Leipzig, 1913. [4] O.Perron , ,,Irrationalzahlen , Goeschens Lehrbuecherei, Band 1, Walter de Gruyter & Co., dritte Auflage, 1946. [5] J.J. Sylvester , ,,On a point in the theeory of vulgar fractions, Amer. J. Math. 3(1980). ================= (,, Ancient books= interesting books ! , e.g. see [4] )Alex/Proposer ============== ==== I'd like to retract what I said about this but I can't because it hasn't appeared. No doubt because I didn't notice that it was also being sent to sci.math.research (if a sci.math.research moderator happens to see this by all means please reject my post - it was wrong wrong wrong, in fact kinda stupid.) >I'm not sure about my news server, which is sometimes erratic. Sorry if >this shows up as a duplicate. >> Consider a linear shift invariant operator A: l1(N) -> l1(N) from a space >of >> real sequences supported on nonnegative integers with bounded sum of >> absolute values into itself. Such an operator can be identified with an >> infinite Toeplitz matrix (it is then a multiplicative operator) or, >> equivalently with a sequence {ai} (it is then a convolution operator). >> What is the induced norm of this operator? Is there any nice expression? >> [...] Stated in terms of convolutions, given a shift-invariant operator T on >l1(N), there is a sequence b = (b0, b1, ...) such that T(a) = b*a, and ||T|| >equals the l1(N) norm of b. I'm not at all sure about this. It seems strange that the norm would always >be attained at the function 1. It's not attained at the function 1, it's attained at the vector (1,0,0,...). Doesn't seem strange at all; in any Banach algebra with identity the norm of the operator y |-> xy is equal to ||x||, and it's attained at the identity. Duh. > I've forgotten too much, and don't have >references handy. But this would jibe with the similar case for L^1(Z), >because the translation-invariant operators on L^1 of a LCA group are just >convolutions with measures on the group, and in the case of Z, L^(Z) and >M(Z) coincide. > ************************ David C. Ullrich ==== > in any Banach algebra > with identity the norm of the operator y |-> xy is equal to ||x||, > and it's attained at the identity. > Dear David, I just want to make sure that I understood that result correctly. If I am not wrong, I can only use this result in this special case of A: X->Y when both the vector spaces X, Y and the set of bounded linear operators A are isomorphic with l_1. With X=Y=l_2 this would not be applicable, right? Since the algebra of bounded linear operators is (isomorphic to) H_infinity (the input and output sequences in H2 after z-transform, of course). With X=Y=l_infinity this would not be applicable too, because the set of all bounded linear operators from X into Y is isomophic with l_1. Am I right? Zdenek ==== > in any Banach algebra >> with identity the norm of the operator y |-> xy is equal to ||x||, >> and it's attained at the identity. >> Dear David, I just want to make sure that I understood that result correctly. If I am >not wrong, I can only use this result in this special case of A: X->Y when >both the vector spaces X, Y and the set of bounded linear operators A are >isomorphic with l_1. I'm not sure exactly what you mean here. >With X=Y=l_2 this would not be applicable, right? Since the algebra of >bounded linear operators is (isomorphic to) H_infinity (the input and >output sequences in H2 after z-transform, of course). With X=Y=l_infinity this would not be applicable too, because the set of all >bounded linear operators from X into Y is isomophic with l_1. Am I right? Zdenek ************************ David C. Ullrich ==== > I'm not sure exactly what you mean here. Well, I admit my comments were confused and confusing. I have to sort the acquired information a bit :-)) Have a nice weekend, Zdenek Hurak ==== > in any Banach algebra >> with identity the norm of the operator y |-> xy is equal to ||x||, >> and it's attained at the identity. >I just want to make sure that I understood that result correctly. If I am >not wrong, I can only use this result in this special case of A: X->Y when >both the vector spaces X, Y and the set of bounded linear operators A are >isomorphic with l_1. No. This result applies in the case where X = Y and is a Banach algebra with identity, and the linear operator happens to be multiplication by an element of the Banach algebra. In this case the Banach algebra is l_1(N) where the multiplication is convolution: (a*b)_n = sum_{j=0}^n a_j b_{n-j} Among the requirements for a Banach algebra is ||a*b|| <= ||a|| ||b|| which says that the norm of the linear operator convolution by a is at most ||a||. Since the Banach algebra has an identity e (e_0 = 1, e_n = 0 otherwise), the norm is exactly ||a||: ||a|| = ||a*e|| = ||a|| ||e||. >With X=Y=l_2 this would not be applicable, right? Since the algebra of >bounded linear operators is (isomorphic to) H_infinity (the input and >output sequences in H2 after z-transform, of course). >With X=Y=l_infinity this would not be applicable too, because the set of all >bounded linear operators from X into Y is isomophic with l_1. l_2 and l_infinity with convolution are not Banach algebras. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== tale: >http://www.ms.uky.edu/~mai/java/stat/brmo.html >First pic is my second go, second pic is more typical Stop posting binaries to non-binary groups. Usenet is a text medium. If you want to show off pretty pictures, get a website and post them there, or find an appropriate binaries group. > -- Douglas Berry gridlore@mindspring.com http://gridlore.home.mindspring.com Atheist #2147, Atheist Vet #5 Ezekiel 13:20 Wherefore thus saith the Lord GOD; Behold, I am against your pillows ==== >Usenet is a text medium. my newsreader begs to differ, get 10 points running aligned with graph like I did (1,000,000 to 1 odds) then you can dispute me. http://www.ms.uky.edu/~mai/java/stat/brmo.html Herc ==== hang on you're one of the answer back and i'll complain to ISP losers. ==== >http://www.ms.uky.edu/~mai/java/stat/brmo.html >First pic is my second go, second pic is more typical The chances of matching an unspecified set of numbers is 100% You can't lose. The illusion that you have that this feat is impressive is described in many books on statistics. My favorite explanation of it is in Richard Dawkins book, Unweaving the Rainbow, Chapter 7, Unweaving the Uncanny He created a term for this, the word is PETWHAC. It stand for Population of Events That Would Have Appeared Coincidental. Read the chapter at your local library and you may begin to understand why your random matching of a few pixels in the graphic you posted is unimpressive. ==== http://www.ms.uky.edu/~mai/java/stat/brmo.html >First pic is my second go, second pic is more typical > The chances of matching an unspecified set of numbers is 100% You can't lose. The illusion that you have that this feat is impressive is described > in many books on statistics. My favorite explanation of it is in > Richard Dawkins book, Unweaving the Rainbow, > Chapter 7, Unweaving the Uncanny > He created a term for this, the word is PETWHAC. It stand for Population of Events That Would Have Appeared > Coincidental. Read the chapter at your local library and you may begin to understand > why your random matching of a few pixels in the graphic you posted is > unimpressive. > Well I'm impressed with your response you fully understand the claim none the less. But there are 2 extraordinary aspects to the graph, the larger square and the smaller square. Actually the fact that 2/3rds of the graph is followed AND that about 8 points were followed WITHIN THAT without a gap. Its the 10,000 to one overall shape multiplied by the 10,000 to one 8 points running. Of all the directions to take some rough mathematics 9*9*9*9*9*9*9*9 TO ONE for the inner sequence. But since 1000s can read this and I guarantee NONE of you can get either 2/3rd of the graph approximated OR a sequence over 6 running where the points match without a gap the claim stands, but only as far as the belief the graph was reproduced honestly. At some points you HAVE to use abduction to form new facts. I have numerous 1000 to 1s. How many coincidences can YOU provide evidence (even circumstantial) for? http://tinyurl.com/fuf8 predict Laurie Holden as the next Truman costar together with this claim I'm the Truman the following week in 2000. http://tinyurl.com/fuf2 http://tinyurl.com/gutr predict the Shuttle Tragedy (I have another with the word preminition in it one year to the day before) I knew that url GUTR off by heart, Grand Unified Theory Religion was the alloted text to the explanation to *christnet* that the shuttle was predicted. 1 in 1000 in itself. http://tinyurl.com/h6uh A reply to me Randi will test you when you properly apply to be tested.... by Rich Shewmaker 100s like this I've formated 50 (FIFTY) from a 2 months sample with a guess the author option list at www.adamskingdom.com Either cover most the graph or get 6 points running, the square root of probability of what I achieved before you dismiss the graph. http://www.ms.uky.edu/~mai/java/stat/brmo.html and who can forget getting 4 (FOUR) multi choice questions all right, on a multi choice psychic test http://tinyurl.com/h6uz the sci.skeptic member insisted I was a private investigator. Do you want 10 other circumstantials, its not a rainbow when its everything i touch. Oh but ofcourse, groups of paranormal claims is called delusion now. Herc ==== >http://www.ms.uky.edu/~mai/java/stat/brmo.html >>First pic is my second go, second pic is more typical >> The chances of matching an unspecified set of numbers is 100% >> You can't lose. >> The illusion that you have that this feat is impressive is described >> in many books on statistics. My favorite explanation of it is in >> Richard Dawkins book, Unweaving the Rainbow, >> Chapter 7, Unweaving the Uncanny >> He created a term for this, the word is PETWHAC. >> It stand for Population of Events That Would Have Appeared >> Coincidental. >> Read the chapter at your local library and you may begin to understand >> why your random matching of a few pixels in the graphic you posted is >> unimpressive. > Well I'm impressed with your response you fully understand the claim none the less. >But there are 2 extraordinary aspects to the graph, the larger square and the >smaller square. Actually the fact that 2/3rds of the graph is followed AND that >about 8 points were followed WITHIN THAT without a gap. Its the 10,000 to one >overall shape multiplied by the 10,000 to one 8 points running. Of all the directions >to take some rough mathematics 9*9*9*9*9*9*9*9 TO ONE for the inner sequence. But since 1000s can read this and I guarantee NONE of you can get either 2/3rd of >the graph approximated OR a sequence over 6 running where the points match without >a gap the claim stands, but only as far as the belief the graph was reproduced honestly. > The graph you posted only showed 2 or 3 pixels repeated. ==== >>But since 1000s can read this and I guarantee NONE of you can get either 2/3rd of >>the graph approximated OR a sequence over 6 running where the points match without >>a gap the claim stands, but only as far as the belief the graph was reproduced honestly. >> The graph you posted only showed 2 or 3 pixels repeated. Relies on me just printing the actual plot honestly ofcourse, but still a decision >can be made, and it would be an odd claim to fake. It would be objective to cite >this as 10 points along the function with no gap between. The worst point is a few >pixels away but citing 'no gap' as the testing parameter is precise. You're welcome to >have 1000s of attempts to get 5 running as specified to claim 10 is ordinary. >http://www.ms.uky.edu/~mai/java/stat/brmo.html you won't get anywhere near this graph > As I said, there are only 2 or 3 pixels which match in succession. They are right in the center of the image. Once it lands on one black pixel, which is guaranteed to happen many times, there is a 1 in 8 chance of landing on another black adjacent pixel. The simulation did that twice, which brings the odds up to 1 in 64. Not very impressive, especially considering how long you ran the simulation, nearly 6000 steps. ==== >>But since 1000s can read this and I guarantee NONE of you can get either 2/3rd of >>the graph approximated OR a sequence over 6 running where the points match without >>a gap the claim stands, but only as far as the belief the graph was reproduced honestly. >> The graph you posted only showed 2 or 3 pixels repeated. Relies on me just printing the actual plot honestly ofcourse, but still a decision >can be made, and it would be an odd claim to fake. It would be objective to cite >this as 10 points along the function with no gap between. The worst point is a few >pixels away but citing 'no gap' as the testing parameter is precise. You're welcome to >have 1000s of attempts to get 5 running as specified to claim 10 is ordinary. >http://www.ms.uky.edu/~mai/java/stat/brmo.html you won't get anywhere near this graph > Here is a challenge. Prevent the simulator from traveling outside the blue square for 10,000 steps. That might impress someone. (I am sure you could find someone that would be impressed) ==== > Here is a challenge. Prevent the simulator from traveling outside the blue square for > 10,000 steps. That might impress someone. > (I am sure you could find someone that would be impressed) > Ask any of 100,000 people who live in my town if there is a truman and find one who says there isn't and I'll be impressed. Its not 64 to 1, if anything it crosses the function a dozen times that is not the heuristic I'm measuring coincidence level on. 10 points _running_ it continues from left to right without_leaving_the_line, it might ALSO be above and ALSO be below but it touches the point at all 10. Measurably 10, get 5 as I prescibed before you dismiss it. SHOW ME ONE point along that 10 that the function is not covered by the random walk! 10 not 2, 10 and its not guaranteed to go anywhere near the function unless you are talking infinite periods. The function only has one point per X value, the random walk has multiple points per X value, the measure is one of those MULTIPLE points covering the function. I didn't say the random walk WAS a function, I said it mimicked the function! Still a measure of closeness is possible. Still waiting for your graph with 5_points_running (that TOUCH the function) Herc Jesus wouldn't want to walk on salt water around you people you'd disqualify him on a bouyency technicality. ==== > my newsreader begs to differ, Your newsreader is wrong then. -- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion ________________________________________________________________ ...lying is central to the survival of nations and to the success of great enterprises, because if our enemies can count on the reliability of everything you say, your vulnerability is enormously increased. - Michael Ledeen, neo-con leader ==== |-|erc warmed at our fire and told this tale: >>Usenet is a text medium. my newsreader begs to differ, What does you ISP say? The standard is that binaries are sent to groups that have the word binaries in their name. Huge files posted to usenet speed the expiration of other postings. It's jut bloody rude. >get 10 points running aligned with graph like I did (1,000,000 to 1 odds) >then you can dispute me. http://www.ms.uky.edu/~mai/java/stat/brmo.html Others are doing that just fine. -- Douglas Berry gridlore@mindspring.com http://gridlore.home.mindspring.com Atheist #2147, Atheist Vet #5 Ezekiel 13:20 Wherefore thus saith the Lord GOD; Behold, I am against your pillows ==== |-|erc warmed at our fire and told this tale: >hang on you're one of the answer back and i'll complain to ISP losers. No, post anytmore huge mucking binary files and I'll complain. As far as I'm cocerned, you can post your rants all you like. -- Douglas Berry gridlore@mindspring.com http://gridlore.home.mindspring.com Atheist #2147, Atheist Vet #5 Ezekiel 13:20 Wherefore thus saith the Lord GOD; Behold, I am against your pillows ==== > Others are doing that just fine. > -- oh naff off, go look up my ISP and tell your friends, and when I argue back in text say right some people don't learn and get me yanked again. worlds full of yank artists and you're #1. Herc ==== in alt.atheism: >>Usenet is a text medium. >my newsreader begs to differ, You use OE, not a newsreader. -- My position concerning God is that of an agnostic. I am convinced that a vivid consciousness of the primary importance of moral principles for the betterment and ennoblement of life does not need the idea of a law-giver, especially a law-giver who works on the basis of reward and punishment. - Letter to M. Berkowitz, October 25, 1950; Einstein Archive 59-215 rukbat at optonline dot net ==== in alt.atheism: >> Others are doing that just fine. >oh naff off, go look up my ISP look up? It's right there in your headers: -- Christians, it is needless to say, utterly detest each other. They slander each other constantly with the vilest forms of abuse and cannot come to any sort of agreement in their teachings. Each sect brands its own, fills the head of its own with deceitful nonsense, and makes perfect little pigs of those it wins over to its side. - Celsus (2nd century C.E.) rukbat at optonline dot net ==== >> Others are doing that just fine. >> -- oh naff off, go look up my ISP and tell your friends, and when I >argue back in text say right some people don't learn and get me yanked again. violation of *anyone's* TOS, including yours. One might also point out that when you signed up with your ISP, you agreed to abide by their TOS, so you have absolutely no cause to whine if you get in trouble for violating it. BTW, again, please seek competent psychiatric help for your schizophrenia. ==== >Usenet is a text medium. my newsreader begs to differ, get 10 points running aligned with graph like I did (1,000,000 to 1 odds) > then you can dispute me. http://www.ms.uky.edu/~mai/java/stat/brmo.html > Herc You crave attention, so I'll give you some. Fuck off and die you asshole! Will this satisfy you? I believe all the bullshit you post. Truman is based on your life, all that other shit, I believe it. Now, I have bills to pay, a shit to take, stuff to study, friends to talk to so excuse me if be believing all the BULLSHIT you post makes absolutely no difference in my life. Keep posting, and maybe you will convince others of the wacked out BULLSHIT you post, and I bet it will make NO difference in their lives either. You pinhead. ==== its an atheism maths post about a graphic, you're blowing the usenet out of the water. I got a good response although negative as expected, with probably 10,000 views, I wouldn't get 20 hits to a web plug. http://www.winternet.com/~mikelr/flame18.html Herc ==== > Keep posting, and maybe you will convince others of the wacked out > BULLSHIT you post, and I bet it will make NO difference in their lives > either. You pinhead. You have no idea of the repurcussions that all your media is taken from my life, I taught you all to speak, every of my desire, taste, favourite music, cars anything is all optimised for my enjoyment. I can't listen to a take 40 music countdown without it spelling out my last weeks adventures and every song tuned into my current status of my love life. You all owe me a life, you've all been programmed by media with raven whine i'm on camera psychology. my goal is by next year I can afford a sound proof house so that i can evade a dozen center anatagonisers. fame without money is torture, they use a satelite like an auxillaury speaker system, constantly harrased over a year now. 100,000 people join in the interrogation, ONE YEAR and its still not public, guess I can log off, be abused by the center another few hours, another few months, into next year, simple fucking house and its all over. couple people admit it then its soon all over. 3 years I chased that girl under intel setting me up, now 2 MORE YEARS into my 30s, haven't seen the girl since she was 21, saw her for 10 minutes 5 YEARS AGO. now the truman NEEDS to be monitored because I wasn't patient enough, 2 YEARS torture every second of the day. 450 days now interrogation from wake up to drop asleep. not 2 seconds running in 2 years to think a private thought, and that includes taking an S Herc ==== > hang on you're one of the answer back and i'll complain to ISP losers. ==== >its an atheism maths post about a graphic, >you're blowing the usenet out of the water. I got a good response although negative as expected, with probably >10,000 views, I wouldn't get 20 hits to a web plug. >http://www.winternet.com/~mikelr/flame18.html >Herc You're listed there too, you know: http://www.winternet.com/~mikelr/flame46.html ==== I'm looking for recommendations for document preparation (e.g., Latex, Word, etc.). Here are my desiderata in priority order: 0. Will run on a Mac 1. Supports mathematical equations 2. Can generate output either in pdf or web (html) 3. Easy to integrate figures and graphs 4. WSIWYG (would be nice, but not absolutely necessary) Antonio ==== > I'm looking for recommendations for document preparation (e.g., Latex, > Word, etc.). Here are my desiderata in priority order: 0. Will run on a Mac > 1. Supports mathematical equations > 2. Can generate output either in pdf or web (html) > 3. Easy to integrate figures and graphs > 4. WSIWYG (would be nice, but not absolutely necessary) > Antonio You might want to have a look at: Scientific Word or it's variants http://www.mackichan.com/ I am not sure about MAC, but you certainly can run in emulation mode. ==== > I'm looking for recommendations for document preparation (e.g., Latex, > Word, etc.). Here are my desiderata in priority order: 0. Will run on a Mac > 1. Supports mathematical equations > 2. Can generate output either in pdf or web (html) > 3. Easy to integrate figures and graphs > 4. WSIWYG (would be nice, but not absolutely necessary) > Mathematicians generally use TeX. On Mac (OS 9 or X) you can try OzTeX... Includes OzTtH for web output For PDF, I guess you would produce Postscript output, then convert that (under Mac OS X you link directly to ps2pdf, dvi2pdf, Distill, Acrobat, etc.) ==== > I'm looking for recommendations for document preparation (e.g., Latex, > Word, etc.). Here are my desiderata in priority order: 0. Will run on a Mac > 1. Supports mathematical equations > 2. Can generate output either in pdf or web (html) > 3. Easy to integrate figures and graphs > 4. WSIWYG (would be nice, but not absolutely necessary) > Antonio Check out OpenOffice.org -- supposedly it now runs on OS X, and it's highly MS Office compatible. I'm sure there are plenty of options to generate PDF output -- one way is to generate output for a PS printer, save that output to a file, and convert the PS to PDF. All the software you need is available for free. ==== > 0. Will run on a Mac > 1. Supports mathematical equations > 2. Can generate output either in pdf or web (html) > 3. Easy to integrate figures and graphs > 4. WSIWYG (would be nice, but not absolutely necessary) > As others have stated, TeX or LaTeX is a good choice. (3) can be a bit difficult though, if you're not making them directly in TeX (ie importing image files) Sam -- If sharing a thing in no way diminishes it, it is not rightly owned if it is not shared. - St Augustine ==== >I'm looking for recommendations for document preparation (e.g., Latex, >>Word, etc.). Here are my desiderata in priority order: >>0. Will run on a Mac >>1. Supports mathematical equations >>2. Can generate output either in pdf or web (html) >>3. Easy to integrate figures and graphs >>4. WSIWYG (would be nice, but not absolutely necessary) >>Antonio > Check out OpenOffice.org -- supposedly it now runs on OS X, and it's highly > MS Office compatible. I'm sure there are plenty of options to generate PDF > output -- one way is to generate output for a PS printer, save that output > to a file, and convert the PS to PDF. All the software you need is > available for free. OpenOffice also uses XML as its native file format, so saving as web will be a snap. -- Will Twentyman ==== > I'm looking for a mathematical foundations of statistics type of > textbook.... The news group sci.stat.math or sci.stat.edu may be a better prospect than this one. Have you asked there? Ken Pledger. ==== I am reading about sets and I have a question. If a set A = { 1,3,5,7 } and B = { 5,15,25,35 } is B considered a power set of A? Since each member of B is 5 times each member of A. If there was a set C that was 5 times each member of B and a set D that is 5 times each member of C and so on. It that a power series? Ernst ==== Ernst Berg scribbled the following: > I am reading about sets and I have a question. > If a set A = { 1,3,5,7 } and B = { 5,15,25,35 } is B considered a > power set of A? No. > Since each member of B is 5 times each member of A. That is not the definition of a power set. The power set of a set is the set of all subsets of the original set, including the empty set and the original set itself. > If there was a set C that was 5 times each member of B and a set D > that is 5 times each member of C and so on. It that a power series? I don't know, I don't know the definition of a power series. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ It's time, it's time, it's time to dump the slime! - Dr. Dante ==== > I am reading about sets and I have a question. If a set A = { 1,3,5,7 } and B = { 5,15,25,35 } is B considered a > power set of A? Since each member of B is 5 times each member of A. If there was a set C that was 5 times each member of B and a set D > that is 5 times each member of C and so on. It that a power series? Ernst The power set of A is the set whose elements are listed here: {} (the empty set) {1} {1, 3} {3} {3, 5} {1, 3, 5} {1, 5} {5} {5, 7} {1, 5, 7} {1, 3, 5, 7} {3, 5, 7} {3, 7} {1, 3, 7} {1, 7} {7} You see that the power set of A has 2^4 = 16 elements. If C is the power set of A, then you can create the power set of C, which has 2^16 = 65536 elements. For example, one of these 65536 elements is { {}, {1, 3}, {5}, {3, 7} } The term power series is not used in this connection, although it is intriguing to think of forming a sequence of sets by starting with a set and forming the power set of it, and then the power set of that, and so on. ==== That is very helpful. Ernst > I am reading about sets and I have a question. If a set A = { 1,3,5,7 } and B = { 5,15,25,35 } is B considered a > power set of A? Since each member of B is 5 times each member of A. If there was a set C that was 5 times each member of B and a set D > that is 5 times each member of C and so on. It that a power series? Ernst The power set of A is the set whose elements are listed here: {} (the empty set) > {1} > {1, 3} > {3} > {3, 5} > {1, 3, 5} > {1, 5} > {5} > {5, 7} > {1, 5, 7} > {1, 3, 5, 7} > {3, 5, 7} > {3, 7} > {1, 3, 7} > {1, 7} > {7} You see that the power set of A has 2^4 = 16 elements. If C is the power set of A, then you can create the power set of C, which > has 2^16 = 65536 elements. For example, one of these 65536 elements is { {}, {1, 3}, {5}, {3, 7} } The term power series is not used in this connection, although it is > intriguing to think of forming a sequence of sets by starting with a set and > forming the power set of it, and then the power set of that, and so on. ==== Hum.. The term power had me. I see the point. Then in this case power means all possible combinations including A and the empty set. Interesting. Ernst > Ernst Berg scribbled the following: > I am reading about sets and I have a question. > > If a set A = { 1,3,5,7 } and B = { 5,15,25,35 } is B considered a > power set of A? No. Since each member of B is 5 times each member of A. That is not the definition of a power set. The power set of a set is > the set of all subsets of the original set, including the empty set and > the original set itself. If there was a set C that was 5 times each member of B and a set D > that is 5 times each member of C and so on. It that a power series? I don't know, I don't know the definition of a power series. ==== There is a well known algorithm for sharing a cake fairly between two children. Let one of the children cut the cake and the other pick his/her piece first. The cutter then has an obvious incentive to cut as fairly as possible. I would like to extend this algorithm to larger groups of children (and hopefully larger cakes). For the sake of the algorithm, I am assuming that all children like cake and they are all greedy and will take as much as they can get. My apologies in advance to parents of nice children who are not greedy. Note my aim is not necessarily to perfectly cut the cake, if it was I would give them a ruler, compass and a course in geometry. My aim is to give them a motive to be fair. Here is my first idea. There are n children numbered 1 to n. Child 1 is asked to cut a slice, then child 2, etc up to child n -1. Now child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left with the remaining slice. So what should child 1 do? If he cuts a slice more than 1 / n of the cake then at least one of the remaining slices will be smaller than 1 / n and he will be left with this or a smaller slice. If he cuts less than 1 / n then he will get either this or an even smaller slice. So his best strategy is to cut as fairly as he can. Now with some simple induction, the other children should cut 1 / n of the diminishing remainder of the cake until n - 1 is presented with a 2 / n piece (and many 1 / n pieces) and cuts it fairly. Can anyone see a flaw? Or suggest a better algorithm? Another interesting question is: If one child gets it wrong. e.g. the first child cuts a slice which is too large, what should the following children do? J ==== > There is a well known algorithm for sharing a cake fairly between two > children. Let one of the children cut the cake and the other pick > his/her piece first. The cutter then has an obvious incentive to cut > as fairly as possible. I would like to extend this algorithm to larger groups of children > (and hopefully larger cakes). For the sake of the algorithm, I am > assuming that all children like cake and they are all greedy and will > take as much as they can get. My apologies in advance to parents of > nice children who are not greedy. Note my aim is not necessarily to perfectly cut the cake, if it was I > would give them a ruler, compass and a course in geometry. My aim is > to give them a motive to be fair. Here is my first idea. There are n children numbered 1 to n. Child 1 > is asked to cut a slice, then child 2, etc up to child n -1. Now > child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left > with the remaining slice. So what should child 1 do? If he cuts a slice more than 1 / n of the > cake then at least one of the remaining slices will be smaller than 1 > / n and he will be left with this or a smaller slice. If he cuts less > than 1 / n then he will get either this or an even smaller slice. So > his best strategy is to cut as fairly as he can. Now with some simple induction, the other children should cut 1 / n of > the diminishing remainder of the cake until n - 1 is presented with a > 2 / n piece (and many 1 / n pieces) and cuts it fairly. Can anyone see a flaw? Or suggest a better algorithm? Another interesting question is: If one child gets it wrong. e.g. the > first child cuts a slice which is too large, what should the following > children do? Often this is described as follows. The Dad moves the knife slowly over the cake, starting at the edge and moving the knife so that it remains parallel to its original orientation. As soon as a child cries stop, the Dad cuts off that piece and gives it to the child. The stop-crier thinks that the piece cut off is worth at least 1/n of the value of the cake, and the other children think otherwise. Notice that it is conceivable that every child could get more than 1/n of the value of the cake, if they each value the cake and frosting differently. This theory is not going to keep some child from whining or even throwing a temper tantrum that he/she has been cheated. There is a big literature on cake-cutting algorithms. You can start here: http://mathworld.wolfram.com/CakeCutting.html ==== > Here is my first idea. There are n children numbered 1 to n. Child 1 > is asked to cut a slice, then child 2, etc up to child n -1. Now > child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left > with the remaining slice. This doesn't quite work. Let n = 3 and suppose the first child cuts fairly. The second child has no motivation to make a fair cut: if he cuts unfairly, he can just take that first piece, leaving the first child with only a crumb, perhaps. There is a famous way to divide a cake fairly into n pieces: the moving knife algorithm. After an initial cut, a parent moves the knife slowly around the cake. Whenever a child yells stop, the knife cuts a piece for her. The moving aspect of this algorithm renders it unappealing though. Apparently there are whole books written on this topic. E.g., Cake- Cutting Algorithms: Be Fair If You Can by Jack Robertson and William Webb. -- | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | ==== > There is a well known algorithm for sharing a cake fairly between two > children. Let one of the children cut the cake and the other pick > his/her piece first. The cutter then has an obvious incentive to cut > as fairly as possible. I would like to extend this algorithm to larger groups of children > (and hopefully larger cakes). For the sake of the algorithm, I am > assuming that all children like cake and they are all greedy and will > take as much as they can get. My apologies in advance to parents of > nice children who are not greedy. Note my aim is not necessarily to perfectly cut the cake, if it was I > would give them a ruler, compass and a course in geometry. My aim is > to give them a motive to be fair. Here is my first idea. There are n children numbered 1 to n. Child 1 > is asked to cut a slice, then child 2, etc up to child n -1. Now > child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left > with the remaining slice. Based on this, all cutters should have a FINAL goal of getting down to slices of size 1/n. There are many ways to achieve this, however, if a piece of cake can be cut a second time. For example, if the first child cuts a slice of size 2/n, the next child could cut it in half. The risk comes if the late children come to an agreement... say there are two pieces of size 2/n left and the rest are size 1/n. Then the last child could cut a 1/n in half and know that there will be a 2/n left. > So what should child 1 do? If he cuts a slice more than 1 / n of the > cake then at least one of the remaining slices will be smaller than 1 > / n and he will be left with this or a smaller slice. If he cuts less > than 1 / n then he will get either this or an even smaller slice. So > his best strategy is to cut as fairly as he can. Now with some simple induction, the other children should cut 1 / n of > the diminishing remainder of the cake until n - 1 is presented with a > 2 / n piece (and many 1 / n pieces) and cuts it fairly. Actually, the last child in this case has no incentive to cut the 2/n slice. It depends on whether the last cutter is willing to screw the last choosers to get the favor of the first picker. > Can anyone see a flaw? Or suggest a better algorithm? Another interesting question is: If one child gets it wrong. e.g. the > first child cuts a slice which is too large, what should the following > children do? If the first child cuts a 2/n, the second should cut it in half or risk getting a 1/(2n). If the first child cuts larger than a 2/n, then they will possibly recieve 1/(2n). The others have no incentive to trim it down to 1/n, only to 2/n. J -- Will Twentyman ==== > There is a well known algorithm for sharing a cake fairly between two > children. Let one of the children cut the cake and the other pick > his/her piece first. The cutter then has an obvious incentive to cut > as fairly as possible. I would like to extend this algorithm to larger groups of children > (and hopefully larger cakes). For the sake of the algorithm, I am > assuming that all children like cake and they are all greedy and will > take as much as they can get. My apologies in advance to parents of > nice children who are not greedy. Well known problem. One solution follows: First child cuts slice. Remaining children each in turn get chance to reject the slice as being too small, to accept that slice as being fair or to shave a bit and accept what remains as being fair. Last child to accept slice as being fair keeps the slice. If all children reject the slice, the first child keeps it. The remaining cake and shavings are then divided recursively among the remaining n-1 children. > Here is my first idea. There are n children numbered 1 to n. Child 1 > is asked to cut a slice, then child 2, etc up to child n -1. Now > child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left > with the remaining slice. 3 child case: Child 1 cuts 1/3 leaving 2/3 Child 2 shaves 0 leaving three slices: 0, 1/3, 2/3 Child 3 chooses 2/3 Child 2 chooses 1/3 Child 1 gets 0 Child 2 has no motive to cut fairly and can negotate with child 3 for a kickback of up to 1/3 of a slice for cutting unfairly. Alternatively: Child 1 cuts 1 leaving 0 Child 2 has no choice but to leave 3 slices: 1, 0, 0 Child 3 chooses 1 Child 2 chooses 0 Child 1 gets 0 Child 1 presumably negotiates with child 3 for a kickback of up to 2/3 of a slice for cutting unfairly. Child 3 is in the driver's seat, being able to guarantee himself at least 1/3 of the cake and being able to negotiate for up to 100% of the cake. (Hey, #1. Give me 99% or I'll work with #2 and leave you with nothing). You need to design a solution so that each child can be assured of getting a fair share in spite of arbitrary collusion on the part of all the other children. John Briggs ==== >There is a well known algorithm for sharing a cake fairly between two >children. Let one of the children cut the cake and the other pick >his/her piece first. The cutter then has an obvious incentive to cut >as fairly as possible. I would like to extend this algorithm to larger groups of children >(and hopefully larger cakes). I can't remember where I read the solution, but here it is. The first child cuts a slice. The second child can choose either to make another cut or to take the piece that was cut by the first child. And so on. Again, at each stage a child has incentive to cut as fairly as possible. However, there's one big problem with this: the unstated assumption that there's no collusion. Obviously if the first two children have formed an alliance, the first child could cut a piece measuring 98%, the second child could take it, and they could go off and share it. The other n-2 children would have to share the remaining tiny sliver of cake. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== >There is a well known algorithm for sharing a cake fairly between two >children. Let one of the children cut the cake and the other pick >his/her piece first. The cutter then has an obvious incentive to cut >as fairly as possible. I would like to extend this algorithm to larger groups of children >(and hopefully larger cakes). I can't remember where I read the solution, but here it is. The first child cuts a slice. The second child can choose either to > make another cut or to take the piece that was cut by the first > child. And so on. Again, at each stage a child has incentive to cut > as fairly as possible. However, there's one big problem with this: the unstated assumption > that there's no collusion. Obviously if the first two children have > formed an alliance, the first child could cut a piece measuring > 98%, the second child could take it, and they could go off and share > it. The other n-2 children would have to share the remaining tiny > sliver of cake. -- > Stan Brown, Oak Road Systems, Cortland County, New York, USA Hmmm. There's the makings of a game of some kind here. Like Diplomacy, where (my son tells me, anyway) you make agreements and then break them when it serves your purpose. Anyway, the other n-2 children wouldn't just share the sliver; they would beat up the first two and take their cake. I was a kid once, and I know how they operate :-) ==== > Here is my first idea. There are n children numbered 1 to n. Child 1 > is asked to cut a slice, then child 2, etc up to child n -1. Now > child n chooses a slice then n - 1, then n - 2 etc. Child 1 is left > with the remaining slice. So what should child 1 do? If he cuts a slice more than 1 / n of the > cake then at least one of the remaining slices will be smaller than 1 > / n and he will be left with this or a smaller slice. If he cuts less > than 1 / n then he will get either this or an even smaller slice. So > his best strategy is to cut as fairly as he can. Now with some simple induction, the other children should cut 1 / n of > the diminishing remainder of the cake until n - 1 is presented with a > 2 / n piece (and many 1 / n pieces) and cuts it fairly. Can anyone see a flaw? Or suggest a better algorithm? > Collusion. First child cuts off 95% of cake, last child takes that 95%, keeps 45% and gives 50% to first child for his cut. > Another interesting question is: If one child gets it wrong. e.g. the > first child cuts a slice which is too large, what should the following > children do? > Make the last child eat all of his slice. ==== this problem had been discussed many times before and had a look before I asked. I would not have guessed to look on Mathworld for cake cutting advice. The possibility of collusion is interesting and had not occurred to me. My children and their friends are still a bit too young for that but I expect that it will come soon enough. Designing algorithms which guard against this possibility adds an extra challenge. Could I just modify the order of choosing (the second phase after the cutting) so that it has a random element or at least one not know to the children? If they are not sure when their turn to choose will come, would it not be harder to effectively collude? J ==== > The possibility of collusion is interesting and had not occurred to > me. My children and their friends are still a bit too young for that > but I expect that it will come soon enough. Designing algorithms > which guard against this possibility adds an extra challenge. Could I just modify the order of choosing (the second phase after the > cutting) so that it has a random element or at least one not know to > the children? If they are not sure when their turn to choose will > come, would it not be harder to effectively collude? > So I'm me first and I get the first chop. I take a wild swing and almost miss, except for 5% which gets whacked off. Now this display of ineptness is really a cover for my premeditatedness. The last kid gets the whopping 95% and while he's gloating, I, my big sister and my mean little brother casually saunder over like a tax collector and inform him of his indebtness. Thus we each have a 3/n chance of getting 1/4 of 95%, presuming we're all fair, which of course we're not as we practice trickle down economics. No matter what the rules, a loop hole exists. ==== The possibility of collusion is interesting and had not occurred to > me. My children and their friends are still a bit too young for that > but I expect that it will come soon enough. Designing algorithms > which guard against this possibility adds an extra challenge. Could I just modify the order of choosing (the second phase after the > cutting) so that it has a random element or at least one not know to > the children? If they are not sure when their turn to choose will > come, would it not be harder to effectively collude? So I'm me first and I get the first chop. I take a wild swing and almost > miss, except for 5% which gets whacked off. Now this display of ineptness > is really a cover for my premeditatedness. The last kid gets the whopping > 95% and while he's gloating, I, my big sister and my mean little brother > casually saunder over like a tax collector and inform him of his > indebtness. Thus we each have a 3/n chance of getting 1/4 of 95%, > presuming we're all fair, which of course we're not as we practice trickle > down economics. No matter what the rules, a loop hole exists. Or, big bully brother simply grabs the whole cake and nobody dares to challenge him. But nice mathematically modelled children would not behave like that, would they? J <386aaf52.0307200158.3ae8579d@posting.google.com> ==== So I'm me first and I get the first chop. I take a wild swing and almost > miss, except for 5% which gets whacked off. Now this display of ineptness > is really a cover for my premeditatedness. The last kid gets the whopping > 95% and while he's gloating, I, my big sister and my mean little brother > casually saunder over like a tax collector and inform him of his > indebtness. Thus we each have a 3/n chance of getting 1/4 of 95%, > presuming we're all fair, which of course we're not as we practice trickle > down economics. No matter what the rules, a loop hole exists. Or, big bully brother simply grabs the whole cake and nobody dares to > challenge him. > But I didn't intend to bring the president into this. > But nice mathematically modelled children would not behave like that, > would they? > Indeed uprighteous Republicans, transcending beyond democracy don't behave like gross plebeian Democrats who don't even recount. ==== I'm trying to find integer coefficients for a polynomial which has sqrt{3} + sqrt{2} as a root. Any ideas? thanks. ==== > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? Start with setting x = (the root you want). Since this expression only has square roots, it should be easy to manipulate (moving terms around, squaring) this expression into an equation between a polynomial in x and zero. -- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman ==== > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? thanks. x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) ==== >I'm trying to find integer coefficients for a polynomial which has >sqrt{3} + sqrt{2} as a root. Any ideas? Let x = sqrt(3) + sqrt(2). Then x^2 = 5 + 2*sqrt(6), so sqrt(6) = (x^2 - 5)/2. Can you take it from there? Brian ==== I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) > x = sqr 3 + sqr 2 x^2 = 5 + 2 sqr 6 (x^2 - 5)^2 = 24 Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3? ==== > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? thanks. Start with x = sqrt(2)+sqrt(3) Rearrange such that one side becomes simpler when squared. Square both sides. Rearrange such that one side becomes simpler when squared. Square both sides. Repeat until there's nothing left to do. Note - you'll be doubling the number of roots to the equation with each square, but that can't be avoided. Phil ==== > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? See pages 5 and 6 of my notes on algebraic number theory for a method of solving all such problems (and an example slightly harder than this). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== >> I'm trying to find integer coefficients for a polynomial which has >> sqrt{3} + sqrt{2} as a root. Any ideas? >> x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) > x = sqr 3 + sqr 2 > x^2 = 5 + 2 sqr 6 > (x^2 - 5)^2 = 24 Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3? That depends... if you require the polynomial to have sqrt 2 and sqrt 3 as _the_ roots, then the answer is no. Clearly one polynomial with these roots is f(x) = (x-sqrt(2))(x-sqrt(3)) = x^2 - (sqrt(2)+sqrt(3))x + sqrt(6). Any other polynomial with exactly sqrt(2) and sqrt(3) as roots is a constant times this polynomial. Now If you require the polynomial to belong to Z[x], then you can only multiply f by integers; but since sqrt(6) is irrational, no polynomial in Z[x] is an integer multiple of f. If you only require that sqrt(2) and sqrt(3) are roots of the polynomial, then the answer is clearly yes. Try g(x) = (x-sqrt(2))(x+sqrt(2))(x-sqrt(3))(x+sqrt(3)) /Rasmus -- ==== > Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3? That depends... if you require the polynomial to have sqrt 2 and sqrt > 3 as _the_ roots, then the answer is no. Clearly one polynomial with > these roots is f(x) = (x-sqrt(2))(x-sqrt(3)) = x^2 - > (sqrt(2)+sqrt(3))x + sqrt(6). Any other polynomial with exactly > sqrt(2) and sqrt(3) as roots is a constant times this polynomial. Now > If you require the polynomial to belong to Z[x], then you can only > multiply f by integers; but since sqrt(6) is irrational, no polynomial > in Z[x] is an integer multiple of f. If you only require that sqrt(2) and sqrt(3) are roots of the > polynomial, then the answer is clearly yes. Try g(x) = > (x-sqrt(2))(x+sqrt(2))(x-sqrt(3))(x+sqrt(3)) > Indeed, however (x^2 - 2)(x^2 - 3) is reducible. Is there a irreducible poly including sqr 2, sqr 3 as roots? I think not. Now Q(sqr 6) subset Q(sqr 2, sqr 3) but equality seems most unlikely. Thus for field F it appears, F(u,v) can't be reduced to F(w) for some w. ==== I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? See pages 5 and 6 of my notes on algebraic number theory for a method > of solving all such problems (and an example slightly harder than this). (For the OP): I ended up there while googling for a proof that the algebraic integers form a ring. Most such proofs (as Prof. Chapman does) establish a connection between an algebraic integer and a matrix for which the algebraic number is an eigenvalue, and they show how to construct that matrix explicitly. Given algebraic integers a and b, the matrixes A and B are constructed so that Av = av and Bv = bv, with a common eigenvector v. Then (a+b) is an eigenvalue of (A+B) with eigenvector v (and ab is an eigenvalue of AB). The construction shows you how to relate the matrix to the minimal polynomial. - Randy ==== I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? See pages 5 and 6 of my notes on algebraic number theory for a method > of solving all such problems (and an example slightly harder than this). Very helpful indeed. I have a related question: It is quite simple to show that 1/x is an algebraic number if x is an algebraic number other than zero. It is also quite easy to show that the square root of a non-integer rational number is an algebraic number, but not an algebraic integer. Now the question: Given an arbitrary algebraic number x, for example by specifying a polynomial P with integer coefficients with P(x) = 0, is there a way to find whether or not x is an algebraic integer? I would guess that x is not an algebraic integer if P(x) = 0 where P is a non-monic polynomial with integer coefficents that cannot be divided by any integer constant or integer polynomial other than +/- 1. But I wouldn't have any idea how to prove this. ==== >Given an arbitrary algebraic number x, for example by specifying a >polynomial P with integer coefficients with P(x) = 0, is there a way to >find whether or not x is an algebraic integer? Factor P over the integers. Exactly one of the factors will have P as a root. The coefficient of the highest power of the variable in this factor is 1 or -1 iff x is an algebraic integer. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Given an arbitrary algebraic number x, for example by specifying a >polynomial P with integer coefficients with P(x) = 0, is there a way to >find whether or not x is an algebraic integer? Factor P over the integers. Exactly one of the factors will have P as a > root. The coefficient of the highest power of the variable in this factor > is 1 or -1 iff x is an algebraic integer. http://mathworld.wolfram.com/AlgebraicNumber.html it says (not in these words): If x is the root of an (irreducible) polynomial with integer coefficents then the other roots are called the conjugates of x. If the same x is the root of _any_ other polynomial with integer coefficients then its conjugates are roots as well (proved by Conway and Guy in 1996. That is quite recent so I guess it must be quite hard to prove? ) ==== http://mathworld.wolfram.com/AlgebraicNumber.html it says (not in these words): If x is the root of an (irreducible) > polynomial with integer coefficents then the other roots are called the > conjugates of x. If the same x is the root of _any_ other polynomial > with integer coefficients then its conjugates are roots as well (proved > by Conway and Guy in 1996. That is quite recent so I guess it must be > quite hard to prove? ) Not so hard, and certainly goes back way beyond 1996. Amongst all polynomials with integer coefficients and x as a root, pick out one, P, with minimal degree. If Q is any other polynomial with integer coefficients and x as a root, divide Q by P and get a remainder R of degree less than the degree of P. But P and Q vanish at x, so R does too; P had minimal degree, so R must be identically zero, so Q is actually a multiple of P. It follows that all the roots of P - all the conjugates of x - are roots of Q. Instead of reading mathworld, try something like Galois Theory by Stewart, or the field theory book by Hadlock. -- ==== > Thus for field F it appears, F(u,v) can't be reduced to F(w) for some w. There is a theorem that if F is a field of characteristic zero (such as the rationals) and u and v are algebraic over F then there does exist w such that F(u, v) = F(w). You just have to be a little careful choosing w - not just any old element of F(u, v) will do. -- ==== > http://mathworld.wolfram.com/AlgebraicNumber.html >it says (not in these words): If x is the root of an (irreducible) >polynomial with integer coefficents then the other roots are called the >conjugates of x. If the same x is the root of _any_ other polynomial >with integer coefficients then its conjugates are roots as well (proved >by Conway and Guy in 1996. That is quite recent so I guess it must be >quite hard to prove? ) No, it's very easy to prove, and very old. The reference to Conway and Guy is only a reference, and not a statement that this was the first time it was proved. >polynomial P with integer coefficients and leading coefficient not +/- >1 and P is primitive, i.e. the gcd of the coefficients is 1 >, then it is indeed impossible to find a different polynomial Q with >leading coefficent +/- 1 and root x, so x is not an algebraic integer. The point is that if P_1 and P_2 are in Q[X] and deg(P_1) >= 1, you can write P_2 = A P_1 + B for A, B in Q[X] with deg(A) < deg(P_1). If P_1(x) = 0 and P_2(x) = 0 then of course B(x) = 0. Take P_1 so that deg(P_1) is minimal among members of Q[X] (other than the constant 0) which have x as a root. Then B must be 0, i.e. P_1 divides P_2. If P_2 is irreducible over the rationals, it's not divisible by any P_1 in Q[X] with 0 < deg P_1 < deg P_2, so deg P_2 = deg P_1 and P_2/P_1 is constant. Thus up to multiplication by a constant, the minimal polynomial P_1 is the only member of Q[X] which has x as a root. The conjugates of x are the other roots of P_1, and since P_1 divides any other polynomial in Q[X] which has x as a root, the conjugates are roots of such a polynomial as well. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== http://mathworld.wolfram.com/AlgebraicNumber.html it says (not in these words): If x is the root of an (irreducible) > polynomial with integer coefficents then the other roots are called the > conjugates of x. If the same x is the root of _any_ other polynomial > with integer coefficients then its conjugates are roots as well (proved > by Conway and Guy in 1996. That is quite recent so I guess it must be > quite hard to prove? ) Not so hard, and certainly goes back way beyond 1996. To be fair to mathworld, the parenthetical Conway and Guy 1996 does not mean that they proved it. It just means that book was the source for the preceding entry. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > Indeed, however (x^2 - 2)(x^2 - 3) is reducible. > Is there a irreducible poly including sqr 2, sqr 3 as roots? No, since Q(sqrt(2)) isn't isomorphic to Q(sqrt(3)). -- Jim Heckman ==== > http://mathworld.wolfram.com/AlgebraicNumber.html >> says (not in these words): If r is the root of an (irreducible) >> polynomial with integer coefficents then the other roots are called the >> conjugates of r. If the same r is the root of _any_ other polynomial >> with integer coefficients then its conjugates are roots as well (proved >> by Conway and Guy in 1996. That is quite recent so I guess it must be >> quite hard to prove? ) Not so hard, and certainly goes back way beyond 1996. The paraphrase is misleading. MathWorld doesn't say proved by but instead merely lists (Conway and Guy, 1996) as a reference. > Amongst all polynomials with integer coefficients and r as a root, > pick out one, P, with minimal degree. If Q is any other polynomial > with integer coefficients and r as a root, divide Q by P and get a > remainder R of degree less than the degree of P. But P and Q vanish > at r, so R does too; P had minimal degree, so R must be identically > zero, so Q is actually a multiple of P. It follows that all the > roots of P - all the conjugates of r - are roots of Q. Ideals provide a conceptually nicer way to view this basic result. It's clear that the set of polys f over Q with r as root comprise an ideal I in Q[x]. Since Q[x] has a Division Algorithm, it is a Euclidean domain, hence a PID. Thus I = (g(x)) for some g in Q[x]. Hence: f(r)=0 <=> f in I <=> g|f, i.e. f = gh, for some h in Q[x] The generator g is called the minimal poly of r (often one first normalizes g by clearing denominators and removing any content). g(x) is simply the gcd of all polys having r as a root. Below we see this is an immediate generalization the Euclidean algorithm. As I emphasized in a prior post [1] on related topics, one should always look for algebraic structures hidden in a problem (here the ideal structure). Once the hidden structure has been revealed, one may immediately apply (reuse!) the entire theory of such structures without having to reinvent the wheel from scratch -- which is essentially what occurs above in Gerry's proof, which is nothing but a specialization of the proof that a Euclidean domain is a PID. The proof in the general case remains the same -- just as easy. Constructively it's just gcd computation via Euclid's algorithm, i.e. F(r)=f(r)=0 => (F mod f)(r)=0 =>...=> g(r)=0, g = gcd(F,f) which results from iterated application of the basic descent step viz. (F, f) -> (f, F mod f), where deg F >= deg f. Note how this has the same form as the Euclidean algorithm for integers. Indeed, both are abstracted in the Euclidean domain structure, and the results from Z generalize immediately to any Euclidean domain. Related instructive examples occur throughout the entire thread containing my prior post [1], which I recommend to Christian. -Bill Dubuque [1] http://groups.google.com/groups?selm=y8zy91ndxs2.fsf%40nestle.ai.mit.edu ==== > http://mathworld.wolfram.com/AlgebraicNumber.html it says (not in these words): If x is the root of an (irreducible) > polynomial with integer coefficents then the other roots are called the > conjugates of x. If the same x is the root of _any_ other polynomial > with integer coefficients then its conjugates are roots as well (proved > by Conway and Guy in 1996. That is quite recent so I guess it must be > quite hard to prove? ) Not so hard, and certainly goes back way beyond 1996. The webpage is really quite misleading there. If I read (quotation of a theorem) (Conway and Guy, 1996) I expect that to mean proved by Conway book with all kinds of popular mathematics inside where the theorem is mentioned, and that book was published in 1996. > Amongst all polynomials with integer coefficients and x as a root, > pick out one, P, with minimal degree. If Q is any other polynomial > with integer coefficients and x as a root, divide Q by P and get a > remainder R of degree less than the degree of P. But P and Q vanish > at x, so R does too; P had minimal degree, so R must be identically > zero, so Q is actually a multiple of P. It follows that all the > roots of P - all the conjugates of x - are roots of Q. So all irreducible polynomials with x as a root are identical except for the sign, so if _one_ irreducible polynomial with root x is not monic then _all_ irreducible polynomials with root x are not monic and then _all_ polynomials with root x, whether reducible or not, are not monic, so x is not an algebraic integer. > Instead of reading mathworld, try something like Galois Theory by > Stewart, or the field theory book by Hadlock. Considering that your proof was just seven lines, and absolutely obvious, it is a shame that proofs of this kind are not included at mathworld. On the positive side, I found Robin Chapman's homepage and learnt some stuff from there, posted two questions on sci.math, and in one day I learnt more about algebraic numbers and algebraic integers than some people did in eight years! ==== >> Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. There is?! How careful must one be? Let's consider u = sqr 2, v = sqr 3, n = degree of w, F = Reals Who's the w with R(w) = R(u,v) ? ---- ==== Indeed, however (x^2 - 2)(x^2 - 3) is reducible. > Is there a irreducible poly including sqr 2, sqr 3 as roots? No, since Q(sqrt(2)) isn't isomorphic to Q(sqrt(3)). > Nifty. ==== > Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. > There is?! How careful must one be? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? > What's wrong with w=sqrt(2)+sqrt(3) ?? Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) . Of course if you really mean F=R, then R(w) = R and R(u,v) = R so just any old element WILL do. Was that your point? ==== > Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. > There is?! How careful must one be? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? What's wrong with w=sqrt(2)+sqrt(3) ?? > Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) . > ww = 5 + 2sqr 6 (w/2)(2sqr6 - 4) = (sqr 6 - 2)(sqr 2 + sqr 3) = sqr 12 - 2.sqr 2 + sqr 18 - 2.sqr 3 = sqr 2 (-w/2)(2.sqr 6 - 6) = -(sqr 6 - 3)(sqr 2 + sqr 3) = -(sqr 12 - 3.sqr 2 + sqr 18 - 3.sqr 3) = sqr 3 Nothing. ww - 5 = 2.sqr 6 (ww - 5)^2 = 24 w^4 - 10w^2 - 1 = 0 > Of course if you really mean F=R, then R(w) = R and R(u,v) = R so just > any old element WILL do. Was that your point? > No, F = Q. Does the general theorem have a name? How complicated is the proof? ==== > >> Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. >There is?! How careful must one be? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals >Who's the w with R(w) = R(u,v) ? > >What's wrong with w=sqrt(2)+sqrt(3) ?? >>Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) . >> >ww = 5 + 2sqr 6 >(w/2)(2sqr6 - 4) = (sqr 6 - 2)(sqr 2 + sqr 3) > = sqr 12 - 2.sqr 2 + sqr 18 - 2.sqr 3 = sqr 2 >(-w/2)(2.sqr 6 - 6) = -(sqr 6 - 3)(sqr 2 + sqr 3) > = -(sqr 12 - 3.sqr 2 + sqr 18 - 3.sqr 3) = sqr 3 >Nothing. ww - 5 = 2.sqr 6 >(ww - 5)^2 = 24 >w^4 - 10w^2 - 1 = 0 >Of course if you really mean F=R, then R(w) = R and R(u,v) = R so just >>any old element WILL do. Was that your point? >> >No, F = Q. Does the general theorem have a name? > Theorem 5.3 (or whatever). >How complicated is the proof? > > It's in the chapter on Galois theory. If the book is conversational, it's easy. If it's formal, it could be pretty hard to understand. (YMMV, some people prefer formal to conversational.) It's not complicated at all, it's straightforward and usually done right where it makes sense. It's just that some books are hard to follow because of the way they're written. (Or the way I process what I read.) I like Van der Waerden, _Algebra_ or _Modern Algebra_, depending on your version. I'm suggesting it because from some other things you seem to like an old-fashioned approach to algebra, so your point of view may agree with his and make it really easy. But any algebra book that covers Galois theory will do. (Not Herstein _Topics in Algebra_, which I consider the best introduction to algebra.) Jon Miller ==== ... >> Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. > There is?! How careful must one be? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? (Let's do it over rationals, yes? u and v are elements of the reals.) If z = sqrt(2) + sqrt(3), (z^3 - 9z)/2 = sqrt(2). So w = z works. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== [.snip.] >>Does the general theorem have a name? >Theorem 5.3 (or whatever). Or the Primitive Element Theorem, since it guarantees the existence of a primitive element for any finite separable extension in those fields. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > So all irreducible polynomials with x as a root are identical except for > the sign, so if _one_ irreducible polynomial with root x is not monic > then _all_ irreducible polynomials with root x are not monic and then > _all_ polynomials with root x, whether reducible or not, are not monic, > so x is not an algebraic integer. Not quite, e.g., x^2 - 2 and 7 x^2 - 14 are both considered irreducible. Sometimes the word primitive is used to describe a polynomial with integer coefficients with no non-trivial common factor. Then there are only two primitive irreducible polynomials with a given x as a root, and each is the negative of the other. -- ==== [in re: the Primitive Element Theorem,] > ...any algebra book that covers Galois theory will do. I think the exception here is Artin's book. If I remember some other discussion correctly, Artin had something against using this theorem in an exposition of Galois Theory, and studiously avoided it. -- ==== > Indeed, however (x^2 - 2)(x^2 - 3) is reducible. > Is there a irreducible poly including sqr 2, sqr 3 as roots? No, since Q(sqrt(2)) isn't isomorphic to Q(sqrt(3)). Nifty. Another approach: Every algebraic element over F is the root of a *unique* monic irreducible polynomial over F. For sqrt(2) it's (x^2 - 2), and ... -- Jim Heckman ==== > > Indeed, however (x^2 - 2)(x^2 - 3) is reducible. > Is there a irreducible poly including sqr 2, sqr 3 as roots? No, since Q(sqrt(2)) isn't isomorphic to Q(sqrt(3)). Nifty. Another approach: Every algebraic element over F is the root of > a *unique* monic irreducible polynomial over F. For sqrt(2) it's > (x^2 - 2), and ... The number z = sqrt(3) + sqrt(2), is a zero of x^4 - 10*x^2 + 1, which shows that z is an algebraic integer and the reciprocal of an algebraic integer. ==== [in re: the Primitive Element Theorem,] ...any algebra book that covers Galois theory will do. I think the exception here is Artin's book. If I remember some other > discussion correctly, Artin had something against using this theorem > in an exposition of Galois Theory, and studiously avoided it. If you're talking about Michael Artin's _Algebra_, he does indeed prove the Primitive Element Theorem. In fact, he then goes on to use it in his proof that For any finite extension K/F, the order |G(K/F)| of the Galois group divides the degree [K:F] of the extension. One thing I find somewhat disconcerting about Artin's exposition of Galois Theory is that he proves several important theorems out of order, i.e., he states them early on, proving them only in later sections. To my mind this makes following the logical development of the ideas a little trickier than need be. -- Jim Heckman ==== >> [in re: the Primitive Element Theorem,] >> ...any algebra book that covers Galois theory will do. >> I think the exception here is Artin's book. If I remember some other >> discussion correctly, Artin had something against using this theorem >> in an exposition of Galois Theory, and studiously avoided it. If you're talking about Michael Artin's _Algebra_, No, he's talking about Emil Artin's book on Galois theory. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== [in re: the Primitive Element Theorem,] ...any algebra book that covers Galois theory will do. I think the exception here is Artin's book. If I remember some other > discussion correctly, Artin had something against using this theorem > in an exposition of Galois Theory, and studiously avoided it. Yes, but Artin still discusses when a field is generated by a single element in his book Galois Theory. It appears on page 64 (of 82 pages) in section M called Simple Extensions. In this section, Artin gives two theorems. The first theorem gives a necessary and sufficient condition for the existence of primitive elements: Theorem 26. A finite extension E of F is primitive over F if and only if there are only a finite number of intermediate fields. Theorem 27 then gives the situation being discussed in this thread. To apply Theorem 27 theorem, we need to note that sqrt(2) and sqrt(3) are separable elements over Q. That is, the irreducible polynomials for sqrt(2) and sqrt(3) do not have repeated roots. I have not yet grasped the impact that separable elements has in Galois theory. I just usually assume that I am working over extensions of Q, where separable follows because the characteristic of Q is 0. To find a couterexample to having a primitive element, I think that you need to have a finite extension E of F, where F is an infinite field of characteristic p. But, I haven't pursued this. -- Bill Hale ==== [in re: the Primitive Element Theorem,] ...any algebra book that covers Galois theory will do. I think the exception here is Artin's book. If I remember some other > discussion correctly, Artin had something against using this theorem > in an exposition of Galois Theory, and studiously avoided it. Yes, but Artin still discusses when a field is generated by a single element in his book Galois Theory. It appears on page 64 (of 82 pages) in section M called Simple Extensions. In this section, Artin gives two theorems. The first theorem gives a necessary and sufficient condition for the existence of primitive elements: Theorem 26. A finite extension E of F is primitive over F if and only if there are only a finite number of intermediate fields. Theorem 27 then gives the situation being discussed in this thread. To apply this theorem, we need to note that sqrt(2) and sqrt(3) are separable elements over Q. That is, the irreducible polynomials for sqrt(2) and sqrt(3) do not have repeated roots. I have not yet grasped the impact that separable elements has in Galois theory. I just usually assume that I am working over extensions of Q, where separable follows because the characteristic of Q is 0. To find a couterexample to having a primitive element, I think that you need to have a non-finite extension E of Z_p, the finite field of order p. But, I haven't pursued this. -- Bill Hale ==== I'm trying to find integer coefficients for a polynomial which has >sqrt{3} + sqrt{2} as a root. Any ideas? > Sure. Set r=sqrt(2)+sqrt(3). Guess there is polynomial of degree 4 meeting your requirements. Set x0[0]=1, x1[0]=r, x2[0]=r^2, x3[0]=r^3, and x4[0]=r^4. Compute the following until x0[i0]=0: x0[i+1]=x1[i]-int(x1[i]/x0[i])*x0[i] . . x3[i+1]=x4[i]-int(x4[i]/x0[i])*x0[i] x4[i+1]=x0[i] At this point you know there are integers a0,..,a4 s.t. a0+..a4*r^4=0. An alternative is to use PSLQ or LLL methods to find an integer relation for 1,r,..,r^4. Kind of overkill in this case though. Rich Burge ==== I'm trying to solve an algebra problem and I came up with four equations (and some side conditions) that I think can be condensed. Here goes: Let G be an abelian group, finite, generated by x and y. Let H be a cyclic subgroup of G. Also, suppose it is known that X = c(X-Y), Y = (c-1)(X-Y), and G/H = , where X is the coset of X, Y is the coset of Y, etc. I've shown that blahblahblahblahblah is true if and only if there exist relatively prime positive integers m_1, n_1 such that 0 <= [G/H:]*n_1 <= |G/H|-1 and m_1*order(X) + n_1*([G/H:]-e_1) = |G/H|-1, where e_1 is the unique integer satisfying 0 <= e_1 <= order(X) and [G/H:]*Y = e_1*X _and_ there exist relatively prime positive integers m_2, n_2 such that 0 <= [G/H:]*n_2 <= |G/H|-1 and m_2*order(Y) + n_2*([G/H:]-e_2) = |G/H|-1, where e_2 is the unique integer satisfying 0 <= e_2 <= order(Y) and [G/H:]*X = e_2*Y. I suspect I can (using relationships between X and Y) reduce the number of equations (and conditions) needed to express this information. Here's what I've tried so far: Since order(X)=|G/H|/gcd(c,|G/H|) and order(Y)=|G/H|/gcd(c-1,|G/H|), we must have order(X)/order(Y) = gcd(c-1,|G/H|)/gcd(c,|G/H|). Call this quotient 1/D. Then we have blahblahblahblah is true iff there exist relatively prime positive integers m_1, n_1 such that 0 <= |G/H|/order(X) * n_1 <= |G/H|-1 and m_1*order(X) + n_1*(|G/H|/order(X) - e_1) = |G/H|-1, where e_1 is the unique integer satisfying 0 <= e_1 <= order(X) and |G/H|/order(X) * Y = e_1*X _and_ there exist relatively prime positive integers m_2, n_2 such that 0 <= |G/H|/(D*order(X)) * n_2 <= |G/H|-1 and m_2*order(X)*D + n_2*(|G/H|/(D*order(X)) - e_2) = |G/H|-1, where e_2 is the unique integer satisfying 0 <= e_2 <= order(X)*D and |G/H|/(D*order(X)) * X = e_2*Y. Even if these equations can't be expressed as a set of fewer equations, is there anything we can say number-theoretically about |G/H|, |H|, or order(X) and order(Y) (besides the _obvious_ inequalities obtained from simply combining these equations)??? Anyone proficient with Mathematica want to give this one a whirl and post any simplifications you discover? (I don't have access to it or any other symbolic algebra package from where I currently work). ==== Let G be an abelian group, finite, generated by x and y. Let H be a > cyclic subgroup of G. Also, suppose it is known that X = c(X-Y), Y = > (c-1)(X-Y), and G/H = , where X is the coset of X, Y is the > coset of Y, etc. I've shown that > I'll presume G is additive as it's Abelian group. H is subgroup. What's 'c'? Some integer? What's meant 'X is coset of X' ? Do you mean 'for x,y in G, let X = x+H, Y = y+H' ? Then -Y = -y+H, X-Y = x-y + H, c(X-Y) = c(x-y) + cH ? > blahblahblahblahblah > What's that? Bush's latest public speach? > is true if and only if there exist relatively prime positive integers > m_1, n_1 such that 0 <= [G/H:]*n_1 <= |G/H|-1 and > m_1*order(X) + n_1*([G/H:]-e_1) = |G/H|-1, where e_1 is the unique integer satisfying 0 <= e_1 <= order(X) and > [G/H:]*Y = e_1*X _and_ there exist relatively prime positive integers m_2, n_2 such > that 0 <= [G/H:]*n_2 <= |G/H|-1 and > m_2*order(Y) + n_2*([G/H:]-e_2) = |G/H|-1, where e_2 is the unique integer satisfying 0 <= e_2 <= order(Y) and > [G/H:]*X = e_2*Y. I suspect I can (using relationships between X and Y) reduce the > number of equations (and conditions) needed to express this > information. Here's what I've tried so far: Since order(X)=|G/H|/gcd(c,|G/H|) and order(Y)=|G/H|/gcd(c-1,|G/H|), > we must have order(X)/order(Y) = gcd(c-1,|G/H|)/gcd(c,|G/H|). Call > this quotient 1/D. Then we have blahblahblahblah is true iff there exist relatively prime positive integers m_1, n_1 such that 0 <= |G/H|/order(X) * n_1 <= |G/H|-1 and > m_1*order(X) + n_1*(|G/H|/order(X) - e_1) = |G/H|-1, where e_1 is the unique integer satisfying 0 <= e_1 <= order(X) and > |G/H|/order(X) * Y = e_1*X _and_ there exist relatively prime positive integers m_2, n_2 such > that 0 <= |G/H|/(D*order(X)) * n_2 <= |G/H|-1 and > m_2*order(X)*D + n_2*(|G/H|/(D*order(X)) - e_2) = |G/H|-1, where e_2 is the unique integer satisfying 0 <= e_2 <= order(X)*D and > |G/H|/(D*order(X)) * X = e_2*Y. Even if these equations can't be expressed as a set of fewer > equations, is there anything we can say number-theoretically about > |G/H|, |H|, or order(X) and order(Y) (besides the _obvious_ > inequalities obtained from simply combining these equations)??? Anyone > proficient with Mathematica want to give this one a whirl and post any > simplifications you discover? (I don't have access to it or any other > symbolic algebra package from where I currently work). > ==== I think the O.P. mistyped (or just miscapitalized). Of course c should be an integer, and X is the coset of H in G containing x, Y is the coset of H in G containing y, etc.... ==== >then x = pi+20 = 23.1415... > >and x1 = ln( x ) = 3.1416315... > >next x2 = sin( x1 ) = 0.05480... Take your calculator out of degree mode. Things are in radians > by default. Actually, since ln(x) is just a bit larger than pi, sin(x1)<0 > and small. Thus sin(x2)<0 and small. --Dan Grubb ====== Let A:= sin(f(pi +20)). Until now I am not sure which inequality A>0 or A<0 is satisfied. Perhap A<0. Please same question regarding B:= sin(sin(A)) , or C=sin(B). Is B positive or B<0 ? Similar question for C . ====== ==== > then x = pi+20 = 23.1415... > > >and x1 = ln( x ) = 3.1416315... > > >next x2 = sin( x1 ) = 0.05480... > >Take your calculator out of degree mode. Things are in radians >>by default. >>Actually, since ln(x) is just a bit larger than pi, sin(x1)<0 >>and small. Thus sin(x2)<0 and small. >>--Dan Grubb >> >====== >Let A:= sin(f(pi +20)). Until now I am not sure which > inequality A>0 or A<0 is satisfied. >Perhap A<0. >Please same question regarding B:= sin(sin(A)) , or C=sin(B). >Is B positive or B<0 ? Similar question for C . > > Why don't you want to calculate? Jon Miller ==== Singapore's math program is being hailed by many as a cure for what ails American math instruction (Strauss, WASHINGTN POST, 3/21). The country's math curriculum is being distributed in the United States. The Singapore curriculum promote a versatility in basic math skills that makes it easier for students to venture later into more difficult problem-solving, reports the paper. Teachers who have used the Singapore text praise it for moving form basic to more advanced math in a logical sequence, according to the paper. - Washington Post. Now you can get the Singapore Testpapers online from just US$6.80. These test papers are set by teachers from top schools in Singapore. A must for anyone who wanted to learn the Singapore math program. Visit our site at http://www.knowledgesoup.com for free sample. ==== Counting pencils and drawing snakes - why couldn't we think of that? ==== > Singapore's math program is being hailed by many as a cure for what > ails American math instruction (Strauss, WASHINGTN POST, 3/21). The > country's math curriculum is being distributed in the United States. The authorized exclusive distributor for most of the best mathematics materials from Singapore in North America is http://www.singaporemath.com/ and I find their product line-up very helpful. (This is not, strictly speaking, an ad. I don't work for the company; I'm just a satisfied customer.) -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ ==== > Singapore's math program is being hailed by many as a cure for what > ails American math instruction (Strauss, WASHINGTN POST, 3/21). [snipped] > Who knew it would be so easy? I'm sure that a math program intended for elite students in a totally different country with a totally different set of educational goals will be the cure to all our education ills. /sarcasm ==== > Singapore's math program is being hailed by many as a cure for what > ails American math instruction (Strauss, WASHINGTN POST, 3/21). The > country's math curriculum is being distributed in the United States. > The Singapore curriculum promote a versatility in basic math skills > that makes it easier for students to venture later into more difficult > problem-solving, reports the paper. Teachers who have used the > Singapore text praise it for moving form basic to more advanced math > in a logical sequence, according to the paper. - Washington Post. Now you can get the Singapore Testpapers online from just US$6.80. > These test papers are set by teachers from top schools in Singapore. > A must for anyone who wanted to learn the Singapore math program. > Visit our site at http://www.knowledgesoup.com for free sample. I worked in Singapore as a Maths teacher in 2000 and I was surprised then that they were talking about the US buying into their methods. The students (young adults) I worked with were a lovely bunch of people, but showed absolutley no independent thought. Give them a problem they had never seen before and there was panic in the room ... but, would you believe it, the pass rate for the courses was generally something like 95% + !! I really hope the US were a little skeptical of the stats they were fed. ==== >> Singapore's math program is being hailed by many as a cure for what >> ails American math instruction (Strauss, WASHINGTN POST, 3/21). The >> country's math curriculum is being distributed in the United States. >> The Singapore curriculum promote a versatility in basic math skills >> that makes it easier for students to venture later into more difficult >> problem-solving, reports the paper. Teachers who have used the >> Singapore text praise it for moving form basic to more advanced math >> in a logical sequence, according to the paper. - Washington Post. >> Now you can get the Singapore Testpapers online from just US$6.80. >> These test papers are set by teachers from top schools in Singapore. >> A must for anyone who wanted to learn the Singapore math program. >> Visit our site at http://www.knowledgesoup.com for free sample. >I worked in Singapore as a Maths teacher in 2000 and I was surprised >then that they were talking about the US buying into their methods. >The students (young adults) I worked with were a lovely bunch of >people, but >showed absolutley no independent thought. Give them a problem they >had never seen before and there was panic in the room ... but, would >you believe it, >the pass rate for the courses was generally something like 95% + !! The difference between training and education; education should prepare for problems of types not seen before. This sounds like the common situation, which is enhanced by teaching facts and methods. Skills can be promoted, but understanding, and even the ability to acquire understanding, is _New York Times_, which pointed out that memorization, beyond the minimum necessary, interferes with the development of language pathways in the brain. >I really hope the US were a little skeptical of the stats they were >fed. The US educationists think in terms of skills, not understanding. -- 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, Deptartment of Statistics, Purdue University ==== > Singapore's math program is being hailed by many as a cure for what > ails American math instruction It was also hailed by the many as worth huge underwriting by national tax monies. A committe has been formed for the purpose of planning a vast number of pilot programs to be administered by an army of nincompoops who will be ordered to busy themselves with yet more empowerment prose and charts and graphs showing the need for yet further funding. All hail. ==== Singapore's math program is being hailed by many as a cure for what > ails American math instruction It was also hailed by the many as worth huge underwriting by national tax > monies. A committe has been formed for the purpose of planning a vast > number of pilot programs to be administered by an army of nincompoops who > will be ordered to busy themselves with yet more empowerment prose and > charts and graphs showing the need for yet further funding. All hail. Just be careful that you don't chew gum in public!!! I don't think that living in SingaClone would be compensated for by the supposed edge in Mathematical prowess. RJ Pease ==== I worked in Singapore as a Maths teacher in 2000 and I was surprised > then that they were talking about the US buying into their methods. The students (young adults) I worked with were a lovely bunch of > people, but > showed absolutley no independent thought. Give them a problem they > had never seen before and there was panic in the room ... but, would > you believe it, > the pass rate for the courses was generally something like 95% + !! I really hope the US were a little skeptical of the stats they were > fed. I've been a teacher (to be exact, TA) at a supposedly elite American university too. Not only do the American students display a stunning inability to produce independent thought, but they can't even solve standard problems which they've seen before. And to think it's one of the top 5 colleges in the country.... ==== School, Advanced, and Challenge. Please visit us at http://math.smsu.edu/~les/POTW.html [I continue to slog through the backlog.] ==== Solutions to (x^p+y^p)/(x+y)=z^p for x,y,z integers and p odd prime. For example: (21^3+7^3)/(21+7)= 7^3 Here's a nice little trick to find integer solutions: rewrite (x^p+y^p)/(x+y)=z^p as (x/z)^p+(y/z)^p=x+y let x=ay/b ((ay)/(bz))^p+(y/z)^p=y(a+b)/b y^p[(a/(bz))^p+(1/z)^p]=y(a+b)/b which simplifies to y^(p-1)=(z^p)(b^(p-1))(a+b)/(a^p+b^p). Close examination of the above will show that choosing a and b to be any integers (except a+b=0) will determine x,y and z such that (x^p+y^p)/(x+y)=z^p. More difficult to show is that x+y can never equal k^p. ==== This looks to be an interesting way to prove the given proposition without using the axiom of choice. I think if one assumes axiom of choice, which is required is the sets are infinite, the following argument can be made. If there exists an injection from X to Y, then card Y >= card X. Similarly, card X >= card Y. Therefore card X = card Y. ==== > This looks to be an interesting way to prove the given proposition > without using the axiom of choice. I think if one assumes axiom of > choice, which is required is the sets are infinite, the following > argument can be made. If there exists an injection from X to Y, then > card Y >= card X. Similarly, > card X >= card Y. Therefore card X = card Y. > Circular reasoning as the Cantor-Bernstein theorem is used to prove that. ==== This looks to be an interesting way to prove the given proposition > without using the axiom of choice. I think if one assumes axiom of > choice, which is required is the sets are infinite, the following > argument can be made. If there exists an injection from X to Y, then > card Y >= card X. Similarly, > card X >= card Y. Therefore card X = card Y. Circular reasoning as the Cantor-Bernstein theorem is used to prove that. I believe Cantor proved the result as a consequence of his well-ordering theorem, which thus relies on the Axiom of Choice. The result proved without such dependence is more often called the Schroeder-Bernstein Theorem. For a concise proof without Choice see: http://www.wikipedia.org/wiki/Cantor-Bernstein-Schroeder_theorem ==== said: > If N is the cardinality of the infinite countable sets and 2^N the > cardinality of the reals (or the power set of N), I think I have found a > more exhaustive way to extract all the 2^N - N reals that are not one to > one and onto with countable sets in the diagonal representation of > Cantor's proof. 2^N - N = 2^N, btw. > As Godel and Cohen proved that this question is independent of the > current set theory axioms, we have to find either an intuitive axiom > that gives us sets of intermediate cardinality or one that prevent us > from finding those sets... Ain't no such thing. Recommended reading: Penelope Maddy, Believing the axioms. I and II, Journal of Symbolic Logic 53 (1988), pp. 481-511, 736-764. Chris Menzel ==== >>You guys seem to define aleph_1 as the >>next biggest cardinal after aleph_0, and for you the CH states >>that |R| = aleph_1. >Another way to describe aleph_1 is the number of distinct ways to well-order >a countable set. (A set is well-ordered if it is totally ordered and every >nonempty subset has a least element.) ... > I guess I wasn't clear enough about what I meant by distinct. > By distinct well-orderings I mean well-orderings that are not > order-isomorphic. An order isomorphism is a one-to-one correspondence > that preserves the ordering (i.e., x < y if and only if their counterparts > f(x) and f(y) satisfy f(x) < f(y)). Is there an obvious proof that this is the same as the smallest cardinal greater than Aleph-0? There must be, otherwise the definitions would not be equivalent. Ralph Hartley ==== | Is there an obvious proof that [the cardinality of the reals] is the | same as the smallest cardinal greater than Aleph-0? There must be, | otherwise the definitions would not be equivalent. No. That's Cantor's Continuum Hypothesis. It's been proved to be independent of the other standard axioms of set theory. In other words, you can assume that it's true, or you can assume that it's false; either way you get a consistent model. -- Jeff Erickson jeffe@cs.uiuc.edu Computer Science Department http://www.uiuc.edu/~jeffe University of Illinois at Urbana-Champaign ==== > | Is there an obvious proof that [the cardinality of the reals] is the I did not say or mean the cardinality of the reals. I meant the cardinality of the countable ordinals, (or of isomorphism classes of well orderings of the integers, which I think is the same thing). > | same as the smallest cardinal greater than Aleph-0? There must be, > | otherwise the definitions would not be equivalent. Ralph Hartley ==== >You guys seem to define aleph_1 as the >next biggest cardinal after aleph_0, and for you the CH states >that |R| = aleph_1. >>Another way to describe aleph_1 is the number of distinct ways to well-order >>a countable set. (A set is well-ordered if it is totally ordered and every >>nonempty subset has a least element.) > ... >> I guess I wasn't clear enough about what I meant by distinct. >> By distinct well-orderings I mean well-orderings that are not >> order-isomorphic. An order isomorphism is a one-to-one correspondence >> that preserves the ordering (i.e., x < y if and only if their counterparts >> f(x) and f(y) satisfy f(x) < f(y)). > Is there an obvious proof that this is the same as the smallest cardinal > greater than Aleph-0? There must be, otherwise the definitions would not be > equivalent. Aleph_1 is the cardinality of omega_1, the first uncountable ordinal. The countable ordinals are all the ordinals less than omega_1. For the ordinals, less than is defined to mean is a member of, as a set. Thus, the members of omega_1 are precisely the countable ordinals, and the cardinality of omega_1 is the cardinality of the set of countable ordinals, which is equivalent to the number of distinct ways to well-order a countable set. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >>Is there an obvious proof that this is the same as the smallest cardinal >>greater than Aleph-0? There must be, otherwise the definitions would not be >>equivalent. Aleph_1 is the cardinality of omega_1, the first uncountable ordinal. ... But you need to prove that there are no uncountable *sets* with cardinality less than any uncountable ordinal. Ralph Hartley ==== >Is there an obvious proof that this is the same as the smallest cardinal >greater than Aleph-0? There must be, otherwise the definitions would not be >equivalent. Suppose S is an uncountable set. By AC, we may assume that S is equipped with a well-ordering. Now we need the fact that every well-ordered set is order-isomorphic to some (von Neumann) ordinal. Let sigma be the ordinal that is order-isomorphic to S. Since S is uncountable, sigma can't be a countable ordinal. So it must be greater than every countable ordinal, and therefore contains every countable ordinal as a member. So the cardinality of S is at least the cardinality of the set of all countable ordinals, i.e., the set of all isomorphism classes of countable well-ordered sets. Conversely, the set of all (at most) countable ordinals is an ordinal, which can't be countable because that would make it a member of itself, contradicting the well-foundedness of the von Neumann ordinals. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== >Is there an obvious proof that this is the same as the smallest cardinal >>greater than Aleph-0? There must be, otherwise the definitions would not be >>equivalent. Suppose S is an uncountable set. By AC, we may assume that S is equipped > with a well-ordering. Actually, I didn't have to think too long to prove it using AC. So what you meant to say was that aleph_1 is the number of distinct (up to isomorphism) ways to well-order a countable set. And that in ZFC this is the smalest cardinal bigger than Aleph_0. Presumably in ZF it might not be (is there a proof in ZF?), so this isn't a good *definition* of Aleph_1. Looking back in the thread I don't think anyone mentioned Choice one way or the other. Ralph Hartley ==== >So what you meant to say was that aleph_1 is the number of distinct (up to >isomorphism) ways to well-order a countable set. >And that in ZFC this is the smalest cardinal bigger than Aleph_0. >Presumably in ZF it might not be (is there a proof in ZF?), so this isn't a >good *definition* of Aleph_1. Aleph_1, more precisely, is usually defined to be the set of all at-most-countable ordinals. If your definition of cardinal automatically requires them to be ordinals, then Aleph_1 will be the smallest cardinal bigger than Aleph_0, with or without AC. If you define the cardinality of x to be the least ordinal a equinumerous to x, if x is well-orderable, and the set of all sets y of least rank which are equinumerous to x, otherwise, and a cardinal to be the cardinality of some set, then I don't think you can prove in ZF that there is a *smallest* cardinal larger than Aleph_0. So I think defining Aleph_1 in terms of countable ordinals rather than as the smallest cardinal larger than Aleph_0 is actually a good thing if you're not assuming AC. >Looking back in the thread I don't think anyone mentioned Choice one way or >the other. In this kind of discussion, usually people assume AC unless otherwise specified. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== >Is there an obvious proof that this is the same as the smallest cardinal >greater than Aleph-0? There must be, otherwise the definitions would not be >equivalent. >> Aleph_1 is the cardinality of omega_1, the first uncountable ordinal. > ... > But you need to prove that there are no uncountable *sets* with cardinality > less than any uncountable ordinal. That follows from the axiom of choice. Every well-ordered set is order-isomorphic to a unique ordinal, and therefore every uncountable well-ordered set has an ordinal type that is >= omega_1, by the trichotomy law for ordinals. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >Is there an obvious proof that this is the same as the smallest cardinal >greater than Aleph-0? There must be, otherwise the definitions would not be >equivalent. >> Suppose S is an uncountable set. By AC, we may assume that S is equipped >> with a well-ordering. > Actually, I didn't have to think too long to prove it using AC. > So what you meant to say was that aleph_1 is the number of distinct (up to > isomorphism) ways to well-order a countable set. > And that in ZFC this is the smalest cardinal bigger than Aleph_0. > Presumably in ZF it might not be (is there a proof in ZF?), so this isn't a > good *definition* of Aleph_1. > Looking back in the thread I don't think anyone mentioned Choice one way or > the other. AC is generally implicit when discussing cardinals. The very definition of a cardinal (= an initial ordinal) depends on AC, since without it there are sets that cannot be well-ordered and therefore don't have a cardinality according to that definition. The usual alternative is to define a cardinal to be an equivalence class of sets. But then we don't get things like the trichotomy law: an uncountable set certainly can't be smaller than N, but it need not be bigger either. It may simply be incomparable with N, in the sense that neither set can be injected into the other. But even without AC, the concept of aleph_1 (= omega_1) still makes sense, and it still holds that no uncountable cardinality can be less than aleph_1. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >> But it is shown later that aleph0 >> < aleph1, a concept that boggles my mind. How can aleph1 actually >> exist?! How can one infinite set be greater than another infinite set? >> This is obviously not clear to me. The set of the integers have no >> end, there is no greatest integer so they go on forever. Same thing >> with the set of irrationals >> Aha. But whereas the integers merely go *on* forever, the irrationals >> also go *in* forever! Think of writing down the decimal representations >> of all the irrational numbers: once you start writing the digits of one >> irrational number, you'll never finish even *that* number. >> (Rational numbers, on the other hand, don't go *in* forever. Eventually, >> every rational number starts repeating, and we can just make up a notation >> that says, this number repeats here, and go on to the next one. That's >> why they're countable. Sort of. ;) >> This is a pretty nice explanation! I also read in another book that in >to arrive. Now, following the logic that irrationals go *in* forever I >hotel has infinitely many rooms, it seems that each room is of finite >space, that's why aleph-1 guests would overflow the hotel. Rooms in >the hotel don't provide infinite space (which would be required to >store guests as large as an irrational number). So, the way I see it, the difference between the infinity of the >integers (aleph-0) and the infinity of the irrationals (aleph-1) seems >to be that the irrationals are infinite in two dimensions (think of a >two dimensional array as opposed to a one dimensional array). The >infinity of the inegers would be the one dimensional array. Hmm... can >we then say that aleph-2 is a three dimensional array, infinite on all >three dimensions?! I would suggest that you read the previous items in this thread, because most of your misconceptions has already been addressed. Larry (this space unintentially left blank ..... ==== > experiences in the military, where I actually had the honor of giving > a lecture on the physics of lasers to the medical personnel at Madigan > Army Medical Center, including the surgeons, other doctors and nurses, > for their medical continuing education credits, I feel like I can > speak confidently on the subject. Radiation Protection Office, Madigan Army Medical Center who posted (which is archived at Vanderbilt). Now the question is: Are you the same James Harris who later posted to RADSAFE as James.Harris@rfets.gov, K-H Manager, 771 Radiological Safety in February 2001? If so, I'd *love* to hear all about the safety violations in Building 771 (including workers being fined $385,000 a few months after you posted that message. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== On Tue, Jul 8, A N Niel inscribed on the eternal scroll: > In simpler words: which char should I use for the square root on my html > pages so that it has a chance to be interpreted/read correctly most of > the time using the most common encodings and fonts? > Here is a table... > With respect, you're referring to a table which was topical in around 1995; the action had moved on considerably by Y2K. Indeed I was using a wide range of mathematical symbols in HTML4 in 1998 and getting very acceptable results, and we're now half a decade beyond that. > If your character is not there, I think you cannot expect most > browsers to show it properly. I can assure you that the majority of installed browser instances have had the capability for quite some time of being able to render a much richer character repertoire. It needed a bit of persuasion with NN4.* versions, but it's certainly possible. There have been some problems with the fact that operating systems (I'm thinking particularly of windows) have installed by default with a rather poor font repertoire, although the wider repertoires have been available to be installed for the asking if wanted. Test pages which might be recommended are http://www.unics.uni-hannover.de/nhtcapri/mathematics.html http://www.alanwood.net/unicode/mathematical_operators.html etc. or indeed my own http://ppewww.ph.gla.ac.uk/~flavell/unicode/unidata22.html All of these display well for me in modern browsers on Windows platforms (Mozilla and derivatives, MSIE5/5.5/6 etc.), even - with minor adjustments - on NN4.* versions. And I understand Macs can also. If not for other Windows users, then they should look to their fonts. Some unix/linux platforms have had some problems, but I think this is getting sorted out now. ==== There is some way to extract the square root (or, in general, the nth root) of a one-variable polynomial? This is elementary for some sorts of polynomials (say, x^2 + 6x + 9 = (x+3)^2), but how about the general case? What is, for instance, the cubic root of P(x) = x+1? I've heard of an algorithm for extracting square roots of polynomials, based on a similar one for numbers, but I don't know any references. I tried using the Newton method for solving equations, and came up with: Let A(x) be the polynomial whose root is to be extracted, P(x) the root, and P_0(x) some initial polynomial. Then, applying the Newton method for the equation F(x) = (P(x))^r - A(x), P_(n+1)(x) = P_n(x) - (F_n(x) / F'_n(x)) (n >= 0) where F_n(x) = (P_n(x))^r - A(x). In F'(x), I'm differentiating in what space, R or the space of functions R -> R? There is something wrong in this method? What? It works? There is any Web references about nth roots of polynomials? Duran Castore (duran_castore@yahoo.com) ==== >There is some way to extract the square root (or, in general, the nth >root) of a one-variable polynomial? This is elementary for some sorts >of polynomials (say, x^2 + 6x + 9 = (x+3)^2), but how about the >general case? What is, for instance, the cubic root of P(x) = x+1? It is (x+1)^(1/3). There is no simpler way to write it (and it's certainly not a polynomial). In general, a polynomial p(x) has a r'th root which is a polynomial if and only if all its roots have multiplicities which are multiples of r. >I've heard of an algorithm for extracting square roots of polynomials, >based on a similar one for numbers, but I don't know any references. I tried using the Newton method for solving equations, and came up >with: Let A(x) be the polynomial whose root is to be extracted, P(x) the >root, and P_0(x) some initial polynomial. Then, applying the Newton >method for the equation F(x) = (P(x))^r - A(x), P_(n+1)(x) = P_n(x) - (F_n(x) / F'_n(x)) (n >= 0) where F_n(x) = (P_n(x))^r - A(x). In F'(x), I'm differentiating in what space, R or the space of >functions R -> R? >There is something wrong in this method? What? It works? There is any >Web references about nth roots of polynomials? You can consider this as just the ordinary Newton's method for solving the equation A = P^r for the variable P, where A happens to involve the parameter x: P_{n+1} = P_n - (P_n^r - A)/(r P_n^(r-1)) = (1-1/r) P_n + A/(r P_n^(r-1)) The result after n iterations will be a rational function of x; if it works (which I think it does if A >= 0 or n is odd), then the convergence will be uniform on compact sets of reals away from the zeros of A. Well, this can't work in the complex plane, because of the branch cut: if A(z) is not the r'th power of a polynomial, it has a root whose multiplicity is not a multiple of r, and it will be impossible to approximate any branch of A(z)^(1/r) by a rational function uniformly on a closed path around that root. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== There is some way to extract the square root (or, in general, the nth > root) of a one-variable polynomial? This is elementary for some sorts > of polynomials (say, x^2 + 6x + 9 = (x+3)^2), but how about the > general case? What is, for instance, the cubic root of P(x) = x+1? I've heard of an algorithm for extracting square roots of polynomials, > based on a similar one for numbers, but I don't know any references. I tried using the Newton method for solving equations, and came up > with: Let A(x) be the polynomial whose root is to be extracted, P(x) the > root, and P_0(x) some initial polynomial. Then, applying the Newton > method for the equation F(x) = (P(x))^r - A(x), P_(n+1)(x) = P_n(x) - (F_n(x) / F'_n(x)) (n >= 0) where F_n(x) = (P_n(x))^r - A(x). In F'(x), I'm differentiating in what space, R or the space of > functions R -> R? There is something wrong in this method? What? It works? There is any > Web references about nth roots of polynomials? > Duran Castore (duran_castore@yahoo.com) Did you learn as a kid that old method for extracting square roots of numbers, where you group the digits in pairs from the decimal point outwards? Well, this does work for polynomials, to give you a series solution at 0. In fact, it works for Taylor series in general. Assume that the Taylor series of the function is 1 + a1 x + a2 x^2 + a3 x^3 + a4 x^4 + ... Group the terms in pairs: 1 + (a1 x + a2 x^2) + (a3 x^3 + a4 x^4) + ... and do what you learned to do for square roots, mutatis mutandis. It's rather fun. My students have not seen the old fashioned way of extracting square roots, and invariably when I show it to them, they are greatly fascinated by it. I think you can expect your Newton's method to work in a neighborhood of any x value where the polynomial is not 0. I looked at f(x) = (1+x)^(1/4). Starting with P_0 = 1, and iterating P0^4 - (1+x) P1 = P0 - ---------------- 4 P0^3 and so forth, you get rational functions. For example, 4 + x P1= ----- 4 1024 + 1024*x + 288*x^2 + 48*x^3 + 3*x^4 P2= ---------------------------------------- 1024 + 768*x + 192*x^2 + 16*x^3 The approximation seems to be pretty good, but not as good as the Pade approximations of the same degree. For example, the degree of the numerator of P3 is 16, that of the denominator, 15. Denoting by R the Pade approximant to (1+x)^(1/4) of degrees 16 and 15, I get f(-.95) = 0.4728708 R(-.95) = 0.4728711 P3(-.95) = 0.5079 Of course, this comparison is unfair because a lot more computation went into R than into P3. ==== >>There is some way to extract the square root (or, in general, the nth >>root) of a one-variable polynomial? This is elementary for some sorts >>of polynomials (say, x^2 + 6x + 9 = (x+3)^2), but how about the >>general case? What is, for instance, the cubic root of P(x) = x+1? It is (x+1)^(1/3). There is no simpler way to write it (and it's >certainly not a polynomial). In general, a polynomial p(x) has >a r'th root which is a polynomial if and only if all its roots have >multiplicities which are multiples of r. On an unrelated basis, the OP's question suggests me another one. A subject often recurring in this ng is that of compositional square root functions. Now I know that in the general case the problem is not easy, but maybe in finite fields, where all functions are in fact polynomials, it may be easier and of a certain combinatorial interest. Are there any studies in this sense? Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc ==== >> There is some way to extract the square root (or, in general, the nth >> root) of a one-variable polynomial? This is elementary for some sorts >> of polynomials (say, x^2 + 6x + 9 = (x+3)^2), but how about the >> general case? What is, for instance, the cubic root of P(x) = x+1? On an unrelated basis, the OP's question suggests me another one. A > subject often recurring in this ng is that of compositional square > root functions. Now I know that in the general case the problem is not easy, but maybe > in finite fields, where all functions are in fact polynomials, it may > be easier and of a certain combinatorial interest. Are there any > studies in this sense? Polynomial decomposition algorithms are much studied in computer algebra since, e.g., they have applications to solving polynomials and simplifying field extensions. A web search on polynomial decomposition along with names like: Ritt, Whaples, Fried, Schinzel, Barton, Zippel, Kozen, Landau, Gutierrez, von zur Gathen, Binder, etc. should discover much of interest, e.g. follow the links below for starters. -Bill Dubuque http://wwwuser.gwdg.de/~cais/CAR/CAR24/node6.html http://www.cs.cornell.edu/kozen/papers/poly.ps http://www.algebra.uni-linz.ac.at/~xbx/pub/DA/DA.ps http://www.eecis.udel.edu/~saunders/papers/sparse-interp2/issac/issac.ps http://math-www.uni-paderborn.de/preprints/preprints_data/Gathen/bivariateDe c.ps.gz ==== >Polynomial decomposition algorithms are much studied in computer algebra >since, e.g., they have applications to solving polynomials and simplifying >field extensions. A web search on polynomial decomposition along with >names like: Ritt, Whaples, Fried, Schinzel, Barton, Zippel, Kozen, Landau, >Gutierrez, von zur Gathen, Binder, etc. should discover much of interest, >e.g. follow the links below for starters. -Bill Dubuque http://wwwuser.gwdg.de/~cais/CAR/CAR24/node6.html >http://www.cs.cornell.edu/kozen/papers/poly.ps >http://www.algebra.uni-linz.ac.at/~xbx/pub/DA/DA.ps >http://www.eecis.udel.edu/~saunders/papers/sparse-interp2/issac/issac.ps >http://math-www.uni-paderborn.de/preprints/preprints_data/Gathen/bivariateD ec.ps.gz I expected that the subject had already been studied, but I couldn't imagine to such an extent. OTOH, surprisingly, it seems not to have been of any particular combinatorial interest (e.g. How many indecomposable polynomials are there in Z_p[X]?) Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc ==== In the conclusion of your excellent Ukraine.doc paper, you mentioned the Brookhaven RHIC project whose goal was to create the cosmic soup that the inflationary model of the Big Bang predicted existed at the start. [note: I don't know how this compares to later Ekopyrotic models which utilize a slow collision of 5-D membranes instead that produces the same results without a need for gravity waves or monopoles] As I recall, many scientists world-wide objected to this project because of conditons that might arise and prove catastrophic to the Earth (if not the Universe). Strangelets was one such possibility. (In fact, didn't someone propose it was a strangelet snaking across the Universe that was responsible for that Siberian explosion?) One of the major questions in the UFO field has always been WHY are they here? Some suggest we are either direct creations of them or have been genetically altered over thousands-of-years by them. A more selfish reason is what your ending paragraphs alluded to -- that being simple self-preservation. In their eyes, we are tinkering with forces that -- under the right situations -- cause them as much harm as ourselves. That would certainly explain why UFOs and atomic energy (especially weapons testing/production) seem to go hand-in- hand. If the latter is minimized, then so are the UFO sightings. Yes, exactly. Before I finish, I appreciate the years of soul-searching, reflection, and debates with colleagues over all these theories. They offer one side of the equation. The other side being of course is the technologies required to create 'exotic matter' and bend/fold spacetime. I realize theorists are immune to this part, but are there any clues as to what form this technology would be in (e.g., rotating magnetic fields, nano-engineered Type III superconductors, quantum lasers, etc.)? Any good books/papers/websites that you could recommend? The basic equations are toward the end of my paper on the Josephson coupling of to <0|e+e-|0>. A big problem is the impedance matching, see Ray Chiao's gravity radio papers on a related problem - it's not identical to what I am talking about but not orthogonal either. There are no papers in the field other than my own because I have only recently created this new approach. It has not sunk in and it may take years. The flying saucers show us that a technology of the /zpf field exists only that US DOD, DOE, NSA, CIA et-al do not have it. I am the best they got at the moment. In fact I am ALL they got that's real pointing the right way IMO of course. Hal Puthoff & Co's PV theory represents what they hope they have and that theory is a joke IMHO. See the recent Mitre Conference on HFGR. Of course what I'm asking for is a Manhattan Project equivalent where you had all the theorists on one side and the engineers coming up with schemes to make the former's equations come to fruition. This has been the most frustrating part for me all along. I think I could understand the physics better if I had some clue as to what physical mechanisms were to be employed to make it happen, captain! That's what Joe Firmage wanted to do with ISSO. He made unwise financial decisions and the project only lasted a year and a half 1999 - 2000. That's what CIA Station Chief Harold Chipman wanted to do in 1985. Those documents are in my book Destiny Matrix. However, the time was not ripe, I did not have the right ideas back then. The right ideas had to wait for the discoveries of both dark matter first and dark energy second. The full experimental picture only was made available to and Beyond II at end of 2002 anticipate the results of NASA WMAP that were confirmed. That's what Hal Puthoff's IAS is supposed to be about but he is barking up the wrong tree, same for Eric Davis's Warp Metrics. That's what Senator Robert Byrd's ISR in W. Va was supposed to be about also with Jim Corum's stuff, but that does not work either as we found out at ISSO with our million dollar contract with SARA of Huntington Beach where Corum came from before he went to ISR. So far The Right Stuff you see in http://qedcorp.com/APS/Ukraine.doc is a One Man Show. WYSYYG. However there are some interesting theory things brewing in Russia, Finland and Kiev in the Ukraine, and possibly some experimental stuff in Beograd with the J.P. Vigier group. My theory does justify Vigier's basic idea for cold fusion i.e. tight atomic states i.e. spatially extended electron as a Bohm hidden variable, that appears isn't in low energy scattering and most importantly in atomic bound states. This is what the Serbs are trying to achieve in a practical way. Will they succeed? I don't know. I am not in touch with them though maybe I will see them in Paris this September. We had Vigier at ISSO and that was one good thing in terms of development of my theory in http://qedcorp.com/APS/Ukraine.doc (1) Making a Wormhole just got easier É => http://www.nature.com/nsu/030527/030527-12.html (MS-Word version archived at http://www.stealthskater.com/Documents/Time_02.doc). This is important. Very good for my program at http://qedcorp.com/APS/Ukraine.doc to appear in Progress in Quantum Physics Research (Nova Scientific Publishers) Good news for time travellers - it just got cheaper. The amount material needed to build a window through time is infinitesimally small, new research shows. To travel through time, all you need to do is open a wormhole in space-time and step through it. And to do that you need a magic ingredient called 'exotic matter', which is repelled rather than attracted by gravity. The hitch is that no one has the remotest idea how to make exotic matter. But don't despair, say Matt Visser, of the Victoria University of Wellington in New Zealand, and his colleagues. They have shown that when we do figure out how to make the stuff, we won't need very much of it (1). The Pundits have not yet realized that the dark energy that is ~ 73% of the large-scale structure of the universe for scale L > 10 megaparsecs, is, on the smaller scales, exactly what they are looking for! They do not understand that the Einstein cosmological constant / in the spatially flat K = 0 FRW post-inflationary bubble metrics is simply a limit of a local field Wigner wavelet /zpf (x,L) when L --> c/Ho ~ 10^26 meters and the x dependence at that scale is very weak. Normal equilibrium vacuum (G.E. Volovik, Universe in a Helium the dominating ODLRO vacuum-polarization coherence <0|e+(x)e-(x)|0>. Upsetting this delicate equilibrium with Josephson links using real superconductors, impedance matched to the physical vacuum. can make /zpf be positive or negative with negative (repulsive) and positive (attractive) zero point energy pressures respectively because w = pressure/(energy density) = -1 and the pressure dominates the energy density by a factor of 3 in Einstein's geometrodynamic generalized source Newtonian-Poisson equation for the effective potential energy per unit test mass of the exotic vacuum volume element at coarse-grained position x and at scale L in the sense of wavelet transform theory. It's nice to know that we do not need a large spacetime region of exotic vacuum /zpf(x,L) to do the tricks that we see flying saucers in fact doing rendering our entire present-day military defense technology essentially impotent and obsolete should such advanced knowledge fall into the hands of small groups of irresponsible individuals. Sir Martin Rees discusses this in Our Final Hour. See also the book The Star Gate Conspiracy by Picknett & Prince and my two books Destiny Matrix and Space-Time and Beyond II for a set of alternate POV's on this topic. Also see http://www.fiu.edu/~mizrachs/techgnosis.html and http://www.techgnosis.com/ As Star Trek: Deep Space 9 ,Quantum Leap and Stargate have taught us, wormholes are the preferred mode of transport for today's fashionable time-traveller. These hypothetical tunnels connect distant parts of space-time, the fabric of our Universe. And despite the philosophical havoc that wormholes wreak with notions of causality, Einstein's theory of general relativity - which describes space-time - allows them to exist. Six years ago, Visser and his colleague David Hochberg showed that in order to stay open, wormholes need exotic matter. It's weird stuff, however - it can be considered to have negative energy, meaning that it has even less than empty space. It's the same as saying that it experiences gravity as a repulsive force, and physicists have never encountered anything of the sort. Nature (i.e., the magazine not GOD(D) in the Quad ;-)) has screwed up here. Their explanation is wrong. What matters is the sign of the pressure. In fact, for w = -1 exotic vacua, our universe is ~ 96% exotic vacuum stuff of both signs in proportion 73% repulsive exotic matter /zpf > 0, and 23% attractive dark matter /zpf < 0 with negative and positive pressures respectively. It's the sign of the pressure that counts! So they imagine it. They key to exotic matter lies in quantum fluctuations, which give empty space a kind of fizziness. Quantum theory continually popping in and out of existence in the vacuum of empty space. Exotic matter might arise by suppressing this fizz, or as a physicist would say, by violating the averaged null energy condition (ANEC). Bingo! Exactly, I show how the macro-quantum coherence <0|e+(x)e-(x)|0> suppresses this fizz in http://qedcorp.com/APS/Ukraine.doc If this were to happen, quantum effects could give rise to tiny amounts of exotic matter. But how much is needed to sustain a > wormhole? That is what Visser and colleagues have now calculated. They find that, if the wormhole is designed carefully, the total quantity of ANEC-violating matter can be made infinitesimally small. This makes a wormhole considerably easier to create. Exactly. That's what the flying saucer have been telling us. That is why what Eric Davis proposes in MUFON 2001 is entirely the wrong approach needing impractical enormous amounts of electromagnetic field stress-energy density to get a very useless small bang for an impossibly large buck since G/c^4 = 10^-33 cm per 10^19 Gev i.e. space-time is too stiff to bend directly with a Tuv source field in Einstein's geometrodynamic field equation Guv = -8pi(G/c^4)Tuv What you need to do the trick is Guv + /zpfguv = -8pi(G/c^4)Tuv ~ 0 where /zpf is controlled by Josephson links tweaking around /zpf = 0. Back to the future again It's not the first time that traversable wormholes have been pulled out of a pit of implausibility. In the 1980s the British astrophysicist Stephen Hawking conjectured that even if you could make a wormhole stabilized by exotic matter, you couldn't go through it to travel in This became known as the Chronology Protection Conjecture. It was a relief for philosophers who were trying to protect the notion of causality. The paradox they envisioned - immortalised in the movie Back to the Future - was that if wormholes could exist it would theoretically be possible to go back in time and prevent your parents from meeting. This would prevent your own existence, and therefore your ability to go back in time. I suspect that Hawking's conjecture here is wrong. We have evidence it is wrong from the flying saucers. It's like saying we could not break the sound barrier. {It's not the first time that traversable wormholes have been pulled out of a pit of implausibility But physicists subsequently thought of a way around this problem - there are, for example, 'time loops' threading through a wormhole along which backwards time travel is possible, but without its being able to alter the future. Sadly, an infinitely small amount of exotic matter is not the same as none at all. This is an example of Sir Michael Berry's singular limit. So until someone figures out how to get hold of it (not to mention how to open up a wormhole in the first place), you can forget about trips to the Jurassic era - or your parents' first date. Don't count on that. The times they are a chang'in. See Time Travel: The Art of the Possible http://www.dvdjournal.com/reviews/s/startrek04voyage.shtml The Star Trek Universe In the mini-documentary Time Travel: The Art of the Possible (11:14), three physicists (mainly me), taped separately, provide their thoughts on the real-world hypotheses behind the concept of time travel. Physics 201 topics are discussed and playfully illustrated, from basic Einsteinian relativity to Carl Sagan's wormhole devised for the novel (and film) Contact . It's an interesting, if only superficially relevant, addition that would be right at home on the Science or Discovery channels. References Visser, M., Kar,S. &Dadhich, N. Traversable wormholes with arbitrarily small energy condition violations. Physical Review Letters ,90, 201102 , (2) New concept of black holes proposes they are a special form of 'dark energy' and not a singularity => Frozen Stars at http://www.sciam.com. What you have been taught in school is almost certainly wrong, because classical black hole spacetimes are inconsistent with quantum mechanics, says physicist George Chapline of Lawrence Livermore National Laboratory. (MS-Word version archived at http://www.stealthskater.com/Documents/BlackHole_01.doc) robbins@math.sfsu.edu, hpw9@pacbell.net, JagdishM@aol.com, swordbuster@earthlink.net, lloomis@bendnet.com, MHoustonx@aol.com, cochran@vancouver.wsu.edu, beenergy@telusplanet.net, pandolfi@zzapp.org, postmaster@fbi.gov, itjobs@nsa.gov, casajobs@nsa.gov, langjobs@nsa.gov, iajobs@nsa.gov, mathjobs@nsa.gov, casajobs@nsa.gov, coopjobs@nsa.gov, genjobs@nsa.gov, vice.president@whitehouse.gov, president@whitehouse.gov, postmaster@vatican.va, president@kremlin.ru, ornet@ossrom.va, investor@newscorp.com, Shareowner-svcs@bankofny.com ==== G'day mate! Had to put down my didgeridoo before finishing my walkabout in the outback to type up this shite, but, Cri-kee mate, WORMHOLES??! The normies will laugh, the crankcases will empathise, the intellects will humidor and roam their luggage, and the bottlenecks will crack their willies against the copy machines, but I and my ilk, the blistering paintlike splotches doubling as airbrushed simulacrum renderings of deep space quasarheaded goo-bots will resonate, uh huh... yes we will; give it up homie! > Making a Wormhole just got easier > http://www.nature.com/nsu/030527/030527-12.html Physical Review Letters _Traversable Wormholes with Arbitrarily Small Energy Condition Violations_ by Matt Visser, Sayan Kar, Naresh Dadhich Traversable wormholes necessarily require violations of the averaged null energy condition, this being the definition of exotic matter. However, the theorems which guarantee the energy condition violation are remarkably silent when it comes to making quantitative statements regarding the total amount of energy condition violating matter in the spacetime. We develop a suitable measure for quantifying this notion and demonstrate the existence of spacetime geometries containing traversable wormholes that are supported by arbitrarily small quantities of exotic URL: http://link.aps.org/abstract/PRL/v90/e201102 Article: http://www.nature.com/nsu/030527/030527-12.html U R U K S T A R G A T E coordinates: ERIDU, where the Annunaki used to be. Latitude: 30.7975000 N Longitude: 45.9777778 E Eridu Thumbnail Images: http://www.oi.uchicago.edu/OI/IS/SANDERS/PHOTOS/MESO/ERIDU/eridu1_1.html Archaeological Site Photography: Mesopotamia http://www.oi.uchicago.edu/OI/IS/SANDERS/PHOTOS/meso_map.html In ancient Mesopotania, music, mathematics, art, science, religion, and poetic fantasy were fused. Around 3000 B.C., the Sumerians simultaneously developed cuneiform writing, in which they recorded their pantheon, and a base-60 number system. Their gods were assigned numbers that encoded the primary ratios of music, with the gods' functions corresponding to their numbers in acoustical theory. Thus the Sumerians created an extensive tonal/arithmetical model for the cosmos. In this far-reaching allegory, the physical world is known by analogy, and the gods give divinity not only to natural forces but also to a 'supernatural,' intuitive understanding of mathematical patterns and psychological forces. -- E.G. McClain (McClain, Ernest G.; Musical Theory and Ancient Cosmology, The World and I, p. 371, February 1994. Cr. L. Ellenberger) <> E BOMB: In the blink of an eye, electromagnetic bombs could throw <> civilization back 200 years. And terrorists can build them for $400. <> BY JIM WILSON <> http://popularmechanics.com/science/military/2001/9/e-bomb/print.phtml <> <> Islamic Studies Pathways <> http://www.lamp.ac.uk/cis/pathways/pathways.html <> HAMAS <> http://www.palestine-info.co.uk/ <> http://www.hizbollah.org <> Azzam Publications <> http://www.azzam.mirrorz.com/ <> Who Is The Mahdi? <> http://islamicweb.com/history/mahdi.htm <> £&$£&$£&$£&$£&$£& $£&$£&$£&$£&$£&$£ &$£&$£&$£&$£&$£&$[Sterling ]&$£&$ [Excerpt from: 'Everything is Under Control' by Robert Anton Wilson - 1998 ISBN 0-06-273417-2 - http://www.rawilson.com] The Con, short for _The Conspiracy_, controls all the other (and lesser) conspiracies you ever heard of, and some you never heard of or even imagined. The indentity of the Con and all its members is known to J.R. _Bob_ Dobbs, founder, Mahatma, Messiah, and CEO of the _Church of the Sub-Genius_/Sub-Genius Foundation. The Con includes the _Bilderbergers, Trilateral Commission Supporters_, the _Illuminati_, communist clones, _Nazi hell creatures_, interstellar bankers, and the leaders of all rival churches and cults. All _Pinks_ (normal or adjusted humans) are indentured servants of the Con. Many think the Con is just a joke or a parody of other conspiracy theories. To such doubters, the Church of the Sub- Genius says that this is the _Time of Pee_--the time foretold, when people would be judged not by works, nor by family, nor even by looks, but by urine. They listen to you through your telephone without its even being off the hook, and record you through satellites that can peer down any street, _anywhere_... They kick your door in anytime they want to. All they have to yell is 'DRUGS!' and your spouse is in jail, your kids are farmed out to the state, your car and house are suddenly theirs... Nobody up there is a friend of yours; nobody up there wants you to have what you would call freedom. The purpose of 'government' is to produce consumers and workers who will keep the cost of labor down, and the profits high for the owners... For this has become so crooked and perverse a nation that your precious bodily fluids are no longer your own, and not even your bladder or bloodstream are private. _There is no place where they may not watch._ According to the _Book of Urinomics_, an ancient sub-genius text recently published by _Relelation X_, And the Beast said: 'By their pee shall ye judge them, and by thy pee shall ye be judged. And all will be divided by their pee. And in the snow shall their names be written. See also: Bob, Mona Charen, Government as Criminal Conspiracy, Corey Hammond, S.O.B. http://www.subgenius.com _Revelation X_, translated from the original tongues by the Sub-Genius Foundation, Simon and Schuster, New York, 1994 1953: http://xray.sai.msu.su/~mystery/images/photo/msu1953/ Actually, I was not part of the official RV unit(s). I was an NCO attached to a Naval R&D unit which was developing the operational parameters for lasar-guided weapons back in the 70's. We've had this conversation before. By association with the work I was doing and a few DIA reports I was able to chart the course of every major event in the Middle East for the next 20 years. I was doing RV in the late 70's on DIA intelligence briefs without realizing I was doing it until the events all occured just as I had predicted them. I made a number of predictions that got me so spooked when they actually happened, I came to the conclusion that many folks who suffer paranoid delusions are actually precognitive and just misinterpret their experiences as CIA mind control, a direct link to God, outrageous ideas of influence, aliens controlling the planet like Courtney Brown's crew believed in Atlanta, etc... I wonder whether or not there was a connection between the Farsight Institute and those poor saps who decided to take a ride on the comet [Hale Bop]??? In 1983 I was referred to SRI by one of their employees when I told her about my impressions of what would happen to the Marine Expeditionary Force in Beirut. I didn't get too far with that due to the sensitive nature of the research studies and their source of funding. I'm also a believer in precognitive remote viewing based upon experience and a review of the literature in the field. If there was nothing to it in terms of statistical results I'm quite certain we would not have wasted our time and effort for so many years in the SRI, SAIC and Ft. Meade studies. Unfortunately, far too many lunatic fringers have jumped on the RV band wagon since Stargate's declassification in 1995, and they have muddied the waters as it were for the genuine talents in the field, and researchers who pursue Theories in Consciousness. I've been experiencing precognition since 1969 and had subjectively given it a lot of thought. Most psi phenomenon can be explained by precognition, and we do it by association. I'm still trying to learn what I can about the subject of consciousness. We're all precognitive because there exists a point where the limits of physical form and spacetime are non-factors relative to our minds. What causes this to be so is up for speculation, but I've been talking about it since 1969. Ken Schlueter Enki and The World Order http://www.earth-history.com/Ancient-texts/Sumer/sumer-enki-worldorder.htm ... When the royal scepter was coming down from heaven, the august crown and the royal throne being already down from heaven, he (the king) regularly performed to perfection the august divine services and offices, laid the bricks of those cities in pure spots. They were named by name and allotted half-bushel baskets. The firstling of those cities, Eridu, she gave to the leader Nudimmud, the second, Bad-Tibira, she gave to the prince and the sacred one, the third, Larak, she gave to Pabilsag, the fourth, Sippar, she gave to the gallant Utu. The fifth, Shuruppak, she gave to Ansud... -- The Eridu Genesis http://www.earth-history.com/Ancient-texts/Sumer/sumer-eridu-genesis.htm SEEDS OF HISTORY IN MYTH AND RELIGION http://www.gatewaystobabylon.com/gods/partnerships/enkienlil.html THE DESTRUCTION OF IRAQI CULTURAL HERITAGE http://www.let.leidenuniv.nl/rencontre/Statement.html[..] We urge all our members to sign the petition published on the 'Threat to World Heritage in Iraq' web site: http://users.ox.ac.uk/~wolf0126, or the petiton on the website: http://www.stoplootersofhistory.babylonians.net Hunt for Stolen Iraqi Antiquities Moves to Cyberspace: THE ORIENTAL INSTITUTE http://www-oi.uchicago.edu/OI/IRAQ/iraq.html http://www-oi.uchicago.edu/OI/IRAQ/Iraqdatabasehome.htm THE ORIENTAL INSTITUTE http://www-oi.uchicago.edu/OI/IRAQ/categories.htm Newly Launched 'Opportunity' Follows Mars-Bound 'Spirit' %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Smart People Believe Weird Things Rarely does anyone weigh facts before deciding what to believe By Michael Shermer ... Smart people believe weird things because they are skilled at defending beliefs they arrived at for nonsmart reasons. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The Schizophrenia Help Home Page http://www.schizophrenia-help.com/index.htm Shocking Secrets Of The Crab Pulsar: http://chandra.harvard.edu/press/02_releases/press_091902.html Rencontre Assyriologique Internationale http://www.let.leidenuniv.nl/rencontre/default.html Project 'The Physical Features of Tablets' Marie-Christine Ludwig (Ludwig.walker@virgin.net) Christopher Walker (Cwalker@thebritishmuseum.ac.uk) http://www.let.leidenuniv.nl/rencontre/Physical_Features.html http://news.nationalgeographic.com/iraq.html Human origins explored through science, archaeology and research http://www.goldenageproject.org.uk/homo.html TREASURES FROM THE ROYAL TOMBS OF UR THE MESOPOTAMIAN TRADITION OF THE FLOOD http://mcclungmuseum.utk.edu/specex/ur/ur-flood.htm Eridu: http://www.oi.uchicago.edu/OI/IS/SANDERS/PHOTOS/MESO/ERIDU/eridu1_1.html Iraq Museum - Baghdad http://info.uibk.ac.at/c/c6/c616/museum/museum.html Sumerian: http://info.uibk.ac.at/c/c6/c616/museum/sumerian.html Basra: http://info.uibk.ac.at/c/c6/c616/museum/basra1.jpg [] M a r t i a n S t a r g a t e [] Why Mars? Because it is humanity's destiny to strive, to seek, to find. And because it is America's destiny to lead. -- George Bush CLICK--> http://www.enterprisemission.com/images/Boeing-M2.jpg [] It is also America's destiny to Fold, to Spindle, to Mutilate. And market exploitation videotapes called: Girls Gone Wild! CLICK--> http://www.girlsgonewild.com/free/payperview/ppv0004.jpg [] Mars Exploration Program http://mars.jpl.nasa.gov/ [] Uruk, home of Gilgamesh.* [] Ziggurat di Uruk: http://www.edicolaweb.net/irak_02g.htm Gilgamesh tomb believed found! Archaeologists in Iraq believe they may have found the lost tomb of King Gilgamesh - the subject of the oldest book in history... http://news.bbc.co.uk/2/hi/science/nature/2982891.stm In the book - actually a set of inscribed clay tablets - Gilgamesh was described as having been buried under the Euphrates, in a tomb apparently constructed when the waters of the ancient river parted following his death. [It was] like Venice in the desert. http://news.bbc.co.uk/2/hi/science/nature/2982891.stm Iraq: Archaeological Expedition Mapping Ancient City Of Uruk http://www.rferl.org/nca/features/2002/05/03052002101632.asp Van Ess: The Euphrates and Tigris through the millennia were always principal trading routes...but when the Sassanians, an Iranian dynasty, conquered Mesopotamia at the end of the third century A.D., they tried to focus on their own trade centers in Iran and to strengthen the [overland] trade routes [that passed] through Iran to China. [] Today, archaeologists are eager to study Uruk and other Mesopotamian cities, but doing so has been severely complicated by the Iraq crisis... http://www.rferl.org/nca/features/2002/05/03052002101632.asp [] * Gilgamesh Epic http://www.wsu.edu/~dee/MESO/GILG.HTM [] The skies roared with thunder and the earth heaved, Then came darkness and a stillness like death. Lightening smashed the ground and fires blazed out; Death flooded from the skies. When the heat died and the fires went out, The plains had turned to ash. --Gilgamesh [] The Epic of Gilgamesh http://www.ancienttexts.org/library/mesopotamian/gilgamesh/ Mesopotamia http://www.mesopotamia.co.uk/index.html Sumerian Language http://www.sumerian.org/sumerian.htm Sumerian Mythology http://www.crystalinks.com/sumermythology.html [] Choctaw wisdomkeeper Sequoyah Trueblood http://www.nancyredstar.com/stan/jose.htm [] The reason we went into Iraq (Sumer, the cradle of civilization) was not because of oil, not because of genocide, not because of Weapons of Mass Destruction, but because there's a Stargate there. - Agent Daniel M. Salter [] [Daniel M. Salter - French Comanche, Former CIA/NRo Agent Agent Daniel M. Salter is a retired counter-intelligence agent for the Scientific and Technical Unit of Interplanetary Phenomena. He was a CONRAD courier for President Eisenhower and a member of the Pilot Air Force, the National Reconnaissance Office (NRO), the CIA, and the Development of Conscious Contact Citizenry Department (DCCCD), with the United States military. He taught courses on electromagnetic anti-gravitational propulsion systems as a professor emeritus at Mountain View College in Texas. Of Comanche and French descent, Daniel is the father of three and grandfather of three.] -- Nancy Red Star [] An Exopolitical Perspective on the Preemptive War against Iraq http://www.exopolitics.org/Study-Paper2.htm by Dr. Michael E. Salla [] And: IRAQ: MESOPOTAMIA DEL SUD, LA TERRA DI SUMER http://www.edicolaweb.net/irak_21g.htm http://www.edicolaweb.net/irak_01g.htm [] M a r t i a n S t a r g a t e [] Qinghai Province Governor: Zhao Leji Capital: Xining Website: http://www.qh.gov.cn Geographical location [][]The province lies on the northeastern part of the Qinghai-Tibet Plateau in west China, bordering Gansu and Sichuan provinces, Xinjiang Uygur Autonomous Region and Tibet Autonomous Region. As the origin of the Yangtze, Yellow, and Lancang rivers, Qinghai has an area of 720,000 square kilometers, the fourth largest in China. Its territory includes 3.86 million hectares of grassland, 590,000 hectares of cultivated land and 266,000 hectares of forest http://www.china.org.cn/e-xibu/2JI/3JI/qinghai/qing-ban.htm [] Mysterious Pipes, Qaidam Basin, Qinghai Province http://english.peopledaily.com.cn/200206/25/eng20020625_98530.shtml The widespread news of mysterious iron pipes at the foot of Mount Baigong, located in the depths of the Qaidam Basin, Qinghai Province of northwest China, has roused concern from related departments. [] Qaidam Basin, in northwest China's Qinghai Province, is expected to have a wide range of industries in the future as a result of a local government plan to tap the rich natural resources in the basin. http://english.peopledaily.com.cn/200207/01/eng20020701_98917.shtml [] Chinese Academy of Sciences CAS astronomers make advance in search of ET http://english.cas.ac.cn/english/news/detailnewsb.asp?infono=24587 [] See also Wingmaker's Inca Tunnel: http://www.wingmakers.com/hakomisite.html [] Scientists Discover Planetary System Similar to Our Own: http://www.nsf.gov/od/lpa/news/03/pr0373.htm [] The Permian http://geology.er.usgs.gov/paleo/geotime.shtml 290 to 248 Million Years Ago [] The Permian period lasted from 290 to 248 million years ago and was the last period of the Paleozoic Era. The distinction between the Paleozoic and the Mesozoic is made at the end of the Permian in recognition of the largest mass extinction recorded in the history of life on Earth. It affected many groups of organisms in many different environments, but it affected marine communities the most by far, causing the extinction of most of the marine invertebrates of the time. [...] http://www.ucmp.berkeley.edu/permian/permian.html [] See also: North Greenland Ice core Project (NGRIP) http://www.glaciology.gfy.ku.dk/ngrip/index_eng.htm And: http://geology.er.usgs.gov/paleo/geotime.shtml An ice core to bedrock will provide samples of each annual layer deposited originally as snow through the last 200,000 years or more. [] Beagle 2 http://www.cnn.com/interactive/space/0305/mars/content.2.1.html Weight: 73 pounds (33 kilograms) Cost: Estimated $60 million Landing site: Isidis Planitia Instruments: Stereo cameras; weather sensors; spectrometers to analyze minerals and search for methane; X-ray detector to determine rock age, arm with rock corer and grinder, microscope and sample analyzer; mole to dig for soil samples; numerous ovens to bake samples to analyze carbon 12. Power: Solar Mission length: Up to 180 days [] Mars Exploration Rover Weight: About 375 pounds (170 kilograms) Cost: $400 million Scheduled landing: Twin rovers in separate missions, both set for January 2004 Landing sites: Gusev Crater and Meridiani Planum Instruments: Cameras providing 360-degree, stereoscopic views; rock abrasion tool; microscopic imager and spectrometer to study rocks and soil samples. Power: Solar Mission length: At least three months, roving on a six-wheeled frame up to 40 meters each day. http://www.cnn.com/interactive/space/0305/mars/frameset.exclude.html [] Mars Global Surveyor The NASA satellite has taken more than 120,000 pictures of the red planet since it went into Mars orbit in 1997, including dramatic shots of eroded gullies and cliffs that suggest modern water activity. [] Mars Odyssey The NASA satellite probe, which arrived in late 2001, is using infrared detectors to look for water-related minerals and a color camera to take panoramic pictures of the surface of Mars. [] Nozomi The Japanese orbiter should reach the red planet telecommunications glitch has put the status of the mission in question. It is designed to study how solar winds interact with the atmosphere, which could help explain the disappearance of surface water. [] Mars Express The European Space Agency mission, its first to another planet, includes an orbiter that will use a radar to search for underground water. Before going into orbit in early 2004, it will jettison Beagle 2 lander to the surface, which will sniff for signs of life. [] Why Mars? It is also America's destiny to Fold, to Spindle, to Mutilate. And market exploitation videotapes called: Girls Gone Wild! CLICK--> http://www.girlsgonewild.com/free/payperview/ppv0004.jpg [] Because it is humanity's destiny to strive, to seek, to find. And because it is America's destiny to lead. -- George Bush CLICK--> http://www.enterprisemission.com/images/Boeing-M2.jpg [] Mars Exploration Program http://mars.jpl.nasa.gov/ [] M a r t i a n S t a r g a t e Long Island got a new state park - which, unlike most of our state parks, comes complete with a fake village, a giant radar tower still glowering out at the Atlantic, and a very eerie legacy. http://www.newsday.com/features/ny-p2cover3345174jun26,0,3729653.story?coll= ny-homepage-promo Fort Hero Air Force Base, Montauk Point, Long Island, New York. http://www.google.com/search?hl=en&ie=ISO-8859-1&q=Fort+Hero+Air+Force+Base% 2C+Montauk UG said: ... All that you do makes it impossible for what already is there to express itself. That is why I call this 'your natural state'. You're always in that state. What prevents what is there from expressing itself in its own way is the search. The search is always in the wrong direction, so all that you consider very profound, all that you consider sacred, is a contamination in that consciousness. You may not like the word 'contamination', but all that you consider sacred, holy and profound is a contamination. So, there's nothing that you can do. It's not in your hands. I don't like to use the word 'grace', because if you use the word 'grace', the grace of whom? You are not a specially chosen individual; you deserve this, I don't know why. If it were possible for me, I would be able to help somebody. This is something which I can't give, because you have it. Why should I give it to you? It is ridiculous to ask for a thing which you already have. [...] The Anti-Krishnamurti: [HAPPY BIRTHDAY UGK!] -- Uppaluri Gopala Krishnamurti (Born 9 July 1918) http://www.well.com/user/jct/mystiq1.htm * * * THE EIGHTFOLD PATH Part Five: ISOLATION by Inanna Arthen Excerpt: http://www.earthspirit.org/fireheart/fhefiso.html ... At the farthest extreme, sensory deprivation and perceptual deprivation experiments in laboratories have explored the influence of isolation, not just from other people, but from the physical world itself. Sensory deprivation, as achieved in flotation tanks like those popular- ized by John Lilly [] [ & ] and the film, Altered States, attempts as far as possible to isolate the very mind, throwing it completely upon its own resources - or upon input from other than the five senses being restricted. By deliberately separating from the consensual web for an extended period (closing our door, turning off the television, walking into the woods), we open ourselves to a profound - and possibly permanent - change. Cutoff from the unending stream of external cues, the mind begins to rely upon its own perceptions to fill in the growing cracks. We find ourselves seeking more and more perceptual material, and doorways that we were taught to close long ago slowly creak open. [...] -- INANNA ARTHEN Scientists Discover Planetary System Similar to Our Own: http://www.nsf.gov/od/lpa/news/03/pr0373.htm S a l v i a d i v i n o r u m ... He described being completely erased from normal consciousness and transported into a vegetable kingdom where sinister mantis-like beings (who he felt had malign intent) were trying to do something to him (he said that as best he could interpret their intent it seemed as though they were trying to 'abduct' him). He was deeply concerned that he would be unable to return, in fact he said that he could not even understand what it was he was trying to return to. He could not remember the normal contents of consciousness at all. Eventually, he latched onto my name and brought himself back. He did not know how long he had been gone and had no memories of what happened physically to him after the second toke. And now the question; if anyone on this list has or has contact with someone who has contacted apparently malign or alien beings such as those desribed above, please address our question: Does anyone know or have experience with 'going-with' such an experience or the beings? Has anyone ever had the presence of will to go with something like this, and if so, what can you tell us about it? Trips - Salvia divinorum http://leda.lycaeum.org/?Table=Trips&Ref_ID=269 ==== Does Jack actually wish to converse with folks in these newsgroups or does he just drop these little tidbits of wisdom (???) in here as (what he sees as) a public service? -- John Zinni > In the conclusion of your excellent Ukraine.doc paper, you mentioned > the Brookhaven RHIC project whose goal was to create the cosmic > soup that the inflationary model of the Big Bang predicted existed > at the start. [note: I don't know how this compares to later > Ekopyrotic models which utilize a slow collision of 5-D membranes > instead that produces the same results without a need for gravity > waves or monopoles] As I recall, many scientists world-wide objected > to this project because of conditons that might arise and prove > catastrophic to the Earth (if not the Universe). Strangelets was > one such possibility. (In fact, didn't someone propose it was a > strangelet snaking across the Universe that was responsible for that > Siberian explosion?) One of the major questions in the UFO field has always been WHY are > they here? Some suggest we are either direct creations of them or > have been genetically altered over thousands-of-years by them. A > more selfish reason is what your ending paragraphs alluded to -- that > being simple self-preservation. In their eyes, we are tinkering with > forces that -- under the right situations -- cause them as much harm > as ourselves. That would certainly explain why UFOs and atomic > energy (especially weapons testing/production) seem to go hand-in- > hand. If the latter is minimized, then so are the UFO sightings. Yes, exactly. Before I finish, I appreciate the years of soul-searching, > reflection, and debates with colleagues over all these theories. > They offer one side of the equation. The other side being of > course is the technologies required to create 'exotic matter' and > bend/fold spacetime. I realize theorists are immune to this part, > but are there any clues as to what form this technology would be in > (e.g., rotating magnetic fields, nano-engineered Type III > superconductors, quantum lasers, etc.)? Any good > books/papers/websites that you could recommend? The basic equations are toward the end of my paper on the Josephson > coupling of > to <0|e+e-|0>. A big problem is the impedance matching, see > Ray Chiao's gravity radio papers on a related problem - it's not identical > to what I am talking about but not orthogonal either. There are no papers in the field other than my own because I have only > recently created this new approach. It has not sunk in and it may take > years. The flying saucers show us that a technology of the /zpf field > exists only that US DOD, DOE, NSA, CIA et-al do not have it. I am the > best they got at the moment. In fact I am ALL they got that's real > pointing the right way IMO of course. Hal Puthoff & Co's PV theory > represents what they hope they have and that theory is a joke IMHO. See > the recent Mitre Conference on HFGR. > Of course what I'm > asking for is a Manhattan Project equivalent where you had all the > theorists on one side and the engineers coming up with schemes to > make the former's equations come to fruition. This has been the most > frustrating part for me all along. I think I could understand the > physics better if I had some clue as to what physical mechanisms were > to be employed to make it happen, captain! That's what Joe Firmage wanted to do with ISSO. He made unwise financial > decisions and the project only lasted a year and a half 1999 - 2000. > That's what CIA Station Chief Harold Chipman wanted to do in 1985. Those > documents are in my book Destiny Matrix. However, the time was not > ripe, I did not have the right ideas back then. The right ideas > had to wait for the discoveries of both dark matter first and dark > energy second. The full experimental picture only was made available to > and Beyond II at end of 2002 anticipate the results of NASA WMAP that > were confirmed. That's what Hal Puthoff's IAS is supposed to > be about but he is barking up the wrong tree, same for Eric Davis's > Warp Metrics. That's what Senator Robert Byrd's ISR in W. Va was > supposed to be about also with Jim Corum's stuff, but that does not work > either as we found out at ISSO with our million dollar > contract with SARA of Huntington Beach where Corum came from before he > went to ISR. So far The Right Stuff you see in > http://qedcorp.com/APS/Ukraine.doc is a One Man Show. > WYSYYG. However there are some interesting theory things brewing in > Russia, Finland and Kiev in the Ukraine, and possibly some experimental > stuff in Beograd with the J.P. Vigier group. My theory does justify > Vigier's basic idea for cold fusion i.e. tight atomic states > i.e. spatially extended electron as a Bohm hidden variable, that appears > isn't in low energy scattering and most importantly in atomic bound > states. This is what the Serbs are trying to achieve in a practical way. > Will they succeed? I don't know. I am not in touch with them though > maybe I will see them in Paris this September. We had Vigier at ISSO and > that was one good thing in terms of development of my theory in > http://qedcorp.com/APS/Ukraine.doc (1) Making a Wormhole just got easier · = http://www.nature.com/nsu/030527/030527-12.html (MS-Word version > archived at http://www.stealthskater.com/Documents/Time_02.doc). This is important. Very good for my program at > http://qedcorp.com/APS/Ukraine.doc > to appear in Progress in Quantum Physics Research (Nova Scientific > Publishers) Good news for time travellers - it just got cheaper. The amount > material needed to build a window through time is infinitesimally small, > new research shows. To travel through time, all you need to do is open a wormhole in > space-time and step through it. And to do that you need a magic > ingredient called 'exotic matter', which is repelled rather than > attracted by gravity. The hitch is that no one has the remotest idea how to make exotic > matter. But don't despair, say Matt Visser, of the Victoria University > of Wellington in New Zealand, and his colleagues. They have shown that > when we do figure out how to make the stuff, we won't > need very much of it (1). The Pundits have not yet realized that the dark energy that is ~ 73% > of the large-scale structure of the universe for scale L > 10 > megaparsecs, is, on the smaller scales, exactly what they are looking > for! They do not understand that the Einstein cosmological constant / > in the spatially flat K = 0 FRW post-inflationary bubble metrics > is simply a limit of a local field Wigner wavelet /zpf (x,L) when L --> c/Ho ~ 10^26 meters and the x dependence at that > scale is very weak. Normal equilibrium vacuum (G.E. Volovik, Universe in a Helium > the dominating ODLRO vacuum-polarization coherence <0|e+(x)e-(x)|0>. > Upsetting this delicate equilibrium with Josephson links using real > superconductors, impedance matched to the physical vacuum. can make > /zpf be positive or negative with negative (repulsive) and positive > (attractive) zero point energy pressures respectively because w = > pressure/(energy density) = -1 and the pressure dominates the energy > density by a factor of 3 in Einstein's geometrodynamic generalized > source Newtonian-Poisson equation for the effective potential energy per > unit test mass of the exotic vacuum volume element at coarse-grained > position x and at scale L in the sense of wavelet transform theory. It's nice to know that we do not need a large spacetime region of exotic > vacuum /zpf(x,L) to do the tricks that we see flying saucers in fact > doing rendering our entire present-day military defense technology > essentially impotent and obsolete should such advanced knowledge fall > into the hands of small groups of irresponsible individuals. Sir Martin > Rees discusses this in Our Final Hour. See also the book The Star > Gate Conspiracy by Picknett & Prince and my two books Destiny Matrix > and Space-Time and Beyond II for a set of alternate POV's on this > topic. Also see http://www.fiu.edu/~mizrachs/techgnosis.html and > http://www.techgnosis.com/ As Star Trek: Deep Space 9 ,Quantum Leap and Stargate have taught us, > wormholes are the preferred mode of transport for today's fashionable > time-traveller. These hypothetical tunnels connect distant parts of > space-time, the fabric of our Universe. And despite the philosophical > havoc that wormholes wreak with notions of causality, > Einstein's theory of general relativity - which describes space-time - > allows them to exist. Six years ago, Visser and his colleague David Hochberg showed that in > order to stay open, wormholes need exotic matter. It's weird stuff, > however - it can be considered to have negative energy, meaning that it > has even less than empty space. It's the same as saying > that it experiences gravity as a repulsive force, and physicists have > never encountered anything of the sort. Nature (i.e., the magazine not GOD(D) in the Quad ;-)) has screwed > up here. Their explanation is wrong. What matters is the sign of the > pressure. In fact, for w = -1 exotic vacua, our universe is ~ 96% exotic > vacuum stuff of both signs in proportion 73% repulsive exotic matter > /zpf > 0, and 23% attractive dark matter /zpf < 0 with negative and > positive pressures respectively. It's the sign of the pressure that counts! So they imagine it. They key to exotic matter lies in quantum > fluctuations, which give empty space a kind of fizziness. Quantum theory > continually popping in and out of existence in the > vacuum of empty space. Exotic matter might arise by suppressing this > fizz, or as a physicist would say, by violating the averaged null energy > condition (ANEC). Bingo! Exactly, I show how the macro-quantum coherence <0|e+(x)e-(x)|0 suppresses this fizz in > http://qedcorp.com/APS/Ukraine.doc > If this were to happen, quantum effects could give rise to tiny amounts > of exotic matter. But how much is needed to sustain a > wormhole? That is what Visser and colleagues have now calculated. They find that, > if the wormhole is designed carefully, the total quantity of > ANEC-violating matter can be made infinitesimally small. This makes a > wormhole considerably easier to create. Exactly. That's what the flying saucer have been telling us. That is > why what Eric Davis proposes in MUFON 2001 is entirely the wrong > approach needing impractical enormous amounts of electromagnetic field > stress-energy density to get a very useless small bang for an impossibly > large buck since G/c^4 = 10^-33 cm per 10^19 Gev i.e. space-time is too > stiff to bend directly with a Tuv source field in Einstein's > geometrodynamic field equation Guv = -8pi(G/c^4)Tuv What you need to do the trick is Guv + /zpfguv = -8pi(G/c^4)Tuv ~ 0 where /zpf is controlled by Josephson links tweaking around /zpf = 0. Back to the future again It's not the first time that traversable wormholes have been pulled out > of a pit of implausibility. In the 1980s the British astrophysicist > Stephen Hawking conjectured that even if you could make a wormhole > stabilized by exotic matter, you couldn't go through it to travel in This became known as the Chronology Protection Conjecture. It was a > relief for philosophers who were trying to protect the notion of > causality. The paradox they envisioned - immortalised in the movie Back > to the Future - was that if wormholes could exist it > would theoretically be possible to go back in time and prevent your > parents from meeting. This would prevent your own existence, and > therefore your ability to go back in time. I suspect that Hawking's conjecture here is wrong. We have evidence it > is wrong from the flying saucers. It's like saying we could not break > the sound barrier. {It's not the first time that traversable wormholes have been pulled > out of a pit of implausibility But physicists subsequently thought of a way around this problem - there > are, for example, 'time loops' threading through a wormhole along which > backwards time travel is possible, but without its being able to alter > the future. Sadly, an infinitely small amount of exotic matter is not the same as > none at all. This is an example of Sir Michael Berry's singular limit. So until someone figures out how to get hold of it (not to mention how > to open up a wormhole in the first place), you can forget about trips to > the Jurassic era - or your parents' first date. Don't count on that. The times they are a chang'in. See Time Travel: The Art of the Possible > http://www.dvdjournal.com/reviews/s/startrek04voyage.shtml The Star Trek Universe In the mini-documentary Time Travel: The Art of the Possible (11:14), > three physicists (mainly me), taped separately, provide their thoughts > on the real-world hypotheses behind the concept of time travel. Physics > 201 topics are discussed and playfully illustrated, from basic > Einsteinian relativity to Carl Sagan's wormhole devised for the novel > (and film) Contact . It's an interesting, if only superficially > relevant, addition that would be right at home on the Science or > Discovery channels. References > Visser, M., Kar,S. &Dadhich, N. Traversable wormholes with arbitrarily > small energy condition violations. Physical Review Letters ,90, 201102 , (2) New concept of black holes proposes they are a special form > of 'dark energy' and not a singularity => Frozen Stars at > http://www.sciam.com. What you have been taught in school is almost > certainly wrong, because classical black hole spacetimes are > inconsistent with quantum mechanics, says physicist George Chapline > of Lawrence Livermore National Laboratory. (MS-Word version archived > at http://www.stealthskater.com/Documents/BlackHole_01.doc) ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Jack is a Write Only entity. Bob Kolker > ==== Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Jack is a Write Only entity. Bob Kolker > Bob, But he is apparently not a Write-Protected entity. He sometimes corrects himself; albeit not with anything resembling any known reality. Tom McDonald ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? I still like the theory I put forward a while ago. It's that Sarfatti has secured funding from kook-brained painfully rich people to develop stargates and stuff so that they can leave the earth. He has to post this nonsense to keep his paying sponsors happy, and he uses his previous pre-kook physics as source for his meaningless word/math salad. Whether he really believes a word of it or not is a harder question. Is he really just a loon, or is he actually dazzlingly clever in having simply found a way to be paid a *mint* for just looning all over usenet? I earn a nice wage for doing clever and difficult things, but if I found I could just dribble as he does and rake it in, I'm damn sure I would. I wasted my youth in poverty trying to unravel the universe. Now I just want to learn to play the piano and grow sunflowers. -- ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Jack is a kook with a degree who likes to spew garbage... he never replies to anyone (since nearly everybody points out that his theories are worthless bullcrap). ==== John Zinni > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Maybe he's out to promote his books. There are various other theories, some of which are, no doubt, of extraterrestrial origin :-) LH ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? -- > John Zinni > [snip] Good news for time travellers - it just got cheaper. The amount > material needed to build a window through time is infinitesimally small, > new research shows. To travel through time, all you need to do is open a wormhole in > space-time and step through it. And to do that you need a magic > ingredient called 'exotic matter', which is repelled rather than > attracted by gravity. > I wonder when NASA will start sinking billions into trying to develop a Star Gate? I wonder if they already have? (Spent the money, that is!) Has anyone heard how their big Anti-Gravity Machine project is going??? Double-A ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? -- > John Zinni [snip] Good news for time travellers - it just got cheaper. The amount > material needed to build a window through time is infinitesimally small, > new research shows. To travel through time, all you need to do is open a wormhole in > space-time and step through it. And to do that you need a magic > ingredient called 'exotic matter', which is repelled rather than > attracted by gravity. > I wonder when NASA will start sinking billions into trying to develop > a Star Gate? I wonder if they already have? (Spent the money, that > is!) Has anyone heard how their big Anti-Gravity Machine project is > going??? > it didn't take off. Herc ==== } } } }> Does Jack actually wish to converse with folks in these newsgroups or does }> he just drop these little tidbits of wisdom (???) in here as (what he sees }> as) a public service? } }Jack is a Write Only entity. } }Bob Kolker Dr H ==== : I wonder when NASA will start sinking billions into trying to develop : a Star Gate? I wonder if they already have? (Spent the money, that : is!) : Has anyone heard how their big Anti-Gravity Machine project is : going??? of _Wired_. Actually, they just debunked the anti-gravity device by putting it in a vacuum... and I remember reading about an ion wind flier in a mechanics magazine of the 1960s in my youth. John Savard ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Jack has changed over the years. He once would respoond to a post now and then. Nearly always with little more than swearing though. Socks ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? Jack has changed over the years. He once would respoond to a post > now and then. Nearly always with little more than swearing though. > Socks Yes it sucks, but I'm not suprised its only that he concentrates on hypothetical science that he is constantly attacked. When I rid the world of skeptics he'll be free to speak again, until then merely broadcasting is suitable for his level of work. Herc ==== > Does Jack actually wish to converse with folks in these newsgroups or does > he just drop these little tidbits of wisdom (???) in here as (what he sees > as) a public service? What a shame that we can't get a dialogue going between Jack and JSH. Gib ==== > What a shame that we can't get a dialogue going between Jack and JSH. > FLT meets FTL? Herc ==== understand the simple considerations required to answer my question regarding the probability that exactly k machines are working just before the (n+1)-th repair is completed, given that exactly j (jI can see that the formula is correct if F is a step function and that >might lead to the general formulation. I don't see how to prove it fully >for an arbitrary F but, apart from that, I have some aesthetic objections >to this way of writing the transition probability. Here is the point, I think. Let (Omega,B,P) be a probability space. We say that two events U,V are independent if the probability of (U and V) is equal to the probability of U times the probability of V. Another way of saying this is that the probability of the cartesian product U x V of U,V in Omega x Omega, with respect to product measure P x P, is equal to the probability in Omega of the preimage of U x V under the diagonal mapping Delta: x |-> (x,x) of Omega into Omega x Omega. In other words, if we denote by Delta_*(P) the pushdown of P to a measure on Omega x Omega via Delta, then U x V is a set for which product measure P x P and Delta_*(P) agree. [Offhand side question: the sets for which the two measures agree form a sigma algebra. Is that sigma algebra generated by sets of the form U x V where U and V are independent?] find it a lot easier to work with for this purpose. I asked earlier for: >a really neat >formula that lives on Omega (or a product of copies of Omega) instead and >which is obviously true and which implies the ones stated above. Is there? The neat formula seems to be that Z_*(P) = X_*(P) x Y_*(P) when X,Y are independent random variables and Z=(X,Y). I'm far from being any kind of expert on probability theory, but I haven't seen things stated in this manner in the probability books I've looked at. What I would like to know is the title and author of a book that does take this slightly more functorial point of view. Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** ==== Can anyone describe the Stone-Cech compactification of the natural numbers topologically and algebraically, for example : is it still a semi-group. is it metrizable and what would be a definition of such a metric ?