mm-463 === Subject: Re: A problem proving a Ring Isomorphism Adjunct Assistant Professor at the University of Montana.>I am using Gallian's book and am trying to complete the proof of>THeorem 15.6 Field of Quotients. I'm trying to prove that the>mapping phi: D -> F given by x -> x/1 is a ring isomorphism from D to>phi (D).>I've proven that it's one-to-one and also that it's a ring>homomorphism. My problem is that I'm not sure how to prove that it's>onto. It would seem awkward to use a/b since I'm given x/1 rather than>any arbitrary a and b.>I appreciate any help completing the theorem.> Reread the problem: you are not asked to show that phi is an> isomorphism from D to F: you are merely asked to show that it is an> isomorphism from D to> phi(D) = { phi(x) : x in D} = {f in F: there exists x in D with f=phi(x)}.> Every function is surjective onto its image.>I knew that I was having to prove that phi is an isomorphism from D to>phi(D). I just did not understand how to prove it. It just seemed>like it 'obviously' had to!In algebra, any homomorphism which is injective is an isomorphism ontoits image. So the problem was just asking you to show that the subringof F given by all fractions of the form x/1 is isomorphic to D. Putanother way: what you had to do was show that phi was an injectivering homomorphism. You did that, so you were done.========================================================= =============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu===Subject: Four Color Theorem Proof> I can't speak for them: but, as I say, I'd love to see your proof.> Will you post it here, please?> Mike.The proof of the FCT is predicated upon the following hypothesis: Every vertex in every maximal planar graph is completely enclosedby a cycle graph; ie, a ring of edges. Essentially, every vertex isthe hub ofits own wheel graph. The validity of this hypothesis is very easy to prove. It is almostself-evident. So I leave it as an exercise for the reader. Next post will discuss the relevance of the hypothesis to the FCT. ===Subject: Re: Four Color Theorem ProofStanding, or are you reasonably tall?Charles RiggsMy email address: chriggs/at/eircom/dot/net===Subject: FW: L00K and feel 20 years y0unger.. boundary=--42224900032991266 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id i3HNPIs12999;X-IP: 18.112.118.229------------------------------------------------ ---------------------As seen on NBC, CBS, and CNN, and even Oprah! The health
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Let x_1, x_2 be linearly independent sols. of a third order homogenous>linear differential equation. Is there a method for finding a third sol.>x_3>such that x_1, x_2 and x_3 are linearly independent?> Yes. See reduction of order in just about any book on differential> equations. (Not a math book, an elementary text - might not be> in a theoretical math book but it's in all the texts.) That's usually> presented as a way to get a second solution to a second-order> equation given one solution, but you should have no trouble> extending it to your problem.>Well, I have trouble extending it. It ends up with a second order linear>equation in an unknown function which I don't know how to solve.Hmm. Sorry - I sort of thought that it was going to be a first-orderequation, but looking at what actually happens with what I hadin mind, of course it's not.===Subject: Re: A question on third order linear differential equations>Let x_1, x_2 be linearly independent sols. of a third order homogenous>linear differential equation. Is there a method for finding a third sol.>x_3>such that x_1, x_2 and x_3 are linearly independent?> Yes. See reduction of order in just about any book on differential> equations. (Not a math book, an elementary text - might not be> in a theoretical math book but it's in all the texts.) That's usually> presented as a way to get a second solution to a second-order> equation given one solution, but you should have no trouble> extending it to your problem.>Well, I have trouble extending it. It ends up with a second order linear>equation in an unknown function which I don't know how to solve. I understand you are gives an ODE resembling(*) x'''(t) + f2(t)x''(t) + f1(t)x'(t) + f0(t)x(t) = 0along with two independent solutions x(t) = x1(t) andx(t) = x2(t). You want a third independent solution, x3(t). Without knowing x3, let W be the Wronskian | x1(t) x1'(t) x1''(t) |W(t) = det | x2(t) x2'(t) x2''(t) | | x3(t) x3'(t) x3''(t) |To form W'(t), separately differentiate the entire first column,entire second column, and entire third column, addingthe three resulting determinants. | x1'(t) x1'(t) x1''(t) |W'(t) = det | x2'(t) x2'(t) x2''(t) | | x3'(t) x3'(t) x3''(t) | | x1(t) x1''(t) x1''(t) | + det | x2(t) x2''(t) x2''(t) | | x3(t) x3''(t) x3''(t) | | x1(t) x1'(t) x1'''(t) | + det | x2(t) x2'(t) x2'''(t) | | x3(t) x3'(t) x3'''(t) | | x1(t) x1'(t) x1'''(t) | = 0 + 0 + det | x2(t) x2'(t) x2'''(t) | | x3(t) x3'(t) x3'''(t) |Use (*) and elementary determinant identitiesto derive W'(t) = -f2(t)W(t). If x1(t), x2(t), x3(t) are linearly dependent (over R)solutions of (*), then W(t) will be the constant zero. If you know an antiderivative for f2(t), you can get thegeneral form for W(t). Substitute the known solutionsx1(t), x2(t) into the definition of W(t) to get a second-order ODE for x3(t), namely(**) (x1(t) x2'(t) - x2(t) x1'(t)) * x3''(t) + (x1''(t) x2(t) - x2''(t) x1'(t)) * x3'(t) + (x1'(t) x2''(t) - x1''(t) x2'(t)) * x3(t) = W(t)Here x1, x2, W are known but x3 is unknown.(**) is linear but non-homogeneous.This may be where you got stuck.The solutions of the corresponding homogeneous equation (when W(t) = 0) are known (what are they?).To get a particular solution, use the variation of parameters. In particular, suppose c1(t) and c2(t) satisfy x3(t) = c1(t) x1(t) + c2(t) x2(t) x3'(t) = c1(t) x1'(t) + c2(t) x2'(t) (these two linear equations can be used to define c1(t) and c2(t)). Then x3''(t) = c1'(t) x1'(t) + c2'(t) x2'(t) + c1(t) x1''(t) + c2(t) x2''(t)Substitute into (**) W(t) = (x1(t) x2'(t) - x2(t) x1'(t)) * (c1'(t) x1'(t) + c2'(t) x2'(t))We also know 0 = c1'(t) x1(t) + c2'(t) x2(t)by differentiating our assumed form for x3(t). This gives us two linear equations in c1'(t) and c2'(t).John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California===Subject: Re: A question on third order linear differential equations>Let x_1, x_2 be linearly independent sols. of a third order homogenous>linear differential equation. Is there a method for finding a thirdsol.>x_3>such that x_1, x_2 and x_3 are linearly independent?> Yes. See reduction of order in just about any book on differential> equations. (Not a math book, an elementary text - might not be> in a theoretical math book but it's in all the texts.) That's usually> presented as a way to get a second solution to a second-order> equation given one solution, but you should have no trouble> extending it to your problem.>Well, I have trouble extending it. It ends up with a second order linear>equation in an unknown function which I don't know how to solve.> I understand you are gives an ODE resembling> (*) x'''(t) + f2(t)x''(t) + f1(t)x'(t) + f0(t)x(t) = 0> along with two independent solutions x(t) = x1(t) and> x(t) = x2(t). You want a third independent solution, x3(t).> Without knowing x3, let W be the Wronskian> | x1(t) x1'(t) x1''(t) |> W(t) = det | x2(t) x2'(t) x2''(t) |> | x3(t) x3'(t) x3''(t) |> To form W'(t), separately differentiate the entire first column,> entire second column, and entire third column, adding> the three resulting determinants.> | x1'(t) x1'(t) x1''(t) |> W'(t) = det | x2'(t) x2'(t) x2''(t) |> | x3'(t) x3'(t) x3''(t) |> | x1(t) x1''(t) x1''(t) |> + det | x2(t) x2''(t) x2''(t) |> | x3(t) x3''(t) x3''(t) |> | x1(t) x1'(t) x1'''(t) |> + det | x2(t) x2'(t) x2'''(t) |> | x3(t) x3'(t) x3'''(t) |> | x1(t) x1'(t) x1'''(t) |> = 0 + 0 + det | x2(t) x2'(t) x2'''(t) |> | x3(t) x3'(t) x3'''(t) |> Use (*) and elementary determinant identities> to derive W'(t) = -f2(t)W(t).> If x1(t), x2(t), x3(t) are linearly dependent (over R)> solutions of (*), then W(t) will be the constant zero.> If you know an antiderivative for f2(t), you can get the> general form for W(t). Substitute the known solutions> x1(t), x2(t) into the definition of W(t) to get> a second-order ODE for x3(t), namely> (**)> (x1(t) x2'(t) - x2(t) x1'(t)) * x3''(t)> + (x1''(t) x2(t) - x2''(t) x1'(t)) * x3'(t)> + (x1'(t) x2''(t) - x1''(t) x2'(t)) * x3(t) = W(t)> Here x1, x2, W are known but x3 is unknown.> (**) is linear but non-homogeneous.> This may be where you got stuck.> The solutions of the corresponding homogeneous equation> (when W(t) = 0) are known (what are they?).> To get a particular solution, use the variation of parameters.> In particular, suppose c1(t) and c2(t) satisfy> x3(t) = c1(t) x1(t) + c2(t) x2(t)> x3'(t) = c1(t) x1'(t) + c2(t) x2'(t)> (these two linear equations can be used to define c1(t) and c2(t)). Then> x3''(t) = c1'(t) x1'(t) + c2'(t) x2'(t)> + c1(t) x1''(t) + c2(t) x2''(t)> Substitute into (**)> W(t) = (x1(t) x2'(t) - x2(t) x1'(t))> * (c1'(t) x1'(t) + c2'(t) x2'(t))> We also know> 0 = c1'(t) x1(t) + c2'(t) x2(t)> by differentiating our assumed form for x3(t).> This gives us two linear equations in c1'(t) and c2'(t).> -- > John Adams served two terms as Vice President and one as President, butlost> reelection. Later his son became President despite losing the popularvote.> That son lost his reelection attempt badly. Now history is repeatingitself.> pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael,California===Subject: Basic sequences question help requiredSum to total by end of nth day=(n/2)(3n+1)Calculate total from end of nth day of 2nth day. Simplify your answerI am stuck on this part of the question the answer is (n/2)(9n+1) butI am unable to reach there any help would be appreciated.===Subject: Re: Basic sequences question help required Adjunct Assistant Professor at the University of Montana.>Sum to total by end of nth day=(n/2)(3n+1)>Calculate total from end of nth day of 2nth day. Simplify your answerThe total at the end of day n is (n/2)(3n+1).The total at the end of day (2n) is ((2n)/2)(3(2n)+1) = n(6n+1).So the total from end of day n to end of day (2n) is equal to(total at end of day (2n) ) - (total at end of day n) = n(6n+1) - (n/2)(3n+1).>I am stuck on this part of the question the answer is (n/2)(9n+1) but>I am unable to reach there any help would be appreciated.Do the operation.==================================================== ==================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu===Subject: Re: Basic sequences question help requiredSorry I meant Calculate the total from the end of the nth day to theend of the 2nth day.===Subject: Re: Basic sequences question help required Adjunct Assistant Professor at the University of Montana.>Sorry I meant Calculate the total from the end of the nth day to the>end of the 2nth day.Sorry what? I gave you the answer.If you have a formula that tells you the total from day 1 to day k,call it T(k), then you can get the total from the end of day n to theend of day 2n by taking the total from day to day 2n, and subtractingthe total from day 1 to day n; that is, calculate T(2n), calculateT(n), and then take T(2n)-T(n). The formula you gave gives the answeryou quoted. (Think of it this way: if you can figure out how much money you getfrom the first of the month to any day after, say M(k) for the k-thday, then how do you figure out how much money you got during the 3rdday? By taking M(3)-M(2): because M(3) gives you all the money you goton days 1, 2, and 3; and M(2) gives you what you got on days 1 and 2;if you take the difference, you get exactly what you got on day 3).=========================================================== ===========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu===Subject: Re: Representations of the circle>Why are the irreducible unitary representations of the circle the same as>the dual of the circle? I can't see any connection here:>Dual of circle = group of continuous homomorphisms from circle to circle,>which are the functions x ---> exp(2Pi*ikx) where k is in Z.>Irreducible unitary representations : We have the regular representation of>S^1 on L^2(S^1) by R(y)f(x) = f(x+y), where f is in L^2(S^1). R : S^1 --->U (L^2(S^1)) where U (L^2(S^1)) is the unitary representations of>L^2(S^1). R is irreducible. Are there other irreducible unitary>representations?No.>Is there something going on here? Are the characters of a locally compact>abelian group the same as the irreducible unitary representations of the>locally compact abelian group? Yes.>Is this true? What if the group is just>locally compact?Then things are different.>I appreciate any insight into this!===Subject: unfamiliar with multinomialsIf I think the probability distribution over three exhaustive events is (p1,p2, p3) and I have observed frequencies of (a,b,c), can I say that theprobability of that probability distribution yielding that frequency is[(a+b+c)!/(a! b! c!)] p1^a * p2^b * p3^c?===Subject: Re: Closest Rank Algorithm?> Hey,> I need an algorithm similar to amazon.com which would give me n items > that were rated highest by the users that rated item x high. i.e. people > who liked this book also liked those 5..Doesn't Amazon just give you the items that were *purchased*? - so they can settle for crosstabulation.I suppose that if there are ratings, you can use Pearsoncorrelation. If it is at all like Amazon, the number ofitems is huge while the number of rating is small.[ snip, rest]Rich Ulrich, wpilib@Pitt.eduhttp://www.pitt.edu/~wpilib/index.html=== Subject: Re: Ability to readnonsense, again.The following might be of some interest for you:Using the theorem prover OTTER Art Quaife has provedfour hundred theorems of von Neumann-Bernays-G.9adel settheory; twelve hundred theorems and definitions ofelementary number theory; dozens of Euclidean geometrytheorems; and G.9adel's incompleteness theorems. It is animpressive achievement. [...][...] In the final chapter, semiautomatic proofs are offered of L.9ab's theorem and of G.9adel's first and second incompletenesstheorems. This is achieved within a formalization of themetatheory of the modal logic K4 -- it is not theorems in logicthat are proved, but theorems about logical theorems -- withheavy use of demodulation. As with the other chapters there isa nice brief presentation of the relevant background.The final hundred pages or so of the book are devoted toan edited list of the theorems proved in NBG set theory andPeano arithmetic, material that should be useful to otherworkers in the area. For them the book may well be asvaluable as its extraordinary price suggests.Automated theorem proving and its prospectsDesmond Fearnley-SanderREVIEW OF: Automated Development of FundamentalMathematical Theories by Art Quaife. (1992: KluwerAcademic Publishers) 271pp. $US123 hbk. ISBN0-7923-2021-2.Source:http://www.maths.utas.edu.au/People/ dfs/QuaifeReview.pdf===Subject: Re: Ability to read> nonsense, again.computation (or program synthesis or database query processing.) Your quote says nothing about these. HA HA HA HA HA HA HAWhat's that about ability to read?This paper also doesn't offer a shred of evidence to substantiate evenwhat it does claim. (You like people who do that, don't you?) On theother hand, I gave the specific 3 axioms, 8 rules of inference,theorems and proofs (remember that P means that P is recursive, notthat P is true):The 3 Axioms (Predicate Calculus wff):TRUE(x) [Peano's 5 Axioms]NIT(I,J,K) [Kleene's T function is recursive]- ~YES(x,x) [Axiom of Foundation]The 8 Rules of Inference (Predicate Calculus wffs => PredicateCalculus wff):NOT: P => ~PAND: P , Q => P ^ QOR: P , Q => P v Q IF: P , Q(x) => P ^ Q(x)DO: P(x) , Q(I,x) => P(x) ^ Q(x,y)UNION: P(x) , Q(x) => P(x) v Q(x)QUIT: P(x,y) => (eA) P(A,x)SUB: P(I) => P()Some of the Theorems (Predicate Calculus wff):NO(x,x) [The set of programs that halt NO on themselves isr.e.]-YES(I,J) [Membership Problem]-HALT(I,I) [Self-applicability Problem]-HALT(I,J) [Halting Problem]-HALT(I,) [Blank Tape Halting Problem]-(eA)HALT(I,A) [Ever-halting Problem]-(aA)HALT(I,A) [Always-halting Problem]etc.Sample Proof:Theorem: -YES(I,J) The Membership Problem is unsolvable. Proof 1. YES(I,J) Given 2. YES(I,I) SUB 1 J=I 3. ~YES(I,I) NOT 2 4. TRUE(x) Axiom 1 We can list the Universal Set. 5. TRUE(x)^~YES(x,x) DO 4,3 I=x 6 . ~YES(x,x) DEF-7 5 Property of Universal Set qed(Notice how I stay entirely within Predicate Calculus syntax, whereasthey have their big fake systems.)Charlie VolkstorfPS It's so much fun watching you scream and squirm when you see howI'm the only person who has ever axiomatized these 3 subjects (asopposed to those dumb, dishonest people whom you admire and think areSorry, you don't get the $100. You gave no axioms, rules ofinference, theorems or proofs. (That proves that you don't have whatyou say.)[Excuse me for acting silly, people, but can't I have fun once in awhile?]> The following might be of some interest for you:Using the theorem prover OTTER Art Quaife has proved> four hundred theorems of von Neumann-Bernays-G.9adel set> theory; twelve hundred theorems and definitions of> elementary number theory; dozens of Euclidean geometry> theorems; and G.9adel's incompleteness theorems. It is an> impressive achievement. [...]No examples! HA HA HA> REVIEW OF: Automated Development of Fundamental> Mathematical Theories by Art Quaife. (1992: Kluwer> Academic Publishers) 271pp. $US123 hbk. ISBN> 0-7923-2021-2.> Source:> http://www.maths.utas.edu.au/People/dfs/QuaifeReview.pdf=== Subject: 2 analysis questionsI need to show that Xn = (tan n) /n is unbounded. Now I clearly see that f(x) = (tan x)/ x is unbounded. Simply let x be near odd multiples of pi/2.Now given, for example, 7pi/2 we can find a sequence of rationals numbersthat approach 7pi/2. That is, I can find integers p and q such that p/q isas close to 7pi/2 as I want. Then the integer p is close to q*pi/2. But q isnot necessary odd! Now my question is, how do I determine which oddmultiples of pi/2 I can approximate with an integer? Or am I correct when Iam leaning toward every odd multiple of pi/2. How do you prove this??Next question: I need to show that Xn = ( n )^(1/n) converges to 1. So, Ineed to show that for e>o there exist N>0 s/t| ( n )^(1/n) - 1| < e whenever n> N.This is as far as I got: |( n )^(1/n) -1| = ( n )^(1/n) - 1 < e <=> (n )^(1/n) < e +1. I just can't seem to find an appropriate bound on (n )^(1/n).Any hints, as always, will be appreciated.Steven===Subject: Re: 2 analysis questions>I need to show that Xn = (tan n) /n is unbounded. Now I clearly see that f>(x) = (tan x)/ x is unbounded. Simply let x be near odd multiples of pi/2.>Now given, for example, 7pi/2 we can find a sequence of rationals numbers>that approach 7pi/2. That is, I can find integers p and q such that p/q is>as close to 7pi/2 as I want. Then the integer p is close to q*pi/2. But q is>not necessary odd! Now my question is, how do I determine which odd>multiples of pi/2 I can approximate with an integer? Or am I correct when I>am leaning toward every odd multiple of pi/2. How do you prove this??Well of course the proof is going to depend on the fact that pi isirrational, so it's really a bit of a fake to say you've actually proved it from first principles, unless you've already _proved_that pi is irrational, which I doubt.Anyway: Let x_n = n mod 2 pi; that is, 0 <= x_n < 2 pi andn = k_n* 2 pi + x_n for some integer k_n. The fact that pi isirrational shows that the x_n are all distinct. Since they'reall distinct and they all lie between 0 and 2 pi, thepigeonhole principle shows that given e > 0 there existn, m with |x_n - x_m| < e. It follows from _this_ that thereexists an integer k within e of an odd multiple of pi/2,because...>Next question: I need to show that Xn = ( n )^(1/n) converges to 1. So, I>need to show that for e>o there exist N>0 s/t>| ( n )^(1/n) - 1| < e whenever n> N.>This is as far as I got: |( n )^(1/n) -1| = ( n )^(1/n) - 1 < e <=> (>n )^(1/n) < e +1. Which is the same as n < (1 + e)^n;you can show that for every e > 0 there exists N such thatthis holds for all n > N from the binomial theorem.>I just can't seem to find an appropriate bound on (>n )^(1/n).>Any hints, as always, will be appreciated.>Steven===Subject: Re: 2 analysis questions>I need to show that Xn = (tan n) /n is unbounded. Now I clearly see that>f (x) = (tan x)/ x is unbounded. Simply let x be near odd multiples of>pi/2. Now given, for example, 7pi/2 we can find a sequence of rationals>numbers that approach 7pi/2. That is, I can find integers p and q such>that p/q is as close to 7pi/2 as I want. Then the integer p is close to>q*pi/2. But q is not necessary odd! Now my question is, how do I>determine which odd multiples of pi/2 I can approximate with an integer?>Or am I correct when I am leaning toward every odd multiple of pi/2. How>do you prove this??> Well of course the proof is going to depend on the fact that pi is> irrational, so it's really a bit of a fake to say you've actually> proved it from first principles, unless you've already _proved_> that pi is irrational, which I doubt.> Anyway: Let x_n = n mod 2 pi; that is, 0 <= x_n < 2 pi and> n = k_n* 2 pi + x_n for some integer k_n. The fact that pi is> irrational shows that the x_n are all distinct. Since they're> all distinct and they all lie between 0 and 2 pi, the> pigeonhole principle shows that given e > 0 there exist> n, m with |x_n - x_m| < e. It follows from _this_ that there> exists an integer k within e of an odd multiple of pi/2,> because...But as best I can tell, this does not then lead us easily to a proof thattan(n)/n is unbounded. Rather, it just tells us that tan(n) is unbounded.Math Tutor might have some of his questions answered at. As far as I know,unboundedness of tan(n)/n has yet to be proven.===Subject: Re: 2 analysis questions>I need to show that Xn = (tan n) /n is unbounded. Now I clearly see that>f (x) = (tan x)/ x is unbounded. Simply let x be near odd multiples of>pi/2. Now given, for example, 7pi/2 we can find a sequence of rationals>numbers that approach 7pi/2. That is, I can find integers p and q such>that p/q is as close to 7pi/2 as I want. Then the integer p is close to>q*pi/2. But q is not necessary odd! Now my question is, how do I>determine which odd multiples of pi/2 I can approximate with an integer?>Or am I correct when I am leaning toward every odd multiple of pi/2. How>do you prove this??> Well of course the proof is going to depend on the fact that pi is> irrational, so it's really a bit of a fake to say you've actually> proved it from first principles, unless you've already _proved_> that pi is irrational, which I doubt.> Anyway: Let x_n = n mod 2 pi; that is, 0 <= x_n < 2 pi and> n = k_n* 2 pi + x_n for some integer k_n. The fact that pi is> irrational shows that the x_n are all distinct. Since they're> all distinct and they all lie between 0 and 2 pi, the> pigeonhole principle shows that given e > 0 there exist> n, m with |x_n - x_m| < e. It follows from _this_ that there> exists an integer k within e of an odd multiple of pi/2,> because...>But as best I can tell, this does not then lead us easily to a proof that>tan(n)/n is unbounded. Rather, it just tells us that tan(n) is unbounded.Hmm, dunno what I was thinking there.>Math Tutor might have some of his questions answered at>. As far as I know,>unboundedness of tan(n)/n has yet to be proven.>David Cantrell===Subject: Re: 2 analysis questions>I need to show that Xn = (tan n) /n is unbounded. Now I clearly see that>f (x) = (tan x)/ x is unbounded. Simply let x be near odd multiples of>pi/2. Now given, for example, 7pi/2 we can find a sequence of rationals>numbers that approach 7pi/2. That is, I can find integers p and q such>that p/q is as close to 7pi/2 as I want. Then the integer p is close to>q*pi/2. But q is not necessary odd! Now my question is, how do I>determine which odd multiples of pi/2 I can approximate with an integer?>Or am I correct when I am leaning toward every odd multiple of pi/2. How>do you prove this??> Well of course the proof is going to depend on the fact that pi is> irrational, so it's really a bit of a fake to say you've actually> proved it from first principles, unless you've already _proved_> that pi is irrational, which I doubt.> Anyway: Let x_n = n mod 2 pi; that is, 0 <= x_n < 2 pi and> n = k_n* 2 pi + x_n for some integer k_n. The fact that pi is> irrational shows that the x_n are all distinct. Since they're> all distinct and they all lie between 0 and 2 pi, the> pigeonhole principle shows that given e > 0 there exist> n, m with |x_n - x_m| < e. It follows from _this_ that there> exists an integer k within e of an odd multiple of pi/2,> because...>But as best I can tell, this does not then lead us easily to a proof that>tan(n)/n is unbounded. Rather, it just tells us that tan(n) is unbounded.>Math Tutor might have some of his questions answered at>. As far as I know,>unboundedness of tan(n)/n has yet to be proven.fraction point of view, it requires one to find, for any given e > 0, apair of natural numbers m and n so that |(2m+1)/n - 2/pi| < e/n^2 [1]That is, we must have an arbitrarily large continuant in the continuedfraction expansion of 2/pi, right after an odd convergent numerator (ife < 1/2, the only possible rational approximations that can satisfy [1]are convergents of the continued fraction for 2/pi). This seems to bequite a difficult problem.Rob Johnson Which is the same as> n < (1 + e)^n;> you can show that for every e > 0 there exists N such that> this holds for all n > N from the binomial theorem.or from Bernoulli's Inequality===Subject: Re: 2 analysis questions> I need to show that Xn = (tan n) /n is unbounded.Well, even to prove that this sequence does not converge is not trivial.See the thread A limit problem with explanation at sci.math.symbolic.===Subject: Re: 2 analysis questions> I need to show that Xn = (tan n) /n is unbounded.> Now I clearly see that f(x) = (tan x)/x is unbounded.> Simply let x be near odd multiples of pi/2. Now given,> for example, 7pi/2 we can find a sequence of rationals> numbers that approach 7pi/2. That is, I can find integers> p and q such that p/q is as close to 7pi/2 as I want. Then> the integer p is close to q*pi/2. But q is not necessary> odd! Now my question is, how do I determine which odd> multiples of pi/2 I can approximate with an integer?> Or am I correct when I am leaning toward every odd> multiple of pi/2. How do you prove this??Maybe it's okay to assume that sin(n) and cos(n) are denseon the interval (-1, 1), because a proof might be difficult.> I need to show that Xn = ( n )^(1/n) converges to 1.Use ln(x(n)) = (1/n)ln(n), which goes to zero (L'Hopital's Rule).Then x(n) goes to e^0 = 1.===Subject: Re: 2 analysis questions> I need to show that Xn = (tan n) /n is unbounded.> Now I clearly see that f(x) = (tan x)/x is unbounded.> Simply let x be near odd multiples of pi/2. Now given,> for example, 7pi/2 we can find a sequence of rationals> numbers that approach 7pi/2. That is, I can find integers> p and q such that p/q is as close to 7pi/2 as I want. Then> the integer p is close to q*pi/2. But q is not necessary> odd! Now my question is, how do I determine which odd> multiples of pi/2 I can approximate with an integer?> Or am I correct when I am leaning toward every odd> multiple of pi/2. How do you prove this??Are you familiar with the fact that if x is irrationalthen there are infinitely many rational numbers p/qwith |x-p/q|<<1/q^2?(in particular, for |qx-p|<<1/q so if x is irrational, there are integral multiples of x which are (arbitrarily) close to integers)===Subject: Re: 2 analysis questions> I need to show that Xn = (tan n) /n is unbounded.> Now I clearly see that f(x) = (tan x)/x is unbounded.> Simply let x be near odd multiples of pi/2. Now given,> for example, 7pi/2 we can find a sequence of rationals> numbers that approach 7pi/2. That is, I can find integers> p and q such that p/q is as close to 7pi/2 as I want. Then> the integer p is close to q*pi/2. But q is not necessary> odd! Now my question is, how do I determine which odd> multiples of pi/2 I can approximate with an integer?> Or am I correct when I am leaning toward every odd> multiple of pi/2. How do you prove this??> Maybe it's okay to assume that sin(n) and cos(n) are dense> on the interval (-1, 1), because a proof might be difficult.> I need to show that Xn = ( n )^(1/n) converges to 1.> Use ln(x(n)) = (1/n)ln(n), which goes to zero (L'Hopital's Rule).> Then x(n) goes to e^0 = 1.Yes!, but this technique would only tell me that n^(1/n) converges to 1, butIS NOT a proof.===Subject: Re: 2 analysis questions>I need to show that Xn = ( n )^(1/n) converges to 1.> Use ln(x(n)) = (1/n)ln(n), which goes to zero (L'Hopital's Rule).> Then x(n) goes to e^0 = 1.> Yes!, but this technique would only tell me that n^(1/n) converges> to 1, but IS NOT a proof.Sure you can restrict yourself to first principles.I'm using theorems, proved in the text or given as exercises.===Subject: Re: Antidiagonal, Infinity> You say there are multiple antidiagonals because you generate an> antidiagonal, and then add that to the list and diagonalize a new> list. If you keep generating antdiagonals that way, they aren't of> the original list, which has one and exactly one binary antidiagonal.If by anti-diagonal you mean a number constructed according to some specific algorithm, that is correct, but if you mean any number constructed so as not to be in the original list, that is false. The method I described will produce one such number not5 in the original list for every iteration. There are at least countably many other algorithms that are guaranteed not to produce numbers in any given list. > I think more about the antidiagonal of an infinite list of> particularly all possible sequences of binary elements.> This concept is that if a set contains every possible permutation,> then no method exists to get a different permutation, because it would> instead be one of the possible permutations instead of different.Thus proving that no list can contain all possible permutations, in whatever sense Ross means by that word.===Subject: Re: Antidiagonal, InfinityMaybe there does not exist those multiple antidiagonals in binary,Virgil. The antidiagonal is different at each of its places from oneof the original list elements, an addended antidiagonal would have todiffer from the antidiagonal at one of those places, and thus notensure exclusivity from whatever element is at the index where theantdiagonal was originally different. There's no end of the listwhere to put the antidiagonal. You could insert the originalantidiagonal somewhere within the list and then some other index, thenext, would apply for each of the following original list elements. Each of the new, unoriginal lists has a different diagonal.It is a different list, given the original list you will generateexactly one given antidiagonal that is always the same given anidentical list. Following addenda will vary based upon where youinsert the original antidiagonal on your new list. I might say thatyour process continues while you have not exhausted duallyrepresentable list items, else your value would not be on the range. The set of list items does not change when you insert an alreadydually represented value, because a set contains no duplicates.So, given a list you can generate one, and exactly and only oneantidiagonal, after which you can modify the list and do as you wish,under most conditions.For these antidiagonals to be able to be generated the list must havea binary sequence with dual representation with not both of itsrepresentations already present on the list, or a sequence withoutdual representation not on the list, and as many can be generated asthere are sequences missing from the list, generating new lists. Theantidiagonal must equal itself for it to be an element of the set.Given a fixed list of all reals, the exact and only antidiagonal willhave a representation different but value equal to a list element. Given a fixed list of all binary sequences, either the antidiagonaldoes not exist or as some say the list does not exist. The setexists, and an ordering can be placed upon the set. Averring that theantidiagonal does not exist does not lead to a contradiction.Here, we encounter some of the differences between a list and a set. The list basically maps each element of a set to an integer andimplies an ordering upon the set elements as a list. The set is thecollection of elements that contains no duplicate elements, elementsare sets, and the empty set, null, or {}, is the only ur-element orleast set element, and from it is constructed other sets.So, I presented a method of sorts to antidiagonalize a set rather thana list, thus making what applies to the list apply to the set, or viceversa.The argument of my previous post that you politely omitted withoutcomment was that the antidiagonal can only be constructed when thereis an element of the original list with not both of its dualrepresentations on the list, else it would not be an element of therange, because it actually has to be one of the dually representeditems else it doesn't exist.If you would, please explain how that the antidiagonal may not existis or is not inconsistent. Also, please present other resultsclaiming no mappings between the integers and reals.===Subject: Re: Yao Ziyuan's Conjecture :-)Originator: richard@cogsci.ed.ac.uk (Richard Tobin)> I did a plot of N against the number of ways can be represented as the> sum of two primes for N <= 1,000,000.What is known (or conjectured) about a lower bound for this function?On log-log axes it looks close to linear, with the power being about0.849.-- Richard===Subject: Re: P vs NP: my proof of P != NPOriginator: dmoews@ccrwest.org (David Moews)|OK, but if more formally, where is the error in (1) <=> (4) ?|(what transition is wrong, or explain please what you mean under local|configuration consistency - I just don't understand).1. Let's take a sample Turing machine, say one which increments its inputby 1, and sample input, say 19. We can write the sequence of tapeconfigurations of the Turing machine down as a matrix: x = 0 1 2 3 4 5 6 7 +-------------------------------------- t | = | +--(alpha)--+ 0 | B 1 0 0 1 | 1# B B | | +---+ +---+ 1 | B 1 0 0 1# 0 | B | B (*) | +---(beta)--+ +---+ 2 | | B 1 0 | 0# 0 0 B B | +---+ +---+ 3 | B | 1 | 0 1# 0 0 B B | +---+ |Here, x denotes the location on the tape, t denotes time, and # marks thehead of the Turing machine. We haven't written down the control state ofthe Turing machine, but let's say it eventually halts successfully. We can consider a local consistency condition corresponding to the box (alpha) above. It will say that the state of cell 6 of the tape at time 1 is what it should be, given the state of cells 5, 6, and 7 of the tape at time 0, the position of the head at time 0, and the control state of the Turing machine at time 0. Similarly, the local consistency condition corresponding to box (beta) says that the state of cell 1 of the tape at time 3 is what it should be, given the state of cells 0, 1, and 2 at time 2, etc. Now, let's look at your formula (1), which I'll rewrite as (E F1)(E F2)...(E Fn)Z, (**) where Z = (A X1)...(A Xn)W, W = (A a)P(X1, ..., Xn, F1(X1), ..., Fn(Xn), a),F1, ..., Fn are unary predicates on bitstrings, and X1, ..., Xn, a arebitstrings. For our example, (**) is satisfied, so there are assignmentsto F1, ..., Fn which satisfy Z. What are these Fi's? They are each basically just the matrix (*), rewritten as predicates of space and time (and, in your construction, input, but let's forget about that---fix the inputto 19.) Z states that (*) is a consistent, possible record of the Turingmachine computation, and it does this by being a conjunction of localconsistency conditions. This conjunction is expressed by the universalquantifiers on X1, ..., Xn, so that for each choice of values of X1, ...,Xn, W is a bounded conjunction of local consistency conditions---in yourconstruction, no more than 2s of them.Formula (4), on the other hand, is (A X1)(A X2)...(A Xn)(E f1)(E f2) ...(E fn)W', (***) where W' = (A a)P(X1, ..., Xn, f1, ..., fn, a),X1, ..., Xn, a are bitstrings, and f1, ..., fn are Boolean variables.This also expresses a conjunction of local consistency conditions, but thereis now nothing corresponding to the matrix (*). (***) therefore says that for each choice of no more than 2s local consistency conditionsexpressible by W', there is some way of locally filling in a few entriesof the matrix (*) that satisfy them, but a different choice of these<=2s conditions might mean filling in contradictory entries in (*).This is why (**) and (***) are not the same.2. I did not point out the error in your derivation of (4) because I didnot understand it. In your formula| ( F1 = (X1 = 0..00)^phi_1 /| (X1 = 0..01)^phi_2 /| ...| (X1 = 1..11)^phi_{2^N}| )^...^(| Fn = (Xn = 0...00)^phi_1 /| (Xn = 0...01)^phi_2 /| ...| (Xn = 1...11)^phi_{2^N}| ) (*)what does (X1 = 0...00)^phi_1 mean? What are the phis? dmoews@xraysgi.ims.uconn.edu===Subject: does this look correct everyone, I was wondering if someone could check this questionfor me and see if I need to add or change things to make my proofsLet G be a group and let G' be the subgroup of G generated by -1 -1the set A = {x y x y | x, y in G. (a) Prove that G is a normal subgroup. It already is a subgroup. moreover if w is in G' and g is in G -1 -1then g w g w is in G' and since w in also in G' -1 -1 g w g = is in G'. -1 -1 -1Replacing g by g and w by w, we get g w g is in G'.Therefore G' is normal.(b) Prove the G/G' is abelian. -1 -1 Since x y x y = e, we get xy = yx(c) If G/N is Abelian, then G' c N. -1 -1 since xy = yx mod N, we get x y x y is in N. Thus A c N and so the subgroup generated by A will be in N.(d) Prove that if H is a subgroup of G and G' c H, then H is normal in G. The same proof as in (a) will work because we can -1 -1 still say g w g w is in H and w is in H. Therefore -1 -1 g w g is in H. Taking the inverses of the elements gives -1 g w g is in H so H is normal.Note: c denotes the symbol for a subroup (sorry only key I could findthat looked decent)Again thank you for the help.===Subject: Complex Function Series Uniform Convergence QuestionI need to prove that sum_{n=0}^infty (frac( z-1, z+1 )) ^ n converges locally uniformly in the half plane Re z > 0, and find the sum.Sorry for the TEx notation, otherwiseoo---- / n | z-1 |/ | --- |---- | z+1 |n=0 /How do I do that? I can't even figure out any sort of partial sums to work from. :(All help and hints greatly appreciated.===Subject: Re: Complex Function Series Uniform Convergence Question> I need to prove that sum_{n=0}^infty (frac( z-1, z+1 )) ^ n converges > locally uniformly in the half plane Re z > 0, and find the sum.> Sorry for the TEx notation, otherwise> oo> ---- / n> | z-1 |> / | --- |> ---- | z+1 |> n=0 /> How do I do that? I can't even figure out any sort of partial sums to > work from. :(Do you mean that you don't know what's the sum of the first n termsof a geometric series?Besides, you should use the fact that the assertion Re z > 0 isequivalent to the assertion |(z - 1)/(z + 1)| < 1.===Subject: Re: Complex Function Series Uniform Convergence Question> I need to prove that sum_{n=0}^infty (frac( z-1, z+1 )) ^ n > converges locally uniformly in the half plane Re z > 0, and find the sum.> Sorry for the TEx notation, otherwise> oo> ---- / n> | z-1 |> / | --- |> ---- | z+1 |> n=0 /> How do I do that? I can't even figure out any sort of partial sums to > work from. :(> Do you mean that you don't know what's the sum of the first n terms> of a geometric series?> Besides, you should use the fact that the assertion Re z > 0 is> equivalent to the assertion |(z - 1)/(z + 1)| < 1.> Jose Carlos SantosDuh !!!.... yikes.===Subject: Re: Haar integralsContent-Length: 1138Originator: rusin@vesuviusThis may help: Write a generating functionZ(x,y) = int dmu(u) exp(- )where x and y are real numbers. We know how to do this integral (Itis a special case of the itzakson-zuber integral).dZ(x,y)/dx |_{x=0} = int dmu(u) exp(- )Then if B is positive definite so teh value at y=inf is zero, we haveint_{-inf}^0 dy (dZ(x,y)/dx) |_{x=0} = int dmu(u) / as you want. > Hi > I am wondering how one might perform the integral> int dmu(u) / where A and B are matrices B>0, |u> (in bra/ket notation) is a > normalized complex vector in C^N, and the measure dmu(u) is the unitarily > invariant uniform measure (Haar measure on unitaries U where |u>=U|0> say). > Any ideas on this or similar integrals? I can see how to do products: > int dmu(u) .... But am not sure how to treat more > complicated ones such as the above.> andrew.===Subject: Re: PROOF that numbers are countable>What real number is greater than all members of S and less than 1/3?>Nonexistent.>What does this tell you?>It tells me you don't have a clue what you're asking.>coming from you..... remember 'whats the difference between>j sharing a digit with all numbers and j sharing the the same digit with all numbers' ?>then a completely unrelated proof, never heard back on that one ghosty,>just stick to the material.>Herc>I answered your questions here. Did you have additional questions?>Just this 1.>Do you understand the difference between:>for any i, there exists a j such that F(i,j) = G(j)>and>there exists a j such that for all i, F(i,j) = G(j)>?>They're quite different.>For that matter, do you understand the difference between:>Ai [F(i,i) /= G(i)]>and>Aj Ei [F(i,j) = G(j)]>The first one is Cantor's argument, the second is what you been trying>to use to disprove it.> Actually I'm using this :> Aj Ei [F(i,j)=G(j) & F(i,j-1)=G(j-1) & F(i,j-2)=G(j-2) ... F(i,1)=G(1)]> Which means whatever number you actually specify computables will match it.> What is your natural language meaning of it? If I did make an infinite list can> you objectively construct a new number? All combinations are present on the> list, your technique is flawed to allow an infinite list and then cite elements at> finite positions in the list. Infinite list by definition means incomplete, not that> you must construct something bigger.> What happens if F is defined as Ai Aj [(i F(i,j)=4) & (i>=j -> F(i,j)=3)], which means for Aj G(j) = 4?> In this case, Ai F_i has only finitely many 4's and infinitely many 3's> in its sequence, but G has infinitely many 4's and no 3's.F itself contains an infinite number of 4s. Remove F(1) and look atthe diagonal, its 0.4..This means when F is mapped to the number line 0.4.. is covered.Therefore the diagonal failed to demonstrate that F does not containthat sequence, and I'm using contain in a certain way here. Atleastwith 0 substituted for the 3.> In any case, you haven't shown that your statement has anything to do> with the diagonalization.All unlimited length finite strings are a subset of all (finite & infinite) strings. Thecomputables contains the former, diagonalisition indicates the latter. The difference betweenthe 2 sets is where unspecifiable irrationals exist, its no mans land, a phantom.It simply DOES NOT PROVE that combinations do not exist on countable lists.The diagonal was supposed to prove a new sequence, it didn't! It proved a selfreferential paradox on handling infinite sequences.No new sequence = Herc is not convinced!Remember what sequences look like?0.12345Thats computable!0.12345.... can be ANYTHING! That's computable!fHerc===Subject: Re: PROOF that numbers are countable>What real number is greater than all members of S and less than 1/3?>None exists.>What does this tell you?>That S does not contain it's least upper bound.>So, S itself has a least upper bound of 1/3.>There is no irrational that you can specify between S and 1/3.>Therefore you cannot specify any 'gap' between S and 1/3.>How are you comparing S and 1/3? < and <= are relational operators>between a real and a real, not between a set of reals and a real.> That is absurd. -3 < N> might take some lateral interpretation but the truth value is easily defined.It can be defined, but it is not, at the moment, defined. More conventional terminology would be to say that -3 is a lower bound of N.> I'm just underlining on the number line what region S covers. The infinite> set S goes right up to 1/3.S doesn't cover a region, it covers discrete points. And, S does not include 1/3.> I don't care what the individual elements of S point to, I'm examining the regions> of the number line.> Your argument is akin to saying the set of all reals of the form 0.xxxxx...> does not include 1. The region 0-1 declares 0.999999.. which includes 1.Let's see: I was talking about finite decimal strings, so you switch to infinite decimal strings. I fail to see the connection.> Why are you incapable of dropping [ELEMENTS OF S] and performing> deductive reasoning on the area of the number line that it maps to?Because you are attempting to discuss things that already have a standard terminology and are abusing it. If you want to use terminology in non-standard ways, you will need to define it first.Also, deductive reasoning is based on logical sequences of statements. You are consistently introducing unrelated statements in an attempt to disprove what you do not agree with.> This is not a direct argument of Cantors proof about numbers existing on computables> R mapped to from N, this is for the seperate proof that Computable numbers contain> all sequences of digits. *This* proof does not require for me to specify ANY> particular N.> This proof only insists that all permutations of digits appear somewhere on the infinite> countable list. The example is S.> S= {> 0.3> 0.33> 0.333> 0.3333> ... }> S is just a different representation of 0.333..Ok, but where is S on your list? Why not simply use .3333....?> This is how it is represented in Functional Programming Languages to handle> infinite streams. S = 1/3.Translation: you are defining non-terminating decimals to be sets of terminating decimals that have the non-terminating decimal as its upper-bound/limit?> There is no disputing S = 1/3.> The argument is not about individual elements of S, only the INFINITE set S.You have presented the discussion as being about the elements of S, then switched to what S represents. It would have been helpful to say so in the first place or, better, simply use what it represents and dispense with talking about S.> It is OBVIOUS that the INFINITE set S contains the point 1/3. No matter> how far you zoom into the number line, the line from the highest lower bound> to the least upper bound draws up to the point 1/3.It is FALSE that 1/3 is an element of S. Or have you redefined the word contains to mean something other than is an element of? If so, a clear definition would have been handy.> Stop thinking that S is finite and can only be represented by ELEMENTS of S.> S is countably infinite. You can't allow S to be defined as countably infinite and> then turn around and declare S never reaches 1/3. Because as surely as> 0.3333.. reaches 1/3 so does S.Have you ever heard of open intervals? These do not contain their endpoints, yet have elements that approach the endpoints in much the same manner that elements of S approach 1/3. Your language is reminiscent of someone claiming that an open interval contains its endpoints, when by definition it does not.>If there is NO irrational between S and 1/3, then there is no missing>number there at all.>If S is adjacent to 1/3 on the number line, then 1/3 is contained by>the SET S. (Numbers cannot be adjacent, if so they are the same point.)>This is evident by examining the infinitely long diagonal of S. Diag_S = 0.333..>By this logic, there are no such things as removable discontinuities in>calculus. Time to start writing the publishers of calculus books and>notifying them of the error.> That is oblivious, there's just as many text books quoting Cantors proof.But does it reflect your oppinion? Do you believe that removable discontinuities exist?>1/3 could be any point, therefore increasingly specified sequences of rationals are sufficient>to map to all of R.>1/3 is a very specific point.>Your argument above boils down to the following:>Given S = {.3, .33, .333, .3333, ... }>(Ae [Ei (i in S & (1/3-i) (1/3 in S)>However, it is also true for S that:>Ai [ (i in S) -> (1/3 > i) ]>So, if your conclusion is true, we can conclude that 1/3 > 1/3. Is this>what you want?> No, I want you to stop limiting the range of S to its indexed members and> consider the INFINITE set S. S = 1/3 in functional programming.I'm not doing functional programming. I'm doing math. Perhaps you hadn't noticed. If you wish to use concepts from functional programming, a few definitions of how you are changing the usual terminology would be convenient.> This is not a (direct) refutation that N cannot map to R, its for reasoning> about the minimal set of numbers required to ->COVER<- the number line,> not ->MAP<- to it.What do you mean by covering the number line?email: wtwentyman at copper dot net===Subject: Re: Can all/everything be in a set, or in a mathematical group (which are two different things in math)?> I just sent this to sci.math.research but it may not show up> there for a few hours since it is moderated. For this> crosspost I have set followups to sci.math so follow> the thread there if you are interested, and also the> thread on sci.math.research if any develops.> -------------------------------------------------------------- -------> I define all/everything (or all as a unit) by sayingIs there anything outside all/everything? No.> But, that's the same way all was defined before> set theory was invented. And probably the > only important thing about set theory is that > it proved that you can't define all.Sorry, dummy, I just defined it.===Subject: Re: Divisibility by 2===>Subject: Divisibility by 2>Message-id: <4073fbb9.0@entanet>If I've got a number of the form (3^k)p-1, is there an easy way to find out>how many times it can be divided by 2. ie. get something like: (3^k)p-1 =>(2^j)q, where 2 and q are coprime.>I know k and p in advance (ie. at the start of calculation), but I need a>standard way of finding j (and therefore q) from them.>More info: p = 1 (mod 2), and both p, k are natural numbers (>=1).>I can construct p if this helps, ie. make p=(2^m)r+1, or something similar>that will help to find j.>Or I can make p=1 (mod 2^z) or something. p has to be odd though.>Any help appreciated.>-- >David VivashSince (3^k)p is always odd, in binary, every number ends in ..11 or ..01. Any that end in ..11 will then become ..10 when you subtract the 1. For these j will always be 1. To find out whether the numbers end in ..11 or ..01, take k (mod 2) and p (mod 4):k (mod 2) p (mod 4) (3^k)p--------- --------- ------------------ 0 1 ends in ..01 0 3 ends in ..11 1 1 ends in ..11 1 3 ends in ..01For the numbers ending in ..01, the actual number of 0sto the left of the least significant bit can be found in the sequence1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 ...If we know how far into the sequence the number is,adding two to the bit position of the least significant 1 ofthe index will tell us j.For k=3 p=3, (3^k)p=81. The index into the sequence is(81-1)/4=20. 20 in binary is 10100. The least significant1 is at bit position 2, so j=4.Unfortunately, in order to find the index, we had to compute(3^k)p-1, so there is really no need to do all that muckingabout with 0 sequences, j would simply be the bit positionof the least significant 1 bit of the binary representationof (3^k)p-1. For k=3 p=3, (3^k)p-1=80. 80 in binary is1010000. The least significant 1 is at bit position 4 (which is the same as the count of least significant 0s, which isthe same as the factors of 2). The GMPY module in Python actually has a function to do this. I use it to remove all factors of two in one fell swoop to avoid having to iteratethrough them.x = 3*x + 1a = scan1(x) # returns bit position of least significant 1x = x/2**a # all factors of 2 removed in one step===Subject: Re: differentiating power series> The series terms are additive, just shorthand notation for a sum of terms,> so..... a proof is trivial, convergence is not required either,> (simple power series k(x-j)^n)> -JSThis is simply wrong. Try differentiating the series Sigma (0 to oo)n!x^n and give me a meaningful derivative.'cid 'ooh===Subject: Re: differentiating power series> The series terms are additive, just shorthand notation for a sum of terms,> so..... a proof is trivial, convergence is not required either,> (simple power series k(x-j)^n)for us.===Subject: Riemann-integrability of a diff'ble functionHi!Let I be a compact interval, X a Banach space of real numbers and f: I -> X be differentiable and such that the derivative is boundedbut not necessarily continuous.Question: Is f' Riemann-integrable?Kai===Subject: computational statisticsWhat is really the difference between: 1. computational statistics 2. statistical computing 3. computer-intensive statistical methods 4. numerical statistics 5. Monte-Carlo simulation methods for statistics?Or is it all the same?4. includes classical methods but lives without randomizationexperiments, whereas 5. is based on experiments with random numbers.In a certain way, 1-3 could be seen as the union of 4+5.Clearly the discussion is only semantic but I am somewhat confusedby that what I found about these concepts in the internet.Any recommendation of a concise introduction available by internet would be welcome.Does anyone know a good German translation of 'computational statistics'?Torsten===Subject: Re: Convergence and term-by-term manipulation of infinite series> Various people made contributions to my original posting, including> Uh, I knew that those were possibilities. There are no> necessary and sufficient conditions there, and the> standard necessary conditions are very different for> differentiation and integration - hence my curiosity what> sort of manipulation he's talking about. There's many> other sorts of manipulations I can imagine he could> be talking about, like reordering the terms, applying> various summability methods...>OK, a few examples:>1 - 1/2 + 1/3 - 1/4 + 1/5... converges.>Let S = 1 - 1/2 + 1/3 - 1/4...>Then 2S = 2(1 - 1/2 + 1/3...) = 2 - 1 + 2/3 - 2/6 - 1/4 + 2/5 ->2/10...> = 1 - 1/2 + 1/3 - 1/4... = S>Thus 2S = S.>In this case, term-by-term manipulation doesn't work because the>original series is not absolutely convergent.I didn't quite follow the manipulation there, but evidentlyyou're reordering the terms - yes, if you want to add up theterms in a different order then absolute convergence isexactly what you need.>sin(x) = x - (x^3)/3! + (x^5)/5!...>i*sin(x) = ix - i(x^3)/3! + i(x^5)/5!... = (ix) + ((ix)^3)/3! +>((ix)^5)/5!...>cos(x) = 1 - (x^2)/2! + (x^4)/4!... = 1 + ((ix)^2)/2! + ((ix)^4)/4!...>cos(x) + i*sin(x) = 1 + (ix) + ((ix)^2)/2! + ((ix)^3)/3!... = exp(ix)>Hence cos(x) + i*sin(x) = exp(ix)>This works because the three series are absolutely and uniformly>convergent over all Z.??? Surely you meant over all R? And by the way, those seriesdo _not_ converge uniformly on R - they converge uniformlyon bounded subsets of R.Anyway no, _those_ manipulations just follow from the factthat the series _converge_; absolute and/or uniform convergencehas nothing to do with it. >Other examples are possible, but you get the idea; multiplication, and>(more importantly) rearrangement were what I principally had in mind.>The book in which I found the conditions I previously alluded to had>pertinent examples for each of the conditions it gave.Well again, there are so many things you might be talking aboutthat it's not possible to really answer your question. If youmention a specific sort of manipulation you're interested inpeople will be able to give you conditions under whichit's valid.>Jabberwocky===Subject: Re: Dual of Frobenius> I find inseparable maps of function fields a bit hard to visualize.> For example, let E be an elliptic curve over a field k of char p>0.> The multiplication-by-p map [p] is then inseparable, so we can factor> it as F^.F, where F is the Frobenius map E->E^ and F^ is its dual.> Thing is, we have deg[p] = p^2, so that deg(F^) = p. Does it follow> that F^ is inseparable?No, a map of degree p does not have to be inseparable. For example,the map f : P^1 --> P^1 f(x) = x^p + xis not inseparable, though it clearly has degree p.For elliptic curves, F^ is inseparable if and only if E issupersingular. You can find a brief discussion of this is my book TheArithmetic of Elliptic Curves, Chapter V, section 3.Joe Silverman===Subject: Natural Transcendental Reflect Asymptotic Exactnesshttp://mathworld.wolfram.com/ TranscendentalNumber.htmlhttp://www.halexandria.org/dward089. htmhttp://sprott.physics.wisc.edu/pickover/trans.htmlEvery transcendental Number is Exactly correlatedwith an Asymptope in the set of all possible Numbers.The stuff that I've been discussing, both in the Proofof Fermat's Last Theorem that I've posted, and in myrecent discussion of Fermat Integers [what I've namedthe new Integers that I've been discussing in the I'veEliminated 'irrational' numbers thread], derives in suchAsymptotic behavior.In a 'natural' transcendental, such as pi or e, the seem-infly 'random' distributions of post-decimal digits is anExact reflection of approaches to an Asymptope, andthe Exactness that I discussed in both my Proof of Fer-mat's Last Theorem, and the Fermat Integers.It's a whole New 'Day' for Mathematics.===Subject: Re: Need some clarification.> This might be a really stupid question to ask but could someone> explain to me what I need to do for these problems. I've read them> through several times and looked through my notes and my book and I> don't know what exactly needs to be done. If someone is willing to> push me in the right direction it would greatly be appreciated.> Question: > If G is a group that acts on a set X, then for any x E X we define the> orbit of x to be O(x) = {g x : g E G} and we also define the> stabilizer> of x to be Gx = {g E G : g x = x}.> Each of the following four parts describes a group action of G on X.> In> each case compute Ox and Gx for every x E X.> (a) Let G = Z and let X = R^2. We define an action of G on X as> follows: For n E Z and (x, y) E R^2 we define n (x, y) to be the> point in R^2 obtained by rotating (x, y) by an angle of pi(n)/2> radians> in a counterclockwise direction about the origin. (Hint: In order> to compute G(x,y) for (x, y) E R^2, consider two cases first when> (x, y) = (0, 0), and then when (x, y) /= (0, 0). )Not sure what you're asking. It's pretty clear what they want. Given the definitions of orbit and stabilizer, they want you to determine these sets for each given action. Which part are you having trouble with? I'll do one just as an example. Gx is the set of elements of G that leave x where it was (stabilize it). In (a), hat's the stabilizer of (0,0)? What rotations leave (0,0) fixed? Clearly all of them. So G(0,0) = Z in this case.If (x,y) is not the origin, what rotations leave it unchanged? Any rotation that's an integer multiple of 2pi, right? So for n a multiple of 4, pi*n/2 is an integer multiple of 2pi, and conversely. So in this case, G(x,y) = {n : n is a multiple of 4}Did that help?===Subject: Erratum for item 4.4.45 in Abramowitz and StegunIt is perhaps surprising that new errors can still be found in thevenerable _Handbook of Mathematical Functions_ by Abramowitz and Stegun(hereafter, A&S).Last month in sci.math.num-analysis, in the process of trying to assistsomeone who had made a request for a computationally cheap approximationfor Arccos, I discovered two problems with their item 4.4.45. Forconvenient reference, see, for example.One of the problems seems to be a simple mistake: Their error bound isincorrect. They should have claimed that |error| < 7*10^(-5), rather than5*10^(-5).The other problem is not so simple. Item 4.4.45 in effect gives anapproximation for Arccos(x) in the form Sqrt(1-x)*(cubic polynomial in x).Surely the vast majority of people who would use 4.4.45 would naturallyassume that the four coefficients given for the polynomial were computed soas to minimize max|error| for the approximation. But that assumption iswrong! In this sense, I consider 4.4.45 to be misleading in an importantway.If one does compute the coefficients so as to minimize max|error| inapproximating Arccos(x), 0<=x<=1, by Sqrt(1-x)*(a + b*x + c*x^2 + d*x^3),one finds that the coefficients should be approximatelya = 1.570758334, b = -0.212875075, c = 0.076897503, d = -0.020892330and that |error| < 3.8*10^(-5).Perhaps you are now wondering how 4.4.45 came to be as it is. Although Iam unsure how their incorrect error bound was obtained, I do know thesource of the other, more significant problem. Notice that A&S give awork by Hastings et al. as their source. Having looked at that work, I cansay that it is not misleading at all. But one must be careful! Hastings etal. in effect determine the coefficients of the polynomial which minimizesmax|error| in approximating Arccos(x)/Sqrt(1-x). Perhaps without payingadequate attention, A&S then gave that same polynomial, multiplied bySqrt(1-x), as the approximation for Arccos(x). The trouble is that, inattempting to minimze max|error|, the optimal polynomial for approximatingArccos(x)/Sqrt(1-x) is _not the same_ as the optimal polynomial to be usedwhen approximating Arccos(x) by Sqrt(1-x)*polynomial.Also note that item 4.4.46 in A&S is misleading in the same way as 4.4.45.[BTW, if we are to approximate Arccos(x), 0<=x<=1, by Sqrt(1-x)*(cubicpolynomial in x), some might prefer that the constant term in thepolynomial be precisely pi/2, and that the three other coefficients thenbe determined so as to minimize max|error|. This givesSqrt(1-x)*(a + b*x + c*x^2 + d*x^3) wherea = pi/2, b = -0.213300989, c = 0.077980478, and d = -0.02164095,with |error| < 4.5*10^(-5).Since the error bound for the prior approximation was slightly smaller, whymight this latter approximation be preferred? Well, if we approximatedArcsin(x) by taking the complement of our approximation for Arccos(x), the_relative_ error would be unbounded at x = 0 unless a = pi/2.]Hmm. What should one do with newly discovered errata for A&S now?===Subject: Re: Erratum for item 4.4.45 in Abramowitz and Stegun< ... on estimation errors concerning Airy fct ... Hmm. What should one do with newly discovered errata for A&S now?> David CantrellI would like to see it at http://dlmf.nist.gov/Contents/or similar as it is the successor of A&S on the www.BTW they claim a (rational) approximation of the cdf normalto be with |error| < 7.5*10(^-8) in (26.2.17) while it isaround 6 digits exact (that formula is quite commonly used),or do i read that a wrong way?---remove the no for mail===Subject: Courant vs. Spivak vs. Apostol> I'm looking for a good book on calculus. I'm interested in a book> that's sort of the equivalent to Feynman Lectures of Physics in the> Calculus world. I currently have books by Stein, Edwards, and> Stewart. This books are good in a practical sense; however, I want> something more insightful. It has to be rigourous, but at the same> time be application oriented. I don't mind if it's wordy. I've heard> Apostol and Spivak's books are a good choice. I've actually seen> Apostol and it looks promising; however, this opinion is based on> reading the preface and a quick look through the chapters (no more> than a half-hour). Any suggestions?> Calculus by Michael Spivak> http://www.amazon.com/exec/obidos/ASIN/0914098896/qid= 1082162810/sr=2-3/ref=sr_2_3/102-1387591-6151359What about Courant's calculus book? How does this book compare toApostol and Spivak's calculus books. How are they similar ordifferent? What is the intended audience (pure mathematics, appliedmathematics, in between)?===Subject: Re: Courant vs. Spivak vs. Apostol>I'm looking for a good book on calculus. I'm interested in a book>that's sort of the equivalent to Feynman Lectures of Physics in the>Calculus world. I currently have books by Stein, Edwards, and>Stewart. This books are good in a practical sense; however, I want>something more insightful. It has to be rigourous, but at the same>time be application oriented. I don't mind if it's wordy. I've heard>Apostol and Spivak's books are a good choice. I've actually seen>Apostol and it looks promising; however, this opinion is based on>reading the preface and a quick look through the chapters (no more>than a half-hour). Any suggestions?>Calculus by Michael Spivak>http://www.amazon.com/exec/obidos/ASIN/0914098896/qid= 1082162810/sr=2-3/ref=sr_2_3/102-1387591-6151359> What about Courant's calculus book? How does this book compare to> Apostol and Spivak's calculus books. How are they similar or> different? What is the intended audience (pure mathematics, applied> mathematics, in between)?Courant is probably the best for someone who wants to /use/ calculusand would like a reasonable understanding of what's going on -- it would be my choice for applied math. For someone interested in mathematicsper se, either Apostol or Spivak might be preferable. I find Apostol===Subject: Courant's Calculus Book Titles?>I'm looking for a good book on calculus. I'm interested in a book>that's sort of the equivalent to Feynman Lectures of Physics in the>Calculus world. I currently have books by Stein, Edwards, and>Stewart. This books are good in a practical sense; however, I want>something more insightful. It has to be rigourous, but at the same>time be application oriented. I don't mind if it's wordy. I've heard>Apostol and Spivak's books are a good choice. I've actually seen>Apostol and it looks promising; however, this opinion is based on>reading the preface and a quick look through the chapters (no more>than a half-hour). Any suggestions?>Calculus by Michael Spivakhttp://www.amazon.com/exec/obidos/ASIN/0914098896/qid= 1082162810/sr=2-3/ref=sr_2_3/102-1387591-6151359> What about Courant's calculus book? How does this book compare to> Apostol and Spivak's calculus books. How are they similar or> different? What is the intended audience (pure mathematics, applied> mathematics, in between)?> Courant is probably the best for someone who wants to /use/ calculus> and would like a reasonable understanding of what's going on -- it would > be my choice for applied math. For someone interested in mathematics> per se, either Apostol or Spivak might be preferable. I find ApostolWhat's the difference between Courant's Introduction to Calculus andAnalysis and Courant's Differential and Integral Calculus. Is it thesame thing?===Subject: Inverse Laplace TransformI have a Laplace transform ofU(x,s) = (q/k) * (alpha/s)^(1/2) * exp[-x * (s/alpha)^(1/2)]and I am trying to find the inverse of it. It's looking like I'll have todo the contour integration, since I can't find anything like this in atable. If I understand it correctly, U(x,s) has a branch point at s=0(from the (alpha/s)^(1/2) term), and that will provide the residue thatshows up in Cauchy's Residue Theorem.A few questions:(1) How do I calculate the reside of the above transform at the branchpoint? (Do I have to go the route of generating the Laurent series for thetransform and then pick out the a_(-1) coefficient, or is there aneasier route? If I do have to generate the Laurent series, what's a goodway to familiarize myself with how to do that?)(2) How do I choose the gamma term in the bounds of the Bromwichintegral? I know that I have to choose a gamma such that all thesingularities are to the left of it in the complex plane, but would anygamma (>0) do?(3) Is this technique guaranteed to give me a closed form solution? Or isthere a possibility I'll get into the integration and realize thenumerical route is my only option?JeffAdd an underscore between 'd' and 's' for email.===Subject: Re: Inverse Laplace Transform> I have a Laplace transform of> U(x,s) = (q/k) * (alpha/s)^(1/2) * exp[-x * (s/alpha)^(1/2)]> and I am trying to find the inverse of it. It's looking like I'll have to> do the contour integration, since I can't find anything like this in a> table. If I understand it correctly, U(x,s) has a branch point at s=0> (from the (alpha/s)^(1/2) term), and that will provide the residue that> shows up in Cauchy's Residue Theorem.> A few questions:> (1) How do I calculate the reside of the above transform at the branch> point? (Do I have to go the route of generating the Laurent series for the> transform and then pick out the a_(-1) coefficient, or is there aneasier route? If I do have to generate the Laurent series, what's a good> way to familiarize myself with how to do that?)> (2) How do I choose the gamma term in the bounds of the Bromwich> integral? I know that I have to choose a gamma such that all the> singularities are to the left of it in the complex plane, but would any> gamma (>0) do?> (3) Is this technique guaranteed to give me a closed form solution? Or is> there a possibility I'll get into the integration and realize the> numerical route is my only option?> JeffYou can find it on the larger tables (say, the Bateman manuscript project)or ask Maple or Mathematica to do it.===Subject: Re: Inverse Laplace Transform>I have a Laplace transform of>U(x,s) = (q/k) * (alpha/s)^(1/2) * exp[-x * (s/alpha)^(1/2)]>and I am trying to find the inverse of it. It's looking like I'll have to>do the contour integration, since I can't find anything like this in a>table.Assuming alpha > 0 and x > 0, Maple gives the answer as 2 1/2 x q alpha exp(- ---------) 4 alpha t u(x, t) = --------------------------- 1/2 1/2 k t PiRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2===Subject: Re: Inverse Laplace Transform>I have a Laplace transform of>U(x,s) = (q/k) * (alpha/s)^(1/2) * exp[-x * (s/alpha)^(1/2)]>and I am trying to find the inverse of it. It's looking like I'll have to>do the contour integration, since I can't find anything like this in a>table.> Assuming alpha > 0 and x > 0, Maple gives the answer as> 2> 1/2 x> q alpha exp(- ---------)> 4 alpha t> u(x, t) = ---------------------------> 1/2 1/2> k t PiAre you familiar with Mathematica? I put the following input intoMathematica: InverseLaplaceTransform[(a/s)^(1/2)*Exp[-x*(s/a)^(1/2)],s,t] and the output I get is: a Sqrt[-] s InverseLaplaceTransform[------------, s, t] Sqrt[s/a] x EIs there a way to specify the same conditions you did (e.g. a > 0 and x 0) to get Mathematica to give me a useable result?JeffAdd an underscore between 'd' and 's' for email.===Subject: Re: Inverse Laplace Transform> Assuming alpha > 0 and x > 0, Maple gives the answer as> 2> 1/2 x> q alpha exp(- ---------)> 4 alpha t> u(x, t) = ---------------------------> 1/2 1/2> k t Pi>Are you familiar with Mathematica? I put the following input into>Mathematica:> InverseLaplaceTransform[(a/s)^(1/2)*Exp[-x*(s/a)^(1/2)],s,t]> and the output I get is:> a> Sqrt[-]> s> InverseLaplaceTransform[------------, s, t]> Sqrt[s/a] x> E>Is there a way to specify the same conditions you did (e.g. a > 0 and x >0) to get Mathematica to give me a useable result?Use the Assumptions option.InverseLaplaceTransform[(a/s)^(1/2)*Exp[-x*(s/a)^(1/2)] ,s,t,Assumptions->{a>0, x>0}]does the trick.Cc: tim@birdsnest.maths.tcd.ie===Subject: Re: Computer based proofs> Surely this did not contain 400 microfiche pages? I don't know, I'm just quoting Appel and Haken from their paper TheFour Color Proof Suffices. An alternative proof is presented athttp://www.math.gatech.edu/~thomas/FC/fourcolor.htmlThe authors comment:There are two reasons why the Appel-Haken proof is not completelysatisfactory. Part of the Appel-Haken proof uses a computer, and cannot be verified by hand, and even the part that is supposedly hand-checkable is extraordinarily complicated and tedious, and as far as we know, no one has verified it in its entirety. We have in fact tried to verify the Appel-Haken proof, but soon gaveup. Checking the computer part would not only require a lot ofprogramming, but also inputing the descriptions of 1476 graphs, andthat was not even the most controversial part of the proof.We decided that it would be more profitable to work out our ownproof. So we did and came up with a proof and an algorithm that aredescribed below.===Subject: Re: Computer based proofs> An alternative proof is presented at> http://www.math.gatech.edu/~thomas/FC/fourcolor.html> The authors comment:> There are two reasons why the Appel-Haken proof is not completely> satisfactory.> Part of the Appel-Haken proof uses a computer, and cannot be> verified by hand, and> even the part that is supposedly hand-checkable is> extraordinarily complicated and tedious, and as far> as we know, no one has verified it in its entirety.I should really shut up until I find the Appel-Haken preprint,which as I recall purported to be a complete proof.But I went to the above URL, and am not entirely convincedthat its status is different to that of the A-H proof.The authors claim to reduce the number of irreducible casesto 600+ in place of 1500+,but their method appears to be the same as A-H (as indeed they say)and it is not clear to me if the reduction is real,or due to several A-H cases being subsumed in one.In any case, verification of 600 cases is not essentially differentfrom verification of 1500 cases;so the issue reduces to the question of whether A-H's proofactually relied on computer output that was not published in a journal(but was available in microfiche).My recollection is that the main use of the computer by A-Hwas to show statistically that it was almost certainly sufficientto consider regions with 14 edges.Although they did use the computer to study these cases,as I recall they then drew all the irreducible regions,and anyone with a lot of time on their handscould have verified that these could all be solved.However, it was a long time ago, and my memory is not goodat the best of times, so I must leave it until I find the preprint.Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, IrelandCc: tim@birdsnest.maths.tcd.ie===Subject: Re: Computer based proofs> But I went to the above URL, and am not entirely convinced> that its status is different to that of the A-H proof. They don't claim that it can be carried out by hand, as far asI could see. The point at issue is whether the Appel-Haken proofcan be carried out by hand. ===Subject: Re: Computer based proofs> [snipped]> The biggest drawback I see to such a tool is the need to standardize> the rigor of virtually all mathematics at a pretty high level. In> particular, some fields (eg, low dimensional topology, some algebraic> geometry) often slink by :-) with a lower level of rigor due to the> intuitive nature of the mathematical concepts involved.> What lower level of rigor? There is a pretty high level of rigor in> low dimensional topology. It's a common mistake for outsiders,> unfamiliar with the standard proof techniques of the field, to> associate known arguments with sloppy reasoning.Good point. By this, I mean the amount of reasoning that the readerneeds to perform between each step of the proof. A proof good enoughfor a few dozen experts in the field who can fill in the gaps may notsuffice under more adverse circumstances (eg, outsiders slandering thefield).Karl Hallowellkhallow@hotmail.com===Subject: Re: Simple differential equation query>Given the following equation eliminate the arbitrary constant:>1. x^3 - 3*(x^2)*y = c>Ok, differentiate both sides giving: 3x^2 - 6xy -3(x^2)y' = 0>DIVIDE THROUGH BY 3X: x - 2y - xy' = 0>Rearrange: (x-2y)dx - x dy = 0 (= answer given in the book)>[Q] But why can we be sure x != 0 ?! (and divide through by it?)> If x can ever take the value 0, then c = 0. So handle c = 0 as a> special case, which makes y = x/3 when x /= 0 and anthing when x = 0.SO you say x^3 = 3(x^2)(y)? Now? What if x = 0? How do you justify thedivision? That's the same problem all over again!> In the event that y is continuous then y = x/3 and again> x - 2y - xy' = 0> which is the desired result for is it not stipulated for the> problem series that y is differentiable, hence continuous?OK, but we don't know that y is differentiable everywhere do we?(Doesn't say it anywhere in the book I'm afraid.)What I'm really asking (now I see this) is how can we show: x^3 - 3(x^2)y = c =(identical) (x-2y) dx - x dy = 0 ?(that represent the same set of points) Take x != 0: and I'm happy since there is no problem about division.Take x = 0: then y can be anything (we have a vertical line - they-axis - (0,b), b element R)Now lets check that this line is represented by the DE/is a solutionto the DE (since it is a solution to the original equation where c =0). DE could be: (x-2y) dx/dy - x = 0, put x=0 to get 0 = 0, so y = celement R (and there's the line if we have x a function y).OR coule be: (x-2y) - x dy/dx = 0, put x=0 to get -2y = 0 => y = 0 ? What about (0,2)? (0,43434371)? What's missing here?>2. cy^2 = x^2 + y>DIVIDE THROUGH BY Y^2: c = (x^2 + y)/y^2>Differentiate both sides and simplify: [ 2xy dx - (y+2x^2) dy ] />y^4 = 0>Now multiply both sides by y^4 and we're done.>[Q] But why can we be sure y != 0 ?!> cy^2 = x^2 + y> 2cyy' = 2x + y'> 2cyyy' = 2xy + yy'Both sides are multiplied by y here. What if y = 0? Problem remains.But I think I got this figured out? What do you think?TO get the DE i need to divide/multiply by y which may be 0. So I sayassume it's not 0: and I get a DE that I arrive at with y != 0.But does/can this represent the point (0,0) like the original equ?Indeed it does since 2(0)(0) dx - (0) dy = 0 whith dx/dy or dy/dx. Sowe have identity and that (0,0) is in the DE.Does the above argument suffice? If it does, where does it fail in myquestion 1?Richard===Subject: Re: Simple differential equation query>Given the following equation eliminate the arbitrary constant:>2. cy^2 = x^2 + y>DIVIDE THROUGH BY Y^2: c = (x^2 + y)/y^2>Differentiate both sides and simplify: [ 2xy dx - (y+2x^2) dy ] />y^4 = 0>Now multiply both sides by y^4 and we're done.>[Q] But why can we be sure y != 0 ?!> cy^2 = x^2 + y> 2cyy' = 2x + y'> 2cyyy' = 2xy + yy'> Both sides are multiplied by y here. What if y = 0? Problem remains.It does not.Replacing cy^2 into last equation 2(x^2 + y) = 2xy + yy'which is the answer expected by the book.So you need not consern yourself about y = 0 as I didn't use division atall. If y = 0 is a problem it will show up when you try to solve thedifferential equation. BTW, notice that to get from cy^2 = x^2 + yto 2cyy' = 2x + y'it has been assumed that y' exists.===Subject: um....mimimum problem.......find mimimum distance from a point on the surfacez = xy to the point (1,1,0)-------------------------um......it's possible??i know second partials test and lagrange multipliers.but i can't find.help me...please...thank you.===Subject: Re: um....mimimum problem.> ...... to you too.===Subject: Re: um....mimimum problem.> ......> find mimimum distance from a point on the surface> z = xy to the point (1,1,0)> -------------------------> um......it's possible??> i know second partials test and lagrange multipliers.> but i can't find.> help me...please...thank you.You want to minimize (x-1)^2+(y-1)^2+(z-0)^2with the constraint z=xy. Substituting, the expression tominimize is (x-1)^2+(y-1)^2+x^2y^2Differentiating wrt x rsp y and setting partial to 0 0 = 2(x-1)+2x(y^2) 0 = 2(y-1)+2(x^2)yand now it's just algebra ...===Subject: Re: um....mimimum problem.> ......> find mimimum distance from a point on the surface> z = xy to the point (1,1,0)> -------------------------> um......it's possible??> i know second partials test and lagrange multipliers.> but i can't find.> help me...please...thank you.Minimize the function f(x,y,z) = (x-1)^2 + (y-1)^2 + z^2subject to the constraint z=xy .Looks like Lagrange multipliers would work.===Subject: Re: um....mimimum problem.> ......> find mimimum distance from a point on the surface> z = xy to the point (1,1,0)> -------------------------> um......it's possible??> i know second partials test and lagrange multipliers.> but i can't find.> help me...please...thank you.the point of minimal distanceon z = x*y should also lie on the plane x = y, for which the squared distance can be expressed as f(x) = (x-1)^2 + (x-1)^2 + (x^2-0)^2.===Subject: Re: um....mimimum problem.>find mimimum distance from a point on the surface>z = xy to the point (1,1,0)> the point of minimal distanceon z = x*y should also lie on the plane > x = y,Why? The problem Find the minimum distance from a point on the surfacez = 1 - (x - y)^2 to the point (0,0,0) is also symmetric in x and y.However, the minimal distance from (0,0,0) to a point of the surfacesuch that x = y is 1, but the distance from (0,0,0) to (1/4,-1/4,3/4) isthe square root of 11/16.===Subject: Re: um....mimimum problem.>find mimimum distance from a point on the surface>z = xy to the point (1,1,0)> the point of minimal distanceon z = x*y should also lie on the plane > x = y,> Why? The problem Find the minimum distance from a point on the surface> z = 1 - (x - y)^2 to the point (0,0,0) is also symmetric in x and y.> However, the minimal distance from (0,0,0) to a point of the surface> such that x = y is 1, but the distance from (0,0,0) to (1/4,-1/4,3/4) is> the square root of 11/16.> Jose Carlos SantosBecause, in addition, for any fixed positive z, the equation z = x*y gives a rectangular hyperbola in a plane parallel to the x-y-plane for which the point closest to (1,1,0) lies on the positive part of the major axis, x = y, of that hyperbola, and is also closer to (1,1,0) than any point on the corresponding rectangular hyperbola -z = x*y.Thus, for any point of z = x*y not on plane x = y, one can find a point on the plane x = y which is closer to (1,1,0).===Subject: Equivalence of algebraic and differential formsSorry that this is a repeat of a question I have posted in anotherthread, but I thought I could state this more succintly and henceexpect a more accurate reply.Why are the equations: x^3 - 3(x^2)y = 0 and: (2x - y) dx - x dy = 0Equivalent? (i.e. Why do they represent exactly the same points? Or inthe very least, why does the differential form contain the 1stequation?).Richard===Subject: Re: Equivalence of algebraic and differential forms> Why are the equations: x^3 - 3(x^2)y = 0> and: (2x - y) dx - x dy = 0> Equivalent? (i.e. Why do they represent exactly the same points? Or in> the very least, why does the differential form contain the 1st> equation?).They aren't. The first has solutions y = x/3 for x /= 0 and y = anything for x = 0.The second 2x - y = xy'has y = x for solutionA third x - 2y = xy'has y = x/3 for solution and doesn't allow y = x/3 for x /= 0 and y = anything /= 0 for x = 0as y' presumes y differentiable, thus continuous.Presuming y is differentiable three times 1 - 2y' = y' + xy 1 = 3y' + xy 0 = 3y + y + xy' xy' = -4y y'/y = -4/x Dlog y = -4.D(log x) log y = -4.log x + c y = e^c x^-4 = cx^-4 y' = bx^-3 + c y = ax^-2 + bx + cResolving constants x - 2y = xy' ?? x - 2ax^-2 - 2bx - 2c = x(-2ax^-3 + b) x - 2ax^-2 - 2bx - 2c = (-2ax^-2 + bx) a = anything, b = 1/3, c = 0y = a/x^2 + x/3, x /= 0 for a /= 0y = x/3 for a = 0Checking x - 2y = xy' ?? x - 2a/x^2 - 2x/3 = x(-2a/x^3 + 1/3) x - 2a/x^2 - 2x/3 = -2a/x^2 + x/3)Checks out.Thus as third equation has many more solutions than the firstequivalence disappears, nay has even vanished.===Subject: Re: sheaf of functions using certain schemes....> I need help on the following algebraic geometry exercise:> Describe the points and the sheaf of functions on the following> schemes:> (1) X = Spec C[x]/(x^2)> (2) X = Spec C[x]/(x^2-x)> (3) X = Spec R[x]/(x^2+1)> I think I can probably handle any two of these given a discussion of> how to solve the other one, but my class is so abstract that I'm> having a hard time applying the definitions and actually computing> these sheaves and points.....Let's look at (2).The points of Spec C[X]/(X^2 - X)are the prime ideals of C[X]/(X^2 - X). These are basicallythe prime ideals of C[X] which contain (X^2 - X).Now the prime ideals of C[X] are (X - z) where z ranges through C.Which of them contain (X^2 - X)?Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_===Subject: Re: sheaf of functions using certain schemes....Actually, determining the prime ideals (for any of those rings) isn'ttoo hard for me, it's writing down the sheaf after those points aredetermined that I'm struggling with...> I need help on the following algebraic geometry exercise:> Describe the points and the sheaf of functions on the following> schemes:> (1) X = Spec C[x]/(x^2)> (2) X = Spec C[x]/(x^2-x)> (3) X = Spec R[x]/(x^2+1)> I think I can probably handle any two of these given a discussion of> how to solve the other one, but my class is so abstract that I'm> having a hard time applying the definitions and actually computing> these sheaves and points.....> Let's look at (2).> The points of Spec C[X]/(X^2 - X)> are the prime ideals of C[X]/(X^2 - X). These are basically> the prime ideals of C[X] which contain (X^2 - X).> Now the prime ideals of C[X] are (X - z) where z ranges through C.> Which of them contain (X^2 - X)?===Subject: Re: Topic for Proof Treatise: Prove/Disprove Manure Really Stinks.> How> Would a mathematician go about proving/disproving Manure really> Stinks. Off-topic. Neither manure nor stinking are mathematical concepts.Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== ===Subject: Re: Weak and strong derivativeIn effect if a function is regular and L^1 on an open subset A of R^n, thenits strong derivative is also the weak derivative. If the function hasn'tthe derivative in one point of the domain, its weak derivative should bealways the strong derivative.===Subject: Re: differential equations teaching>Ok, so I'm finally able to settle down enough to take some>conventional classes, and one of them is basic differential equations.>Halfway through the class, I see that it has rapidly degenerated into:>1. Spend 2 minutes doing differential equations>2. Spend 3 hours doing basic algebra to clean up the mess>3. Repeat over and over and over>And looking at the trend so far, one can only conclude it will become>more so before the end.>Please god tell me all math isn't this way.>Or if it is, put a bullet in my head.>Also, when people write math problems, do they just randomly throw>whatever numbers pop in their heads into a differential operator? >Until now, it's usually seemed like they've taken efforts to minimize>the above nonsense, ie by having strategic things cancel out,>radicands conveniently being powers, etc.Use the very nice differential equations method to change this>differential equation into a system of 20 equations in 20 unknowns,>all quadratic. Then solve. NOOOO!!!You're absolutely right. That's mindless equation-crunching. Have alook at Hirsch, Smale: Ordinary differential equations and linearalgebra to see what differential equations is *really* about.BTW: Linear algebra is a very important part of differential equationstheory. So you need to learn a lot of that too.===Subject: Re: differential equations teaching> Spend 3 hours doing basic algebra to clean up the messSkill in integration is a necessary pre-requisite.Course startsusually asking students to eliminate arbitrary constant/s to set updifferential equation,to appreciate the two way process.For example,y= C e^(-x^2/2) sets up dy/dx= -x*y , that integrates back to givenrelation by separable variables. etc..===Subject: Re: differential equations teaching> Ok, so I'm finally able to settle down enough to take some> conventional classes, and one of them is basic differential equations.> Halfway through the class, I see that it has rapidly degenerated into:> 1. Spend 2 minutes doing differential equations> 2. Spend 3 hours doing basic algebra to clean up the mess> 3. Repeat over and over and over> And looking at the trend so far, one can only conclude it will become> more so before the end.> Please god tell me all math isn't this way.Some math isn't this way. In fact, most math isn't this way. Get away fromthe service courses and into the math courses.What you do have to do is make a choice between proving that a solutionexists and is unique, and actually finding a solution. You know the jokeabout the mathematician who wakes up in the middle of the night to find hisroom filled with smoke, wanders into the bathroom and runs the water in thesink, then goes back to bed because he's proved the existence of a solutionto his problem. Physicists and engineers are unimpressed.> Or if it is, put a bullet in my head.> Also, when people write math problems, do they just randomly throw> whatever numbers pop in their heads into a differential operator?> Until now, it's usually seemed like they've taken efforts to minimize> the above nonsense, ie by having strategic things cancel out,> radicands conveniently being powers, etc.Use the very nice differential equations method to change this> differential equation into a system of 20 equations in 20 unknowns,> all quadratic. Then solve. NOOOO!!!1. Why pure mathematicians hate applied mathematics. After all, ifyou're concerned with casting brass cannons so that they won't crack underrepeated firing, you're stuck with whatever numbers the physics gives you(both before and after the Fourier transform) and you just have to calculatewith what the world gives you.2. Why mathematicians hate applied courses. The students just grind awayat the calculations without understanding what they're doing. Yuck!3. What's the purpose of math (and science in general)? Understanding. Ifyou just calculate, you're just getting an answer, you're not understandinganything. So uncomprehending calculation is really just a big waste oftime.The problem is, that nice numbers in diff eqs tend to make the problemstrivial. It's really, really hard to strike the balance between too easy tosee if you know what you're doing and too hard to see if you know whatyou're doing. Besides, we've got other stuff to do too. And we have atendency to procrastinate, especially with unpleasant tasks like looking fornice diff eq problems.I attended a seminar in the early 80s by one of the big names in fluid flow(sorry, I've already taken my bedtime meds and I'm fading fast). He had areally messy algebraic calculation that he decided he'd declare done when hegot the same answer twice in one day (took him about two hours to do ittwice). Took him two weeks. (Turbulence can be shock to the system.)Jon Miller===Subject: Re: Descending sets> Let { A_xi } be a descending transfinite series of subsets of> A = A_0 with the property> for all eta, xi < eta, A_eta subset A_xi> Let |A| = omega_nu and mu = omega_(nu+1)So mu = |A|^+.> Without loss of generality we can not have> A_eta = /{ A_xi | xi < eta }> for limit ordinals eta? / = cap = intersectionI'm not sure what you're asking here. A_eta could be a proper subset of the intersection, so you can't just assume they're equal.> Show A_mu = A_(mu+1).What if A_xi = A for each xi <= mu, and A_xi = 0 for xi > mu?> Often there's some eta < mu with> A_eta = A_(eta+1).> Dare one conjecture that will always be possible?Yes. Because if there wasn't such an eta, then you could get an injection from mu into A by mapping each xi in mu to some element of A_xi A_(xi+1).===Subject: Re: Descending sets===Subject: Re: Descending sets > Let { A_xi } be a descending transfinite series of subsets of > A = A_0 with the property > for all eta, xi < eta, A_eta subset A_xi > Let |A| = omega_nu and mu = omega_(nu+1) >So mu = |A|^+. > Without loss of generality we can not have > A_eta = /{ A_xi | xi < eta } > for limit ordinals eta? / = cap = intersection >I'm not sure what you're asking here. A_eta could be a proper >subset of the intersection, so you can't just assume they're equal.Indeed, however it that case shift A_(eta+n) -> A_(eta+n+1)for all n in omega_0 and set A_eta = /{ A_xi | xi < eta } > Show A_mu = A_(mu+1). >What if A_xi = A for each xi <= mu, and A_xi = 0 for xi > mu?Pest. > Often there's some eta < mu with > A_eta = A_(eta+1). > Dare one conjecture that will always be possible? >Yes. Because if there wasn't such an eta, then you could get an >injection from mu into A by mapping each xi in mu to some element of >A_xi A_(xi+1).Nice.Let |A| = omega_nu, mu = omega_(nu+1) = |A|^+, A = A_0 and{ A_xi | xi < mu } be a descending series of subsets, ie for all eta, xi < eta, A_eta subset A_xiThen there's some eta < mu with A_eta = A_(eta+1)In addition, is there some beta < mu with for all xi in [beta,mu) A_beta = A_xi ?----Supersedes: x0, and the fact thecircle center didn't change wrt the circle, means thatthe relationship of x0' with x0 is the same as that ofany x' on the circle with the corresponding x: x'=x-vt;x0'=x0-vt.This is to prepare you for the True Believer crackpots thatsay 'constant' coordinates can't be transformed; some evensay they aren't coordinates. These crackpots include somethat brag about how they were childhood geniuses, btw.QED: The galilean transformation for any law ongeneralized Cartesian coordinates is invariant underthe Galilean transform.The use of the privileged form explains HOW the transformedequation can be messed up, the next Subject explains whatthe screwed up effect of the transform is, and how useof the generalized form corrects the screwup.------------------------------===Subject: 6. The data scale degradation absurdity.The SR transforms and the Galilean transforms bothconvert good, ratio scale data to inferior intervalscale data. The effect is corrected, allowed for,when the transforms are conducted on the generalizedcoordinate forms specified by analytic geometry andvector algebra.Both sets of transforms are 'translations' - lateralmovements of an axis, increasing over time in thesecases - but with the SR transform also involving arescaling. It is the translation term, -vt in the xtransform to x', and -xv/cc in the t transform to t',that degrades the ratio scale data to interval scaledata. In general, rescaling does not effect scalequality in the size-of-units sense we have here.SR likes to consider its transforms just rotations,however - in spite of the fact Einstein correctly saidthey were 'translations' (movements) - and in the caseof 'good' rotations, ratio scale data quality is indeedpreserved, but SR violates the conditions of good ro-tations; they are not rigid rotations and they don'tappropriately rescale all the axes that must be rescaledto preserve compatibility.The proof is in the pudding, and the pudding is thecombination of simple tests of the transformations.We can tell if the transformed data are ratio scaleor interval.Ratio scale data are like absolute Kelvin. A measure-ment of zero means there is zero quantity of thestuff being measured. Ratio scale data support add-ition, subtraction, multiplication, and division.The test of a ratio scale is that if one measurelooks like twice as much as another, the stuffbeing measured is actually twice as much. Withabsolute Kelvin, 100 degrees really is twice theheat as 50 degrees. 200 degrees really is twiceas much as 100.Interval scale data are like relative Celsius, whichis why your science teacher wouldn't let you use itin gas law problems. There is only one mathematicaloperation interval scales support, and that has tobe between two measures on the same scale: subtraction.100 degrees relative (household) Celsius is not twiceas much as 50; we have to convert the data to absoluteKelvin to tell us what the real ratio of temperaturesis.However, whether we use absolute Kelvin or relativeCelsius, the difference in the two temperature readingsis the same: 50 degrees.Thus, if we know the real quantities of the 'stuff'being measured, we can tell if two measures are ona ratio scale by seeing if the ratio of the twomeasures is the same as the ratio of the known quant-ities.If a scale passes the ratio test, the interval scale testis automatically a pass.If the scale fails the ratio test, the interval scaletest becomes the next in line.It isn't just the bare differences on an intervalscale that provides the test, however. Differencesin two interval scale measures are ratio scale, soit is ratios of two differences that tell the tale.Let's do some testing, and remember as we do that ourconcern is for whether or not the data are messed up,not with 'reasons', excuses, or avoidance.---------------------------------------------------- --Are we going to take a transformed length (difference)and see whether that length fits ratio or interval scaledefinitions?Of course, not. Interval scale data are ratio afterone measure is subtracted from another. That is themajor reason the SR transforms can be used in science.Let there be three rods, A, B, C, of length 10, 20, 40,respectively. These lengths are on a known ratio scale,our original x-axis, with one end of each rod at theorigin, where x=0, and the other end at the coordinatethat tells us the correct lengths.Note that these x-values are ratio scale only becauseone end of each rod is at x=0. That may remind you ofthe correct way to use a ruler or yard/meter-stick:put the zero end at one end of the thing you aremeasuring. Put the 1.00 mark there instead of the zero,and you have interval scale measures.Let A,B,C, be 10, 20, 40.Let a,b,c be x' at v=.5, t=10.x'=x-vt.A B C a b c---------------- --------------------10 20 40 5 15 35---------------- --------------------B/A = 2 b/a = 3C/A = 4 c/a = 7C/B = 2 c/b = 2.333 Obviously, the transformed values are no longer ratio scale. The effect is less on the greater values.C-A = 10 b-a = 10C-A = 30 c-a = 30C-B = 20 c-b = 20 Obviously, the transformed values are now interval scale. This will hold true for any value of time or velocity.(C-A)/(B-A) = 3 (c-a)/(b-a) = 3(C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the differences are ratio scale, being identical to the ratios of the corresponding original - ratio scale - differences.The main difference between these results and the SRresults is that the differences do not correspond soneatly to the original, ratio scale, differences.This is due only to the rescaling by 1/sqrt(1-(v/c)^2).The ratios of the differences on the transformed valuesdo correspond neatly and exactly to the ratio scaleresults.Using the generalized coordinate form, such as (x-x0),the transform produces an interval scale x' and aninterval scale x0'. That gives us a ratio scale (x'-x0'),just like - and equal to - (x-x0).------------------------------===Subject: 7. The Crackpots' Version of the Transforms.It has become apparent - whether misleading or not -that the crackpot responses to the obvious derive froma common source, whether it be bandwagoning or theirSR instructors.Below, in the sci.math subject, we see that all sci.mathrespondents agree with the basic controversial positionof this faq: every coordinate is transformed, whether asupposed constant or not.Think about it, the generalized coordinate of a circlecenter, x0, applies to infinities upon infinities ofcircle locations (given y and z, too); it is a constantonly for a given circle, and even then only on a givencoordinate axis.And even variables are often held 'constant' duringeither integration or differentiation.The utility of a variable is that you can discuss allpossible particular values without having to single outjust one. That utility does not make particular - singledout - values on the variable's axis not values of thevariable just because they have become named values.In any case, all that is preamble to the incompetent ideathey have proposed for a transform of coordinates. It isbased on the idea that the circle center, point of emission,whatever, has coordinates that cannot be transformed.Let there be an equation, say (x)^2 - (ict)^2 = 0.What is the transformed version of that equation?Answer: (x')^2 - (ict')^2 = 0. That's the one thing theBrittanica got right. Note that the leading crackpot justcriticized this faq for presuming to correct the Britt-anica, but it then and before poses the incompetent pseudo-transform we analyze here in this section.x to x' and t to t' are obviously coordinate transforms;the x and t coordinates have been replaced by the coord-inates in the primed system.A tranform of an equation from one coordinate system toanother is NOT a substitution of the/a definition of xfor itself; that is not a coordinate transformation.The most that can said for such a substitution is thatit is a change of variable.But the crackpots are calling this a coordinate trans-form of the original equation: (x'+vt)^2 - (ict')^2 = 0.It is not a coordinate transform, of course, exceptaccidentally. (x'+vt) is not the primed systemcoordinate, it is another form/expression of x. Theyget that substitution by solving x'=x-vt for x; x=x'+vt.So, by incompetent misnomer, they accomplish what theyhave been railing against all along.It has been the generalized coordinate form in question allthis time: (x-x0)^2 - (ict)^2 = 0.Here they substitute for x instead of transforming to theprimed frame: (x'+vt-x0)^2 - (ict')^2. ----- ^ | ^ |It is still x ^ but see what they have accomplishedby their mis/malfeasance: [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)]. =[x'-x0']The crackpots have been bragging about how you don'thave to transform the circle center's coordinate totransform the circle center's coordinate. Braggingthat what they were doing was not what they saidthey were doing.This does give us insight as to some of the crackpotvariations on their x0'<>x0-vt theme, which in all thevariations will be discussed in later sections..They are used to seeing the mixed coordinate form,(x'+vt-x0) without realizing what it respresented,so - accompanied with a lack of understanding ofthe term 'dependent' - they are used to seeing justthe one vt term, and not the one hidden in the defi-nition of x' and are used to imagining it makes thewhole expression time dependent and thus not invariant.About which, let x=10, let, x0=20, v=10, and tvariously 10 and 23:(x-x0)=-10. Using their (x'+vt-x0):For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0)For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230 - 20 = -10 = (x-x0)The result depends in no way on the value of time;we showed the obvious for a couple of instances of tjust so you can see that the crackpots not only donot understand the obvious logic of the algebra{ (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which showsthat the transform has no possible time term effect -but they don't understand even a simple arithmeticdemonstration of the facts.Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the sameway since t'=t: (x-vt+vt-x0)=(x-x0).Their process, which says (x'+vt') is the transformof x, says that (x'+vt') is the moving system locationof x, but it can't be because x is moving further inthe negative direction from the moving viewpoint.That formula will only work out with v<0 which is indeedthe velocity the primed system sees the other moving at.However, that formula cannot be derived from x'=x-vt,the formula for transformation of the coordinates fromthe unprimed to the primed,------------------------------===Subject: 8. What does sci.math have to say about x0'=x0-vt?The crackpots' positions/arguments were put to sci.mathin such a way that at least two or three who posted re-sponses thought it was your faq-er who was on the idiot'sside of the questions.Their responses:---------------------------------------------------- ------I. x0' = x0. In other words: x0' <> x0-vt, or constant values on the x-axis are not subject to the transform.AA: ============================================================== ====== No. x0' = x0 - vt. Well, if you want, you could define constant values on the x-axis, butin the context of the question that is not relevant. The relevant fact isthat if the unprimed observer holds an object at point x0, then theprimed observer assigns to that object a coordinate x0' which isnumerically related to x0 by x0'= x0 -vt.AA: ============================================================== ======EE: ============================================================== ======What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0'is to give the coordinate in the (x',t',)-system, it will be given byx0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest(constant x-coordinate) is a coordinate-dependent concept.EE: ============================================================== ======GG: ============================================================== ======Sounds very false. We can say that the representation of the point X0 isthe number x0 in the unprimed system, and x0' in the primed system.Clearly x0 and x0' are different, if vt is not zero. However one may saythat (though it sounds/is stupid) the point X0 itself is the samethroughout the transformation. However that expression soundsmeaningless, since a transform (ok, maybe we should call it a change ofbasis) is only a function that takes the point's representation in onesystem into the same point's representation in another system. It ispreferrable to use three notations: X0 for the point itself and x0 andx0' for the points' representations in some coordinate systems.GG: ============================================================== ======------------------------------===Subject: 9. But Doesn't x.c'=x.c?That idea is one of the most idiotic to come up, and it doesso frequently. And in a number of guises.The idea being that x.c' <> x.c-vt, with x.c being whatwe have called x0 above; the notation makes no difference.Some crackpots have managed to maintain that position evenafter graphs have illustrated that such an idea means thatafter a while a circle center represented by x.c' could beoutside the circle.The leading crackpot just make that explicit, as far asone can tell from his befuddled post in response to a lineabout active transforms, which are actually moving bodysituations, not coordinate transformations:---------------------------------------------- ----------------------e>An active transform is not a coordinate transform, ... Right, it is a transform of the center (in the opposite direction) done to effect the change of coordinates without a coordinate transform. ...E: Transform of the center? Center of a circle? He really is saying a circle center moves in the opposite direction of the circle! Right?-------------------------------------------------------- ------------If r=10 and x.c was at x.c=0, then the points on the circle(10,0), (-10,0), (0,10) and (0,-10) could at some time become(-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c,the circle center would be at (0,0) still! The circle is herebut its center is way, way over there! Indeed, although a changeof coordinate systems is not movement of any object described inthe coordinates, the x.c'=x.c crackpottery is tantamount to thecircle staying put but the center moving away. Or vice versa.------------------------------===Subject: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationshiplike this: (x-vt+vt - x.c).See, he says, that is transforming x (with x-vt - x.c) and thenreversing the transform (x-vt+vt - x.c).That's just another crackpot form of the idiocy thatx.c' <> x.c-vt. You'll have noticed the implicationis that there is no transform vt term relating to x.c.------------------------------===Subject: 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent?That particular crackpottery is perhaps more corrupt thanmoronic, since it includes deliberately hiding a vt term fromview, and pretending it isn't there. [However, we have seenabove that it is a familiar incompetency, and not likely anoriginal.]Look, the crackpots say, there is a time term in thetransformed (x' - x.c+vt). The transform isn't invariant!It's time dependent!Just put x' in its original axis form, also, which revealsthe other time term, the one they hide: (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c).So, at any and all times, the transform reduces to theoriginal expression, with no time term on which to bedependent.Then there is the fact that if you leave the equationin any of the various notation forms - with or withoutreducing them algebraicly - the arithmetic always comesdown to the same as (x-x.c). That means nothing to crack-pots, but may mean something to you.------------------------------===Subject: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology?My dictionary relates 'tautology' to needless repetition.That's another form of the x.c' <> x.c-vt idiocy.The repetition involved is the vt transformation term.Apply the -vt term to the x term, and it is needlessrepetition to apply it anywhere again? The 'again' isto the x.c term. The x.c' = x.c crackpot idiocy.The repetition of the vt terms is required by the presenceof two x values to be transformed.Be sure to note the next section.------------------------------===Subject: 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform?Now, how on earth can we relate a tautology to a basicdefinition in math?we get this definition:--------------------------------------------------- -----------A linear transformation, A, on the space is a method of corr-esponding to each vector of the space another vector of thespace such that for any vectors U and V, and any scalarsa and b, A(aU+bV) = aAU + bAV.---------------------------------------------------------- ---Let points on the sphere satisfy the vector X={x,y,z,1},and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1,and b=-1.Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0 1 )A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ).The left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c , y-y.c, z-z.c, 0 ).The right hand side: aAX= ( x-ut, y-vt, z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ).Need it be said?Sure: QED. On the galilean transform thedefinition of a linear transform, A(aU+bV)=aAU + bAV,is completely satisfied.The generalized form transforms exactly andnon-redundantly - with ONE TRANSFORM, not atransform and reverse transform - and non-tautologically, just as the very definitionof a linear transform says it should.And does so with absolute invariance, with thisgalilean transformation.------------------------------===Subject: 14. But The Transform Won't Work On Time Dependent Equations?The main crackpot that has asserted such a thing was referringto equations such as in Subject 4, above. The Light Sphereequation; for which we have shown repeatedly elsewhere that thenumerical calculations are identical for any primed values asfor the unprimed values.The presence - before transformation - of a velocity termseems to confuse the crackpots. It turns out there is ex-treme historical reason for this, as you will see in thesubject on Maxwell's equations.------------------------------===Subject: 15. But The Transform Won't Work On Wave Equations?See Subject 17, below, for a discussion of Second Derivativeforms and the galilean transforms.------------------------------===Subject: 16. But Maxwell's Equations Aren't Galilean Invariant?Oh? Just what is the magical term in them that prevents(x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true?It turns out not to be magic, but reality, that interfereswith the application of the galilean transforms to the gen-eralized coordinate form(s) of Maxwell: there are no coordi-nates to transform!When True Believer crackpots are shown the simpledemonstration that the galilean transform ongeneralized cartesian coordinates is invariant,their first defense is usually an incredibly stupidx0'=x0, because the coordinate of a circle center,or point of emission, etc, is a constant and can'tbe transformed.The last defense is but Maxwell's equations are notinvariant under that coordinate transform. Whenasked just what magic occurs in Maxwell that wouldprevent the simple algebra (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0)from working, and when asked them for a demonstration,they will never do so, however many hundreds oftimes their defense is asserted.The reason may help you understand part of Einstein's1905 paper in which he gave us his absurd SpecialRelativity derivation:THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.Einstein gave the electric force vector as E=(X,Y,Z)and the magnetic force vector as B=(L,M,N), where theforce components in the direction of the x axis areX and L, Y and M are in the y direction, Z and N inthe z direction.Those values are not, however, coordinates, but valuesvery much like acceleration values.BTW, the current fad is that E and B are 'fields', havingbeen 'force fields' for a while, after being 'forces'.So, when Einstein says he is applying his coordinatetransforms to the Maxwell form he presented, he iseither delusive or lying.(a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1-(v/c)^2): X'=X. L'=L. Y'=B(Y-(v/c)N). M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N-(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the x-axis, but force magnitudes in that direction. Similarly for Y and M and y, Z and N and z.(b) the v of the coordinate transforms is in Maxwell before any transform is imposed; Einstein's transform v is the velocity of a coordinate axis, not the velocity he touched it.(c) if they were honest Einsteinian transforms, they'd be x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the x-axis as the v for a y-axis and z-axis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway.(d) as everyone knows, the effect of E and B are on the direction. Both the speed and direction are changed by E and B, but v - the speed - is a constant in SR.As absurd as are the previously demonstrated Einsteinianblunders, this one transcends error and is an incredibleexample of True Believer delusion propagating over decades.The components of E and B do differ from point to point,and in the variations that are not coordinate free,they are subject to the usual invariant galilean trans-formation when put in the generalized coordinate form.--------------------------------------------------------- ----The SR crackpots don't know what coordinates are. Thevarious things they call coordinates include coordin-nates, but also include a variety of other quantities.--------------------------------------------------- ---1. One may express coordinates in a one-axis-at-a-time manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are so-called basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each.2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel- erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively.Einstein's - and Maxwell's - E and B arenot coordinate vectors.====================================================== ======There is another variety of intellectual befuddlement thatmisinforms the idea that Maxwell isn't invariant under thegalilean transform: confusions about velocities.Velocities With Respect to Coordinate Systems.-----------------------------------------------Aaron Bergman supplied the background in a post to a sci.physics.*newsgroup:======================================= ========================Imagine two wires next to each other with a current I in each.Now, according to simple E&M, each current generates a magneticfield and this causes either a repulsion or attraction betweenthe wires due to the interaction of the magnetic field and thecurrent. Let's just use the case where the currents are parallel.Now, suppose you are running at the speed of the current betweenthe wires. If you simply use a galilean transform, each wire,having an equal number of protons and electrons is neutral. So,in this frame, there is no force between the wires. But this is acontradiction.=============================================== =================First of all, the invariance of the galilean transform (x'-x.c')=(x-x.c), insures that it is an error to imagine there is anydifference between the data and law in one frame and in another;the usual, convenient rest frame is the best frame and only framerequired for universal analysis. [Well, (x'<>x, x,c'<>x.c, but(x'-x.c')=(x-x.c).]Second, given that you decide unnecessarily to adapt a law toa moving frame, don't confuse coordinate systems with meaningfulphysical objects, like the velocity relative to a coordinatesystem instead of relative to a physical body or field.In other words, what does current velocity with respect to acoordinate system have to do with physics?Nothing. Certainly not anything in the example Bergman gave.What is relevant is not current velocity with respect to acoordinate system, but current velocity with respect to wiresand/or a medium. The velocity of an imaginary coordinate sys-tem has absolutely nothing to do with meaningful physical vel-ocity. You can - if you are insightful enough and don't violateitem (e) - identify a coordinate system and a relevant physicalobject, but where some v term in the pre-transformed law isin use, don't confuse it with the velocity of the coordinatetransform.Velocities With Respect to ... What?-----------------------------------------------Albert Einstein opened his 1905 paper on Special Relativitywith this ancient incompetency:================================================= ==============The equations of the day had a velocity term that was takenas meaning that moving a magnet near a conductor would createa current in the conductor, but moving a conductor near awire would not. This was belied by fact, of course.The important velocity quantity is the velocity of themagnet and conductor with respect to each other, not tosome absolute coordinate frame (as far as we know) andnot to an arbitrary coordinate system.One possible cause was the idea: but the equation says the magnetmust be moving wrt the coordinate system or ... the absoluterest frame.There not being anything in the equation(s) to say either ofthose, it is amazing that folk will still insist the velocityterm has nothing to do with velocity of the two bodies wrteach other.-------------------------------------------------------- ---------------------------------===Subject: 17. First and Second Derivative differential equations.One of the intellectually corrupt ways ofdenying the very simple demonstration ofgalilean invariance of all laws expressedin the generalized coordinate form demandedby analytic geometry, vector analysis, andmeasurement theory [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ]is the assertion that those equations 'over there'(usually Maxwell or wave) are somehow immune tothe elementary laws of algebra used to demon-strate the invariance. [Unfortunately, theassertions are never accompanied by referenceto the magical math that makes elementary al-gebra invalid. Wonder why that is?]Part of the time it is based on the old lorebased on the incompetent transformation ofthe privileged form of an equation insteadof the correct form. [Evidence of this isany reference to an effect due to the velocityof the transform; it falls out algebraicly- as you see above - and cancels out arith-metically - as you can see above.]But usually it is just whistling in the dark,waving the cross (zwastika, I'd say) atthe mean old vampire.The most general equation that could be conjuredup is a differential with either First or SecondDerivatives.Let's examine the plausibility of such magicalmagical, non-invariance assertions.(a) to get a Second Derivative you must have a First Derivative.(b) to get a First Derivative you must have a function to differentiate.(c) to get a Second Derivative you must have a function in the second degree.So, let us examine the question as to whetherany such common Maxwell/wave equation willdiffer for(a) the common, privileged form, represented as ax^2, with a being an unknown constant function.(b) the generalized cartesian form, represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a being an unknown constant function.(c) the transformed generalized cartesian form, represented as a(x-vt -x.c+vt)^2, same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a being an unknown constant function.I. for (a), remembering that x.c is a constant, and that this version is only correct because x.c=0, otherwise (b) is the correct form: d/dx ax^2 = 2ax (d/dx)^2 ax^2 = 2aII. for (b), remembering that x.c is a constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2aIII. for (c); same as for (b).So, what we have seen so far is(1) differential equations in the second degree- the wave equations - must clearly be the same forall forms: the privileged form in x, the generalizedcartesian form in x and the centroid, x.c, or thetransformed generalized cartesian form.That is, anyone who imagines that correct usagegives different results for galilean transformedframes is at first showing his ignorance, and inthe end showing his intellectual corruption.(2) As far as the First Derivatives are concerned, theonly cases in which there really is a difference betweenthe two forms is where x.c <> 0, and in that case, theuse of the privileged form is obviously incompetent.So, how do you correctly use the differential equations?If you are using rest frame data with the centroidat x=0, etc, you can't go wrong without trying togo wrong.If you are using rest frame data with the centroidnot at x=0, you must use (x-x.c) anyplace x appearsin the equation.If you are using moving frame data, you must use themoving frame centroid as well as the light front(or whatever) moving frame data itself, perhaps firstcalculating (x'-x.c'), which equals (x-x.c) which isobviously correct, and which is obviously the plain oldcorrect x of the privileged form.Unless, of course, there really is some magical termor expression that invalidates the obvious and elemen-tary algebra of the invariance demonstration.Or maybe you just whistle when you don't want basicalgebra to hold true.Eleaticus!---?---!---?---!---?---!---?---!---?---!---?--- !---?---!---?---!---?! Eleaticus Oren C. Webster ThnkTank@concentric.net ?! Anything and everything that requires or encourages systematic ?! examination of premises, logic, and conclusions ?!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---?===Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws===> Subject: 1. Purpose> The purpose of this document is to provide the student of Physics,> especially Relativity and Electromagnetism, the most basic princ-> iples and logic with which to evaluate the historic justification> of Relativity Theory as a necessary alternative to the classical> physics of Newton and Galileo.Galilean invariance has been empirically falsified. See -Special Theoryof Relativity- by A.P. French pp 10-11. If velocities added an electroncould be pushed to move faster than light in an inertial frame. This isnot possible.Bob KolkerSupersedes: 540-180. We know that the two clocks are related by t' = t/1 since both are using the same second, hour, etc units. But 870 (14:30 in minutes) is not 540/1-180, so once again we know something is wrong. However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 (midnite) - 180, so t.z' = -180, and t' = -180 + 540/1 = 360.Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes.We see thus that like distance measures/coordinates, timeaxis origins (zero points) must either be 'lined up' or adjusted for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving system elapsed time interval since the time axes were both at a common zero. There is no t.z' adjustment: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)Make sure you understand that in the clock case, if theEST is showing a good number for elapsed time since thetravelling observer passed NYC, then the PST clock issilliness. t.z' must be zero or must be taken out oftime lapse calculations for the PST clock to be usedintelligently, just as was true for x.z'.What is lacking as yet for Lorentz t' is the -vx/cc term thatcorresponds to the x' formula -vt term.Break it up into two parts: v/c and x/c. v/c is a scaling factor that changes velocity from whatever kind of unit you are using over to fractions of c.x/c is distance divided by velocity, which is time. x/cis thus the time interval since the two time axeshad a common zero point - which they have to have in theLorentz transforms which do not have the t.z' term welearned to use above.Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving system clock has been changed - since the common/adjusted time - over and beyond the elapsed time intervalrepresented by x/sqrt(1-vv/cc).We have discovered that the only way for t' to be t/gis for t' and t to have a common zero point, just asfor x' and x. It would be otherwise if the t' formulacontained an adjustment t.z' under some name or other,but the necessity to include such a term correlates100% with t' numbers that aren't directly usable.As for x and x', our knowledge of how to setup a properformula relating t and t' is of no use unless we usethe knowledge in scientific formulas; (t'-t.z'+xv/gcc)gives us the only directly useful value: t/g.------------------------------ ===Subject: 9. Einstein's (1905) derivations.When we return to Einstein's derivations of the transformformulas with a well-focused eye, we find he was a wee bitconfused - or at least self-contradictory.When he set up his (at first unknown) tau=moving systemtime formulas, he created three particular instances of tau.Tau.0 is the time at which light is emitted at the movingorigin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t,a constant, the stationary system starting time.Tau.1 is the time at which the light is reflected. Helets the stationary time be t+x'/(c-v); t is still aconstant and x'/(c-v) is the time interval since t.Tau.2 is the time at which the light gets (back) to themoving origin. The stationary time value is put as t +x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v)+ x'/(c+v) is the time interval since t.On the thesis that the moving observer sees the time tothe mirror as the same as the time back to the origin,he sets .5[ tau.0 + tau.2 ] = tau.1.Tau.0 completely drops out of the analysis and leavesno trace, and has no effect.Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what you'd get if you had explicilty said t' is an interval and so is t.What doesn't drop out in the stationary time values isx'/(c-v) and x'/(c+v), the time interval it takes forlight to get to the fleeing mirror, and the time intervalit takes for light to get back to the approaching origin.Thus, his resultant t' formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals.There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments.Let's look at the x dimension; it is x'=x-vt [as x increasesby vt, the effect over time is x'=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance.He uses that x' not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reflected. x' can't be the stationary system coordinate of the mirror at that time. That value is x'+vt.x' is explicitly an interval, distance. Thus, the whole tau derivation of the t' formula is fully andexplicitly based on x' - a spatial length/distance/interval -and the two time interals x'/(c-v) and x'/(c+v).While we're at it, if the starting t is not zero, his x'=x-vt formula is complete nonsense also. Given thatthere was some L that was the mirror x-location and lengthwhen the light is emitted, if t was already, say, 500, thenx'=L-vt could have been a very negative length.------------------------------ ===Subject: 10. A word about intervals.There are intervals, and there are intervals.If we put our yard stick zero point at one endof a piece of paper and read off the coordinateat the other end of the paper, we have a goodmeasure of the paper's length, a Ratio Scalemeasure. [Absolute temperature scales are ratioscale.]If instead we put the one end of the paper at theone inch mark (or the zero end of the stick oneinch 'into' the length of the paper) we get measuresthat are one inch off the true, ratio scale length.The two messed up measures are still intervals,but they are Interval Scale measures. [Householdtemperature scales are interval scale, which iswhy your physics and chemistry professors won'tlet you use them without first converting to theratio scale absolute temperatures.)t'=t/g and x'=x/g represent ratio scale measures,given that t and x were ratio scalae to start with.t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale measures, even given a good ratio scale tand a good ratio scale x.x'=x.z'+x/g and x'=x/g-vt/g are both interval scale measures, even given a good ratio scale xand a good ratio scale t.Look for the (SR) Lorentz t', x' = degraded measuresdocument soon at a newsgroup near you.------------------------------ ===Subject: 11. Intervals versus the Twins Paradox.t'=(t-vx/cc)/g shows t' being greater than t.The reason Special Relativity will not allow theuse of its basic time equation in determining whatSR has to say about the twins' ages, is that t' andx' are supposedly just coordinates, and they say you have to take the coordinate pairs (t',x') and (x,t)into consideration in both the time and place the twins' separation started and the time and place the twins reunited.Since t' and x' are actually both intervals, notjust coordinates, the 'excuse' is spurious, and is so even without use of the obvious (x_b-x_a) and(t_b-t_a) usages.However, SR is right to be embarrassed by theirtransformation formulas.Look for the (SR) Lorentz t', x' = degraded measuresdocument at a newsgroup near you. ------------------------------ ===Subject: 12. SummaryA. t'=t/g and x'=x/g can be almost 'just coordinates' in the sense that the values obtained may not be of much use except in the most primal and useless way: how long and how far since/from the time/ place they were zero. Even here, however, the zero points within each of the two scale pairs (t',t) and (x'.x) must have been lined up. If the zero points have been intelligently selected (such as at the starting point and time of a trip) they can be rationally used 'as is' in any valid sci- entific equation.B. Even the interval scale t'=t.z' - xv/gcc + t/g and x'=x.z' - vt/g + x/g are not 'just coordinates'. They can be used to good effect by establishing the relevant starting times/points and using (t'-t.z'+xv/gcc) and (x'-x.z'+vt/g), as the situation may require.C. When you see vx/gcc or vt/g in use in any guise with non-zero values, you know the resultant t' or x' is a degraded, interval scale value.E-X: Anytime you do not see what amounts to t.z' and xv/gcc in the time case, or x.z' and vt/g in the distance case, you know that the t' and/or x' in use are intervals. Period.Y: Either set your clock to zero at the start of the relevant time interval, or use (t-t0), with both being readings on the same clock. Either move your x-axis origin to the starting end or point, or use (x-x0), with both being readings on the same axis.Z: In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see that t' and x' satisfy the mathematical tests for/of interval scales when -vt and -vx/cc are not zero; thus, they must be intervals. When -vt and -vx/cc are zero, t' and x' satisfy the much better mathematical definition of ratio scales, and are thus not just mere intervals, but (rescaled) good ones.Eleaticus!---?---!---?---!---?---!---?---!---?---!---?--- !---?---!---?---!---?! Eleaticus Oren C. Webster ThnkTank@concentric.net ?! Anything and everything that requires or encourages systematic ?! examination of premises, logic, and conclusions ?!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---?===Subject: Projected Newtonian OrbitEL where FP sweeps out equal areas in equal time intervals.This motionis projected onto another plane ELP arbitrarily inclined to EL asdescribes another projected ellipse orbit EP on ELP. F is mapped to apseudofocus FP which is no more the focus of EP. However, does theFP-PP radial line still sweep out equal areas on ELP in equal timeintervals?===Subject: Re: Projected Newtonian Orbit> EL where FP sweeps out equal areas in equal time intervals.This motion> is projected onto another plane ELP arbitrarily inclined to EL as> describes another projected ellipse orbit EP on ELP. F is mapped to a> pseudofocus FP which is no more the focus of EP. However, does the> FP-PP radial line still sweep out equal areas on ELP in equal time> intervals?This should be true, since the equal areas-equal time law (Kepler II),is just a consquence of angular momentum being a strict constant ofthe motion. And this is valid in any central force problem. Itshould also be true in any projection, since the projection of aconstant vector will still be a constant (though not necessarilyidentical to the original). So the rule should follow.===Subject: Re: Projected Newtonian Orbit>EL where FP sweeps out equal areas in equal time intervals.This motion>is projected onto another plane ELP arbitrarily inclined to EL as>describes another projected ellipse orbit EP on ELP. F is mapped to a>pseudofocus FP which is no more the focus of EP. However, does the>FP-PP radial line still sweep out equal areas on ELP in equal time>intervals?Yes, if you mean an orthogonal projection. The mapping from EL to ELP is affine, and therefore multiplies all areas by the same constant. No if it's a projective transformation (i.e. the observer is at a finite distance): areas closer to the observer will appear larger.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ===Subject: looking for best numerical integration methodf(x) = exp(-P(x))P(x) is a polynomial in xP(x) >= 0 for all xThe coefficients of P(x) are all knownasked: int(exp(-P(x)),x=-1..1)What is the best (computer based) integration method to solve this? Theproblem is that f(x) can be 0 nearly everywhere except near a specificlocation, for example with P(x) = (100*x-55)^4So if you measure f(x) at a number of points without looking to P(x) itself,it can easily go wrong...I think one should look to P(x) itself, where the minimum value is and soon, but you cannot calculate the roots for degree 5 or higher, so...?===Subject: Re: looking for best numerical integration method> f(x) = exp(-P(x))> P(x) is a polynomial in x> P(x) >= 0 for all x> The coefficients of P(x) are all known> asked: int(exp(-P(x)),x=-1..1)> What is the best (computer based) integration method to solve this? The> problem is that f(x) can be 0 nearly everywhere except near a specific> location, for example with P(x) = (100*x-55)^4> So if you measure f(x) at a number of points without looking to P(x) itself,> it can easily go wrong...> I think one should look to P(x) itself, where the minimum value is and so> on, but you cannot calculate the roots for degree 5 or higher, so...?You state you are looking for a 'computer based' method of integration. I assumeyou mean a numerical method. If so, note that you *can* calculate the roots of apolynomial for degree 5 or higher using, for example, the Jenkins-Traub method.There are two things you must never attempt to prove: the unprovable -- and theobvious.Democracy: The triumph of popularity over principle.http://www.crbond.com===Subject: Re: Limit Points===Subject: Re: Limit Points >Unfortunately, different texts have different definitions for limit >point.Alas. >Personally, I prefer to use accumulation point, boundary point, >or cluster point (depending on what I mean) and avoid the >ambiguous term.----===Subject: 2 problems on complex numbersLately, my friend has given to me few problems on complex numbers,there are 2 that I can't solve. Could anybody give me some hint?the 1st:Let for k = 1,2,...,2^n (where n is a natural number) x_k =cos(2k*pi/(2^n +1)) + i sin(2k*pi/(2^n +1)) denote the correspondingroot of binomial equation x^(2^n + 1) = 1. The task: For given k findpolynomials f(x) and g(x) with integer coefficients such that f(x_k)^2+ g(x_k)^2 = -1.and the 2nd:Let n be a natural number and A, z_1, z_2,..., z_n be some complexnumbers such that |A|=1, |z_k|>=1 for each k = 1,2,...,n. How many(maximally) complex numbers z (according to the values of n and A)does it exist such that |z|<1 andn/(1-A) = z/(z-z_1) + z/(z-z_2) + ... + z/(z-z_(n-1)) + z/(z-z_n). my thoughts on it:If we adjust the equation above, we obtain: (@) (n-1+A).[z^n + b_(n-1)*z^(n-1) + ... + b_1*z] + n*b_0 = 0 where according to Viet's formulas: b_(n-1) = -(z_1 + z_2 + ... + z_n) b_(n-2) = z_1*z_2 + ... + z_(n-1)*z_n ..................................... b_1 = (-1)^(n-1)*[z_1*z_2*...*z_(n-1) + ... + z_2*z_3*...*z_n] b_0 = (-1)^n*z_1*z_2*...*z_n The coefficient of the member of degree 0 in the polynomial (@) isexactly n*b_0/(n-1+A). If we denote the roots of (@) asx_1,x_2,...,x_n, according to Viet's formulas (-1)^n*x_1*x_2*...*x_n =n*b_0/(n-1+A). Now it's clear that |b_0|>=1, so|n*b_0/(n-1+A)|>=|n/(n-1+A)|. If we discuss the values of |n/(n-1+A)|over all possible A, we obtain that 1<|n/(n-1+A)|1. But this signifies that|(-1)^n*x_1*x_2*...*x_n|>1, so for at least one value of x_k, theinequality |x_k|>1 must be held. This signifies that the equation (@)can have maximally n-1 roots z such that |z|<1. And that's all fromme. My question is, whether the problem has really n-1 suitablesolutions z. If it does, could you show me any example, if itdoesn't - what is the correct solution?===Subject: Integral of a Vector over Distance Integral of a Vector over Distance -------------------If f=ma (mr/tt) is integrated over distance r the result is a scalar.If p=mv (mr/t) is integrated over distance r the result is a scalar.Is there any mathematical necessity for this? In other words is thereany mathematical principle dictating that the integration of a vectorover distance result in a scalar - apart from physics conventions?===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?Only if it's a dot product. An example is work. Force anddisplacement are both vectors, so the integral over their dot productwill be a scalar called work.But you can also integrate over a cross product. A good example isangular momentum, which is the integral of a displacement crossed intoa differential of momentum.===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?>Only if it's a dot product. An example is work. Force and>displacement are both vectors, so the integral over their dot product>will be a scalar called work.I understand that. What I'm asking is whether there is any fundamentalmathematical principle at work for colinear vector multiplication notat work in the case of cross products for right angle vectors.>But you can also integrate over a cross product. A good example is>angular momentum, which is the integral of a displacement crossed into>a differential of momentum.But in the case of a cross product respective vectors are notcolinear. With a dot product you're multiplying colinear vectorcomponents whereas in cross products you're multiplying rightangle vector components.===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?Neither of these assertions is correct.The integral is of the same type, scalar, vector, tensor, cohomologyclass, what have you, as the integrand, once you take the differentialnature of the dx factor into account.For instance, if I integrate the vector f = ma, along an interval onthe x-axis: integral(f(x,y,z,t) dx; x= 0, 1)the result is a vector; its components are the individual integrals of the components of f(x,y,z,t) over that integral.Does it mean anything? Who cares? After all, the integral isindependent of the origin of the integrand. It's up to the personwriting an integral to write something that makes sense. Mathematicsis not required to make sense out of nonsense.I think you're thinking of the definition of work as the line integral integral(f(x,y,z,t).ds, s_o, s_1)in which the integrand is the dot product of the vector f and thevector ds, tangent to the path of integration. This is the integralof a scalar over the parameter interval, and therefore a scalar.Dale===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?>Neither of these assertions is correct.>The integral is of the same type, scalar, vector, tensor, cohomology>class, what have you, as the integrand, once you take the differential>nature of the dx factor into account.>For instance, if I integrate the vector f = ma, along an interval on>the x-axis:> integral(f(x,y,z,t) dx; x= 0, 1)>the result is a vector; its components are the individual integrals of >the components of f(x,y,z,t) over that integral.>Does it mean anything? Who cares? After all, the integral is>independent of the origin of the integrand. It's up to the person>writing an integral to write something that makes sense. Mathematics>is not required to make sense out of nonsense.>I think you're thinking of the definition of work as the line integral> integral(f(x,y,z,t).ds, s_o, s_1)>in which the integrand is the dot product of the vector f and the>vector ds, tangent to the path of integration. This is the integral>of a scalar over the parameter interval, and therefore a scalar.In the above you refer to vector f and vector ds. So where does eitherbecome scalar for the integral of a scalar over a parameter interval?===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?Yes, look at any engineering text where it talks about what things like foot-pounds or gram centimetres mean. Basically, if the same measure appears both above and below the fraction bar, then the result is simple ratio, which has no measure. When you integrate a vector over distance, you eseentially do distance/distance, which is a ratio (or scalar, in vector-speak.)NB that (3/5 lbs) is _not_ the same as (3 lbs / 5 lbs). In decimal notation, the first fraction is 0.6lb, and the second is just 0.6.===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?>Yes, look at any engineering text where it talks about what things like >foot-pounds or gram centimetres mean. Basically, if the same measure >appears both above and below the fraction bar, then the result is simple >ratio, which has no measure. When you integrate a vector over distance, >you eseentially do distance/distance, which is a ratio (or scalar, in >vector-speak.)>NB that (3/5 lbs) is _not_ the same as (3 lbs / 5 lbs). In decimal >notation, the first fraction is 0.6lb, and the second is just 0.6.I think I agree with your analysis, Wolf. But I'm unclear why the samerationale wouldn't also apply to cross products.===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?Dimensions2-->1 projection The world is flat it's pi that's round!There is only one number.about me http://cparkes.actewagl.net.au===Subject: Re: Integral of a Vector over Distance> Integral of a Vector over Distance> -------------------> If f=ma (mr/tt) is integrated over distance r the result is a scalar.> If p=mv (mr/t) is integrated over distance r the result is a scalar.> Is there any mathematical necessity for this? In other words is there> any mathematical principle dictating that the integration of a vector> over distance result in a scalar - apart from physics conventions?>Dimensions>2-->1 projection Could the result not be represented in another dimension analogous tor x p even though it is a scalar product?>-- >The world is flat it's pi that's round!>There is only one number.>about me http://cparkes.actewagl.net.au===Subject: Characters, Representations basics questionI posted this in an earlier thread, and was wondering if someone couldexplain or prove the following (is it easy to prove?) :Why are the irreducible unitary representations of the circle the same asthe dual of the circle?Why is this true in general for a locally compact abelian groups and nottrue for just locally compact groups? (What's the proof for Locally compactabelian?)Why are the only unitary irreducible representations of the circle given bythe right regular representations?===Subject: Re: Characters, Representations basics question>I posted this in an earlier thread, and was wondering if someone could>explain or prove the following (is it easy to prove?) :>Why are the irreducible unitary representations of the circle the same as>the dual of the circle?>Why is this true in general for a locally compact abelian groupsIf F is a family of nomal operators on C^n and AB = BA for all A, B inF then F is simultaneously diagonalizable (hence if n > 1 then Fhas a non-trivial invariant subspace.) This is one of the manyfacts known as the spectral theorem.> and not>true for just locally compact groups? (What's the proof for Locally compact>abelian?)>Why are the only unitary irreducible representations of the circle given by>the right regular representations?===Subject: Re: Characters, Representations basics question> I posted this in an earlier thread, and was wondering if someone could> explain or prove the following (is it easy to prove?) :> Why are the irreducible unitary representations of the circle the same as> the dual of the circle?Assuming that by circle you mean the torus group T = R/Z = U(1) = SO(2)then the dual of this group, as a locally compact abelian group, is Z.If A is a locally compact abelian group (LCAG)then by definition the elements of the dual LCAG A*are the irreducible (ie 1-dimensional) unitary representations of A.So the irreducible unitary representations of Tcorrespond to the elements of Z, and vice versa.Since one of these groups is enumerable, and the other is not,they cannot be the same in any sense.Perhaps I have misunderstood? > Why is this true in general for a locally compact abelian groups and not> true for just locally compact groups? The irreducible representations of a non-abelian group do not form a group.Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland===Subject: Re: Characters, Representations basics question> I posted this in an earlier thread, and was wondering if someone could> explain or prove the following (is it easy to prove?) :> Why are the irreducible unitary representations of the circle the same as> the dual of the circle?>Assuming that by circle you mean the torus group> T = R/Z = U(1) = SO(2)>then the dual of this group, as a locally compact abelian group, is Z.>If A is a locally compact abelian group (LCAG)>then by definition the elements of the dual LCAG A*>are the irreducible (ie 1-dimensional) unitary representations of A.I think the question was _why_ the irreducible representationsare one-dimensional. The ie makes it sound like that's justtrue by definition.>So the irreducible unitary representations of T>correspond to the elements of Z, and vice versa.>Since one of these groups is enumerable, and the other is not,>they cannot be the same in any sense.>Perhaps I have misunderstood?One of us has misunderstood something: looking at thequestion and your reply it seems that you're claimingthat one of Z and Z is not enumerable (I'm not surewhich one...) > Why is this true in general for a locally compact abelian groups and not> true for just locally compact groups? >The irreducible representations of a non-abelian group do not form a group.===Subject: On the Cosmological Constant problemOn the derivation of my basic zero point field formula for /zpf below.Some new points:state must also be positive.Therefore the over-all sign is a matter of convention determined by the 3 GR conventions described in Ch. 1 of John Peacock's Cosmological Physics.2. Coarse graining. My /zpf field is a local field, which on the cosmological scale beginning at L = 10^2 megaparsecs out to 10^4 megaparsecs limits to Einstein's cosmological constant. We know from statistical mechanics, from Bohr-Rosenfeld in the 1930's to Ken Wilson in the 1960's Renormalization Group on a lattice, that we need to coarse grain to get to different scales, i.e. integrate out smaller wave lengths and higher frequencies. Look at my formula that has the generic structure/zpf(x) = (Area)^-1[(Volume)|Vacuum Coherence|^2 - 1]Vacuum Coherence is a giant quantum BIT pilot wave, i.e. a separated by distance L. The quantum wave is normalized to|psi| ~ 1/L^3/2In macro-quantum physics |Vacuum Coherence|^2 is the condensate number density./zpf is a local field that is coarse-grained to a ball of radius L at scale L. That is, the point is replaced by the ball! For example, in the FRW cosmology metric the ball has a radius ~ 10^2 megaparsecs! An entire galaxy is a point.So that in general at scale L/zpf(x,L) = (Quantum of Area)^-1[aL^3|Vacuum Coherence|^2 - 1]a is a dimensionless number.Using the world hologram idea of Susskind et-al note thatL^2/(Quantum of Area) ~ number of Bekenstein-Hawking bits that can be packed into the point i.e. coarse-grained sphere whose center is x.PS Ibison's PV Cosmology paper is well written and he proves that PV is a wrong theory. It does not at all fit the now established facts of precision cosmology whose data is in http://supernova.lbl.gov/~evlinder/linderteachin1.pdf The fact is that there is missing mass and it is 96% of the Universe. PV theory is a wrong theory in Feynman's sense, even if the math is nice, and Ibison's math is nice, if it does not fit the facts, and it doesn't it should be rejected and it is by all competent physicists in the field.Collins Lets get started; 1) This interview is over 4 years old. Allot has changed in ZPE thinking since that time. OK, tell us exactly what has changed.Collins 2) Virtual photons have positive zero point energy density and virtual electron-positron pair only make up a very small fraction of the total Vacumm Energy Density. That fraction was at one time all anybody thought the vacuum was comprised of. Now, it's considered a drop in the bucket compared to 10^114 ergs/cm^3. Show us your calculations. What are your assumptions? How are your assumptions justified by experiment? It is true that virtual photons have positive zero point energy density. I have written that many times. So what is your point? So what? Justify your pronouncement that virtual electron-positron pair only make up a very small fraction of the total Vacumm Energy Density.Collins 1.57 under, The Quantum Vacuum: Introduction to Quantum Electrodynamics: Peter W. Milonni; Los Alamos, New Mexico. at; http://ufoconspiracy.com/misc/Energy%20Density%20of%20the%20ZPE %20Quantum%20Vacuum.doc There is nothing relevant in the document showing that |zero point energy density of virtual electron positron pairs| << |zero point energy density of virtual photons| One would hope they were equal in absolute value since they are of opposite sign! virtual photons have negative pressure. virtual electron-positron pairs NOT in the vacuum condensate have positive pressure.Collins 3) There is no reason you can't have vacuum coherence down to the Planck Scale: vacuum coherence = 0 If you have vacuum coherence down at the Planck scale then vacuum coherence =/= 0Collins 4) /zpf = (Quantum of Area)^-1[(Quantum of Volume)|Vacuum Coherence|^2 - 1] : What is your ref? Actually it comes directly from the two-fluid model of Tiza, London et-al, which has a very general significance as shown by Oliver Penrose, Gorkov, P.W. Anderson and others. When the reduced quantum density level for fermions), in the diagonal limit of say rho(x,x') Total density = Superfluid Density + Normal Fluid Density In my model In Hagen-Kleinert's world crystal lattice intuitive picture of discrete spacetime. As usual Superfluid Density = |Ground State Coherence|^2 In the vacuum Normal Fluid = Zero Point Fluctuations Therefore (Quantum of Volume)^-1 = |Vacuum Coherence|^2 + (Zero Point Density of I assume for reasons of parsimony AKA Occam's Razor, i.e. More with less = Big Bang for small buck = Over-Unity Creative Imagineering! (Quantum of Volume) = (Quantum of Area)^3/2 i.e. in the Penrose spin-network pre-geometry, there is only one independent quantized parameter not two or three. (Quantum of Volume)^-1 = |Vacuum Coherence|^2 + (Zero Point Density of transforms by my above conceptual poetic tech-gnostic substitutions of the real Alchemy to (Quantum of Area)^-3/2 = |Vacuum Coherence|^2 + /zpf(Quantum of Area)^-1/2 Hence by middle school algebra that you can no longer do in your reduced state of cognitive virility (Quantum of Area)^-1 = (Quantum of Area)^1/2|Vacuum Coherence|^2 + /zpf /zpf = (Quantum of Area)^-1 - (Quantum of Area)^1/2|Vacuum Coherence|^2 or /zpf = (Quantum of Area)^-1[1 - (Quantum of Area)^3/2|Vacuum Coherence|^2] The overall sign is not determined until one decides on the 3 GR conventions (ch 1 of John Peacock's Cosmological Physics) The way it's written above needs Guv - /zpfguv = 0 So that when |Vacuum Coherence| -> 0 you get gravity attraction i.e. dark matter exotic vacuum You also have to know that Vacuum Coherence = Giant Quantum Wavefunction, i.e. macroscopically [psi] = (Volume)^-1/2 consistent with Born probability interpretation. 7) Jack: Hal's idea here violated Einstein's equivalence principle + Heisenberg's uncertainty principle:Collins: PV enhances GR by taking the next step up. And, the intrinsic reason for Uncertainty is the ZPE Vacuum fluctuations. Jack: Show us. Your second statement is wrong. You put the cart before the horse. Heisenberg's principle applies in different ways both to real and virtual quanta. Also, Heisenberg's principle may only be a limiting case of dx = h/dp + alpha' dp/hCollins That doesn't show the physical reason behind the expression which is the Vacuum Fluctuations which is already quantized.....Rmccalled IR/UV duality. alpha' is the Witten string parameter. Collins And, in case you didn't get it K is the dielectric constant and NOT a Tensor but it does have a exponential form in PV. When K is made a Tensor PV will most likely account for the full output from a binary pulsar and agree with GR. References; Hal has not been able to solve this problem in more than 10 years. He is near 70 and I am not holding my breath that he will ever solve it. This shows that problems hard in PV are easy in GR. If it ain't broke don't fix it with something not as good.http://xxx.lanl.gov/abs/gr-qc/9909037http://xxx.lanl.gov/ abs/astro-ph/0302273 RmcInterview is athttp://www.pbs.org/safarchive/3_ask/archive/qna/3282_ hpsweinberg.html===Subject: Re: Rubik's cube operators> rupertmccallum@yahoo.com says...> I have a good collection of edge operators for my Rubik's cube, but I> don't know any corner operators. Can anyone give me any of the> following:> [...]> A meson (twist one corner cubie one way, another corner cubie the> other way)> Rupert,> This type of move (or operator) you can easily create yourself> through the use of the group-theoretic concept of _commutation_.> The idea is very simple. Say you have a move sequence P that> twists one corner cubelet, while scrambling large portions> of the cube, _except_ for the one of the face layers the> cubelet lies in. For example, the sequence R'DRFDF' twists> the FRU cubelet CW and leaves the U layer untouched.> If you now perform the move sequence P' (P backwards) the> cube will be restored. However, consider what happens if you> perform the move sequence Q consisting of the single move U> before you do P'. You'll (1) move the twisted cubelet at FRU> to FLU. You'll twist the new cubelet at FRU ccw, leaving> the top layer otherwise intact and you'll unscramble the> bottom two layers into their original state. Finish off> with Q' to restore the top layer back into position and> you'll have performed your meson move!> The sequence PQP'Q' (= R'DRFDF' U FD'F'R'D'R U') which you> just performed is called a commutator of P and Q.> A baryon (twist three corner cubies the same way)> Can you see how you can create a move sequence for this> using a commutation of the sequence for the meson?> Here's a good page talking about the relevance of group> theory to puzzles like the Rubik's Cube:> http://www.geocities.com/jaapsch/puzzles/theory.htm> Enjoy!> Christer Ericson> Sony Computer Entertainment, Santa MonicaYes indeed. My corner swap isFSF'S' B SFS'F' B'which is rather longer winded than the others that have been mentionedbut I discovered because FSF'S is the commutator of Front and Sideand now the the full sequence is the commutator of this element withBottom.===Subject: Where to download 60 millions of HEXADECIMAL digits of Pi ?it seems that many sites allow to download millions or even billionsof decimal digits, but is there one where HEXADECIMAL (only 60-100millions...) digits can be found ? nous devons agir comme si la chose qui peut-.90tre ne sera pas devait.90tre (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel 90 (except for the 0/90cases).Where a,b are the equatorial, polar radii and Oz--equalingacos{b/a}--is the angle of oblate ellipticity, or angulareccentricity: e^2 = sin{Oz}^2, e'^2 = tan{Oz}^2 andf = ver{Oz} = 2 * sin{.5*Oz}^2 = 1 - cos{Oz}.Denoting the spherical/graticular TLat, AP as TLats, APs and theelliptically correct as TLate, APe; on a spheroid with minorellipticity (such as Earth), TLate ~=~ atan{tan{TLats}*sec{Oz}} andAPe ~=~ atan{tan{APs}*sec{Oz}}.If, using TLats, one finds APe (via the standard geodetic distancecalculation, letting Lat1 = 0, Lat2 = TLats and Long2 - Long1 = 90)and labels the results TLatm:L1 = TLats, APm:L1 = APe, then findsTLate using APs (this is a little more complicated--Lat1, Long1 = 0;Azm1 = APe; Distance_0 = 10000; IC = 10; Distance = Distance + IC,recurse until Long2 = 90: When Long2 > 90, thenDistance = Distance - IC; IC = .1 * IC, resume recursion) and labelsthese results as TLatm:U1 = TLate, APm:U1 = APs, one will see that(when TLats and APs are integers) the decimal of TLate will nearlymatch that of APe's.Using the average of one of the elements, find the second generationof that set and do the same for the other element (using the aboveroutines): TLatm:L2 = .5 * [TLatm:L1 + TLatm:U1] = .5 * [TLats +TLate], get APm:L2; APm:U2 = .5 * [APm:L1 + APm:U1] =.5 * [APs + APe], get TLatm:U2. Repeat the process until likeelements converge to a common value: TLatm = TLatm:L = TLatm:U,APm = APm:L = APm:U.Again with minor ellipticities, TLatm ~=~ atan{tan{TLats}*sec{Oz}^.5}and APm ~=~ atan{tan{APs}*sec{Oz}^.5}.If one changed the TLat/AP relationship from APs + TLats = 90 toAPm + TLatm = 90 + [2 * ES] = [APs + ES] + [TLats + ES], thenTLatm = TLats + ES, APm = APs + ES and the ELLIPTIC SUPPLEMENT (ES)is a single value, providing a common, offset pseudospherical mergebetween TLat and AP, hence creating a SPHEROIDAL ISOPATHICMEDIAN--?: http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid= 68282&fname=isopathicmedian_g.gifWhile either the direct solution (finding a second set of geodeticcoordinates, given one set, a distance and azimuth) or inversesolution (finding the distance and direction between two givengeodetic coordinate sets) can be utilized, the inverse approach isconsiderably faster and easier:--- Lat1 = 0: Long1 = 0: Lat2 = atan{tan{TLats}*sec{Oz}^.5}: Long2 = 90: APe = atan{cot{TLats}*sec{Oz}^.5}: ES = 0------- ------- ----------ES_o = ES: ES = .5 * [APe + Lat2 - 90]---Lat2 = TLats + ES---Calculate APe---Recurse until ES = ES_o------- ------- ----------TLatm = Lat2: APm = APeIs this a known concept and, if so, does it have a more formalname/identity? Is there a more direct equation for ES (such asan--elliptic?--integral or series expansion)?A likely question someone may have at this point is yeah, so what doyou do with TLatm, APm once you have them?.The immediate use this writer has, is as the object of differentiation(the differentiant or differentiand?): Differentiating byAPs/TLate produces a different result than by APe/TLats.With APm, one can find the theoretical, pseudospherical distance--or,as is this writer's quest, the quadrantal arcradius--for a given APs,from the equator out to TLats (remembering that APs --> TLate andTLats --> APe).Applying it further, APm can be used in finding the transverse surfacearea--though it would seem that, as APe = APs at the equator, Qs goesto Qe only in the complete case (i.e., 0 to 90), hence ES needs tobe reduced as Lats goes from TLats to 0 (i.e., at the equator),likely by the sine (squared?) of the geographic/spherical transverselatitude (TvL): Lats = asin{cos{APs}*sin{TvL}}; Latm = Lats + [sin{TvL} * ES] or [sin{TvL}^2 * ES]; (TLats = TLats + [sin{90} * ES]) APm = APs + [sin{TvL} * ES] or [sin{TvL}^2 * ES];Likewise, Longs = atan{sin{APs}*tan{TvL}}, so Long1 = Longs1,Long2 = Longs2 and--instead of 90--Long2 - Long1 = Longs2 - Longs1,which, if finding Latm and APm via the direct solution, would be thecomparative measure.This is presuming TvL is used exclusively, and not the parametricTvL (TpL)--or even TvLe or TpLe--in which case formulationwould likely become considerably more complex.So, is there such a thing recognized as a pseudospherical, spheroidalisopathic median (though likely identified under another name/term)? ~Kaimbridge~----- WantedKaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ----------DigitologyThe Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. *****===Subject: Re: Difficult Algebra Problem> I have a problem that I'm having a tough time with. Let R be a ring (perhaps without 1).> Prove that (S,+,.) is a ring with unity> when S=R X Z, (r,m)+(s,n)=(r+s,m+n), and (r,m)(s,n)=(rs+ms+nr,mn).>[...]> If R has a unity 1, how does that unity relate to the unity of S?> Are they the same? Would it contradict the uniqueness of unity?> Since R is not even a subset of S, the two unities are certainly not equal. > Even if you regard R as a subset of S via the identification> r --> (r,0) (for r in R)> you can check, that (r,0)(0,n) = (nr,0).> Hence no element of the form (r,0) can be the unity of S.> So if you start with a ring unity, the new ring S will> have another unity.> MarcThe unity is S is (0,1). So let's take r-->(r,0). If you take(r,0)(1,0) you get (r,0) using the multiplication defined. If youtake (r,0)(0,1) you get (r,0) too but this is assuming that you canuse a (0,1). Can you assume that every element must be in (r,0) form? If not, wouldn't that contradict the uniqueness of identity for R?Maybe I am missing something here. Jacob===Subject: Re: Difficult Algebra Problem>I have a problem that I'm having a tough time with.Let R be a ring (perhaps without 1). Prove that (S,+,.) is a ring with> unity>when S=R X Z, (r,m)+(s,n)=(r+s,m+n), and (r,m)(s,n)=(rs+ms+nr,mn).> SNIP> We want a (s,n) such that (r,m)(s,n)=(r,m) therefore r=rs+ms+nr and m=mn> clearly the second part implies n=1 which is feasible since n e Z> so now we know r = rs+ms + r = s(r+m) + r> if we let s=0 (since R is a ring, 0 has to be in it), the above equation> becomes r = r> So unity in S is (0,1). Notice this does NOT require unity in R.> -TralfazI remember the problem from algebra. It is a formal trick to embed aring without a unity into a ring with a unity. My book didn't presentany applications of it though. I suppose, then, that the problem isuseful for demonstrating what algebra has always been about: assumingthat some particular element exists and giving it an undetermined form(which in this case is an element of the big ring with undeterminedcomponents) and then solving for the exact form of the element givenwhatever constraints it has to satisfy (which in this case is thatthose components are determined by the requirement that the elementact as a unity in the big ring S).Patrick===Subject: Maximum GCD-Sum PermutationsIf we have the permutation of {1,2,3,...,m}, where the sum of the(m-1) GCDs of adjacent elements in the permutation is maximized, weget (perhaps, found by hand) the sums equal to:(starting at m=2)1, 2, 4, 5, 9, 10, 14, 17, 23, 24, 32,...(For example, for m = 12, one of the maximum-GCD-sum permutations{there are a few which add up to 32} is{unless I erred}5,10,2,8,4,12,6,9,3,7,11,1which gives the GCD-sum of5+2+2+4+4+6+3+3+1+1+1 = 32)Now, the sequence of maximum sums matches, for the terms given,sequence A063985 of the EIS:http://www.research.att.com/projects/OEIS?Anum=A063985This sequence is defined assum{k=1 to m} (k - phi(k)),where phi(k) is the Euler totient function.Are these two sequences the same for all terms?By the way, in another post I am making today, GCD-game (and puzzlequestion), I ask about the 2-dimensional analogy to this sequence,where the first m^2 positive integers are placed somehow in a grid,and we want to maximize the sum of the GCDs of adjacent elements inthe grid.Leroy Quet===Subject: GCD-game (and puzzle question)Here is a simple game (which may have some teaching possibilities, into it.As in many of my games, the board is an n-by-n grid drawn on paper.(See below regarding what n should be.)2-person game.One player plays columns, the other plays rows.You have a deck of n^2 cards (perhaps just slips of paper), where eachcard has a different one of the integers from 1 to n^2 written on it.The deck of cards is placed face-down after shuffling.Players take turns drawing one card per turn.one of the grid's squares which does not yet have any integer writtenin it.Scoring is as follows:For row-player, the player's score is the sum of the greatest commondivisors between every pair (among the {n^2 -n} left/right-pairs inthe grid) of immediately left/right-adjacent integers.(See example below.)For the column-player, as you might have guessed, the player's scoreis the sum of the GCDs taken over all the pairs of integers which areimmediately up/down-adjacent.-As for n,I felt at first n should be even so each player places the same number(n^2/2) of integers into the grid. But then I realized that the secondplayer to write down an integer in the game will be forced to make thegame's final move in a particular square.But if n is odd, the second player has a choice for their last movebetween 2 squares where to place the last integer. Then the remainingcard in the deck is simply written in the last empty square.Still, the first player might have an advantage, which I bet vanishesas n gets bigger.So, I would suggest an odd n which is as big as the players bothbelieve they can handle.-Example: n = 5:Board at game's completion:2 4 3 6 110 8 14 9 75 11 12 20 2125 13 22 18 1617 19 24 23 15Row-player scoring:2+1+3+1 +2+2+1+1 +1+1+4+1 +1+1+2+2 +1+1+1+1 =30Column-player scoring:2+4+1+3+1 +5+1+2+1+7 +5+1+2+2+1 +1+1+2+1+1 =44Column wins.-Questions/puzzle:What is the highest possible score a player could theoreticallyachieve for a given n?What is the highest sum of both players' scores which could possiblybe achieved for a given n?(I ask about the 1-dimension version of this problem in another post Iam making today, Maximum GCD-Sum Permutations.)-And it might be fun to have variations on this game which use othercriteria besides GCD.Maybe players can decide to use the LCM of, the number of primedivisors in the sum of, the product of (as some examples) the pairs ofadjacent integersas a criteria for scoring.Leroy Quet===Subject: Re: Q: Orbital Intercept> Is there any chance you could help me out with this problem?> AndyThis is a 2-body problem, right? I mean, it's only the nuclei that are gravitationally active, so to speak. I've always understood that the term n-body problem means that all the bodies participate in the dynamics, not just the kinematics. In your description, all those test So, solve the 2-body problem in CM coordinates. Surely that's been done. Then apply the disc structures to the result, making sure to conserve angular momentum.BTW, I'm confused by your remark The masses are in a parabolic orbit.If you meant elliptic and this was a typo, I am less confused, but if not, why on earth would they be in parabolic orbits (let alone a [sic] parabolic orbit)?=== ===Subject: Re: Generating Matrices with for-loop in Matlab > I have a problem and am about to flip out, because I am a Matlab > beginner. You may also be a beginner at spelling in English. That is certainlyno crime, but worth correcting. Please take this as friendly correction: > I have say an 12x3 matrice named M and want to seperate 4 3x3 ^^^^^^^ ^^^^^^^^Spelling: MATRIX SEPARATE > Matrices from that named S1, S2, S3, and S4. How do I solve this in > Matlab? Later, I would like to have an 3*nx3 matrice and seperate ^^^^^^^ ^^^^^^^Spelling: MATRIX SEPARATE > that into single labled 3x3 matrices S1,S2,...,Sn. Is there anybody > who can help me out? Two remarks:1. There is a Matlab group that has many experts as regular contributors: comp.soft-sys.matlab2. It is almost always a good idea to avoid for-loops in Matlab. Manythings that other language appear to force into a loop construction can and should be done some other way in Matlab. This particular case is a good example.Your partitioning of the matrix M can be done in many ways, andyou will need to be *explicit* in terms of how you care to do yourseparation. For example, you may wish to have the S-matrices formedfrom blocks of M, so that M looks like this: [ S1 ] [ S2 ] M = [... ] [ Sn ]On the other hand, I can imagine you could want to take every fourth row to produce S1, S2, S3, and S4 (if M represented a data stream, and you wished to cause a burst of errors to be spread out over a range of locations so it could be handled by error-correction coding, for instance). As I said, it's essential to be explicit in how you wish to partition your matrix. Matlab needs to be told WHICH elements of M getmapped to WHICH elements of WHICH matrix S.Now, just for laughs, let's suppose you have decided that the rows of each S-matrix will be rows of M, so all you care to do is to write an expression of the sort: > S( row,: ) = M(row', :);where row and row' represent expressions to be determined.Let's suppose you decided that row 1 of S1 would be row R(1,1) of M, row 2 of S1 R(2,1) row 3 of S1 R(3,1) row 1 of S2 R(1,2) row 2 of S2 R(2,2) row 3 of S2 R(3,2)and so forth, down to row 3 of S4 R(3,4).Then, you could write this series of statements: S1 = M(R(:,1),:); S2 = M(R(:,2),:); S3 = M(R(:,3),:); S4 = M(R(:,4),:);to fill out S1 through S4.For example: the block format I mentioned earlier give us this: S1 = M([1:3],:); S2 = M([4:6],:); S3 = M([7:9],:); S4 = M([10:12],:);Check it out: here's the index matrix R that I mentioned above: +------------------ rows of S1 | +--------------- rows of S2 | | +------------ rows of S3 | | | +-------- rows of S4 | | | | V V V V [ 1 4 7 10] R = [ 2 5 8 11] [ 3 6 9 12]The above technique has the obvious generalization to your expanded case, where the matrix is of size (3n) x 3 .===Subject: Moebius-Function-Based Sequencesmu(k) is the Moebius (Mobius) function,defined by:sum{k=1 to oo} mu(k)/k^r = 1/zeta(r).http://www.research.att.com/projects/OEIS?Anum= A008683I do not believe this (interesting, in my opinion) sequence is in theEISyet. (Although it might be in the process of being submitted.):a(n) = smallest integer > n wheremu(n) = mu(a(n)).6, 3, 5, 8, 7, 10, 11, 9, 12, 14, 13, 16, ...-Also, a related sequence:b(n) = a(n) - n:5, 1, 2, 4, 2, 4, 4, 1, 3, 4, 2, 4,...average?ie. does x =limit{m->oo} (1/m) sum{n=1 to m} b(n)exist?Does x have a closed form?Robert G. Wilson v of the seq.fan email group determined that x seemsto be approaching 3.Now, the sequences of n's wheremu(n) = m(n+1) IS in the EIS.http://www.research.att.com/projects/OEIS?Anum=A064148But the actual values of mu(c(n)), where c(n) is the n_th term of A064148, is not in the EIS.-1, 0, 1, 1, 0, 0, -1, -1 ,...Finally, I do not believe the following is in the EIS either:d(n) = Number of positive divisors k of n, wheremu(k) = 1 and mu(n/k) = -1.0, 1, 1, 0, 1, 0, 1, 0, 0, 0,...I get the relation (hopefully correct):4*d(n) + sum{k|n} mu(k)*mu(n/k) =product{p|n} e(p,n),where the product is over the distinct primes dividing n;e(p,n) = 2 if p|n but p^2 does not divide n;e(p,n) = 1 if p^2|n but p^3 does not divide n;e(p,n) = 0 if p^3|n.Leroy Quet===Subject: When R=*R in a Simplified Bonner Sphere FunctionNeutron Spectra By Bonner Sphere Inverted and Solved for the Extendedor Point Source. When R=*RA set of varying sized plastic spheres called Bonner spheres are usedto moderate the neutron field. And the interior detector is a neutronsensitive one. Each variable is solved and the spectra calculated onlyafter the response coefficient of the linear set is fitted . Agaussian least squares assumption fit is commonly utilized.R- neutron/mev A bin system is not used in this analysis, only thepure response function.To determine this function, R, for a measurement using the Bonnersphere is the experimenter’s goal. Many simplifying assumptionsare utilized to allow the function to be solved.1. Adding plastic diameter to the sphere always lowers the spectra..2. A pure thermal detector in the exact center of each sphere existswithout perturbing the plastic’s moderating properties.3. Ambient source distribution is invertable with point sourcedistribution.Ambient inversion is the meaning of pure spherical detectors. In manyapplications in radiation dosimetry the small mass of tissue is to bereplacing the system. A dosimetry calculation for real usage.And inverting the response function exactly is the goal of thisanalysis. This would allow a single sphere to measure the knownsource’s absolute emission rate, total neutrons per unit time.Remember the least square fitting is commonly required because of thedifficulty in solving the system. This is dispensed with.Inverting the best fit function is not the same as using each datapoint to cause the solution, given the solution’s formula.r- mev :this is the sphere response to any energy of neutron.c- count :this is the counting efficieny for any given energy.Two differential constants are given. Each determines the rate of theneutron. Simple sets of linear differentials.A time partial is added giving:dr/dt= r dtdc/dt= c dtAnd of course for each equation, the set is caused by the diametersselected, making:C = c1 + c2 + c3 + c4 +c5 + c6 +c7 +c8R = r1 + r2 + r3 +r4 + r5 + r6 + r7 + r8Making the field responded to, the calibration fieldset(R1….Rn). And here the coefficients are simultaneously solvedmatrix solutions.The inversion to allow an exact single partial solution for a singlesphere is possible!Possible without graph fitting solution. The mistaken assumption hereis the set of example differentials in relation to the abstractdifferential set. We have assigned a particular diameter sphere as thecause of (1…..8). Mathematically, this is a fault, but itallows the old style Bonner set solution.And here the true cause is the differential constants, r and c. Iselected the units of these constants to allow the right solution.*********The search was nontrivial.************15 days for the givenhint of a dilemma.Why does the example sphere dislocate mathematical logic? Because thelimit concept of the sphere diameter to the neutron bin bandwidth wasnot discussed and is atrocious. A horrible cheat of the width of theneutron is allowed mathematically.A dispensed with, bin width, makes common Bonner set solutionpossible. Now function’s coeffiecients solve.And the pure function is an assumed power function here. The spectrawill always be a simple power function, because the assumptions causethis to be the caused differential. Others could be substituted andonly the matrix solution altered.Also, spikes of monoenergetic neutrons on the power function will beunseen. In fact the capacity to resolve this system is in doubtexperimentally. Only nice gentle spectra of a single monoenergeticsource could be determined. Playing with the detector differentialswould allow a gamma spectra like unfolding.The response of the Sphere detector to the entire spectra energy rangeis supposed to be large? Making it small with many differentexperimental system examples allows the different sources tocorrespond to different detectors?!Here is the inversion of the detector bin width removed. We no longeruse the last conception of functional response. The arbitraryfunction is sufficient. And experimental discovery would define thesuccessful set of Bonner spheres, or maybe just atoms. A set ofatoms, fluorescing, could replace the set of spheres.And so the selected constants are left to pure logical inversion toallow the single set example to measure the total neutron emissionrate experimentally!!!! A big dilemma resolved.R1 = dc r1/dtR2- same...Rn- sameEach differential constant is independent for the integral!!!! ::::J)))))))))))))I chose them independent. ************************************************************** ************************Remember all the calibration work? Each coefficient must be resolvedwith respect to a total energy.And this last set is of course the matrix differential eigenfunction.JA one dimension matrix.R=C |*R1| |*R2| |*R3| |*Rn|Why yes R is a matrix function!!!!!!!! We can do them correctly now adays… HurahaaR= *R here *R is the unknown experimental data, while R is the abstractmatrix of the calibration experimental data set.Selecting the correct eigenfunctions for the abstract attenuatingradiation detector is this discovery. A real nice function to rememberbecause it is nowhere intuitive.Gaithersburg, MD USA