mm-3339 === Subject: : linear algebra with diagonalizable. Hello sir~ Matrix A = |cos@ -sin@| |sin@ cos@ | Show that A is diagonalizable. Find P such that P^-1.A.P = D -------------------------------- I think.... |xA - I| = x^2 - 2xcos@ + 1 = 0. so, x = cos@ +-sqrt[(cos@)^2 - 1] Hm...I have a trouble. How do you solve this? === === Subject: : Re: linear algebra with diagonalizable. ... > so, x = cos@ +-sqrt[(cos@)^2 - 1] what is '1 - (cos y)^2'? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === === Subject: : Re: linear algebra with diagonalizable. > Hello sir~ Matrix A = |cos@ -sin@| > |sin@ cos@ | Show that A is diagonalizable. > Find P such that P^-1.A.P = D -------------------------------- > I think.... > |xA - I| = x^2 - 2xcos@ + 1 = 0. so, x = cos@ +-sqrt[(cos@)^2 - 1] Hm...I have a trouble. > How do you solve this? x = cos@ +-i sin@ = exp(i@) or exp(-i@). === === Subject: : Re: linear algebra with diagonalizable. > Hello sir~ > Matrix A = > |cos@ -sin@| > |sin@ cos@ | > Show that A is diagonalizable. > Find P such that P^-1.A.P = D > -------------------------------- > I think.... > |xA - I| = x^2 - 2xcos@ + 1 = 0. > so, x = cos@ +-sqrt[(cos@)^2 - 1] > Hm...I have a trouble. > How do you solve this? > > x = cos@ +-i sin@ = exp(i@) or exp(-i@). Yes, thank you very much. === === Subject: : =?windows-1256?B?x93KysfNINTI38kg09/TIN3m0SDa0cguLi4uLiAgd3d3LjY0YXIuY29t?= === === Subject: : =?windows-1256?B?UmU6IMfdysrHzSDUyN/JINPf0yDd5tEg2tHILi4uLi4gd3d3LjY0YXIuY29 t ?= === === Subject: : =?windows-1256?B?x93KysfNINTI38kg09/TIN3m0SDa0cguLi4uLiAgd3d3LjY0YXIuY29t?= === === Subject: : algorithm with a,b Hello sir~ I showed algorithm. a is non-negative integer. b is positive integer. start with a, b. (*) if a < b, output a. End if a >= b, a = a-b. returen to (*). for given a, b, what's output ?? hint) with definition of sequence. ---------------------------------- How do you think about it ? === === Subject: : Re: algorithm with a,b mina_world .95.8f.96.88.91(.88): > Hello sir~ I showed algorithm. a is non-negative integer. > b is positive integer. start with a, b. (*) > if a < b, output a. End if a >= b, a = a-b. > returen to (*). for given a, b, > what's output ?? hint) with definition of sequence. ---------------------------------- > How do you think about it ? output is x = b*((a/b) - [a/b]) - residue of division . What is sequence about you say ? === === Subject: : Re: algorithm with a,b mina_world .86.83.94.8d.92(.8d): Hello sir~ > I showed algorithm. > a is non-negative integer. > b is positive integer. > start with a, b. > (*) > if a < b, output a. End > if a >= b, a = a-b. > returen to (*). > for given a, b, > what's output ?? > hint) with definition of sequence. > ---------------------------------- > How do you think about it ? output is x = b*((a/b) - [a/b]) - residue of division . Oh, good idea. thank you very much. > What is sequence about you say ? Maybe, I have written the meaning of hint uncertainly. This is a problem with mathematics education. so, I translated hint wrongly. Anyway, thank you very much. === === Subject: : Re: algorithm with a,b > Hello sir~ I showed algorithm. a is non-negative integer. > b is positive integer. start with a, b. (*) > if a < b, output a. End if a >= b, a = a-b. > returen to (*). for given a, b, > what's output ?? hint) with definition of sequence. ---------------------------------- > How do you think about it ? Try some nontrivial examples. --- Christopher Heckman === === Subject: : Re: algorithm with a,b one integer x = 0...b. === === Subject: : Re: algorithm with a,b > one integer x = 0...b. x = 0,...,b-1. I want to know the hint. with definition of sequence. I can't understand this hint. in this problem, there is a sentence that This algorithm sometimes used in the definition of sequence. I can't know it's meaning. How do you think about it ? === === Subject: : =?windows-1256?B?x93KysfNINTI38kg09/TIN3m0SDa0cguLi4uLiAgd3d3LjY0YXIuY29t?= === === Subject: : time complexity for positive definiteness For a non-symmetric square matrix of size n x n. What is the time complexity of finding out whether the matrix is positive definite or not. Also, which would be the fastest, in terms of time complexity, to compute the same. I want to know the answer for finding singularity of a matrix too. Which would be the fastest in terms of the time complexity of n and how fast will it be? Dilip === === Subject: : Re: time complexity for positive definiteness For a non-symmetric square matrix of size n x n. > What is the time complexity of finding out whether the matrix is > positive definite or not. > Also, which would be the fastest, in terms of time complexity, to > compute the same. I want to know the answer for finding singularity of a matrix too. > Which would be the fastest in terms of the time complexity of n and how > fast will it be? Dilip > I can offer you n^3 algorithms for both problems. 1. Do Choleski factorization of your symmetric matrix. If the factorization succeeds, you know it is positive definite. 2. Do Gaussian elimination (i.e. LU factorization). If it succeeds, it is not singular. But I am not an expert, so there could be better ways. === === Subject: : Re: time complexity for positive definiteness But I would like to know an algorithm which can work for 1) in best time for symmetric matrices 2) for non-symmetric matrices please enlighten upon the non-symmetric matrices as well dilip > For a non-symmetric square matrix of size n x n. > What is the time complexity of finding out whether the matrix is > positive definite or not. > Also, which would be the fastest, in terms of time complexity, to > compute the same. > I want to know the answer for finding singularity of a matrix too. > Which would be the fastest in terms of the time complexity of n and how > fast will it be? > Dilip > I can offer you n^3 algorithms for both problems. 1. Do Choleski factorization of your symmetric matrix. If the > factorization succeeds, you know it is positive definite. 2. Do Gaussian elimination (i.e. LU factorization). If it succeeds, it > is not singular. But I am not an expert, so there could be better ways. === === Subject: : Re: time complexity for positive definiteness But I would like to know an algorithm which can work for > 1) in best time for symmetric matrices > 2) for non-symmetric matrices please enlighten upon the non-symmetric matrices as well > dilip Positive definiteness doesn't really make much sense for non-symmetric matrices. But if it is non-symmetric, symmetrize it. That is a short n^2 operation. As for Gaussian elimination, I think you could do it about twice as fast for symmetric matrices (by using a mild variant of Choleski factorization). === === Subject: : Re: inner product space question > Final exams coming up and I found this question on an old exam (no > solutions posted though): Let V be an innner product space, and let W be a finite-dimensional > subspace of V. If x is not /in W, prove that there exists a y /in W > perp (space orthogonal to W) that is not orthogonal to x. This is intuitively clear, but I can't get a (decent looking) proof. This is what I've tried so far - < * , * > denotes the inner product: By a prev. theorem, we know x decomposes uniquely into x = u + z > such that u is /in W and z is /in W perp. So pick y = z. < x, y > = < u+z, z > = + = < z , z > (since u & y > orthogonal) = ||z||^2 Now I have to show that z <> 0 > By contradiction: assume z = 0. Then x = u + z = u > Since u is /in W, x=u also /in W. But this contradicts hypothesis. > Thus z <> 0. Thus = ||z||^2 > 0 which means x and y are not orthogonal. > For some funny reason, I think I've complicated the issue or made some > error somewhere. > Can anyone point out any mistakes? What does this need besides the definition of W^perp? Suppose Ez Ay in W^perp = 0. Then z in W. -- Michael Press === === Subject: : Re: inner product space question On 2006-12-03 01:35:11 -0500, Michael Press said: Suppose Ez Ay in W^perp = 0. Then z in W. By definition, this shows z is in W^perp^perp. We always know W is contained in W^perp^perp. But how did you conclude that W=W^perp^perp? -- -kira === === Subject: : Re: inner product space question > On 2006-12-03 01:35:11 -0500, Michael Press said: > Suppose Ez Ay in W^perp = 0. Then z in W. By definition, this shows z is in W^perp^perp. We always know W is > contained in W^perp^perp. But how did you conclude that W=W^perp^perp? I am trying to prove Let V be an innner product space, and let W be a finite-dimensional subspace of V. If x is not /in W, prove that there exists a y /in W perp (space orthogonal to W) that is not orthogonal to x. I proved that given z if for all y in W^perp = 0, then z in W. The contrapositive is if x not in W then not (for all y in W^perp = 0) ==> Exists y in W^perp <> 0. -- Michael Press === === Subject: : Re: inner product space question On 2006-12-02 21:29:29 -0500, Paul Sperry said: > >> Final exams coming up and I found this question on an old exam (no >> solutions posted though): >> >> Let V be an innner product space, and let W be a finite-dimensional >> subspace of V. If x is not /in W, prove that there exists a y /in W >> perp (space orthogonal to W) that is not orthogonal to x. (W perp)perp = ? [...] I recall myself doing this very same exercise from the textbook Linear Algebra 4th ed. by Friedberg. The result of this problem is then used to solve a later exercise that asks what you hinted: show that W^perp^perp = W. And the original poster's proof looks just fine to me. -- -kira === === Subject: : the error in Godels proof Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. TP ( x <-> ~Eproof(x) ) = NONSENSE ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The theorem prover is trying to tell us that a statement that asserts its own unprovability is not a valid formula. In godels proof it is 1st shown that there is no proof of x. Then it is noticed that x asserts exactly that, hence the dubious conclusion that x is true. But asserting x is TRUE is itself an action that relies on a proof of x, so its really nonsense! The conventional argument follows up on the paradox that x is proven in some other system. This is a consistent arrangement but is not necessary as G is not strictly true. Herc === === Subject: : Re: the error in Godels proof > [Junk, as usual]. Whoa! The jerk is back, and he still has it. Keep it coming, lad! === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ In what sense (if any) is your TP a theorem-prover? An ACTUAL theorem prover DOESN'T output true or false. Theorems BY DEFINITION are NOT merely true or false: they are PROVEN. Because of Godel's COMpleteness theorem (as OPPOSED the Incompleteness one that cranks like you obsess about), we know that theorems that can be proved from axioms will come up true IN ALL MODELS of the axioms. But the point is that under the standard classical first- order paradigm, NO statement, BY ITSELF, IS EVER merely true or false: MODELS make a statement true or false. Statements that are PROVEN THEOREMS are true in ALL models-of-the-axioms-they-were-proven-from, NOT in just one or some. Denials of theorems are false in ALL Models, not just one or some. Statements that are true in some models of the axioms and false in others are simply not DECIDED by those axioms. So IF your TP is ACTUALLY a theorem-prover, then its outputs NEED to be Proved, Refuted, and Undecided. And after you correctly figure out what you are TRYING to do, you will STILL have the problem that you can't write a TM or any other finitary algorithm that ACTUALLY DOES this correctly. === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ In what sense (if any) is your TP a theorem-prover? > An ACTUAL theorem prover DOESN'T output > true or false. Theorems BY DEFINITION are NOT > merely true or false: they are PROVEN. In the context of a Universal Theorem Prover, true and proven are synonymous. It could output, PROVEN, DISPROVEN but it means the same as TRUE, FALSE, coming from a theorem prover. > Because of Godel's COMpleteness theorem (as OPPOSED > the Incompleteness one that cranks like you obsess about), > we know that theorems that can be proved from axioms > will come up true IN ALL MODELS of the axioms. > But the point is that under the standard classical first- > order paradigm, NO statement, BY ITSELF, IS EVER > merely true or false: MODELS make a statement true or > false. Statements that are PROVEN THEOREMS are true > in ALL models-of-the-axioms-they-were-proven-from, > NOT in just one or some. Denials of > theorems are false in ALL Models, not just one or some. > Statements that are true in some models of the axioms > and false in others are simply not DECIDED by those axioms. Models were introduced to fit Godels proof, by allowing G statements to be true. There's no need to use models with a Universal Theorem Prover. Just say G is NONSENSE then all true formula will be encapsulated by logic. So IF your TP is ACTUALLY a theorem-prover, then its > outputs NEED to be Proved, Refuted, and Undecided. > And after you correctly figure out what you are TRYING to > do, you will STILL have the problem that you can't write a TM > or any other finitary algorithm that ACTUALLY DOES this correctly. > ...because this has no proof defeats the system?? get real. Herc === === Subject: : Re: the error in Godels proof <4573c915$0$97217$892e7fe2@authen.yellow.readfreenews.net In the context of a Universal Theorem Prover, true and proven are > synonymous. It could output, PROVEN, DISPROVEN but it > means the same as TRUE, FALSE, coming from a theorem prover. You should not talk about a universal theorem-prover. Theorems follow from axioms. Every collection of axioms is different. If you are going to make it universal then you have to provide the axioms as an input, separate from the theorem. Or, you could conjoin the axioms and the DENIAL of the theorem into one grand conjunction and try to derive a contradiction from it (that is usually called a refutational theorem-prover). You can confirm, algorithmically, that a 1st-order conjunction entails a contradiction. But you can't always confirm that it doesn't. > But the point is that under the standard classical first- > order paradigm, NO statement, BY ITSELF, IS EVER > merely true or false: MODELS make a statement true or > false. > Models were introduced to fit Godels proof, No, they weren't. Models just are what they are. Deal with it. > There's no need to use models with a Universal Theorem Prover. That doesn't change the fact that the models exist and that they assign truth-values to the statements -- something that the theorem- prover, ESPECIALLY if it is universal, does NOT do. > Just say G is NONSENSE then all > true formula will be encapsulated > by logic. That really is not possible. The unprovable statement that you are dismissing as nonsense IS TRUE. So if you have called it nonsense, YOU ARE JUST WRONG. I am not provable as a theorem is the natural-language translation of that sentence, but BEFORE it was translated, it was JUST a statement ABOUT NUMBERS, with addition and multiplication. IT HAS a truth-value. In the standard model, anyway (where all the things in the model are natural numbers). The reason why the theorem-prover cannot output true for it, EVEN THOUGH IT IS TRUE, is because it is NOT a provable theorem. Truth and provability ARE DIFFERENT. The completeness theorem implies that the reason why this true unprovability statement is not provable is that its truth-value is DIFFERENT in OTHER (non-standard) models of the axioms. So you cannot say that models were invented for Godel's proof. It has ALWAYS been the truth about the axioms of Peano Arithmetic that they have DIFFERENT models, and that some sentences that are true in one model are not true in others. > And after you correctly figure out what you are TRYING to > do, you will STILL have the problem that you can't write a TM > or any other finitary algorithm that ACTUALLY DOES this correctly. > ...because this has no proof defeats the system?? No, NOT because of that. But rather because SOMEthing must ALWAYS defeat the system. It would be easy to make the system UNdefeated by this has no proof from PA. You could just ADD this as an axiom. Then it would be very trivially provable. But then there would be another truth that wasn't provable from the newer bigger system. There will ALWAYS be some. The fact that you want to dismiss them all as nonsense is beside the point: they are all just numerical facts about addition and multiplication, so they CAN'T be nonsense. The other (Main) point is that YOU CAN'T TELL whether a given statement is or is not EQUIVALENT to nonsense. We don't === === Subject: : Re: the error in Godels proof [...] > I am not provable as a theorem is the natural-language translation > of that > sentence, but BEFORE it was translated, it was JUST a statement > ABOUT NUMBERS, with addition and multiplication. IT HAS > a truth-value. In the standard model, anyway (where all the things > in the model are natural numbers). The reason why the theorem-prover > cannot output true for it, EVEN THOUGH IT IS TRUE, is because it is > NOT a provable theorem. Truth and provability ARE DIFFERENT. The completeness theorem implies that the reason why this true > unprovability statement is not provable is that its truth-value is > DIFFERENT False? > in OTHER (non-standard) models of the axioms. What I don't get is this: Either there is a finite sequence of formulae that proves G_T, or there isn't, and if the theory T is consistent, then there isn't. How does interpreting the formulae of T in a non-standard way change the syntactic fact of the matter? How can G_T be false in some non-standard model of the formulae of T? -- hz === === Subject: : Re: the error in Godels proof herbzet says... >What I don't get is this: Either there is a finite sequence of formulae that proves G_T, or >there isn't, and if the theory T is consistent, then there isn't. >How does interpreting the formulae of T in a non-standard way change >the syntactic fact of the matter? How can G_T be false in some >non-standard model of the formulae of T? A proof consists of a sequence of formulas such that each formula is either an axiom, a logical tautology, or follows from previous formulas. If you allow *infinite* sequences, then it's possible for *everything* to be provable. To prove a formula B, just construct the sequence B or not B B or not B B or not B ... B B B B This is meant to suggest an infinite (nonstandard, actually) sequence with the following characteristics: (Let F_n be the nth formula in the sequence). If n is a standard natural, then F_n = the tautology B or not B. If n is a nonstandard natural, then F_n = B. Since there is no smallest nonstandard natural, if n is nonstandard, then formula F_n follows from formula F_{n-1}. So if you allow nonstandard naturals, and correspondingly allow nonstandard sequences of formulas, then nonstandard proofs become possible, as well. -- Daryl McCullough Ithaca, NY === === Subject: : Re: the error in Godels proof herbzet says... > >What I don't get is this: >Either there is a finite sequence of formulae that proves G_T, or >there isn't, and if the theory T is consistent, then there isn't. >How does interpreting the formulae of T in a non-standard way change >the syntactic fact of the matter? How can G_T be false in some >non-standard model of the formulae of T? A proof consists of a sequence of formulas such that > each formula is either an axiom, a logical tautology, > or follows from previous formulas. If you allow *infinite* > sequences, then it's possible for *everything* to be > provable. To prove a formula B, just construct the sequence B or not B > B or not B > B or not B > ... > B > B > B > B This is meant to suggest an infinite (nonstandard, actually) > sequence with the following characteristics: (Let F_n be the > nth formula in the sequence). If n is a standard natural, then F_n = the tautology B or not B. > If n is a nonstandard natural, then F_n = B. Since there is no smallest nonstandard natural, if n is nonstandard, > then formula F_n follows from formula F_{n-1}. So if you allow nonstandard naturals, and correspondingly allow > nonstandard sequences of formulas, then nonstandard proofs become > possible, as well. -- > Daryl McCullough > Ithaca, NY just tickles me. -- hz === === Subject: : Re: the error in Godels proof herbzet a .8ecrit : > [...] > >> I am not provable as a theorem is the natural-language translation >> of that >> sentence, but BEFORE it was translated, it was JUST a statement >> ABOUT NUMBERS, with addition and multiplication. IT HAS >> a truth-value. In the standard model, anyway (where all the things >> in the model are natural numbers). The reason why the theorem-prover >> cannot output true for it, EVEN THOUGH IT IS TRUE, is because it is >> NOT a provable theorem. Truth and provability ARE DIFFERENT. >> The completeness theorem implies that the reason why this true >> unprovability statement is not provable is that its truth-value is >> DIFFERENT False? > >> in OTHER (non-standard) models of the axioms. What I don't get is this: Either there is a finite sequence of formulae that proves G_T, or > there isn't, and if the theory T is consistent, then there isn't. Right > How does interpreting the formulae of T in a non-standard way change > the syntactic fact of the matter? It doesn't. A model never speaks about provability, but about truth. How can G_T be false in some > non-standard model of the formulae of T? Compare with non-euclidean geometry. Te euclid axiom (the parallel postulate) is false in the model of the Poincare plane, true in R^2, and not provable *precisely* because if it was (in a consistent theory, of course) it would be true in all models. > -- > hz === === Subject: : Re: the error in Godels proof herbzet a Ì©crit : [...] > What I don't get is this: > Either there is a finite sequence of formulae that proves G_T, or > there isn't, and if the theory T is consistent, then there isn't. Right > How does interpreting the formulae of T in a non-standard way change > the syntactic fact of the matter? It doesn't. A model never speaks about provability, but about truth. How can G_T be false in some > non-standard model of the formulae of T? Compare with non-euclidean geometry. The euclid axiom (the parallel > postulate) is false in the model of the Poincare plane, true in R^2, and > not provable *precisely* because if it was (in a consistent theory, of > course) it would be true in all models. no real grip on what a non-standard model of arithmetic would look like. -- hz === === Subject: : Re: the error in Godels proof > >> herbzet a Ì©crit : [...] > > What I don't get is this: >> Either there is a finite sequence of formulae that proves G_T, or > there isn't, and if the theory T is consistent, then there isn't. >> Right > How does interpreting the formulae of T in a non-standard way change > the syntactic fact of the matter? >> It doesn't. A model never speaks about provability, but about truth. > How can G_T be false in some > non-standard model of the formulae of T? >> Compare with non-euclidean geometry. The euclid axiom (the parallel >> postulate) is false in the model of the Poincare plane, true in R^2, and >> not provable *precisely* because if it was (in a consistent theory, of >> course) it would be true in all models. no real grip on what a non-standard model of arithmetic would look like. I might add, given that GC's (Goldbach Conjecture's) truth has been, and still is, in limbo for more than 200 years, we don't seem to have a real grip on what the (so-called) standard model of arithmetic would look like (in its entirety). -- > hz -- ----------------------------------------------------- What we call 'I' is just a swinging door which moves when we inhale and exhale. Shunryu Suzuki ---------------------------------------------------- === === Subject: : Re: the error in Godels proof > Either there is a finite sequence of formulae that proves G_T, or > there isn't, I mean, of course, a finite sequence of formulae in T. -- hz === === Subject: : Re: the error in Godels proof > Just say G is NONSENSE then all > true formula will be encapsulated > by logic. That really is not possible. The unprovable statement > that you are dismissing as nonsense IS TRUE. > So if you have called it nonsense, YOU ARE JUST WRONG. No, it looks wrong but its quite consistant in a single universal system. this has no proof cannot be false and cannot be true according to the axioms. Now you'd like to think that it really has no proof then doesn't it?? Making it TRUE? But that's the trap you have all fallen into, that's just one of the cyclic deductions that make the statement nonsense, it has to be nonsense as it refutes its own provability. We class the statement in the same class as this is false. > I am not provable as a theorem is the natural-language translation of > that > sentence, but BEFORE it was translated, it was JUST a statement > ABOUT NUMBERS, with addition and multiplication. IT HAS > a truth-value. In the standard model, anyway (where all the things > in the model are natural numbers). The reason why the theorem-prover > cannot output true for it, EVEN THOUGH IT IS TRUE, is because it is > NOT a provable theorem. Truth and provability ARE DIFFERENT. The completeness theorem implies that the reason why this true > unprovability statement is not provable is that its truth-value is > DIFFERENT > in OTHER (non-standard) models of the axioms. So you cannot say that > models were invented for Godel's proof. It has ALWAYS been the > truth > about the axioms of Peano Arithmetic that they have DIFFERENT models, > and that some sentences that are true in one model are not true in > others. > And after you correctly figure out what you are TRYING to > do, you will STILL have the problem that you can't write a TM > or any other finitary algorithm that ACTUALLY DOES this correctly. > ...because this has no proof defeats the system?? No, NOT because of that. But rather because SOMEthing must ALWAYS > defeat the system. It would be easy to make the system UNdefeated > by this has no proof from PA. You could just ADD this as an > axiom. > Then it would be very trivially provable. But then there would be > another truth > that wasn't provable from the newer bigger system. There will ALWAYS > be > some. The fact that you want to dismiss them all as nonsense is beside > the point: they are all just numerical facts about addition and > multiplication, > so they CAN'T be nonsense. The other (Main) point is that YOU CAN'T > TELL > whether a given statement is or is not EQUIVALENT to nonsense. We > don't > You mean, Yes because of that. These self denying statementS are part of knowledge and have to be classified correctly. Herc === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all > of mathematics. > for every input to the TP, it outputs TRUE, > FALSE, or NONSENSE. > when it inputs Godel's statement, the output is > NONSENSE. > TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > In what sense (if any) is your TP a > theorem-prover? > An ACTUAL theorem prover DOESN'T output > true or false. Theorems BY DEFINITION are NOT > merely true or false: they are PROVEN. In the context of a Universal Theorem Prover, true > and proven are > synonymous. It could output, PROVEN, DISPROVEN but > it > means the same as TRUE, FALSE, coming from a theorem > prover. Because of Godel's COMpleteness theorem (as OPPOSED > the Incompleteness one that cranks like you obsess > about), > we know that theorems that can be proved from > axioms > will come up true IN ALL MODELS of the axioms. > But the point is that under the standard classical > first- > order paradigm, NO statement, BY ITSELF, IS EVER > merely true or false: MODELS make a statement true > or > false. Statements that are PROVEN THEOREMS are > true > in ALL models-of-the-axioms-they-were-proven-from, > NOT in just one or some. Denials of > theorems are false in ALL Models, not just one or > some. > Statements that are true in some models of the > axioms > and false in others are simply not DECIDED by those > axioms. Models were introduced to fit Godels proof, by > allowing G statements > to be true. There's no need to use models with a > Universal Theorem > Prover. Just say G is NONSENSE then all true formula > will be encapsulated > by logic. > So IF your TP is ACTUALLY a theorem-prover, then > its > outputs NEED to be Proved, Refuted, and > Undecided. > And after you correctly figure out what you are > TRYING to > do, you will STILL have the problem that you can't > write a TM > or any other finitary algorithm that ACTUALLY DOES > this correctly. > > ...because this has no proof defeats the system?? > get real. Herc > I think you miss the point. There is no technical equivalence, where proofs are concerned, between true-false-nonsense and proven-unproven-undecided. The first set of conclusions (yours) assumes application of the logic of excluded middle and is strictly two- valued. The second set is three-valued, allowing future proof techniques to admit a constructive proof not in your hypothetical theorem-prover's canon of knowledge. Fact is, there is no class of conjecture that can bear the label nonsense. A nonsensical statement is not subject to proof method. Your idea of a universal theorem prover seems merely to be another version of the hoped-for formalist program (Russell, Hilbert Frege, et al)in which all mathematics could be formulaicly packaged. That program failed long ago. Why? One word: Godel. Tom === === Subject: : Re: the error in Godels proof |-|erc kirjoitti: > Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! The conventional argument follows up on the paradox that x is proven > in some other system. This is a consistent arrangement but is not necessary > as G is not strictly true. Herc All of the mathematics is not recursevily axiomatible, because we can always diagonalize by G?dels`s way a theory which contains basic truths about arihmetic. So your theorem prover doens`t exist. === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. > Well, tell us more about this theorem prover. However you go about defining your theorem prover, this contention of yours will be a philosophical one. Mathematically, it will leave Goedel's proof untouched. === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. Well, tell us more about this theorem prover. However you go about defining your theorem prover, this contention of > yours will be a philosophical one. Mathematically, it will leave > Goedel's proof untouched. > I could say the same about you, that Godels proof achieves nothing, nothing can be calculated or anything useful come about of it, so it is also philosophical. Godels proof is the belief that this has no proof is true and the accompanying ideas that this petty utterence will forever defeat a total formal mathematics. You have it the wrong way around, moving TOWARDS a solution is mathematical. Herc === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all > of mathematics. > for every input to the TP, it outputs TRUE, > FALSE, or NONSENSE. > when it inputs Godel's statement, the output is > NONSENSE. > Well, tell us more about this theorem prover. > However you go about defining your theorem prover, > this contention of > yours will be a philosophical one. Mathematically, > it will leave > Goedel's proof untouched. > > I could say the same about you, that Godels proof > achieves nothing, nothing > can be calculated or anything useful come about of > it, so it is also philosophical. Godels proof is the belief that this has no proof > is true and the accompanying > ideas that this petty utterence will forever defeat a > total formal mathematics. You have it the wrong way around, moving TOWARDS a > solution is mathematical. Herc Moving mathematics forward is exactly why Godel's theorem is significant, and why your philosophy is not significant to mathematics. Mathematics is a language. Mathematics is not identical to calculation, as you would have it. As a first order description of some activity, a mathematical result consistent with some set of assumptions cannot explain itself. That explanation has to be bracketed by a higher (meta) set of assumptions. Mathematicians doing everyday work of calculating results by standards of reducing problems to those previously solved, whether with pencil and paper or in your vision of automated theorem proving, do not have to be worried about the antinomies with which Godel (and Russell before) concerned himself. Researchers breaking new ground in correlating theory with result, however, have to know that they are not creating contradictions in the process of proving theorems. If mathematics is about anything, it is about truth, and in the making of true statements, a theorem offers its complete meaning, no part of which may oppose the meaning of any other part. This, then, briefly, is the usefulness of Godel's theorem -- the knowledge of the existence of a systematized means to demarcate known results from new definitions. An example of this phenomenon might be the rule which disallows division by zero in arithmetic. So now that we have identified the importance of Godel's Theorem with your own standard of usefulness, what is your argument left with? The output I read is: NONSENSE. Tom === === Subject: : Re: the error in Godels proof <4572925c$0$97260$892e7fe2@authen.yellow.readfreenews.net > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > Well, tell us more about this theorem prover. > However you go about defining your theorem prover, this contention of > yours will be a philosophical one. Mathematically, it will leave > Goedel's proof untouched. I could say the same about you, that Godels proof achieves nothing, nothing > can be calculated or anything useful come about of it, so it is also philosophical. > Well, it's a proof. If you accept the axioms and rules of inference which it uses, then you must accept that the result it states is true. That's what it achieves. Tell me more about this theorem prover of yours. What is the language in which the sentences which are fed to it as input are? For which sentences does it output TRUE? For which FALSE? For which NONSENSE? > Godels proof is the belief that this has no proof is true and the accompanying > ideas that this petty utterence will forever defeat a total formal mathematics. You have it the wrong way around, moving TOWARDS a solution is mathematical. Herc === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > Well, tell us more about this theorem prover. > However you go about defining your theorem prover, this contention of > yours will be a philosophical one. Mathematically, it will leave > Goedel's proof untouched. > I could say the same about you, that Godels proof achieves nothing, nothing > can be calculated or anything useful come about of it, so it is also philosophical. Well, it's a proof. If you accept the axioms and rules of inference > which it uses, then you must accept that the result it states is true. > That's what it achieves. Well if you've proven something then apply that knowledge, what can you use this fact to do? Herc === === Subject: : Re: the error in Godels proof <4572925c$0$97260$892e7fe2@authen.yellow.readfreenews.net> <45729f72$0$97276$892e7fe2@authen.yellow.readfreenews.net > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > Well, tell us more about this theorem prover. > However you go about defining your theorem prover, this contention of > yours will be a philosophical one. Mathematically, it will leave > Goedel's proof untouched. > I could say the same about you, that Godels proof achieves nothing, nothing > can be calculated or anything useful come about of it, so it is also philosophical. > Well, it's a proof. If you accept the axioms and rules of inference > which it uses, then you must accept that the result it states is true. > That's what it achieves. Well if you've proven something then apply that knowledge, what can > you use this fact to do? > It depends on the application. Can you give a specific example? > Herc === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! No, it isn't. The proof that x has no proof (in a particular consistent system) is as follows: Assume that x is false. Then there _is_ a proof of x. But then we have a proof of a false result. Contradiction, so x must be true. Since x is true, what it states is truth, and there is in fact no proof of x. You're putting the horse before the cart. --- Christopher Heckman === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! No, it isn't. The proof that x has no proof (in a particular consistent system) is as > follows: Assume that x is false. Then there _is_ a proof of x. But then > we have a proof of a false result. Contradiction, so x must be true. You have just proven x. > Since x is true, what it states is truth, and there is in fact no proof > of x. You're putting the horse before the cart. --- Christopher Heckman === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! > No, it isn't. > The proof that x has no proof (in a particular consistent system) is as > follows: Assume that x is false. Then there _is_ a proof of x. But then > we have a proof of a false result. Contradiction, so x must be true. You have just proven x. But not using that system of axioms. What I didn't state (which I should really have) is that when I talk about whether x is provable, it was in the context of that system S. A more thought-out argument is: > The proof that x has no proof (in a particular consistent system S) is as > follows: Assume that x is false. Then there _is_ a proof of x in S. But then > we have a proof of a false result in S. Contradiction, so x must be true ... and there is no proof of x in S. |-|erc seems to be saying that x could also have the value of NONSENSE, but this is disproven while setting up the Godel numbering of statements in S. --- Christopher Heckman > Since x is true, what it states is truth, and there is in fact no proof > of x. > You're putting the horse before the cart. === === Subject: : Re: the error in Godels proof > |-|erc seems to be saying that x could also have the value of NONSENSE, > but this is disproven while setting up the Godel numbering of > statements in S. > why is this? lots of godel numbers will be NONSENSE as every sequence of symbols has a godel number. Herc === === Subject: : Re: the error in Godels proof <4573c5ec$0$97218$892e7fe2@authen.yellow.readfreenews.net > |-|erc seems to be saying that x could also have the value of NONSENSE, > but this is disproven while setting up the Godel numbering of > statements in S. why is this? lots of godel numbers will be NONSENSE as every sequence > of symbols has a godel number. True, but it can be determined which numbers are nonsense. Read _Godel, Escher, Bach_ by Douglas Hofstadter in order to see how to do Godel numbering. A more thorough treatment is given in the book _Computability and Logic_, by Boolos and Jeffrey. --- Christopher Heckman === === Subject: : Re: the error in Godels proof > The proof that x has no proof (in a particular consistent system) is as > follows: Assume that x is false. Then there _is_ a proof of x. But then > we have a proof of a false result. Contradiction, so x must be true. You have just proven x. exactly! > Since x is true, what it states is truth, and there is in fact no proof > of x. contradiction also. x cannot be true or false. Herc === === Subject: : Re: the error in Godels proof contradiction also. x cannot be true or false. The well known Goedel sentence is true, but unprovable. Which is exactly the point. Bob Kolker === === Subject: : Re: the error in Godels proof <45736e31$0$97234$892e7fe2@authen.yellow.readfreenews.net> <4th9fuF13sp4pU1@mid.individual.net> Bob Kolker kirjoitti: > contradiction also. x cannot be true or false. The well known Goedel sentence is true, but unprovable. Which is exactly > the point. There is no THE goedel`s sentence. Every theory containg enoug arithmetic has it`s own. > Bob Kolker === === Subject: : Re: the error in Godels proof > The well known Goedel sentence is true, but unprovable. What sentence is that? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: the error in Godels proof <45736e31$0$97234$892e7fe2@authen.yellow.readfreenews.net> <4th9fuF13sp4pU1@mid.individual.net> The well known Goedel sentence is true, but unprovable. What sentence is that? > You are just as aware as Mr. Kolker of the simple mathematical fact that we can prove mathematically the existence of true sentences in the language of PA which are not provable in PA. Any of those can be used as Mr. Kolker's Goedel sentence. Why not suggest one yourself, rather than pretend that you've never heard of such an idea before? > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: the error in Godels proof > You are just as aware as Mr. Kolker of the simple mathematical fact > that we can prove mathematically the existence of true sentences in the > language of PA which are not provable in PA. Any of those can be used > as Mr. Kolker's Goedel sentence. So the famed well known G.9adel sentence is a random arithmetical truth unprovable in PA? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: the error in Godels proof >> The well known Goedel sentence is true, but unprovable. What sentence is that? It has been suppressed by the evil Cantorians. Lee Rudolph === === Subject: : Re: the error in Godels proof <45736e31$0$97234$892e7fe2@authen.yellow.readfreenews.net> <4th9fuF13sp4pU1@mid.individual.net> > The well known Goedel sentence is true, but unprovable. >What sentence is that? It has been suppressed by the evil Cantorians. > Lee Rudolph Tell me; do you see all those who use the term 'Goedel sentence' as True Believers in a Cantorian Conspiracy? That sounds like a rather ideological approach to the study of mathematics. === === Subject: : Re: the error in Godels proof > The well known Goedel sentence is true, but unprovable. >>What sentence is that? >> It has been suppressed by the evil Cantorians. >> Lee Rudolph Tell me; do you see all those who use the term 'Goedel sentence' as >True Believers in a Cantorian Conspiracy? Tell *me*: do you see all True Believers as Provable Believers? >That sounds like a rather >ideological approach to the study of mathematics. Evil is not a term from ideological analysis, it is a term from moral analysis. Lee Rudolph === === Subject: : Re: the error in Godels proof > It has been suppressed by the evil Cantorians. They're evil alright, sneaky little scoundrels! -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: the error in Godels proof > The proof that x has no proof (in a particular consistent system) is as > follows: Assume that x is false. Then there _is_ a proof of x. But then > we have a proof of a false result. Contradiction, so x must be true. > Since x is true, what it states is truth, and there is in fact no proof > of x. > And here your error is obvious. Where you write Contradiction, so x must be true IS the proving of x, therefore x has a proof, or is nonsense. Infact your final sentence, so x MUST BE TRUE.... and there is in fact no proof of x is a contradiction in terms. Herc === === Subject: : Re: the error in Godels proof >> The proof that x has no proof (in a particular consistent system) is as >> follows: Assume that x is false. Then there _is_ a proof of x. But then >> we have a proof of a false result. Contradiction, so x must be true. >> Since x is true, what it states is truth, and there is in fact no proof >> of x. And here your error is obvious. Where you write > Contradiction, so x must be true IS the proving of x, > therefore x has a proof, or is nonsense. Infact your final sentence, > so x MUST BE TRUE.... and there is in fact no proof of x is a contradiction in terms. There is a disambiguation of proof that could be helpful. There are proofs that can be done in first order Peano arithmetic, a formal system; and then there are just plain proofs, or logical arguments. Every Peano arithmetic proof can be turned into a plain proof. But not necessarily vice versa. What Proginoskes could have inserted between and there is in fact no proof of x and '.' is: (in a particular consistent system). Also, the proof that x has no proof (in a particular consistent system) is not a proof (in a particular consistent system). It is a plain proof. David Bernier === === Subject: : Re: the error in Godels proof <457290fe$0$97244$892e7fe2@authen.yellow.readfreenews.net> > The proof that x has no proof (in a particular consistent system) is as >> follows: Assume that x is false. Then there _is_ a proof of x. But then >> we have a proof of a false result. Contradiction, so x must be true. >> Since x is true, what it states is truth, and there is in fact no proof >> of x. > And here your error is obvious. Where you write > Contradiction, so x must be true IS the proving of x, > therefore x has a proof, or is nonsense. > Infact your final sentence, > so x MUST BE TRUE.... and there is in fact no proof of x is a contradiction in terms. > There is a disambiguation of proof that could be helpful. > There are proofs that can be done in first order Peano arithmetic, > a formal system; and then there are just plain proofs, or > logical arguments. > What you call plain proofs can really be formalized in ZF, since proof theory itself can be interpreted in ZF. |-|erc needs to understand that the Godel statement does not encode This statement is unprovable, but rather This statement is unprovable in PA, or whatever other sufficiently strong arithmetic you want. It can still be proved in other systems, and ZF is generally accepted as a good formalization of mathematics. > Every Peano arithmetic proof can be turned into a plain proof. > But not necessarily vice versa. What Proginoskes could have inserted between > and there is in fact no proof of x and '.' is: > (in a particular consistent system). Also, the proof > that x has no proof (in a particular consistent system) is not > a proof (in a particular consistent system). It is a plain proof. David Bernier === === Subject: : Re: the error in Godels proof >> The proof that x has no proof (in a particular consistent system) is as >> follows: Assume that x is false. Then there _is_ a proof of x. But then >> we have a proof of a false result. Contradiction, so x must be true. >> Since x is true, what it states is truth, and there is in fact no proof >> of x. >> So you agree the above has a giant logical flaw as its written? Herc === === Subject: : Re: the error in Godels proof >> The proof that x has no proof (in a particular consistent system) is as >> follows: Assume that x is false. Then there _is_ a proof of x. But then >> we have a proof of a false result. Contradiction, so x must be true. >> Since x is true, what it states is truth, and there is in fact no proof >> of x. > So you agree the above has a giant logical flaw as its written? There is no logical flaw, but the details can be tricky. -- David Marcus === === Subject: : Re: the error in Godels proof <457290fe$0$97244$892e7fe2@authen.yellow.readfreenews.net> <45729fc0$0$97251$892e7fe2@authen.yellow.readfreenews.net >> The proof that x has no proof (in a particular consistent system) is as >> follows: Assume that x is false. Then there _is_ a proof of x. But then >> we have a proof of a false result. Contradiction, so x must be true. >> Since x is true, what it states is truth, and there is in fact no proof >> of x. So you agree the above has a giant logical flaw as its written? Herc You should ask Boucher, who has a much better theoretical refutation of Godel's damnation of you to inconsistency, or: incompleteness, and both. There is only one, theory, of everything. Ross === === Subject: : Re: the error in Godels proof > Suppose a theorem prover exists that mimics all of mathematics. for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. when it inputs Godel's statement, the output is NONSENSE. TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. The actual conclusion is: If the system is consistent, then there is no proof of x. > Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! > The conventional argument follows up on the paradox that x is proven > in some other system. This is a consistent arrangement but is not necessary > as G is not strictly true. Herc > -- David Marcus === === Subject: : Re: the error in Godels proof Suppose a theorem prover exists that mimics all of mathematics. > for every input to the TP, it outputs TRUE, FALSE, or NONSENSE. > when it inputs Godel's statement, the output is NONSENSE. > TP ( x <-> ~Eproof(x) ) = NONSENSE > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The theorem prover is trying to tell us that a statement that asserts its > own unprovability is not a valid formula. In godels proof it is 1st shown > that there is no proof of x. The actual conclusion is: If the system is consistent, then there is no > proof of x. So we have: If T is consistent then G_T is unprovable and G_T is true. Did I get it right? > Then it is noticed that x asserts exactly that, > hence the dubious conclusion that x is true. But asserting x is TRUE > is itself an action that relies on a proof of x, so its really nonsense! > The conventional argument follows up on the paradox that x is proven > in some other system. This is a consistent arrangement but is not necessary > as G is not strictly true. Herc > -- > David Marcus === === Subject: : Re: the error in Godels proof If T is consistent then G_T is unprovable and G_T is true. > Did I get it right? If T is a consistent, recursively axiomatized theory strong enough to express the axioms of, say, PA, then there is a sentence G in the language of T that is not a theorem of T but G is true in the standard model of PA. And usually(?) mathematicians take 'true in the standard model of PA' to be longwinded for 'true'. It is not claimed that there is a true G that is not provable in ANY consistent, recursively axiomatized theory strong enough to express the axioms of, say, PA, but rather it is shown that for any consistent, recursively axiomatized theory strong enough to express the axioms of, say, PA, there is a true G not provable in T. I.e., G may be a different for different T's. MoeBlee === === Subject: : Re: JSH: Ever talk to a mathematician? <4564df20$0$97270$892e7fe2@authen.yellow.readfreenews.net> > Well I have a B.Sc. in physics so I'm pondering why in the hell > polish matters when the result is so dramatic, and you know, sitting > here now I think that professor was just doing things by the rules of > his society. My polish means so much in an academic world where polish > is part of the rules, like the social rules that govern human behavior > in many other areas. > Polish? Are you playing the race card? > JSH does not have a degree, he only took some courses, he admitted that >> about 6 months ago. >> He does hate Polish people by his racist statement, and any other race that >> is good at math. > I am going to contact his employers and get him fired. Then I'm going > to contact Princeton. Then I'm going to go to Kinkos, make a giant > banner that says James Harris Hates Polish People and glue it outside > of his apartment. > Haven't we Polish people suffered enough!?!!? LOL! With a little p, polish is a word meaning a shiny surface, > (or related concepts) Yes, that was the joke. BTW, this is the only word in the English language whose pronunciation changes when you capitalize it. --- Christopher Heckman === === Subject: : Re: JSH: Ever talk to a mathematician? | Well I have a B.Sc. in physics so I'm pondering why in the hell | polish matters when the result is so dramatic, and you know, sitting | here now I think that professor was just doing things by the rules of | his society. My polish means so much in an academic world where polish | is part of the rules, like the social rules that govern human behavior | in many other areas. |>> Polish? Are you playing the race card? |> JSH does not have a degree, he only took some courses, he admitted that |> about 6 months ago. |> He does hate Polish people by his racist statement, and any other race that |> is good at math. |>> I am going to contact his employers and get him fired. Then I'm going |>> to contact Princeton. Then I'm going to go to Kinkos, make a giant |>> banner that says James Harris Hates Polish People and glue it outside |>> of his apartment. |>> |>> Haven't we Polish people suffered enough!?!!? |> LOL! |> |> With a little p, polish is a word meaning a shiny surface, |> (or related concepts) | Yes, that was the joke. | | BTW, this is the only word in the English language whose pronunciation | changes when you capitalize it. How about reading & Reading ? ____________________Gerard S. === === Subject: : Re: JSH: Ever talk to a mathematician? > | Well I have a B.Sc. in physics so I'm pondering why in the hell > | polish matters when the result is so dramatic, and you know, sitting > | here now I think that professor was just doing things by the rules of > | his society. My polish means so much in an academic world where polish > | is part of the rules, like the social rules that govern human behavior > | in many other areas. |>> Polish? Are you playing the race card? |> JSH does not have a degree, he only took some courses, he admitted that > |> about 6 months ago. > |> He does hate Polish people by his racist statement, and any other race > |> that > |> is good at math. |>> I am going to contact his employers and get him fired. Then I'm going > |>> to contact Princeton. Then I'm going to go to Kinkos, make a giant > |>> banner that says James Harris Hates Polish People and glue it outside > |>> of his apartment. > |> |>> Haven't we Polish people suffered enough!?!!? |> LOL! > | |> With a little p, polish is a word meaning a shiny surface, > |> (or related concepts) | Yes, that was the joke. > | > | BTW, this is the only word in the English language whose pronunciation > | changes when you capitalize it. How about reading & Reading ? ____________________Gerard S. > Alos job and Job. -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === === Subject: : Re: JSH: Ever talk to a mathematician? <4564df20$0$97270$892e7fe2@authen.yellow.readfreenews.net> <_rudnTuEepLq4u_YnZ2dnUVZ_uWdnZ2d@onvoy.com> <031220061654108047%chenrich@monmouth.com > | Well I have a B.Sc. in physics so I'm pondering why in the hell > | polish matters when the result is so dramatic, and you know, sitting > | here now I think that professor was just doing things by the rules of > | his society. My polish means so much in an academic world where polish > | is part of the rules, like the social rules that govern human behavior > | in many other areas. > |>> Polish? Are you playing the race card? > |> JSH does not have a degree, he only took some courses, he admitted that > |> about 6 months ago. > |> He does hate Polish people by his racist statement, and any other race > |> that > |> is good at math. > |>> I am going to contact his employers and get him fired. Then I'm going > |>> to contact Princeton. Then I'm going to go to Kinkos, make a giant > |>> banner that says James Harris Hates Polish People and glue it outside > |>> of his apartment. > |> |>> Haven't we Polish people suffered enough!?!!? > |> LOL! > | > |> With a little p, polish is a word meaning a shiny surface, > |> (or related concepts) > | Yes, that was the joke. > | > | BTW, this is the only word in the English language whose pronunciation > | changes when you capitalize it. > How about reading & Reading ? ____________________Gerard S. > Alos job and Job. If we're going to delve into mythology, we may as well use Cthulhu, the only word that changes pronunciation when you look at it. (Clearly I need to amend my statement, which had taken on urban legend proportions. Alas, I don't remember what the original statement was; outlawing proper names clearly doesn't work.) --- Christopher Heckman === === Subject: : Re: JSH: Ever talk to a mathematician? <4564df20$0$97270$892e7fe2@authen.yellow.readfreenews.net> <_rudnTuEepLq4u_YnZ2dnUVZ_uWdnZ2d@onvoy.com | Well I have a B.Sc. in physics so I'm pondering why in the hell > | polish matters when the result is so dramatic, and you know, sitting > | here now I think that professor was just doing things by the rules of > | his society. My polish means so much in an academic world where polish > | is part of the rules, like the social rules that govern human behavior > | in many other areas. |>> Polish? Are you playing the race card? |> JSH does not have a degree, he only took some courses, he admitted that > |> about 6 months ago. > |> He does hate Polish people by his racist statement, and any other race that > |> is good at math. |>> I am going to contact his employers and get him fired. Then I'm going > |>> to contact Princeton. Then I'm going to go to Kinkos, make a giant > |>> banner that says James Harris Hates Polish People and glue it outside > |>> of his apartment. > |> |>> Haven't we Polish people suffered enough!?!!? |> LOL! > | |> With a little p, polish is a word meaning a shiny surface, > |> (or related concepts) | Yes, that was the joke. > | > | BTW, this is the only word in the English language whose pronunciation > | changes when you capitalize it. How about reading & Reading ? Only in England. --- Christopher Heckman === === Subject: : Re: JSH: Ever talk to a mathematician? |>| Well I have a B.Sc. in physics so I'm pondering why in the hell |>| polish matters when the result is so dramatic, and you know, sitting |>| here now I think that professor was just doing things by the rules of |>| his society. My polish means so much in an academic world where polish |>| is part of the rules, like the social rules that govern human behavior |>| in many other areas. |>|> |>> Polish? Are you playing the race card? |> |>|> JSH does not have a degree, he only took some courses, he admitted that |>|> about 6 months ago. |>|> He does hate Polish people by his racist statement, and any other race that |>|> is good at math. |>|> |>> I am going to contact his employers and get him fired. Then I'm going |>|>> to contact Princeton. Then I'm going to go to Kinkos, make a giant |>|>> banner that says James Harris Hates Polish People and glue it outside |>|>> of his apartment. |>|>> |>|>> Haven't we Polish people suffered enough!?!!? |>|> |> LOL! |>|> |>|> With a little p, polish is a word meaning a shiny surface, |>|> (or related concepts) |>|> | Yes, that was the joke. |>| |>| BTW, this is the only word in the English language whose pronunciation |>| changes when you capitalize it. |> How about reading & Reading ? | Only in England. --- and in America (whoever has played Monopoly: take a ride on the Reading Railroad. And also residents of Pennsylvania and Massachusetts, and I'm sure, other states as well. _______________________________Gerard S. === === Subject: : Re: JSH: Ever talk to a mathematician? <4564df20$0$97270$892e7fe2@authen.yellow.readfreenews.net> <_rudnTuEepLq4u_YnZ2dnUVZ_uWdnZ2d@onvoy.com> | Well I have a B.Sc. in physics so I'm pondering why in the hell > |>| polish matters when the result is so dramatic, and you know, sitting > |>| here now I think that professor was just doing things by the rules of > |>| his society. My polish means so much in an academic world where polish > |>| is part of the rules, like the social rules that govern human behavior > |>| in many other areas. > |>|> |>> Polish? Are you playing the race card? > | |>|> JSH does not have a degree, he only took some courses, he admitted that > |>|> about 6 months ago. > |>|> He does hate Polish people by his racist statement, and any other race that > |>|> is good at math. > |>|> |>> I am going to contact his employers and get him fired. Then I'm going > |>|>> to contact Princeton. Then I'm going to go to Kinkos, make a giant > |>|>> banner that says James Harris Hates Polish People and glue it outside > |>|>> of his apartment. > |>|> |>|>> Haven't we Polish people suffered enough!?!!? > |>|> |> LOL! > |>| |>|> With a little p, polish is a word meaning a shiny surface, > |>|> (or related concepts) > |>|> | Yes, that was the joke. > |>| > |>| BTW, this is the only word in the English language whose pronunciation > |>| changes when you capitalize it. |> How about reading & Reading ? | Only in England. --- and in America (whoever has played Monopoly: > take a ride on the Reading Railroad. I've always pronounced it the same as reading. > And also residents of Pennsylvania and Massachusetts, and I'm > sure, other states as well. Yes, but only YANKEES! --- Christopher Heckman === === Subject: : Re: JSH: Ever talk to a mathematician? |>| Well I have a B.Sc. in physics so I'm > pondering why in the hell > |>| polish matters when the result is so > dramatic, and you know, sitting > |>| here now I think that professor was just > doing things by the rules of > |>| his society. My polish means so much in > an academic world where polish > |>| is part of the rules, like the social > rules that govern human behavior > |>| in many other areas. > |>|> |>> Polish? Are you playing the race card? > | > |>|> JSH does not have a degree, he only took > some courses, he admitted that > |>|> about 6 months ago. > |>|> He does hate Polish people by his racist > statement, and any other race that > |>|> is good at math. > |>|> |>> I am going to contact his employers and > get him fired. Then I'm going > |>|>> to contact Princeton. Then I'm going to go to > Kinkos, make a giant > |>|>> banner that says James Harris Hates Polish > People and glue it outside > |>|>> of his apartment. > |>|> |>|>> Haven't we Polish people suffered enough!?!!? > |>|> |> LOL! > |>| > |>|> With a little p, polish is a word meaning > a shiny surface, > |>|> (or related concepts) > |>|> | Yes, that was the joke. > |>| > |>| BTW, this is the only word in the English > language whose pronunciation > |>| changes when you capitalize it. > |> How about reading & Reading ? > | Only in England. > --- and in America (whoever has played Monopoly: > take a ride on the Reading Railroad. I've always pronounced it the same as reading. And also residents of Pennsylvania and > Massachusetts, and I'm > sure, other states as well. Yes, but only YANKEES! --- Christopher Heckman > If you're going to be proper, the railroad and the town are pronounced redding no matter where you live. At least one reason to capitalize is to be proper. Tom === === Subject: : Re: Coweb diagram in matlab I'm interested too ha scritto nel messaggio > Is there any suggestion about how to build a coweb diagram of a > function (describing a dynamical system) in matlab? A. > === === Subject: : Re: Parametric solutions of Euler Brick? <4tdi46F134ja8U1@mid.dfncis.de > Well? Comments? > I guess there's nothing obviously wrong (or someone would have said > so right away). So what *has* been proved? No Euler Brick if Proposition 7 > is true, but we still lack a proof of that? One direction is obvious (to me); if the variables have the special values, you end up with a perfect square. What I can't see is the reverse (only if), that a perfect square leads to one of the cases. Here is the statement of Proposition 7, for those of you who have forgotten it: Proposition 7: The expression (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) is a perfect square iff one of the following holds: (a) P^2 = n^2 m, (b) P^2 = n m^2, (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR (d) m = t, n = a^2, P = a^3 (where a and t are integers). --- Christopher Heckman === === Subject: : Re: Parametric solutions of Euler Brick? [Proginoskes] > ... > One direction is obvious (to me); if the variables have the special > values, you end up with a perfect square. What I can't see is the > reverse (only if), that a perfect square leads to one of the cases. Here is the statement of Proposition 7, for those of you who have > forgotten it: Proposition 7: The expression (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) is a perfect square iff one of the following holds: (a) P^2 = n^2 m, > (b) P^2 = n m^2, > (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR > (d) m = t, n = a^2, P = a^3 (where a and t are integers). I haven't followed this, so don't know whether I'm missing hidden assumptions. Looks like: m=15 n=20 P=60 makes the first expression a perfect square (104976000000000000 = 324000000^2) but doesn't satisfy any of #a through #d. It's easy to find others. It seems to get substantially harder if the conditions: P^4 > n^4 m^2 and P^4 > n^2 m^4 are added. Then, e.g., m=21 n=52 P=273 is the first I found with all of {m, n, P} > 0. === === Subject: : Re: Parametric solutions of Euler Brick? <4tdi46F134ja8U1@mid.dfncis.de> <5e-dnWeiMdbwHe_YnZ2dnUVZ_o6dnZ2d@comcast.com [Proginoskes] > ... > One direction is obvious (to me); if the variables have the special > values, you end up with a perfect square. What I can't see is the > reverse (only if), that a perfect square leads to one of the cases. > Here is the statement of Proposition 7, for those of you who have > forgotten it: > Proposition 7: The expression > (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) > is a perfect square iff one of the following holds: > (a) P^2 = n^2 m, > (b) P^2 = n m^2, > (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR > (d) m = t, n = a^2, P = a^3 (where a and t are integers). I haven't followed this, so don't know whether I'm missing hidden > assumptions. Looks like: m=15 n=20 P=60 makes the first expression a perfect square (104976000000000000 = > 324000000^2) but doesn't satisfy any of #a through #d. A quick look through the paper only suggests that P has to divide evenly into m*n*k. Your examples do this, even if k=1. I'll have to take a closer look then. (The existence of such a small counterexample suggests there's something strange going on.) --- Christopher Heckman > It's easy to find others. It seems to get substantially harder if the > conditions: P^4 > n^4 m^2 > and > P^4 > n^2 m^4 are added. Then, e.g., m=21 n=52 P=273 is the first I found with all of {m, n, P} > 0. === === Subject: : Re: Parametric solutions of Euler Brick? [Proginoskes] > ... > One direction is obvious (to me); if the variables have the special > values, you end up with a perfect square. What I can't see is the > reverse (only if), that a perfect square leads to one of the cases. >> Here is the statement of Proposition 7, for those of you who have > forgotten it: >> Proposition 7: The expression >> (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) >> is a perfect square iff one of the following holds: >> (a) P^2 = n^2 m, > (b) P^2 = n m^2, > (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR > (d) m = t, n = a^2, P = a^3 (where a and t are integers). [Tim Peters] >> I haven't followed this, so don't know whether I'm missing hidden >> assumptions. Looks like: >> m=15 n=20 P=60 >> makes the first expression a perfect square (104976000000000000 = >> 324000000^2) but doesn't satisfy any of #a through #d. [Proginoskes] > A quick look through the paper only suggests that P has to divide > evenly into m*n*k. Your examples do this, even if k=1. Not sure I follow. If there's no restriction on k, then it's trivial that k exist s.t. P | mnk, no matter what m, n, and P (=/= 0) may be. It's true anyway that /most/ counterexamples I found had P | mn directly. Here's one where it's more of a stretch: m = 98 = 7^2 * 2 n = 259 = 7 * 37 P = 1127 = 7^2 * 23 That is, P|mnk here iff 23|k. > I'll have to take a closer look then. (The existence of such a small > counterexample suggests there's something strange going on.) Seems that way. OTOH, while the integers are small, the prop 7 expression grows quickly. In the <98, 259, 1127> example, it's 14593054812133845870592414670025 = 3820085707432995^2 That is, the arithmetic is tedious enough that it's almost tempting to use a computer ;-) >> ... === === Subject: : Re: Parametric solutions of Euler Brick? <4tdi46F134ja8U1@mid.dfncis.de> <5e-dnWeiMdbwHe_YnZ2dnUVZ_o6dnZ2d@comcast.com> ... > One direction is obvious (to me); if the variables have the special > values, you end up with a perfect square. What I can't see is the > reverse (only if), that a perfect square leads to one of the cases. >> Here is the statement of Proposition 7, for those of you who have > forgotten it: >> Proposition 7: The expression >> (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) >> is a perfect square iff one of the following holds: >> (a) P^2 = n^2 m, > (b) P^2 = n m^2, > (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR > (d) m = t, n = a^2, P = a^3 (where a and t are integers). [Tim Peters] >> I haven't followed this, so don't know whether I'm missing hidden >> assumptions. Looks like: >> m=15 n=20 P=60 >> makes the first expression a perfect square (104976000000000000 = >> 324000000^2) but doesn't satisfy any of #a through #d. [Proginoskes] > A quick look through the paper only suggests that P has to divide > evenly into m*n*k. Your examples do this, even if k=1. Not sure I follow. If there's no restriction on k, then it's trivial that k > exist s.t. P | mnk, no matter what m, n, and P (=/= 0) may be. There is a requirement that there is an integer Q such that P Q = m n k; this is so that a particular Pythagorean triple can be constructed. > It's true anyway that /most/ counterexamples I found had P | mn directly. > Here's one where it's more of a stretch: m = 98 = 7^2 * 2 > n = 259 = 7 * 37 > P = 1127 = 7^2 * 23 That is, P|mnk here iff 23|k. > I'll have to take a closer look then. (The existence of such a small > counterexample suggests there's something strange going on.) Seems that way. OTOH, while the integers are small, the prop 7 > expression grows quickly. In the <98, 259, 1127> example, it's 14593054812133845870592414670025 = 3820085707432995^2 That is, the arithmetic is tedious enough that it's almost tempting to use a > computer ;-) I don't know when I'll be able to get back to checking the paper, so you don't need to find any more counterexamples until I find out how they affect the other variables in the paper. 8-) (OTOH, if the paper has to fail, at least it's failing where I suspected it would. And after only one reading, at that. 8-).) --- Christopher Heckman >> ... === === Subject: : Re: hodge decomposition harmonic vector field relax your boundary conditions some. For example, on R^2 you have the harmonic function x^2-y^2. On R^2 with > the negative real line removed, you have the harmonic function arctan(y/x). you'll have to please pardon my ignorance, I am trying blindly to teach this subject to myself, but I thought a hodge decomposition would break a smooth space filling vector field into three vector fields whose sum equals the original vector field. You say on R^2, x^2-y^2 is a harmonic function. x^2-y^2 would seem to me to return a scalar field, not a vector field. Perhaps I have misunderstood the hodge decomposition altogether? === === Subject: : Re: hodge decomposition harmonic vector field >>You can have harmonic functions that vary in space, but you have to >>relax your boundary conditions some. >>For example, on R^2 you have the harmonic function x^2-y^2. On R^2 with >>the negative real line removed, you have the harmonic function arctan(y/x). > you'll have to please pardon my ignorance, I am trying blindly to teach > this subject to myself, but I thought a hodge decomposition would break > a smooth space filling vector field into three vector fields whose sum > equals the original vector field. > You say on R^2, x^2-y^2 is a harmonic function. x^2-y^2 would seem to > me to return a scalar field, not a vector field. Perhaps I have > misunderstood the hodge decomposition altogether? > No, it is my fault - I am giving harmonic functions, not harmonic vector fields. On R^2 try (x^2-y^2,-2xy). For the other one, try ((1/2)log(x^2+y^2), arctan(y/x)) (or perhaps a minus sign here or there). In general, pick (u,-v) where u+iv is analytic. I.e. the condition the both the divergence and curl are zero becomes the Cauchy-Riemann conditions. (It is equivalent to saying that d^2u/dx^2+d^2u/dy^2=0 which is the more common usage of the word harmonic.) Stephen === === Subject: : Re: hodge decomposition harmonic vector field <97vch.257771$FQ1.64764@attbi_s71 On R^2 try (x^2-y^2,-2xy). For the other one, try ((1/2)log(x^2+y^2), > arctan(y/x)) (or perhaps a minus sign here or there). oh! (x^2-y^2,-2xy) is a harmonic vector field? hm, visully it looks like there is curl... I guess maybe I need to learn more about curl. http://i78.photobucket.com/albums/j102/ivandewolf/harmonic.jpg === === Subject: : Student resource site.. We are looking for your help. Hi all, After getting jack of wading through the internet trying to find useful sites while studying, me and a few friends have decided to try and establish a site to coordinate study material. This may yet prove to be a huge disaster.. We've put together a site where you can suggest useful links, with keywords, sorted by category. The site allows you to search these. We also have a forum that allows posting of attachments and a chat room (which is a tad ordinary though). The idea is to get a consolidated site of quality student resources. So once you've spent two hours finding something, you can share it with like minded people. Since its a new site we have no users and no content right now (We're getting there). We haven't even added out own graphics yet (but they will be spectacular).. What we want are some start up users to post suggested links, and resources to get it going (we even want comments and suggestions). If you want to help, the site is at www.pandanotes.net It's new and rough, and a work in progress, but none-the-less I think a good idea.. P === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved Has anybody managed to successfully used Tom White's tiling program > Alhambra? There doesn't seem to be any user's guide or description > of how to use it. I was able to run it bringing up a graphics window > but I couldn't much of anything with it. There appears to be some > tiling functions (e.g. Manual Tiler, Sequential Tiler, etc.) but they > aren't available from the menus. If I could figure out how to use this > program, it might save me months of work. I would appreciate any help > I could get. Have you checked out the website? (Google is your friend ... Google is your friend ...) http://tiling.org/alhambra/ --- Christopher Heckman === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved > Has anybody managed to successfully used Tom White's tiling program > Alhambra? There doesn't seem to be any user's guide or description > of how to use it. I was able to run it bringing up a graphics window > but I couldn't much of anything with it. There appears to be some > tiling functions (e.g. Manual Tiler, Sequential Tiler, etc.) but they > aren't available from the menus. If I could figure out how to use this > program, it might save me months of work. I would appreciate any help > I could get. Have you checked out the website? (Google is your friend ... Google is > your friend ...) http://tiling.org/alhambra/ Yes, that is the site for the program but as I said there doesn't seem to be any documentation on how to use the program. d=scan'208; a=567208759:sNHT15203536 === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved > Yes, that is the site for the program but as I said there doesn't seem to > be any documentation on how to use the program. No documentation is a sign that the author simply is not serious about his or her work. I'd not waste your time on such a product (or money) and would have a look elsewhere. I'm not sure what you mean by tiling but if it's just tiling graphics there are plenty of programs out there - if it is for game tiling then again there are plenty of options out there. Scott Kane Moderator comp.software.shareware.* === === Subject: : Re: Alhambra Tiling Program <457587a1$0$28953$5a62ac22@per-qv1-newsreader-01.iinet.net.au> Approved-Poster: approved > Yes, that is the site for the program but as I said there doesn't seem to > be any documentation on how to use the program. No documentation is a sign that the author simply is not serious about his > or her work. I'd not waste your time on such a product (or money) and would > have a look elsewhere. I'm not sure what you mean by tiling but if it's > just tiling graphics there are plenty of programs out there - if it is for > game tiling then again there are plenty of options out there. By tiling I mean tiling the euclidean plane. In my case the tiles will be polygons. A particular tile is said to tile the plane if you can fit copies of tile together without gaps or overlap so that they cover the entire plane -- like a jigsaw puzzle. What I want is a program that takes an arbitary polygon as input and then starts trying all possible ways of putting copies of the tile together until it either (a) proves that it cannot tile the plane or (b) stops at some point and displays the partial tilings it found that might possibly be extended into a planar tiling. Unfortunately, Alhambra seems to be the only available program that has an automated tiling capability for an arbitrary polygon. Furthermore, the source code is available so that I could modify it which is something that I would want to do no matter what tiling program I used (there are some unique things I would need to add). The only other option I have is write such a program myself from scratch. I would rather not spend months doing so if I don't have to. d=scan'208; a=594580794:sNHT451664002 === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved > The only other option I have is write such a program myself from > scratch. I would rather not spend months doing so if I don't have to. I hope you find a solution and sorry I could not offer something more substantial for you. Scott Kane Moderator comp.software.shareware.* d=scan'208; a=63149145:sNHT31858911 === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved Hi Glenn, > Has anybody managed to successfully used Tom White's tiling program > Alhambra? There doesn't seem to be any user's guide or description > of how to use it. I was able to run it bringing up a graphics window > but I couldn't much of anything with it. There appears to be some > tiling functions (e.g. Manual Tiler, Sequential Tiler, etc.) but they > aren't available from the menus. If I could figure out how to use this > program, it might save me months of work. I would appreciate any help > I could get. Have you contacted the author for help? Most authors are more than happy to answer questions from potential customers. Scott Kane Moderator comp.software.shareware.* === === Subject: : Re: Alhambra Tiling Program <457251ec$0$28912$5a62ac22@per-qv1-newsreader-01.iinet.net.au> Approved-Poster: approved Hi Glenn, > Has anybody managed to successfully used Tom White's tiling program > Alhambra? There doesn't seem to be any user's guide or description > of how to use it. I was able to run it bringing up a graphics window > but I couldn't much of anything with it. There appears to be some > tiling functions (e.g. Manual Tiler, Sequential Tiler, etc.) but they > aren't available from the menus. If I could figure out how to use this > program, it might save me months of work. I would appreciate any help > I could get. Have you contacted the author for help? Most authors are more than happy to > answer questions from potential customers. Scott Kane > Moderator comp.software.shareware.* I emailed the author but have not received any response. d=scan'208; a=68263430:sNHT14459940 === === Subject: : Re: Alhambra Tiling Program Approved-Poster: approved > I emailed the author but have not received any response. Well. I'd say that's the answer to your question. Email once more and if you get no response then you're either dealing with an author with a spam filter switched so high he can't communicate with users or he couldn't care less to respond. Either way I'd not be looking at investing time or money in unsupported software. Scott Kane Moderator comp.software.shareware.* === === Subject: : the AIM of LIFE What is your purpose in life? What is the rationale behind our life? Why do we live in this life? These questions frequently intrigue people who try to find accurate answers. People provide different answers to these questions. Some people believe the purpose of life is to accumulate wealth. But one may wonder: What is the purpose of life after one has collected colossal amounts of money? What then? What will the purpose be once money is gathered? If the purpose of life is to gain money, there will be no purpose after becoming wealthy. And in fact, here lies the problem of some disbelievers or misbelievers at some stage of their life, when collecting money is the target of their life. When they have collected the money they dreamt of, their life loses its purpose. They suffer from the panic of nothingness and they live in tension and restlessness. Can Wealth Be an Aim? We often hear of a millionaire committing suicide, sometimes, not the millionaire himself but his wife, son, or daughter. The question that poses itself is: Can wealth bring happiness to one's life? In most cases the answer is NO. Is the purpose of collecting wealth a standing purpose? As we know, the five-year old child does not look for wealth: a toy for him is equal to a million dollars. The eighteen-year old adolescent does not dream of wealth because he is busy with more important things. The ninety-year old man does not care about money; he is worried more about his health. This proves that wealth cannot be a standing purpose in all the stages of the individual's life. Wealth can do little to bring happiness to a disbeliever, because he/she is not sure about his fate. A disbeliever does not know the purpose of life. And if he has a purpose, this purpose is doomed to be temporary or self destructive. What is the use of wealth to a disbeliever if he feels scared of the end and skeptical of everything. A disbeliever may gain a lot of money, but will surely lose himself. Worshipping Allah as an Aim On the contrary, faith in Allah gives the believer the purpose of life that he needs. In Islam, the purpose of life is to worship Allah. The term Worship covers all acts of obedience to Allah. The Islamic purpose of life is a standing purpose. The true Muslim sticks to this purpose throughout all the stages of his life, whether he is a child, adolescent, adult, or an old man. Worshipping Allah makes life purposeful and meaningful, especially within the framework of Islam. According to Islam this worldly life is just a short stage of our life. Then there is the other life. The boundary between the first and second life is the death stage, which is a transitory stage to the second life. The type of life in the second stage a person deserves depends on his deeds in the first life. At the end of the death stage comes the day of judgment. On this day, Allah rewards or punishes people according to their deeds in the first life. The First Life as an Examination So, Islam looks at the first life as an examination of man. The death stage is similar to a rest period after the test, i. e. after the first life. The Day of Judgment is similar to the day of announcing the results of the examinees. The second life is the time when each examinee enjoys or suffers from the outcome of his behavior during the test period. In Islam, the line of life is clear, simple, and logical: the first life, death, the Day of Judgment, and then the second life. With this clear line of life, the Muslim has a clear purpose in life. The Muslim knows he is created by Allah. Muslims know they are going to spend some years in this first life, during which they have to obey God, because God will question them and hold them responsible for their public or private deeds, because Allah knows about all the deeds of all people. The Muslim knows that his deeds in the first life will determine the type of second life they will live in. The Muslim knows that this first life is a very short one, one hundred years, more or less, whereas the second life is an eternal one. The Eternity of the Second Life The concept of the eternity of the second life has a tremendous effect on a Muslims during their first life, because Muslims believe that their first life determines the shape of their second life. In addition, this determines the shape of their second life and this determination will be through the Judgment of Allah, the All just and Almighty. With this belief in the second life and the Day of Judgment, the Muslim's life becomes purposeful and meaningful. Moreover, the Muslim's standing purpose is to go to Paradise in the second life. In other words, the Muslim's permanent purpose is to obey Allah, to submit to Allah, to carry out His orders, and to keep in continues contact with Him through prayers (five times a day), through fasting (one month a year), through charity (as often as possible), and through pilgrimage (once in one's life). The Need for a Permanent Purpose Disbelievers have purposes in their lives such as collecting money and property, indulging in sex, eating, and dancing. But all these purposes are transient and passing ones. All these purposes come and go, go up and down. Money comes and goes. Health comes and goes. Sexual activities cannot continue forever. All these lusts for money, food and sex cannot answer the individual's questions: so what? Then What? However, Islam saves Muslims from the trouble of asking the question, because Islam makes it clear, from the very beginning, that the permanent purpose of the Muslim in this life is to obey Allah in order to go to Paradise in the second life. We should know that the only way for our salvation in this life and in the hereafter is to know our Lord who created us, believe in Him, and worship Him alone. We should also know our Prophet whom Allah had sent to all mankind, believe in Him and follow Him. We should, know the religion of truth which our Lord has commanded us to believe in, and practice it ... Those in search of truth Who have an open mind and heart, Islamic Education Foundation Welcome You. Objectives: - To Convey the message of Islam To Educate Muslims about Islam To keep in close contact with new Muslims. Activities: Offering Courses and presenting lectures about Islam in several languages. Teaching Islam and Arabic. Teaching new Muslims to receive the Holy Quran. Helping Non- Muslims embrace Islam and complete the required procedures === === Subject: : algebra with D_4~S_3 Hello sir~ D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. S_3 = <(1,2), (1,2,3)> Show that D_3 and S_3 are isomorphic. ------------------------------------ I think... If (1,2) = a and (1,2,3) = b, I can show that (1,2)^2 = (1,2,3)^3 = e and (1,2)^-1 = (1,2) (1,2,3)^-1 = (1,2,3) so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. from this, can I say that D_3 and S_3 are isomorphic ? === === Subject: : Re: algebra with D_4~S_3 days. My association with the Department is that of an alumnus. >Hello sir~ D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. >S_3 = <(1,2), (1,2,3) >Show that D_3 and S_3 are isomorphic. ------------------------------------ >I think... If (1,2) = a and (1,2,3) = b, >I can show that (1,2)^2 = (1,2,3)^3 = e and >(1,2)^-1 = (1,2) >(1,2,3)^-1 = (1,2,3) so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. from this, can I say that D_3 and S_3 are isomorphic ? Almost immediately. Remember what it means to give a group by generators and relations. If G = < X | R> is a group, where X is a set of generators and R is the set of relations, then given any group H and any subset Y of elements of H, if you can find a surjection f:X->Y such that the elements f(x) satisfy the relations in R, then f extends to a unique group homomorphism from G to H. So here you have X = {a,b}, and R = {a^2=1, b^3=1, a^{-1}ba=b^{-1} }. You have mapped a to (1,2) and b to (1,2,3), and shown that the corresponding relations hold, that is, that (1,2)^2=1, (1,2,3)^3 = 1, and (1,2)^{-1}(1,2,3)(1,2) = (1,2,3)^{-1}. That means that there is a group homomorphism from D_3 to S_3 mapping a to (1,2) and b to (1,2,3). Now, this is all you get from your work so far. You still don't have that the homomorphism is an isomorphism, though. For that, in this case, you can note that: (i) (1,2) and (1,2,3) generate S_3, so the map is surjective; (ii) The orders of S_3 and D_3 are the same, and we have a surjective group homomorphism between then, hence an isomorphism. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: algebra with D_4~S_3 >Hello sir~ >D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. >S_3 = <(1,2), (1,2,3) >Show that D_3 and S_3 are isomorphic. >------------------------------------ >I think... >If (1,2) = a and (1,2,3) = b, >I can show that (1,2)^2 = (1,2,3)^3 = e and >(1,2)^-1 = (1,2) >(1,2,3)^-1 = (1,2,3) >so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. >from this, can I say that D_3 and S_3 are isomorphic ? Almost immediately. Remember what it means to give a group by generators and relations. If G = < X | R> is a group, where X is a set of generators and R is > the set of relations, then given any group H and any subset Y of > elements of H, if you can find a surjection f:X->Y such that the > elements f(x) satisfy the relations in R, then f extends to a unique > group homomorphism from G to H. So here you have X = {a,b}, and R = {a^2=1, b^3=1, a^{-1}ba=b^{-1} }. You have mapped a to (1,2) and b to (1,2,3), and shown that the > corresponding relations hold, that is, that (1,2)^2=1, (1,2,3)^3 = 1, > and (1,2)^{-1}(1,2,3)(1,2) = (1,2,3)^{-1}. That means that there is a > group homomorphism from D_3 to S_3 mapping a to (1,2) and b to > (1,2,3). Now, this is all you get from your work so far. You still don't have > that the homomorphism is an isomorphism, though. For that, in this > case, you can note that: (i) (1,2) and (1,2,3) generate S_3, so the map is surjective; > (ii) The orders of S_3 and D_3 are the same, and we have a surjective > group homomorphism between then, hence an isomorphism. Oh, I can understand my problem certainly from this. thank you very much for your detailed advice. === === Subject: : Re: algebra with D_4~S_3 On 2006-12-03 02:57:06 -0500, mina_world said: > Hello sir~ D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. > S_3 = <(1,2), (1,2,3) > Show that D_3 and S_3 are isomorphic. ------------------------------------ > I think... If (1,2) = a and (1,2,3) = b, > I can show that (1,2)^2 = (1,2,3)^3 = e and > (1,2)^-1 = (1,2) > (1,2,3)^-1 = (1,2,3) so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. from this, can I say that D_3 and S_3 are isomorphic ? So, you've shown S_3 holds all of D_3's relations, but S_3 may hold additional relations that D_3 doesn't. I think you need a bit more work. -- -kira === === Subject: : Re: algebra with D_4~S_3 > On 2006-12-03 02:57:06 -0500, mina_world said: > Hello sir~ > D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. > S_3 = <(1,2), (1,2,3) > Show that D_3 and S_3 are isomorphic. > ------------------------------------ > I think... > If (1,2) = a and (1,2,3) = b, > I can show that (1,2)^2 = (1,2,3)^3 = e and > (1,2)^-1 = (1,2) > (1,2,3)^-1 = (1,2,3) > so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. > from this, can I say that D_3 and S_3 are isomorphic ? So, you've shown S_3 holds all of D_3's relations, but S_3 may hold > additional relations that D_3 doesn't. I think you need a bit more work. What's additional relations ? === === Subject: : Re: algebra with D_4~S_3 On 2006-12-03 04:48:08 -0500, mina_world said: > >> On 2006-12-03 02:57:06 -0500, mina_world said: >> > Hello sir~ > > D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. > S_3 = <(1,2), (1,2,3)>> > Show that D_3 and S_3 are isomorphic. > > ------------------------------------ > I think... > > If (1,2) = a and (1,2,3) = b, > I can show that (1,2)^2 = (1,2,3)^3 = e and > (1,2)^-1 = (1,2) > (1,2,3)^-1 = (1,2,3) > > so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. > > from this, can I say that D_3 and S_3 are isomorphic ? >> >> So, you've shown S_3 holds all of D_3's relations, but S_3 may hold >> additional relations that D_3 doesn't. >> >> I think you need a bit more work. What's additional relations ? Well, suppose you have a group G. It is conceivable to add a new relation to the generators of G to get a new group H. In fact, H will be a quotient group of G with the additional relation. Certainly, H will not be isomorphic to G, but H will still satisfy all the relations of G. In your case, I suppose you can count the size of the two groups. If they're the same, then there cannot be any additional relations on S_3 or else |S_3| < |D_3|. -- -kira === === Subject: : Re: algebra with D_4~S_3 <2006120303210416807-kirakun@earthlinknet> <2006120305010950073-kirakun@earthlinknet On 2006-12-03 04:48:08 -0500, mina_world said: >> On 2006-12-03 02:57:06 -0500, mina_world said: > Hello sir~ >> D_3 = , a^2 = b^3 = 1, {a^(-1)}.b.a = b^-1. > S_3 = <(1,2), (1,2,3) >> Show that D_3 and S_3 are isomorphic. >> ------------------------------------ > I think... >> If (1,2) = a and (1,2,3) = b, > I can show that (1,2)^2 = (1,2,3)^3 = e and > (1,2)^-1 = (1,2) > (1,2,3)^-1 = (1,2,3) >> so, {a^(-1)}.b.a = {(1,2)^-1}.(1,2,3).(1,2) = (1,2,3) = b^-1. >> from this, can I say that D_3 and S_3 are isomorphic ? >> So, you've shown S_3 holds all of D_3's relations, but S_3 may hold >> additional relations that D_3 doesn't. >> I think you need a bit more work. > What's additional relations ? Well, suppose you have a group G. It is conceivable to add a new > relation to the generators of G to get a new group H. In fact, H will > be a quotient group of G with the additional relation. Certainly, H > will not be isomorphic to G, but H will still satisfy all the relations > of G. In your case, I suppose you can count the size of the two groups. If > they're the same, then there cannot be any additional relations on S_3 > or else |S_3| < |D_3|. Yes, so to complete the proof that S_3 and D_3 are isomorphi, mina_world still needs to prove that |D_3| <= 6, which is not difficult. Derek Holt. === === Subject: : topology with continuous. Hello sir~ Basis B = {V in X | V and V^c are open set} This is a basis of X. F : closed set. p in X and p in F^c. Show that there exists continuous function f : X->R such that f(p) = 0 and f(F) = {1}. --------------------------------------------------------------- How do you show it ? === === Subject: : Re: topology with continuous. > Basis B = {V in X | V and V^c are open set} This is a basis of X. > This is absurdity. Do you mean B = { V subset X | V and V^c are open }? How am I to understant this? That X has a topology and that a base for X is B? Hm. You are defining zero-dimensional space. A space S is zero-dimensional when S has a base of clopen sets. So you are considering a space X that is zero dimensional. Have I understood you correctly? > F : closed set. p in X and p in F^c. Show that there exists continuous function f : X->R > such that f(p) = 0 and f(F) = {1}. How do you show it ? > Some clopen base set U with p in U subset F^c. Define f(U) = 0, f(XU) = 1. The rest falls into place. === === Subject: : Re: topology with continuous. > Basis B = {V in X | V and V^c are open set} > This is a basis of X. > This is absurdity. Do you mean > B = { V subset X | V and V^c are open }? Yes, right. > How am I to understant this? > That X has a topology and that a base for X is B? > Hm. You are defining zero-dimensional space. A space S is zero-dimensional when S has a base of clopen sets. So you are considering a space X that is zero dimensional. > Have I understood you correctly? Zero-dimensional space ? I am unfamiliar with zero-dimensional space. If X is a zero-dimensional space, is my problem wrong fundamentally ? === === Subject: : Re: topology with continuous. Basis B = {V in X | V and V^c are open set} > This is a basis of X. > B = { V subset X | V and V^c are open }? > How am I to understant this? > That X has a topology and that a base for X is B? > Hm. You are defining zero-dimensional space. > A space S is zero-dimensional when S has a base of clopen sets. > So you are considering a space X that is zero dimensional. > Have I understood you correctly? Zero-dimensional space ? > I am unfamiliar with zero-dimensional space. > I defined it directly above. > If X is a zero-dimensional space, > is my problem wrong fundamentally ? > No. What I was asking is your space X a zero-dimensional space, ie a space with a base of clopen sets. That's what you appeared to be saying. (clopen means open and closed) === === Subject: : Re: topology with continuous. > Basis B = {V in X | V and V^c are open set} > This is a basis of X. > B = { V subset X | V and V^c are open }? > How am I to understant this? > That X has a topology and that a base for X is B? > Hm. You are defining zero-dimensional space. > A space S is zero-dimensional when S has a base of clopen sets. > So you are considering a space X that is zero dimensional. > Have I understood you correctly? > Zero-dimensional space ? > I am unfamiliar with zero-dimensional space. > I defined it directly above. > If X is a zero-dimensional space, > is my problem wrong fundamentally ? > No. What I was asking is your space X a zero-dimensional space, > ie a space with a base of clopen sets. That's what you appeared > to be saying. (clopen means open and closed) Yes, thank you very much. === === Subject: : Re: topology with continuous. > Hello sir~ Basis B = {V in X | V and V^c are open set} This is a basis of X. F : closed set. p in X and p in F^c. Show that there exists continuous function f : X->R > such that f(p) = 0 and f(F) = {1}. --------------------------------------------------------------- Oh, I can... because, X is normal. because, X is normal <=> given a closed set A and an open set U containing A, there is an open set V containing A such that cl(V) in U. namely, let A be a closed set. so, A^c is a closed set by basis B. so, A = V = cl(V) in U for open set U containing A. thus, X is normal. By Urysohn lemma, F and F^c are disjoint closed subsets of X. so, there exists a continuous map f : X -> [0,1] such that f(x) = 0 for every x in F^c, and f(x) = 1 for every x in F. since p in F^c, f(p) = 0 === === Subject: : Re: topology with continuous. > Basis B = {V in X | V and V^c are open set} Do not use in to mean subset. It makes nonsense, for example: 2 in N and 2 in P(N), thus 2 in two disjoint subsets. > This is a basis of X. > F : closed set. > p in X and p in F^c. > Show that there exists continuous function f : X->R > such that f(p) = 0 and f(F) = {1}. Oh, I can... because, > X is normal. because, You cannot assume X is normal. Steens's Counter examples in Topology give examples of not normal zero dimensional spaces such as space #64 the Sorgenfrey plane with base of half open rectangles. > X is normal <=> given a closed set A and an open set U > containing A, there is an open set V containing A such that > cl(V) in U. > namely, > let A be a closed set. > so, A^c is a closed set by basis B. > so, A = V = cl(V) in U for open set U containing A. thus, X is normal. By Urysohn lemma, > F and F^c are disjoint closed subsets of X. > so, there exists a continuous map f : X -> [0,1] > such that f(x) = 0 for every x in F^c, > and f(x) = 1 for every x in F. since p in F^c, f(p) = 0 > === === Subject: : topology with closure and boundary. Hello sir~ X_1 = (R, U) X_2 = (R, D) U : usual topology. D : discrete topology. X = X_1 x X_2 (product space) A = (-1,1) x (-1,1) in X. Find the closure of A and boundary of A without proof. -------------------------------------- I think... cl(A) = [-1,1]x(-1,1) b(A) = {{-1}x(-1,1)} U {{1}x(-1,1)} is this right thinking ? === === Subject: : Re: topology with closure and boundary. > X_1 = (R, U) > X_2 = (R, D) U : usual topology. D : discrete topology. X = X_1 x X_2 (product space) > A = (-1,1) x (-1,1) in X. Find the closure of A and boundary of A cl(A) = [-1,1]x(-1,1) > cl(AxB) = cl A x cl B > b(A) = {{-1}x(-1,1)} U {{1}x(-1,1)} > X - A = (-oo,-1]xR / [1,oo)xR / Rx(-oo,-1] / Rx[1,oo) > is this right thinking ? Correct. === === Subject: : Re: topology with closure and boundary. > X_1 = (R, U) > X_2 = (R, D) U : usual topology. D : discrete topology. X = X_1 x X_2 (product space) A = (-1,1) x (-1,1) in X. Find the closure of A and boundary of A > without proof. > -------------------------------------- I think... cl(A) = [-1,1]x(-1,1) b(A) = {{-1}x(-1,1)} U {{1}x(-1,1)} is this right thinking ? Well, I don't know what you were _thinking_, but it is the right _answer_, all right. :-) Jose Carlos Santos === === Subject: : Re: Function with a local maximum in each point === === Subject: : Function with a local maximum in each point > It can be proved that a continuous function f from a closed segment > (a,b), having real values and a local maximum in each point of (a,b) > is constant. Correct notation for closed interval is [a,b]. Wlog a <= b. Let r = inf f([a,b]). Some c in [a,b] with f(c) = r. Let A = f^-1(r). A is nonnul closed set. Let d = inf A. d in A, since A nonnul closed. Assume a < d. Now since d is a local maximum and f(d) = r there's an open interval U about d, for which f(U) = r. This contradicts d = inf A. Thus a = d. Let B = [a,b] - A, assume B nonnul and let t = inf B. Show a < t, t in A and since t is a local maximum, there's some open U nhood t with f(U) = r. This cointradicts t = inf B. Thus [a,b] = A = f^-1(c) and f is constant. Now assume the same for f on the nonnul open interval (a,b). Then for all sufficienlty large n in N, f is contant on [a+1/n, b-1/n]. Thus f is constant on (a,b). > What if f is not supposed continuous ? f(x) = 1, if 0 <= x f(x) = 0, if x < 0 ---- === === Subject: : Re: locally proper map (?) <4572484F.7020503@netscape.net >>What does a locally proper map mean? >f:X -> Y is proper when for all K, f^-1(K) is compact. >Ok, I'll make an educated guess. >f:X -> Y is locally proper when for all y, some open U nhood y >with for all compact K subset U, f^-1(K) is compact. >If perchance Y is Hausdorff, then >f:X -> Y is locally proper iff for all y, >some open U nhood y with compact f^-1(cl U). How about this guess? For all x in X, there exists a neighborhood U > of x such that the restriction of f to U is proper. > Educated. Does either imply the other? === === Subject: : Re: locally proper map (?) >>What does a locally proper map mean? >f:X -> Y is proper when for all K, f^-1(K) is compact. >Ok, I'll make an educated guess. >f:X -> Y is locally proper when for all y, some open U nhood y >with for all compact K subset U, f^-1(K) is compact. >If perchance Y is Hausdorff, then >f:X -> Y is locally proper iff for all y, >some open U nhood y with compact f^-1(cl U). > How about this guess? For all x in X, there exists a neighborhood U > of x such that the restriction of f to U is proper. > Educated. Does either imply the other? Well, (WLP) implies (SLP): for x in X just take an open nbhd U of f(x) givenh by (WLP) and then consider the restriction of f to f^1(U). But (WLP) is stronger than (SLP) in general: consider the special case of a map f: X --> 1 to the one point space. Then (WLP) <==> X is compact but (SLP) <==> every point of x has a compact open neighbourhood. That last condition is one of the two competing definitions for locally compact. So (SLP) would fit a little better into the naming scheme. Marc === === Subject: : Re: locally proper map (?) <4572484F.7020503@netscape.net> === === Subject: : Re: locally proper map (?) >>What does a locally proper map mean? >f:X -> Y is proper when for all K, f^-1(K) is compact. >f:X -> Y is locally proper when for all y, some open U nhood y >with for all compact K subset U, f^-1(K) is compact. > For all x in X, there exists a neighborhood U > of x such that the restriction of f to U is proper. > Does either imply the other? > Well, (WLP) implies (SLP): for x in X just take an open nbhd U of > f(x) givenh by (WLP) and then consider the restriction of f to > f^1(U). > But (WLP) is stronger than (SLP) in general: consider the special > case of a map f: X --> 1 to the one point space. > Then (WLP) <==> X is compact > but (SLP) <==> every point of x has a compact open neighbourhood. Open neighborhoods within a Hausdorff space cannot be compact unless they're also closed. Thus for Hausdorff X, that f is SLP iff every point has a compact clopen neighborhood. > That last condition is one of the two competing definitions > for locally compact. So (SLP) would fit a little better into > the naming scheme. Your last condition should read every point has a compact neighborhood, ie every point is in of the interior of some compact set. Stephen's definition fits in with a general notion of local map property. For example (f|U is f restricted to U) f:X -> Y local homeomorphism when for all x in X, some open U nhood x with f|U:U -> f(U) homeomorphism f:X -> Y locally constant when for all x in X, some open U nhood x with f(U) = { f(x) } Does Stephen mean f:X -> Y locally proper when for all x in X, some open U nhood x with f|U:U -> Y is proper or f:X -> Y locally proper when for all x in X, some open U nhood x with f|U:U -> f(U) is proper? The former because proper maps are not required to be surjective? For local homeomorphism it's f|U:U -> f(U) because homeomorphisms are surjective. For locally constant, the two are equivalent, whether f|U:U -> Y or f|U:U -> f(U). ---- === === Subject: : Re: locally proper map (?) === > === Subject: : Re: locally proper map (?) >>What does a locally proper map mean? >f:X -> Y is proper when for all K, f^-1(K) is compact. >f:X -> Y is locally proper when for all y, some open U nhood y >with for all compact K subset U, f^-1(K) is compact. > For all x in X, there exists a neighborhood U > of x such that the restriction of f to U is proper. > Does either imply the other? Well, (WLP) implies (SLP): for x in X just take an open nbhd U of > f(x) givenh by (WLP) and then consider the restriction of f to > f^1(U). But (WLP) is stronger than (SLP) in general: consider the special > case of a map f: X --> 1 to the one point space. Then (WLP) <==> X is compact > but (SLP) <==> every point of x has a compact open neighbourhood. ___________________________________________________^^^^__ forget the open. I had misinterpreted Stephens condition. >[...] > Does Stephen mean > f:X -> Y locally proper when for all x in X, > some open U nhood x with f|U:U -> Y is proper or > f:X -> Y locally proper when for all x in X, > some open U nhood x with f|U:U -> f(U) is proper? Good question. I do not know. Marc === === Subject: : Re: Axiom of choice and the three spheres. <242651.1165087821592.JavaMail.jakarta@nitrogen.mathforum.org> Yes, that there are non-measurable sets in the reals is attributed to basically that line of argument of Vitali. Thus you might understand that for quite some time I have been suggesting to re-Vitali-ize that consideration. That there could be a partition of the unit interval into infinitely many (disjoint) subsets, where for a given subset of the partition any other partition's elements are either all greater or all less than any of the elements of the subset, with each subset represented thusly a line segment with some constant measure, labelled c, leads to the standard outcry that then there would be a least positive real number, or rather the partition containing zero would have some infinitesimal yet positive value of measure while the sum of the measures of those partitions is the whole and unity. Partitioning the unit interval in that way, where the measure could not be finite so would be infinitesimal, yet would sum over the partitions to unity, leads to that partition looking just like the differential, eg, dx, from the integral calculus: the infinitesimal analysis. The sum of dx from zero to one is one. That there would be some least or sufficiently small positive real (less than 1/n for finite natural n) that as well sums over an index of each displacement by that value until it reaches exactly one, to equal exactly one, contradicts a variety of closely held notions of standard real analysis, for example that there is no least positive real. While that may be so, standard real analysis itself sees counterexamples to that there is not such a thing. Standar theory contradicts: itself. So; forgo, forgive, integration or non-measurable sets. Pick one or the other, exclusice, integration, or non-measurable sets, until everyone is less naive. There can be only one, theory, of everything. Ross === === Subject: : Re: Need German equivalent of crank >> I have a wide knowledge of English - but not German - in which you are >> more expert than me. Than I You are wackelig - take a running jump. Nick === === Subject: : Re: MACSYMA and AXIOM - the same failure pattern <45221DC3.9030100@tid.es> Human-driven quality assurance process gets inefficient for large scale projects like AXIOM and MAXIMA, and in the long run, destroys the initial efforts. The solution is automated environments like the VM machine. === === Subject: : Re: large ordinals, help! > An ordinal a little larger than epislon_0 is omega_1. A little larger? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! An ordinal a little larger than epsilon_0 is omega_1. A little larger? > Yup, a little larger. Not a tiny bit larger like epsilon_1 but just little larger, like omega_2 which is a little more larger. There's lots of these pretentious little more larger's, but the biggest one of all, that really does try to be larger is omega_(omega_0). This last number is ample chicken feed to keep most mathematicians happy. To keep most set theorists happy, one may have to actually find a larger ordinal, viz omega_omega_omega_omega_... Beyond that is but quibbling of my thing is bigger than yours, a very popular sport both for esoteric set theorists, preachy preaching preachers and deranged leaders who deign to out trump Borali-Forte's mother of all ordinals. === === Subject: : Re: large ordinals, help! > > An ordinal a little larger than epsilon_0 is omega_1. >> A little larger? > Yup, a little larger. Would you also consider the Howard ordinal or the Church-Kleene ordinal just a little larger than epsilon-0? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net > An ordinal a little larger than epsilon_0 is omega_1. >> A little larger? > Yup, a little larger. Would you also consider the Howard ordinal or the Church-Kleene ordinal > just a little larger than epsilon-0? > No, I conside all intangible untouchables too big for their britches. === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> > Would you also consider the Howard ordinal or the Church-Kleene ordinal >> just a little larger than epsilon-0? > No, I consider all intangible untouchables too big for their britches. I'm glad that's all sorted out, then. Do you consider the Howard ordinal > and the Church-Kleene ordinal intangible untouchables? > Incidentally notice how tiny winy they are as compared to Borali-Forte's mother of all ordinals, the last big ordinal, the biggest ordinal of all, the only fixed point of f(xi) = xi + 1. The mother of all ordinals is the smallest and only known incomprehensible ordinal. === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> I'm glad that's all sorted out, then. Do you consider the Howard ordinal > and the Church-Kleene ordinal intangible untouchables? > ??? Huh? These ordinals are countable, and thus barely infinite and are a long way from being inaccessible. Perhaps you are using a different definition of inaccessible? > Incidentally notice how tiny winy they are as compared to Borali-Forte's > mother of all ordinals, the last big ordinal, the biggest ordinal of all, > the only fixed point of f(xi) = xi + 1. There is no last big ordinal. Perhaps you are loosely applying the word ordinal to On, the class of all ordinals? I suppoose you could, since On has many properties of ordinals, but then you couldn't call On the class of all ordinals since classes must contain only sets, never proper classes. In any case, it is clearly non-standard usage. However, your use of the only fixed point of f(xi) = xi + 1 is certainly false regardless of what you meant. For no ordinal is x + 1 = x, since no ordinal is equal to its successor. As for On, it has no successor since it's not even a set, and thus +1 is not defined. > The mother of all ordinals is the smallest and only known > incomprehensible ordinal. Please define what you mean by incomprehensible. Are you alluding to Cantor's Absolute Infinity? If so, it is strange to refer to it as the smallest and only known one, since there cannot be a larger one than Absolute Infinity. If you simply mean On, this is not really even an ordinal, so it is already larger than any known (or unknown) ordinal. On, being a proper class, cannot be an element of any other class, and thus *known* to be the biggest. Your terminology is strange and rather sloppy, and I've had to do a lot of educated guessing. Please define what you mean by the terms inaccessible, last big ordinal, and incomprehensible, so that you can be more coherent. Jonathan Hoyle Eastman Kodak === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> Incidentally notice how tiny winy they are as compared to Borali-Forte's > mother of all ordinals, the last big ordinal, the biggest ordinal of all, > the only fixed point of f(xi) = xi + 1. There is no last big ordinal. Perhaps you are loosely applying the > word ordinal to On, the class of all ordinals? I suppoose you could, > since On has many properties of ordinals, but then you couldn't call On > the class of all ordinals since classes must contain only sets, never > proper classes. In any case, it is clearly non-standard usage. > Oh yea I know, all mathematicians are supposed to be married to logic. > However, your use of the only fixed point of f(xi) = xi + 1 is > certainly false regardless of what you meant. For no ordinal is x + 1 > = x, since no ordinal is equal to its successor. As for On, it has no > successor since it's not even a set, and thus +1 is not defined. > Exactly why it does get any bigger. > The mother of all ordinals is the smallest and only known > incomprehensible ordinal. Please define what you mean by incomprehensible. Are you alluding to > Cantor's Absolute Infinity? If so, it is strange to refer to it as > the smallest and only known one, since there cannot be a larger one > than Absolute Infinity. > I've given you the only know example of an incomprehensible ordinal and you dare to complain that you don't comprehend it? You're not supposed to, that's why it's incomprehensible. Never heard of Cantor's Absolute Infinity. What's that? The opposite of a perfect vaccuum? > If you simply mean On, this is not really even an ordinal, so it is > already larger than any known (or unknown) ordinal. On, being a > proper class, cannot be an element of any other class, and thus *known* > to be the biggest. Your terminology is strange and rather sloppy, and I've had to do a lot > of educated guessing. Please define what you mean by the terms > inaccessible, last big ordinal, and incomprehensible, so that you > can be more coherent. > Borali-Forte Now get you up and go just one more, just one little more, beyond the mother of all ordinals. === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> Incidentally notice how tiny winy they are as compared to Borali-Forte's > mother of all ordinals, the last big ordinal, the biggest ordinal of all, > the only fixed point of f(xi) = xi + 1. > There is no last big ordinal. Perhaps you are loosely applying the > word ordinal to On, the class of all ordinals? I suppoose you could, > since On has many properties of ordinals, but then you couldn't call On > the class of all ordinals since classes must contain only sets, never > proper classes. In any case, it is clearly non-standard usage. > Oh yea I know, all mathematicians are supposed to be married to logic. > However, your use of the only fixed point of f(xi) = xi + 1 is > certainly false regardless of what you meant. For no ordinal is x + 1 > = x, since no ordinal is equal to its successor. As for On, it has no > successor since it's not even a set, and thus +1 is not defined. > Exactly why it does get any bigger. ** Exactly why it doesn't get any bigger. > The mother of all ordinals is the smallest and only known > incomprehensible ordinal. > Please define what you mean by incomprehensible. Are you alluding to > Cantor's Absolute Infinity? If so, it is strange to refer to it as > the smallest and only known one, since there cannot be a larger one > than Absolute Infinity. > I've given you the only know example of an incomprehensible ordinal and > you dare to complain that you don't comprehend it? You're not supposed > to, that's why it's incomprehensible. Never heard of Cantor's Absolute Infinity. What's that? The opposite of > a perfect vaccuum? > If you simply mean On, this is not really even an ordinal, so it is > already larger than any known (or unknown) ordinal. On, being a > proper class, cannot be an element of any other class, and thus *known* > to be the biggest. > Your terminology is strange and rather sloppy, and I've had to do a lot > of educated guessing. Please define what you mean by the terms > inaccessible, last big ordinal, and incomprehensible, so that you > can be more coherent. > Borali-Forte Now get you up and go just one more, > just one little more, > beyond the mother of all ordinals. === === Subject: : Re: large ordinals, help! >> I'm glad that's all sorted out, then. Do you consider the Howard ordinal >> and the Church-Kleene ordinal intangible untouchables? >> In what sense is the Howard ordinal or the Church-Kleene ordinal inaccessible? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> <4jcdh.179$4J6.68@reader1.news.saunalahti.fi >> I'm glad that's all sorted out, then. Do you consider the Howard ordinal >> and the Church-Kleene ordinal intangible untouchables? In what sense is the Howard ordinal or the Church-Kleene ordinal > inaccessible? > They aren't??? Of course they are. === === Subject: : Re: large ordinals, help! >> In what sense is the Howard ordinal or the Church-Kleene ordinal >> inaccessible? They aren't??? Of course they are. They aren't inaccessible in the technical sense. It's somewhat obscure what you have in mind. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> <4jcdh.179$4J6.68@reader1.news.saunalahti.fi> <%Lcdh.187$4J6.90@reader1.news.saunalahti.fi >> In what sense is the Howard ordinal or the Church-Kleene ordinal >> inaccessible? > They aren't??? Of course they are. They aren't inaccessible in the technical sense. It's somewhat obscure > what you have in mind. > They aren't? Then what small underlings are they claiming the privilege of flouting pretensions names? === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> <4jcdh.179$4J6.68@reader1.news.saunalahti.fi> <%Lcdh.187$4J6.90@reader1.news.saunalahti.fi> They aren't inaccessible in the technical sense. It's somewhat obscure > what you have in mind. They aren't? No, the Howard ordinal and the Church-Kleene ordinal are relatively large countable ordinals. > Then what small underlings are they claiming the privilege of flouting pretensions names? They have their significance in proof theory and recursion theory. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! <1GSch.65784$6x2.31079@reader1.news.jippii.net> <4jcdh.179$4J6.68@reader1.news.saunalahti.fi> <%Lcdh.187$4J6.90@reader1.news.saunalahti.fi> They aren't inaccessible in the technical sense. It's somewhat obscure > what you have in mind. > They aren't? No, the Howard ordinal and the Church-Kleene ordinal are relatively > large countable ordinals. > Then what small underlings are they claiming the privilege of flouting pretensions names? They have their significance in proof theory and recursion theory. > Bigger or smaller than the first non-recursive ordinal? I'll suppose the smaller one is the first fixed point of epsilon_xi = xi === === Subject: : Re: large ordinals, help! > No, I conside all intangible untouchables too big for their britches. I'm glad that's all sorted out, then. Do you think the Howard ordinal and the Church-Kleene ordinal are intangible untouchables? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: large ordinals, help! > Can anybody give me much information about the large ordinals, > especially about $epsilon_0$ ? John Baez posted a description of ordinals, along with a story about driving through South Dakota. You can find it here: --- Christopher Heckman === === Subject: : Re: large ordinals, help! > >Can anybody give me much information about the large ordinals, >>especially about $epsilon_0$ ? >> John Baez posted a description of ordinals, along with a story about >driving through South Dakota. You can find it here: > --- Christopher Heckman > Infinite ordinals always remind me of that Dr. Seuss book about the the letters beyond Z. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: large ordinals, help! Infinite ordinals always remind me of that Dr. Seuss book about the the > letters beyond Z. > Missed that. Please do enlighten us what he said other than they go beyond. === === Subject: : Re: large ordinals, help! > >Infinite ordinals always remind me of that Dr. Seuss book about the the >>letters beyond Z. >> >Missed that. Please do enlighten us what he said other than they go >beyond. > > The book is On Beyond Zebra. See http://www-unix.oit.umass.edu/~poetsma/zebra.html , though it is better with pictures, of course. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: large ordinals, help! <4572E30F.9040700@netscape.net> <45737E2B.5020407@netscape.net >>Infinite ordinals always remind me of that Dr. Seuss book about the the >>letters beyond Z. >Missed that. Please do enlighten us what he said other than they go >beyond. The book is On Beyond Zebra. See > http://www-unix.oit.umass.edu/~poetsma/zebra.html , though it is better > with pictures, of course. http://www.evertype.com/standards/csur/seuss.html has the glyphs themselves. (Unicode includes spaces for Seuss's letters, btw!) A relative of mine had that book, and after my grad school advisor said, There are not enough letters in the alphabet, I was reminded of it and spent a few years looking for a (used) copy. Interestingly enough (while we're digressing), I referenced this page a few days ago, in Archimedes Plutonium's thread about typesetting math and science equations. --- Christopher Heckman === === Subject: : Re: large ordinals, help! <45737E2B.5020407@netscape.net >>Infinite ordinals always remind me of that Dr. Seuss book about the the >>letters beyond Z. >Missed that. Please do enlighten us what he said other than they go >beyond. The book is On Beyond Zebra. See > http://www-unix.oit.umass.edu/~poetsma/zebra.html , though it is better > with pictures, of course. > LOL ;-) Borali-Forte Now get you up and go just one more, just one little more, beyond the mother of all ordinals. === === Subject: : Compactness I heard that the defn of compactness by filter basis of closed sets, or the convergence of ultrafilters, is related to the following in some way, which I don't know: Compactness of anything, not only of topological spaces, means it is without holes, where holes are infinite process without end. Kindly explain if there is at all such concepts, and if there is, explain it. === === Subject: : Re: Compactness > I heard that the defn of compactness by filter basis of closed sets, > or the convergence of ultrafilters, is related to the following in some > way, which > I don't know: > Compactness of anything, not only of topological spaces, means it is > without holes, > where holes are infinite process without end. Kindly explain if there is at all such concepts, and if there is, > explain it. The concept of a hole can be misleading to one's intuition. There is nothing contradictory about a compact torus (donut) having a hole. Perhaps the absence of missing nests: in the sense that in a non-compact space, one can find a nested system of closed sets with empty intersection. (I assume your text has defined nested system of closed sets.) === === Subject: : Re: Compactness > I heard that the defn of compactness by filter basis of closed sets, > or the convergence of ultrafilters, is related to the following in some > way, which > I don't know: > Compactness of anything, not only of topological spaces, means it is > without holes, where holes are infinite process without end. Kindly explain if there is at all such concepts, and if there is, > explain it. > [0,1] / [2,3] is compact and has a big hole, (1,2). (1,2) is an infinite process, ie (1,2) = /_n (1+1/n, 1-1/n) === === Subject: : Re: Compactness (1,2) is an infinite process, ie (1,2) = /_n (1+1/n, 1-1/n) You missed it! After a stage your process skips off [0,1]/[2,3]; and I couldn't make out why you have considered it an infinite process. === === Subject: : Re: Compactness [0,1] / [2,3] is compact and has a big hole, (1,2). > (1,2) is an infinite process, ie (1,2) = /_n (1+1/n, 1-1/n) > Whoops. (1,2) is an infinite process, ie (1,2) = /_n (1+1/n, 2-1/n) > You missed it! After a stage your process skips off [0,1]/[2,3]; and I > couldn't make out why you have considered it an infinite process. > It's an infinite process because you have to take the union of an infinite sequence of sets. It's a short infinite process as compared to long infinite processes of dealing with unjust political corruptions. === === Subject: : Re: Compactness > I heard that the defn of compactness by filter basis of closed sets, > or the convergence of ultrafilters, is related to the following in some > way, which > I don't know: > Compactness of anything, not only of topological spaces, means it is > without holes, > where holes are infinite process without end. Kindly explain if there is at all such concepts, and if there is, > explain it. > A space S is compact iff every open cover of S has a finite subcover every filter on S has a cluster point every ultra filter on S converges. === === Subject: : Re: Compactness every open cover of S has a finite subcover > every filter on S has a cluster point > every ultra filter on S converges. Well, there is another point: In a topological space (X,T), a subset A is compact iff for any filter basis F consisting of closed sets s.t. A clusters F( i.e., A intersects every member of F), A also cuts the core of F(i.e., /F). === === Subject: : Re: Compactness > Compactness of anything, not only of topological spaces, means it is > without holes, > where holes are infinite process without end. As a mathematical concept, compact is a term only used in topology - at least I never heard of it in any other (mathematical) context. But a topological space can very well have holes (another concept making sense only in topology), once you specify what a hole is. For example, a torus is compact but has a hole. Should infinite process without end rather mean something like converging sequence without limit (in the space considered)? === === Subject: : Re: Compactness As a mathematical concept, compact is a term only used in topology - > at least I never heard of it in any other (mathematical) context. No, not at all, there is a theorem of propositional calculus, named Compactness Theorem, which states that if S be a set of propositional formulas, then S is satisfiable iff every finite subset of S is satisfiable. === === Subject: : Re: Compactness > >>As a mathematical concept, compact is a term only used in topology - >>at least I never heard of it in any other (mathematical) context. > No, not at all, there is a theorem of propositional calculus, named > Compactness Theorem, which states that if S be a set of propositional > formulas, then S is satisfiable iff every finite subset of S is > satisfiable. While the two concepts are not identical there is a loose relation. The Heine-Borel property says that if a set of open sets covers a subset of the space then a -finite- subset of that set will also cover the subset of the space. There is an inference running from the whole thing to a finite subset of the thing. Bob Kolker > === === Subject: : Re: Compactness <4tidp3F13upc8U1@mid.individual.net While the two concepts are not identical there is a loose relation. The > Heine-Borel property says that if a set of open sets covers a subset of > the space then a -finite- subset of that set will also cover the subset > of the space. There is an inference running from the whole thing to a > finite subset of the thing. Bob Kolker > What at all did you say? Is it by any means related to whether the without hole concept is right? I can't understand your statement; === === Subject: : Re: Compactness > >>While the two concepts are not identical there is a loose relation. The >>Heine-Borel property says that if a set of open sets covers a subset of >>the space then a -finite- subset of that set will also cover the subset >>of the space. There is an inference running from the whole thing to a >>finite subset of the thing. >>Bob Kolker > What at all did you say? Is it by any means related to whether the > without hole concept is right? I can't understand your statement; Google A subset of a topological space is compact if and only if it has the Heine-Borel Property. The Heine-Borel property. Every open covering of a set contains a finite open covering. Bob Kolker > === === Subject: : Re: Compactness <4tidp3F13upc8U1@mid.individual.net> <4tin3sF14av8vU3@mid.individual.net> There is nothing weird about Heine Borel's covering criterion of compactness; what I asked you to explain is the line There is an inference running from the whole thing to a > finite subset of the thing. === === Subject: : Re: Compactness > As a mathematical concept, compact is a term only used in topology - >> at least I never heard of it in any other (mathematical) context. No, not at all, there is a theorem of propositional calculus, named >Compactness Theorem, which states that if S be a set of propositional >formulas, then S is satisfiable iff every finite subset of S is >satisfiable. But this also comes from topology. Consider the configuration space K = {true,false}^V, where V is the set of variables of S, and give it the product topology. Each member A of S corresponds to the subset K(A) of K consisting of those assignments to the variables that satisfy it. Since this depends on only finitely many variables, it is a closed set. Tychonoff's theorem says K is compact, and then the finite intersection property of compact spaces says if every finite subfamily of these closed sets has nonempty intersection, then so does the whole family. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : =?iso-8859-9?B?a2FyZWv2a2xlcmluIHRvcGxhbf0=?= 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) i.8din genel bir form.9fl bulunacak === === Subject: : Re: karek.9aklerin toplam.88 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) i.8din genel bir form.9fl bulunacak (boyle bir formul oldugunu sanmiyorum) Translation: He wants to know the general expression for this sum: 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) === === Subject: : Re: karek.9aklerin toplam.88 > 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) i.8din genel bir form.9fl bulunacak (boyle bir formul oldugunu sanmiyorum) Translation: He wants to know the general expression for this sum: > 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) There is no closed form in terms of elementary functions, but it could be expressed in terms of a type of zeta function. Perhaps an approximation might be adequate. When k is large, a simple approximation is Sqrt(k) (2k/3 + 1/2) + zeta(-1/2) where the constant term, zeta(-1/2), is -0.20788... David Cantrell === === Subject: : Re: karek.9aklerin toplam.88 -- Translation to Turkish: Genel olarak boyle bir formul yokmus. k buyuk ise toplam yaklasik olarak: Sqrt(k) (2k/3 + 1/2) + zeta(-1/2) Burada, zeta(-1/2) = -0.20788... Menderes >> 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) i.8din genel bir form.9fl bulunacak >> (boyle bir formul oldugunu sanmiyorum) >> Translation: He wants to know the general expression for this sum: >> 1+2^(1/2)+3^(1/2)+4^(1/2)....k^(1/2) There is no closed form in terms of elementary functions, but it could be > expressed in terms of a type of zeta function. Perhaps an approximation > might be adequate. When k is large, a simple approximation is Sqrt(k) (2k/3 + 1/2) + zeta(-1/2) where the constant term, zeta(-1/2), is -0.20788... David Cantrell === === Subject: : Liptschitz function approximation Let f be a Liptschitz function on [0,1]. Show that there exists f_n in C^1([0,1]) and M such that (i) |f_n'| <= M (ii) lim f_n(x) = f(x), for all x in [0,1]. === === Subject: : Re: Liptschitz function approximation >Let f be a Liptschitz function on [0,1]. >Show that there exists f_n in C^1([0,1]) and M such that >(i) |f_n'| <= M >(ii) lim f_n(x) = f(x), for all x in [0,1]. Are you familar with the notion of convolution with an approximate identity? ************************ David C. Ullrich === === Subject: : Re: Liptschitz function approximation > Let f be a Liptschitz function on [0,1]. > Show that there exists f_n in C^1([0,1]) and M such that > (i) |f_n'| <= M > (ii) lim f_n(x) = f(x), for all x in [0,1]. In fact there are polynomials that do the above; what class is this, and what material are you covering at the moment? === === Subject: : Re: Liptschitz function approximation This is an exercise in a real analysis textbook. On Dec 4, 5:03 am, The World Wide Wade > Let f be a Liptschitz function on [0,1]. > Show that there exists f_n in C^1([0,1]) and M such that > (i) |f_n'| <= M > (ii) lim f_n(x) = f(x), for all x in [0,1].In fact there are polynomials that do the above; what class is > this, and what material are you covering at the moment? === === Subject: : a real analysis problem f is Lebesgue integrable over R. if it holds int_G f = 0 for any open set G with m(G)=1. Show that f = 0 a.e. === === Subject: : Re: a real analysis problem >f is Lebesgue integrable over R. if it holds int_G f = 0 for any open >set G with m(G)=1. Show that f = 0 a.e. > Show first that f(x+1)=f(x) a.e. by considering int f on open intervals (a,a+1) and (a+epsilon,a+1+epsilon). Then show that f=0 a.e. on (0,1). To do this, suppose that e.g. f>0 on a subset S of (0,1) having measure 1/n. Then, consider int f over S+1 union S+2 union ... union S+n. -- Daniel Mayost === === Subject: : Re: a real analysis problem > f is Lebesgue integrable over R. if it holds int_G f = 0 for any open > set G with m(G)=1. Show that f = 0 a.e. Many contributors to this group will know how to do this question because they have already done their homework. Now it is your turn to learn-by-doing. R.G. Vickson === === Subject: : Re: ? variable coef DEs > Hi: J H Hubbard's DE book provides good understandings for constant linear ODEs. In his book, the linear DEs with variable coeffs will be described in book III, but I cannot find that anywhere. Does that even be published? Any good book for linear DE with variable coefficients? > Just recalled one thing: let t as another dependent variable, the linear variable ODE system is then transformed to a nonlinear ODE system. So does that mean that one only needs to understand NL ODE system, i.e. nothing special for system of ODEs with variable coefficients? System of ODEs with constant coefficients is highly related to linear algebra but nothing interesting to say for variable coefficients? by Cheng Cosine Dec/02/2k6 NC === === Subject: : Re: ? variable coef DEs > Hi: >> J H Hubbard's DE book provides good understandings for >> constant linear ODEs. In his book, the linear DEs with variable >> coeffs will be described in book III, but I cannot find that >anywhere. >> Does that even be published? Any good book for linear DE with >> variable coefficients? Just recalled one thing: let t as another dependent variable, >the linear variable ODE system is then transformed to a nonlinear ODE system. So does that mean that one only needs to understand NL ODE >system, i.e. nothing special for system of ODEs with variable coefficients? No it doesn't. >System of ODEs with constant coefficients is highly related to linear >algebra but nothing interesting to say for variable coefficients? Linear algebra says a lot for variable coefficient systems: fundamental matrices, variation of parameters, etc. The main difference is that it may be difficult or impossible to find a fundamental set of solutions in closed form. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: ? variable coef DEs > Hi: >> J H Hubbard's DE book provides good understandings for >> constant linear ODEs. In his book, the linear DEs with variable >> coeffs will be described in book III, but I cannot find that >>anywhere. >> Does that even be published? Any good book for linear DE with >> variable coefficients? > Just recalled one thing: let t as another dependent variable, >>the linear >> variable ODE system is then transformed to a nonlinear ODE system. >> So does that mean that one only needs to understand NL ODE >>system, i.e. >>nothing special for system of ODEs with variable coefficients? No it doesn't. >System of ODEs with constant coefficients is highly related to linear >>algebra but nothing interesting to say for variable coefficients? Linear algebra says a lot for variable coefficient systems: > fundamental matrices, variation of parameters, etc. The > main difference is that it may be difficult or impossible > to find a fundamental set of solutions in closed form. > Well, the most impressive idea learnt from const coef ODE system is that the system dynamics could be investigated by separated eigenmodes that evolve with time independently. But when the coefficients are time-dependent, how does one investigate the system's transient behaviors or to trace some special modes? Or think in another direction, given an initial vector for a const coef ODE system, one can separately trace different modes' transient behavior. When coefs are time dependent, so long as it remains linear, one can decompose the initial vector into a given basis. But then how does one trace the dynamics of different modes for a given initial vector for a system of ODEs with time-dependent coefs? by Cheng Cosine Dec/03/2k6 NC === === Subject: : Re: ? variable coef DEs >> Hi: >> J H Hubbard's DE book provides good understandings for >> constant linear ODEs. In his book, the linear DEs with variable >> coeffs will be described in book III, but I cannot find that >anywhere. >> Does that even be published? Any good book for linear DE with >> variable coefficients? >> Just recalled one thing: let t as another dependent variable, >the linear >> variable ODE system is then transformed to a nonlinear ODE system. >> So does that mean that one only needs to understand NL ODE >system, i.e. >>nothing special for system of ODEs with variable coefficients? >> No it doesn't. >System of ODEs with constant coefficients is highly related to linear >algebra but nothing interesting to say for variable coefficients? >> Linear algebra says a lot for variable coefficient systems: >> fundamental matrices, variation of parameters, etc. The >> main difference is that it may be difficult or impossible >> to find a fundamental set of solutions in closed form. Well, the most impressive idea learnt from const coef ODE system is that the system dynamics could be investigated by separated eigenmodes that evolve with time independently. But when the coefficients are time-dependent, how does one investigate the system's transient behaviors or to trace some special modes? Or think in another direction, given an initial vector for a const coef ODE system, one can separately trace different modes' transient behavior. When coefs are time dependent, so long as it remains linear, one can decompose the initial vector into a given basis. But then how does one trace the dynamics of different modes for a given initial vector for a system of ODEs with time-dependent coefs? In general there is no such thing as a mode for the time-dependent case. You can take any n solutions with linearly independent initial conditions as your basis. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: ? variable coef DEs >>... >> Well, the most impressive idea learnt from const coef ODE system >>is that the system dynamics could be investigated by separated >>eigenmodes that evolve with time independently. But when the >>coefficients are time-dependent, how does one investigate the >>system's transient behaviors or to trace some special modes? >>Or think in another direction, given an initial vector for a const >>coef ODE system, one can separately trace different modes' transient >>behavior. When coefs are time dependent, so long as it remains >>linear, one can decompose the initial vector into a given basis. But then >>how does one trace the dynamics of different modes for a given >>initial vector for a system of ODEs with time-dependent coefs? In general there is no such thing as a mode for the > time-dependent case. You can take any n solutions with > linearly independent initial conditions as your basis. > Okay, taking any n linear indep unit vectors as the basis and project the IC onto it. Could one trace dynamics of certain individual components? by Cheng Cosine Dec/04/2k6 NC === === Subject: : linear algebra problem Define T in L(C^2) by T(w,z) = (z,0) Find all generalized eigenvectors of T. I proceeded as follows: T(w,0)= (0,0)= 0(w,0) so, 0 is an eigenvalue with eigenvector (1,0) Also, note that T^2(1,1)= T(1,0)= (0,0) So, (T- OI)^2(1,1)= T^2(1,1)= (0,0) for j=2 Therefore, (1,0) and (1,1) are generalized eigenvectors. Are there any more is my question? === === Subject: : Re: linear algebra problem > Define T in L(C^2) by T(w,z) = (z,0) Find all generalized eigenvectors of T. I proceeded as follows: T(w,0)= (0,0)= 0(w,0) so, 0 is an eigenvalue with eigenvector (1,0) Also, note that T^2(1,1)= T(1,0)= (0,0) So, (T- OI)^2(1,1)= T^2(1,1)= (0,0) for j=2 Therefore, (1,0) and (1,1) are generalized eigenvectors. Are there any more is my question If v is an eigenvector then so is a*v if a != 0. === === Subject: : Re: linear algebra problem > Define T in L(C^2) by > T(w,z) = (z,0) > Find all generalized eigenvectors of T. > I proceeded as follows: > T(w,0)= (0,0)= 0(w,0) > so, 0 is an eigenvalue with eigenvector (1,0) > Also, note that T^2(1,1)= T(1,0)= (0,0) > So, (T- OI)^2(1,1)= T^2(1,1)= (0,0) for j=2 > Therefore, (1,0) and (1,1) are generalized eigenvectors. > Are there any more is my question If v is an eigenvector then so is a*v if a != 0. what are you saying? === === Subject: : [newbieQ] Normal group Let me denote the inverse of g in G with the writing g'. Moreover, K <= G means that K is a subgroup of G, A Kg < gK => K < gKg', and since gKg', K are groups, then we have (A). Note that I had to switch <= into basic set inclusion <, that's why we do not know if Kg and gK are groups. Is the previous proof correct? === === Subject: : Re: Normal group > Let me denote the inverse of g in G with the writing g'. Moreover, K <= G > means that K is a subgroup of G, A necessarily strict). > Let K <= G.If we have gKg' <= K for all g in G, then we have g'Kg <= K > too, which leads to the inverse group inclusion K <= gKg' (A) and then K is normal in G. > I want to know if the following method to get (A) from hypotheses is > correct. > If we pick a particular g in G, we know that g'Kg is a group itself, and is > a subgroup of K for every g in G (hypothesis). We have that g'Kg < K => Kg < gK => K < gKg', and since gKg', K are groups, then we have (A). > Note that I had to switch <= into basic set inclusion <, that's why we do > not know if Kg and gK are groups. Is the previous proof correct? You are using that left multiplication by g and right multiplication by g' are set-theoretic bijections, hence conjugation is a set-theoretic bijection. This is valid. You more or less re-proved that conjugation is an isomorphism. -hagman === === Subject: : finsler structure Can you suggest me where to study definition and main properties of Finsler structures? === === Subject: : Re: finsler structure > Can you suggest me where to study definition and main properties of > Finsler structures? > there's a 190 page book by s.s. chern & z. shen: riemann-finsler geometry, world scientific, 2005. there's a 440 page book by d. bao, s.s. chern and z. shen: an introduction to riemann-finsler geometry, springer, 2000. vale, rip -- NB eddress is r i p 1 AT c o m c a s t DOT n e t === === Subject: : Re: Humans Lack Genetics to Understand Physics > The right cerebral cortex is a picture/imagination > on mind , > but it uses the pixels to add value judgements to > o the picture . > This is far more productive in science than a left > t brained person > trying to use > a textual method . No scientist can work from the L side ! They'd be > be refering to the > book every second ! So now what is a professional scientist ?! No school teaches R side . All teach from a book . > All the worlds top scientists were R sided , and > nd had to teach > themselves > science . They had to translate text to pictures > es then memorise it . > It is impossible for any female to develope a R > a R side in science . > Every female ever born has a R side NOT twice as > as larg as her L side . > Its hormonal .. > The connection is that no female has ever > ver created new science . > Only white males ..... > ______________________________________________________ The names Marie Curie and Emmy Noether, among others, must be unknown to you. As well as the convention of not top-posting in this forum. Tom > ________ > Even professional physicists go home at > night, pop a beer, and watch the > game on TV. > Did Dick Feynman ever do that? > No, Since Fetmann was by his own > admission, a mathematician, not a phycist, > Not so, he said he worked through intuition. > Einstein > commented similarly, that his strength was insight > into > nature, and that math was a means to an end. > It would be interesting to survey professional > scientists, > to find out how many regard themselves primarily as > mathematicians. > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than > physics. OK > -- > Rich > === === Subject: : Re: Humans Lack Genetics to Understand Physics <18319467.1165162323062.JavaMail.jakarta@nitrogen.mathforum.org > The right cerebral cortex is a picture/imagination > on mind , > but it uses the pixels to add value judgements to > o the picture . > This is far more productive in science than a left > t brained person > trying to use > a textual method . > No scientist can work from the L side ! They'd be > be refering to the > book every second ! > So now what is a professional scientist ?! > No school teaches R side . All teach from a book . > All the worlds top scientists were R sided , and > nd had to teach > themselves > science . They had to translate text to pictures > es then memorise it . > It is impossible for any female to develope a R > a R side in science . > Every female ever born has a R side NOT twice as > as larg as her L side . > Its hormonal .. > The connection is that no female has ever > ver created new science . > Only white males ..... > ______________________________________________________ The names Marie Curie and Emmy Noether, among others, > must be unknown to you. Everybody's heard of Madame Curie. oitherwise the physics cranks wouldn't even know what alpha particels are. But that is obvious to most non math cranks, since alpha decay is a chemical reaction rather than a physical reaction. And Noether's theorem is the crutch maximus for all Quantum Mechanics cranks, since Hamilton discovered everything about it, except Euler.and Von Neumann. As well as the convention of not top-posting in this > forum. Tom > ________ > Even professional physicists go home at > night, pop a beer, and watch the > game on TV. > Did Dick Feynman ever do that? > No, Since Fetmann was by his own > admission, a mathematician, not a phycist, > Not so, he said he worked through intuition. > Einstein > commented similarly, that his strength was insight > into > nature, and that math was a means to an end. > It would be interesting to survey professional > scientists, > to find out how many regard themselves primarily as > mathematicians. > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than > physics. > OK > > -- > Rich > === === Subject: : Re: Humans Lack Genetics to Understand Physics The right cerebral cortex is a > picture/imagination > on mind , > but it uses the pixels to add value judgements > to > o the picture . > This is far more productive in science than a > left > t brained person > trying to use > a textual method . > No scientist can work from the L side ! They'd be > be refering to the > book every second ! > So now what is a professional scientist ?! > No school teaches R side . All teach from a > book . > All the worlds top scientists were R sided , and > nd had to teach > themselves > science . They had to translate text to pictures > es then memorise it . > It is impossible for any female to develope a R > a R side in science . > Every female ever born has a R side NOT twice as > as larg as her L side . > Its hormonal .. > The connection is that no female has ever > ver created new science . > Only white males ..... > ______________________________________________________ > The names Marie Curie and Emmy Noether, among > others, > must be unknown to you. > zz bunker replied: > Everybody's heard of Madame Curie. > oitherwise the physics cranks > wouldn't even know what alpha particels are. But that is obvious to most non math cranks, > since alpha decay is a chemical reaction > rather than a physical reaction. And Noether's theorem is the crutch maximus > for all Quantum Mechanics cranks, since > Hamilton discovered everything > about it, except Euler.and Von Neumann. Apparently, you possess little actual knowledge of either Hamilton's work, or Noether's. Tom > As well as the convention of not top-posting in > this > forum. > Tom > > ________ > Even professional physicists go home at > night, pop a beer, and watch the > game on TV. > Did Dick Feynman ever do that? > No, Since Fetmann was by his own > admission, a mathematician, not a phycist, > Not so, he said he worked through intuition. > Einstein > commented similarly, that his strength was > insight > into > nature, and that math was a means to an end. > It would be interesting to survey professional > scientists, > to find out how many regard themselves > primarily as > mathematicians. > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than > physics. > OK > > -- > Rich > === === Subject: : Re: Humans Lack Genetics to Understand Physics <15791023.1165187313941.JavaMail.jakarta@nitrogen.mathforum.org > The right cerebral cortex is a > picture/imagination > on mind , > but it uses the pixels to add value judgements > to > o the picture . > This is far more productive in science than a > left > t brained person > trying to use > a textual method . > No scientist can work from the L side ! They'd be > be refering to the > book every second ! > So now what is a professional scientist ?! > No school teaches R side . All teach from a > book . > All the worlds top scientists were R sided , and > nd had to teach > themselves > science . They had to translate text to pictures > es then memorise it . > It is impossible for any female to develope a R > a R side in science . > Every female ever born has a R side NOT twice as > as larg as her L side . > Its hormonal .. > The connection is that no female has ever > ver created new science . > Only white males ..... > ______________________________________________________ > The names Marie Curie and Emmy Noether, among > others, > must be unknown to you. > zz bunker replied: > Everybody's heard of Madame Curie. > oitherwise the physics cranks > wouldn't even know what alpha particels are. > But that is obvious to most non math cranks, > since alpha decay is a chemical reaction > rather than a physical reaction. > And Noether's theorem is the crutch maximus > for all Quantum Mechanics cranks, since > Hamilton discovered everything > about it, except Euler.and Von Neumann. > > Apparently, you possess little actual knowledge of > either Hamilton's work, or Noether's. Hamiton's is simple to understand, since he was the first string theorist, since the rest of them are misunderstood idiots, Noether' theorem is self explanitory, since Eistein was the first to claim GTR doesn't work without it, Tom > As well as the convention of not top-posting in > this > forum. > Tom > ________ > Even professional physicists go home at > night, pop a beer, and watch the > game on TV. > Did Dick Feynman ever do that? > No, Since Fetmann was by his own > admission, a mathematician, not a phycist, > Not so, he said he worked through intuition. > Einstein > commented similarly, that his strength was > insight > into > nature, and that math was a means to an end. > It would be interesting to survey professional > scientists, > to find out how many regard themselves > primarily as > mathematicians. > So he did want all mathematicians do, > he camped out in the Himalayas > looking for NASA Mirror grants, rather than > physics. > OK > > -- > Rich > === === Subject: : Re: nice,erdos,can. math bull === === Subject: : 2006 Putnam problems Would someone please be kind enough to post this year's Putnam Ray Steiner === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam For Brits, what are the Putnam problems? Nick === === Subject: : Re: 2006 Putnam problems Would someone please be kind enough to post this year's Putnam For Brits, what are the Putnam problems? The same problems as for Americans. 8-) The Putnam Exam is a continent-wide contest held every year where students are given 12 problems (in all) and have 6 hours (in all) to work on them. These problems require creativity; they are (usually) not regular problems which appear, for instance, in Calculus. Homepage: http://math.scu.edu/putnam/ Archive (old tests & solutions): http://www.unl.edu/amc/a-activities/a7-problems/putnam/ --- Christopher Heckman === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam >> For Brits, what are the Putnam problems? The same problems as for Americans. 8-) The Putnam Exam is a continent-wide contest held every year where > students are given 12 problems (in all) and have 6 hours (in all) to > work on them. These problems require creativity; they are (usually) not > regular problems which appear, for instance, in Calculus. Sounds like the Cambridge (UK) Entrance exam. They weren't problems found in maths books - and would demonstrate the ability to think outside the box. Nick === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner I am wondering if anybody has solutions to any of the Putnam questions that they would like to discuss. I was unable to take the Putnam this year (the GRE subject test was on the same day), but I have been looking over some of the problems. I was surprised by some of the later problems on the test, specifically A4 and B5. Both had very short solutions, which is usually only the case for the first couple of problems in each section. I haven't yet attempted A1 or B1, but at first glance, they both look to be more difficult than A4 or B5. === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam >> Ray Steiner I am wondering if anybody has solutions to any of the Putnam questions >that they would like to discuss. I was unable to take the Putnam this >year (the GRE subject test was on the same day), but I have been >looking over some of the problems. I was surprised by some of the >later problems on the test, specifically A4 and B5. Both had very >short solutions, which is usually only the case for the first couple of >problems in each section. I haven't yet attempted A1 or B1, but at >first glance, they both look to be more difficult than A4 or B5. For B1, use the identity x^3 + y^3 + z^3 - 3xyz = (x+y+z)(wx+w^2y+z)(w^2x+wy+z) where w is a cube root of unity. Mike Guy === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. B6 is a push-over or I have missed something. Did I do this right? B6) Find lim_{n->oo} {x_n^{k+1}}/n^k where k is an integer greater than one, x_0 is positive, and x_{n+1} is defined recursively as x_n + x_n^{-1/k}. Prove by induction that x_n ge x_0 + n * x_0^{-1/k}. x_0 = x_0 + 0 * x_0^{-1/k}. Suppose for r = 0,1,2, ..., n we have x_r ge x_0 + r*x_0^{-1/k}, then x_{n+1} = x_n + x_n^{-1/k} = x_0 + n * x_0^{-1/k} + (x_0 + n * x_0^{-1/k})^{-1/k} ge x_0 + (n+1) * x_0^{-1/k}. Now lim x_n^{k+1} / n^k ge lim n * x_0^{(k+1)/k} = oo. -- Michael Press === === Subject: : Re: 2006 Putnam problems Would someone please be kind enough to post this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. B6 is a push-over or I have missed something. Did I do this right? > B6) Find lim_{n->oo} {x_n^{k+1}}/n^k where k is an > integer greater than one, x_0 is positive, and x_{n+1} is defined > recursively as x_n + x_n^{-1/k}. Prove by induction that x_n ge x_0 + n * x_0^{-1/k}. > x_0 = x_0 + 0 * x_0^{-1/k}. > Suppose for r = 0,1,2, ..., n we have > x_r ge x_0 + r*x_0^{-1/k}, then > x_{n+1} = x_n + x_n^{-1/k} > = x_0 + n * x_0^{-1/k} + (x_0 + n * x_0^{-1/k})^{-1/k} > ge x_0 + (n+1) * x_0^{-1/k}. Now > lim x_n^{k+1} / n^k ge lim n * x_0^{(k+1)/k} = oo. -- > Michael Press Your mistake is on the last equation you give. You are trying to use the fact that (x_0 + n * x_0^{-1/k})^{-1/k} is greater than or equal to x_0^{-1/k}. However, this is incorrect as the exponent is negative. === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post > this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of > the Putnam questions > that they would like to discuss. I was unable to > take the Putnam this > year (the GRE subject test was on the same day), > but I have been > looking over some of the problems. I was > surprised by some of the > later problems on the test, specifically A4 and > B5. Both had very > short solutions, which is usually only the case > for the first couple of > problems in each section. I haven't yet > attempted A1 or B1, but at > first glance, they both look to be more difficult > than A4 or B5. > B6 is a push-over or I have missed something. Did I > do this right? > B6) Find lim_{n->oo} {x_n^{k+1}}/n^k where k is an > integer greater than one, x_0 is positive, and > x_{n+1} is defined > recursively as x_n + x_n^{-1/k}. > Prove by induction that x_n ge x_0 + n * > x_0^{-1/k}. > x_0 = x_0 + 0 * x_0^{-1/k}. > Suppose for r = 0,1,2, ..., n we have > x_r ge x_0 + r*x_0^{-1/k}, then > x_{n+1} = x_n + x_n^{-1/k} > = x_0 + n * x_0^{-1/k} + (x_0 + n * > x_0^{-1/k})^{-1/k} > ge x_0 + (n+1) * x_0^{-1/k}. > Now > lim x_n^{k+1} / n^k ge lim n * x_0^{(k+1)/k} = oo. > -- > Michael Press Your mistake is on the last equation you give. You > are trying to use > the fact that (x_0 + n * x_0^{-1/k})^{-1/k} is > greater than or equal to > x_0^{-1/k}. However, this is incorrect as the > exponent is negative. > Here is a very simple solution using Stolz-Cesaro. a_n clearly goes increasingly to infinity. Denoting f(x) = x + x^(-1/k), the limit L of (a_n)^((k+1)/k) / n can be computed easily using Stolz-Cesaro: L = lim_(n-->oo) ( f(a_n) ^ (1+1/k) - (a_n) ^ (1+1/k) = lim_(x-->oo) ( f(x)^(1+1/k) - x^(1+1/k) ) = lim_(x-->oo) x^(1+1/k) [ ( 1 + x^(-1-1/k) )^(1+1/k) - 1 ] = lim(y-->0) [ (1 + y)^(1+1/k) - 1 ]/y = 1+1/k Hence the original limit is (1+1/k)^k. V. Anisiu === === Subject: : Re: 2006 Putnam problems <2654414.1165398674325.JavaMail.jakarta@nitrogen.mathforum.org > Would someone please be kind enough to post > this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of > the Putnam questions > that they would like to discuss. I was unable to > take the Putnam this > year (the GRE subject test was on the same day), > but I have been > looking over some of the problems. I was > surprised by some of the > later problems on the test, specifically A4 and > B5. Both had very > short solutions, which is usually only the case > for the first couple of > problems in each section. I haven't yet > attempted A1 or B1, but at > first glance, they both look to be more difficult > than A4 or B5. > B6 is a push-over or I have missed something. Did I > do this right? > B6) Find lim_{n->oo} {x_n^{k+1}}/n^k where k is an > integer greater than one, x_0 is positive, and > x_{n+1} is defined > recursively as x_n + x_n^{-1/k}. > Prove by induction that x_n ge x_0 + n * > x_0^{-1/k}. > x_0 = x_0 + 0 * x_0^{-1/k}. > Suppose for r = 0,1,2, ..., n we have > x_r ge x_0 + r*x_0^{-1/k}, then > x_{n+1} = x_n + x_n^{-1/k} > = x_0 + n * x_0^{-1/k} + (x_0 + n * > x_0^{-1/k})^{-1/k} > ge x_0 + (n+1) * x_0^{-1/k}. > Now > lim x_n^{k+1} / n^k ge lim n * x_0^{(k+1)/k} = oo. > -- > Michael Press > Your mistake is on the last equation you give. You > are trying to use > the fact that (x_0 + n * x_0^{-1/k})^{-1/k} is > greater than or equal to > x_0^{-1/k}. However, this is incorrect as the > exponent is negative. > Here is a very simple solution using Stolz-Cesaro. a_n clearly goes increasingly to infinity. > Denoting f(x) = x + x^(-1/k), the limit L of (a_n)^((k+1)/k) / n > can be computed easily using Stolz-Cesaro: L = lim_(n-->oo) ( f(a_n) ^ (1+1/k) - (a_n) ^ (1+1/k) > = lim_(x-->oo) ( f(x)^(1+1/k) - x^(1+1/k) ) > = lim_(x-->oo) x^(1+1/k) [ ( 1 + x^(-1-1/k) )^(1+1/k) - 1 ] > = lim(y-->0) [ (1 + y)^(1+1/k) - 1 ]/y = 1+1/k Hence the original limit is (1+1/k)^k. V. Anisiu That's a very clever argument. I actually hadn't seen the Stolz-Cesaro theorem before, but it seems like a natural discrete analog of l'Hopital. I was trying to use something similar to solve this problem, but I couldn't see to pass from the limit of a sequence to the limit a continuous variable goes to infinity. === === Subject: : Re: 2006 Putnam problems > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. Question B6 was a quick one for me, surprisingly enough. However, my solution is not air tight. The idea is to assume that sequence a_n has a Laurent series expansion in 1/n. Then the leading power of n as well as the coefficient fall immediately fall out by matching coefficients on either side of the recurrent relation. It is then trivial to take the desired limit. Of course the Laurent series expansion is much stronger than what is necessary, as evidence by other solutions posted on the Putnam Directory site. However, it's a very nice assumption to work with. So, I wonder whether it's actually valid. One way to prove it is to somehow show that the defining recurrence relation fixes the coefficients of the Laurent expansion up to one free parameter and that the Taylor series tail has a finite radius of converence. Then the free parameter can be fixed from the some a_n (computed by recurrence from the initial condition) within this radius of convergence. I haven't tried estimating the coefficients as it superficially looks like a lot of work. But I wonder if there's a quicker way to see that a Laurent expansion would be valid. Igor === === Subject: : Re: 2006 Putnam problems > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. Question B6 was a quick one for me, surprisingly enough. However, my > solution is not air tight. The idea is to assume that sequence a_n has > a Laurent series expansion in 1/n. [...] That should have been Puiseux series, not Laurent. The leading power of 1/n is non-integral. Igor === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. And what about B5? The question basically asks for a function f that maximizes the integral from 0 to 1 of (f(x) * x ^ 2 - x * (f(x)) ^ 2). It is clearly sufficient to maximize (f(x) * x ^ 2 - x * (f(x)) ^ 2) for each x. By a simple computation, (f(x) * x ^ 2 - x * (f(x)) ^ 2) is maximal when f(x) = x / 2. === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. And what about B5? The question basically asks for a function f that > maximizes the integral from 0 to 1 of (f(x) * x ^ 2 - x * (f(x)) ^ 2). > It is clearly sufficient to maximize (f(x) * x ^ 2 - x * (f(x)) ^ 2) > for each x. By a simple computation, (f(x) * x ^ 2 - x * (f(x)) ^ 2) > is maximal when f(x) = x / 2. Agh! I was _minimizing_ this expression. --- Christopher Heckman === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. > And what about B5? The question basically asks for a function f that > maximizes the integral from 0 to 1 of (f(x) * x ^ 2 - x * (f(x)) ^ 2). > It is clearly sufficient to maximize (f(x) * x ^ 2 - x * (f(x)) ^ 2) > for each x. > By a simple computation, (f(x) * x ^ 2 - x * (f(x)) ^ 2) > is maximal when f(x) = x / 2. Agh! I was _minimizing_ this expression. --- Christopher Heckman Can't you make the expression arbitrarily small by choosing f to be a large, negative constant function? === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. I was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. I was surprised by some of the > later problems on the test, specifically A4 and B5. Both had very > short solutions, which is usually only the case for the first couple of > problems in each section. I haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. > And what about B5? The question basically asks for a function f that > maximizes the integral from 0 to 1 of (f(x) * x ^ 2 - x * (f(x)) ^ 2). > It is clearly sufficient to maximize (f(x) * x ^ 2 - x * (f(x)) ^ 2) > for each x. > By a simple computation, (f(x) * x ^ 2 - x * (f(x)) ^ 2) > is maximal when f(x) = x / 2. > Agh! I was _minimizing_ this expression. Can't you make the expression arbitrarily small by choosing f to be a > large, negative constant function? Yes; I made my minimize-instead-of-maximize mistake after I realized that. --- Christopher Heckman === === Subject: : Re: 2006 Putnam problems > æ æWould someone please be kind enough to post this year's Putnam > Ray Steiner > I am wondering if anybody has solutions to any of the Putnam questions > that they would like to discuss. æI was unable to take the Putnam this > year (the GRE subject test was on the same day), but I have been > looking over some of the problems. æI was surprised by some of the > later problems on the test, specifically A4 and B5. æBoth had very > short solutions, which is usually only the case for the first couple of > problems in each section. æI haven't yet attempted A1 or B1, but at > first glance, they both look to be more difficult than A4 or B5. > And what about B5? æThe question basically asks for a function f that > maximizes the integral from 0 to 1 of (f(x) * x ^ 2 - x * (f(x)) ^ 2). > It is clearly sufficient to maximize (f(x) * x ^ 2 - x * (f(x)) ^ 2) > for each x. > By a simple computation, (f(x) * x ^ 2 - x * (f(x)) ^ 2) > is maximal when f(x) = x / 2. > Agh! I was minimizing this expression. > Can't you make the expression arbitrarily small by choosing f to be a > large, negative constant function?Yes; I made my minimize-instead-of-maximize mistake after I realized > that. I just set $f(x) = frac{x}{2}-g(x)$, whence $I(f)-J(f) = int 0^1 frac{x^3}{4} dx - int 0^1 x [g(x)]^2 dx$. æ æ æ --- Christopher Heckman- Hide quoted text -- Show quoted text - === === Subject: : Re: 2006 Putnam problems The solution to A4 is only short if you know about random variables. > (And since it requires knowledge of this advanced topic, it got put > later in the test.) It would be horrible to solve using, say, > inclusion-exclusion. I agree that it is natural and intuitive to think about such a problem in terms of expectation of random variables, but I don't agree that the short solution requires knowledge of that advanced topic. You don't really need the language or conceptual framework of random variables; you merely need to recognize that the problem calls for the evaluation of a double sum, and then to observe that interchanging the order of summation simplifies the task. That is, the sum, over all permutations pi, of the number of local maxima of pi, is equal to the sum, over all k, of the number of permutations having a local maximum at k. I think this little trick is within reach of a clever student, even if he has not had a formal course in probability theory. === === Subject: : Re: 2006 Putnam problems > Would someone please be kind enough to post this > his year's Putnam > Ray Steiner > See: http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml V. Anisiu === === Subject: : Re: 2006 Putnam problems On Dec 3, 11:37 am, stei...@math.bgsu.edu Ray Steiner They are available in the October, 2006, issue of The American Mathematical Monthly. === === Subject: : Re: 2006 Putnam problems > On Dec 3, 11:37 am, stei...@math.bgsu.edu Would someone please be kind enough to post this year's Putnam > Ray Steiner They are available in the October, 2006, issue of The American > Mathematical Monthly. > No, that issue has the problems from a year ago. Gowan wants the problems from the exam of yesterday. === === Subject: : Re: 2006 Putnam problems I can post the ones I didn't submit (slight notational modifications): A3) Consider the sequence defined by $a_k = k$ for $k le 2006$, and recursively as $a_{k+1} = a_k + a_{k-2005}$ for $k ge 2006$ (the sequence appears as 1, 2, 3, ..., 2005, 2006, 2007, 2009, 2012, 2016, ....) Show that there exist 2005 consecutive terms which each have a factor of 2006. A4) Define a local maximum of a permutation of a finite set (of magnitude greater than 1) if any of the following conditions hold: i) the first element of the permutation is greater than the second element, ii) an element is greater than the elements on either side (assuming the element is not at the end) iii) the last element is greater than the preceding element. (For example, if we take a set of five elements, then the permutation arranging $Z_5$ as $2,1,4,5,3$ has a local maximum at the first and fourth elements.) Averaging over all permutations of a finite set, what is the average number of local maxima? B1) Show that there exists only one solution set to $x^3+3xy+y^3 = 1$ which form the vertices of a non-trivial equilateral triangle, and find the area. B6) Find $$lim_{ntoinfty} frac{x_n^{k+1}}{n^k}$$ where $k$ is an integer greater than one, $x_0$ is positive, and $x_{n+1}$ is defined recursively as $x_n + x_n^{-1/k}$. === === Subject: : Re: 2006 Putnam problems >B6) Find $$lim_{ntoinfty} frac{x_n^{k+1}}{n^k}$$ where $k$ is an >integer greater than one, $x_0$ is positive, and $x_{n+1}$ is defined >recursively as $x_n + x_n^{-1/k}$. Let y_n = x_n^(1+1/k). Then y_{n+1} = y_n (1 + 1/y_n)^(k/(k+1)). Using the binomial theorem (and some thought about signs of terms) y_n + 1 + 1/k <= y_{n+1} <= y_n + 1 + 1/k + (k+1)/(2 k^2 y_n). Hence there A, B > 0 such that A + n(1+1/k) <= y_n <= B + n(1+1/k) + log(n)/(2 k^2). [ Note that the second inequality is much tighter than the first. This shows that y_n has a logarithmic singularity as n -> oo, so x_n can't be expressed as a Laurent series, as suggested by another poster. ] Since x_n^{1+k} = y_n^k, thye required limit is (1 + 1/k)^k. Mike Guy === === Subject: : Re: 2006 Putnam problems > I can post the ones I didn't submit (slight notational modifications): > [...] > A4) Define a local maximum of a permutation of a finite set (of > magnitude greater than 1) if any of the following conditions hold: > i) the first element of the permutation is greater than the second > element, > ii) an element is greater than the elements on either side (assuming > the element is not at the end) > iii) the last element is greater than the preceding element. > (For example, if we take a set of five elements, then the permutation > arranging $Z_5$ as $2,1,4,5,3$ has a local maximum at the first and > fourth elements.) > Averaging over all permutations of a finite set, what is the average > number of local maxima? Hmm. This one I knew how to do right off. The answer is (n+1)/3 (if n >= 3), and the idea is to use random variables. Define X(p,k) to be 1 if k is a local max of the permutation p, and 0 otherwise. Then the average number of local maxes in a permutation is the expected value of X(p,1) + X(p,2) + ... + X(p,n), which is the sum of the expected values of each X(p,k). The expected value of X(p,k) is the probability that there is a local max at k. If k = 1 or n, then this value is 1/2 (choose two distinct numbers from {1,2,..,n}, one at a time; what is the probability that the first one is larger?), otherwise it is 1/3. The sum of the expected values thus becomes 1/2 + 1/3 * (n-2) + 1/2 = (n+1)/3. GENERALIZATION: Choose a positive integer m. If p is a permutation, and k is an integer, then k is a local m-max if p(k) > p(i) for all i such that 0 < abs(i-k) <= m. (Hence a local 1-max is a local max from the original problem.) What is the average number of m-maxes, averaged over all permutations? --- Christopher Heckman === === Subject: : Local k-Maxes in Latin Squares: Was: Re: 2006 Putnam problems > I can post the ones I didn't submit (slight notational modifications): > [...] > A4) Define a local maximum of a permutation of a finite set (of > magnitude greater than 1) if any of the following conditions hold: > i) the first element of the permutation is greater than the second > element, > ii) an element is greater than the elements on either side (assuming > the element is not at the end) > iii) the last element is greater than the preceding element. > (For example, if we take a set of five elements, then the permutation > arranging $Z_5$ as $2,1,4,5,3$ has a local maximum at the first and > fourth elements.) > Averaging over all permutations of a finite set, what is the average > number of local maxima? Hmm. This one I knew how to do right off. The answer is (n+1)/3 (if n >= 3), and the idea is to use random variables. Define X(p,k) to be 1 if k is a local max of the permutation p, and 0 > otherwise. Then the average number of local maxes in a permutation is > the expected value of X(p,1) + X(p,2) + ... + X(p,n), which is the sum of the expected values of each X(p,k). The expected > value of X(p,k) is the probability that there is a local max at k. If k > = 1 or n, then this value is 1/2 (choose two distinct numbers from > {1,2,..,n}, one at a time; what is the probability that the first one > is larger?), otherwise it is 1/3. The sum of the expected values thus > becomes 1/2 + 1/3 * (n-2) + 1/2 = (n+1)/3. GENERALIZATION: Choose a positive integer m. If p is a permutation, and > k is an integer, then k is a local m-max if p(k) > p(i) for all i such > that 0 < abs(i-k) <= m. (Hence a local 1-max is a local max from the original problem.) What is > the average number of m-maxes, averaged over all permutations? Better generalization: A Latin Square (see http://mathworld.wolfram.com/LatinSquare.html ) is a 2-dimensional version of a permutation. Define a local k-max of a Latin Square to be an entry (i,j) such that L(i,j) > L(i+a, j+b), for all (a,b) where the infinity norm of (a,b) is positive and <= k (wherever L(i+a,j+b) makes sense). What is the expected number of k-maxes in a Latin Square? (The local k-max definition could be relaxed to L(i,j) > L(i+m,j) and L(i,j) > L(i, j+m), where m is a nonzero integer between -k and k; this only considers the orthogonal neighbors and should be about as difficult as (A4) on the 2006 Putnam.) --- Christopher Heckman === Hey, I have a few copies of papers in Romanian that I wouldn't >> mind having translated, if you're really hard up! (Most by >> Solomon Marcus, such as MR 26 #3849, if you're curious.) > I just looked up the review. So, X is a complete separable > metric space with a sigma finite complete borel measure such > that singletons have measure 0 and there is a perfect set > which is rare (defn?) and such that whenever it has nonempty > intersection with a sphere of X, the intersection has positive > measure. The paper proves constructs a Frink set which is > G-delta and proves there is no Frink set which is F-sigma and > asks whether there exists a Frink set which is non-Borel or > non-analytic in Lusin's sense. By a Frink set, one means a > set of first category and of measure 0 which cannot be written > as a countable union of sets B_n such that such that > mu(Fr B_n) = 0. Does Fr B_n mean the boundary of B_n? Yes, and so they're talking about sets that are the analog of the Jordan measure zero sets in R^n, where for bounded subsets in R^n this means that a finite covering by open rectangles can have an arbitrarily small positive value for the sum of the volumes of the rectangles in the covering. Recall that Lebesgue measure zero is less restrictive, allowing for countable coverings. For bounded subsets of R^n, Jordan measure zero is equivalent to the closure having Lebesgue measure zero. For arbitrary subsets of R^n, the collection of countable unions of Jordan measure zero sets forms a sigma-ideal of small sets that is properly contained in the sigma-ideal of sets simultaneously Lebesgue measure zero and first category (Frink sets, a term that never caught on, are sets that verify properly contained in), and identical to the collection of subsets of F_sigma Lebesgue measure zero sets. [I think rare in the review means nowhere dense.] For much more about this, see the first post below and the comments under PROOFS OF 4 & 4': in the second post below. HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS References for Continuity Sets I mention some things about the Marcus paper that was written in Romanian in the first post above, which can be found be doing a page search for Marcus [14]. Yes, I realize none of this is anywhere near your mathematical interests. The details are in case someone besides us reads this post and is interested. Dave L. Renfro === > Yes, and so they're talking about sets that are the analog of > Yes, I realize none of this is anywhere near your mathematical > interests. The details are in case someone besides us reads > this post and is interested. Are there any independence results in set theory known or conjectured in connection with these strange ideals? For example, various kinds of sets are known to satisfy the axiom of determinacy (I think Martin proved Borel determinacy and I don't know what the story is with analytic sets). How about Frink sets? In Sierpinski's book, Hypothese du Continu, his proposition C25 asserts (assuming the continuum hypothesis) that there is a permutation f of the set of real numbers such that f induces a bijection of the set of sets of measure 0 onto the set of sets of first category. So, the power set of the reals modulo the ideal of sets of measure 0 is isomorphic to the power set of the reals modulo the ideal of sets of first category. So, C25 is a result that relates ideals of the sigma-algebra of measurable sets under the action of automorphisms of the power set of the set of all real numbers. Do you think there might be results like this for these other ideals you mentioned? Also, if B is the algebra of borel sets and I is its ideal of sets of measure 0 and J is its ideal of sets of first category, then the algebras B/I and B/J are complete algebras, even though B is only a sigma algebra and I,J are only sigma-ideals, and B/I and B/J are not isomorphic. Is B/K complete for K any of the other ideals you mentioned and, if so, are all these quotient algebras non-isomorphic? -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === Yes, I realize none of this is anywhere near your mathematical > interests. I think I mentioned that I do like stuff like this. is that I'm more intersted in getting books, especially for free. Books opportunities to make local sense of the document. So, from the standpoint of language acquisition (given that I'm going to be doing it in this way) is inevitably on the particular mathematics, while with a book, even though the medium is mathematics, I'm under less pressure to accomplish anything in particular. Since I'm already trying to accomplish a lot of particular things, I'm trying not to take on any more until I finish some of them. differential equations, which I don't know a lot about, and I've been incredibly side tracked trying to acquire the necessasry background and to read the bibliographical references. Reading mathematics in another language has the advantage of being on familiar ground, compared with reading a novel, but it has the disadvantage that I have certain standards that I bring to reading mathematics, no matter what language it is in, and then the reading takes on a life of its own. At the moment, that life of its own has now led me to work slowly through one of Ladyzhenskaya's books in Russian with a dictionary. It is a lot of fun. In the case of Romanian, what caught my eye was the existence of particular books in English that had been translated from books in Romanian and which I already know I would find useful. I guess I'm quibbling with the word interests which has one meaning in common parlance and another in the newspeak that professionals speak. I'm a big fan of common parlance, when it already does the job. Anyway, if there was actually a book on this stuff in Romanian and I could have it for free, I'd be more inclined to go for it. Meanwhile, I think I've made a little progress. At the website http://www.theta.ro/Right.html there are translations into Romanian of various books, including Rudin's Real and Complex Analysis, Rockafellar's Convex Analysis and Narasimhan's Analysis on Manifolds. I don't really understand the prices or ordering procedures (the following is for Narasimhan's book): 218 pagini -----------------> 218 pages 2001 Theta, Bucuresti --------> Bucharest: Theta, 2001 ISBN 973-99097-9-5 120 000 lei (pret de librarie) ------> ? (interlibrary loan?) 15% reducere la distributie directa ---> 15 percent discount? To whom? I'll keep looking. -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === through one of Ladyzhenskaya's books in Russian with a dictionary. > It is a lot of fun. I don't know if this would be of interest to you, but I thought I'd mention the following book. Russian for Mathematicians by Olga Glazunova, St. Petersburg University Press (1997). http://tinyurl.com/yhtbca [book information] http://tinyurl.com/ajrgx [preface of book] http://tinyurl.com/yf9jok [Olga's web page] Dave L. Renfro === > At the moment, that life of its own has now led me to work slowly > through one of Ladyzhenskaya's books in Russian with a dictionary. > It is a lot of fun. I don't know if this would be of interest to you, but I thought > I'd mention the following book. > Russian for Mathematicians by Olga Glazunova, St. Petersburg > University Press (1997). > http://tinyurl.com/yhtbca [book information] > http://tinyurl.com/ajrgx [preface of book] > http://tinyurl.com/yf9jok [Olga's web page] scientific Russian. Also a German book entitled Russisch fuer Mathematiker. What would be really useful would be something like a Russian equivalent of Macintyre and Witte's German-English Mathematical Vocabulary. But actually I seem to be managing quite well with just a dictionary, even though I don't know much about the language. In the right context, a vague word that seems to mean not broken pops out as continuous. I now recall that when I was in Paris a couple of decades ago, I passed the Sorbonne and saw some kind of event taking place inside. One of the features was a table that had, among other things, some Romanian mathematics books. I shouldn't have resisted my impulse to get one of them. -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === ... > 218 pagini -----------------> 218 pages > 2001 Theta, Bucuresti --------> Bucharest: Theta, 2001 > ISBN 973-99097-9-5 > 120 000 lei (pret de librarie) ------> ? (interlibrary loan?) I think you've been taken in by false friends. That surely means price for bookstores. Lee Rudolph === conjectured in connection with these strange ideals? For > example, various kinds of sets are known to satisfy the axiom > of determinacy (I think Martin proved Borel determinacy and > I don't know what the story is with analytic sets). How > about Frink sets? Oh, I forgot to mention that there are some cardinal invariants results involving F_sigma measure zero sets which, depending on the model of ZFC, can be different than for the collection of sets simultaneously measure zero and first category. See the references listed in my post below where cardinal invariants first appears (do a search in page for this phrase). HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS Dave L. Renfro === conjectured in connection with these strange ideals? For > example, various kinds of sets are known to satisfy the > axiom of determinacy (I think Martin proved Borel determinacy > and I don't know what the story is with analytic sets). > How about Frink sets? In Sierpinski's book, Hypothese du Continu, his proposition > C25 asserts (assuming the continuum hypothesis) that there > is a permutation f of the set of real numbers such that f > induces a bijection of the set of sets of measure 0 onto > the set of sets of first category. So, the power set of > the reals modulo the ideal of sets of measure 0 is isomorphic > to the power set of the reals modulo the ideal of sets of > first category. So, C25 is a result that relates ideals of the sigma-algebra > of measurable sets under the action of automorphisms of the > power set of the set of all real numbers. Do you think there > might be results like this for these other ideals you mentioned? Also, if B is the algebra of borel sets and I is its ideal > of sets of measure 0 and J is its ideal of sets of first > category, then the algebras B/I and B/J are complete algebras, > even though B is only a sigma algebra and I,J are only > sigma-ideals, and B/I and B/J are not isomorphic. Is B/K > complete for K any of the other ideals you mentioned and, > if so, are all these quotient algebras non-isomorphic? I don't know much about this aspect of sigma-algebras and sigma-ideals, but I do know that F_sigma measure zero sets (Frink sets are measure zero and first category sets that are too big to be covered by an F_sigma measure zero set) make an appearance in Oxtoby's book Measure and Category in a few places, in particular, I think in Chapter 13 Transforming Linear Sets into Null Sets. (I don't have Oxtoby's book with me now, however.) If anyone was really interested in pursuing this, a good person to ask is Marek Balcerzak (Technical University of Lodz). He's written a number of papers extending in various ways (to generalized F_sigma null notions -- what are now usually called F_sigma supported sigma-ideals, arbitrary sigma-ideal/algebra notions, etc.) related work by R. Daniel Mauldin, Dorothy Maharam Stone, Alexander S. Kechris, and others on these sorts of issues. Marek's Habilitation Thesis was on this topic, for example (first URL below). Marek's e-mail address is at the second URL below. http://bazy.opi.org.pl/raporty/opisy/dokhab/19000/d19802.htm http://www.ii.uj.edu.pl/EMIS/journals/JAA/vol1i2/7.html Dave L. Renfro === === Subject: : Re: The proof of the Riemann hypothesis Why would you have to say things like Do you have a reason or should I start thinking that perhaps this is why the paper is ignored? The last sentence is particularly hurting. But, of course, if the idea is proved to be wrong, I will recommend you to ignore it. Well, I have revised the paper over the past week. In the revised version, you will see more clearly that we 1. first calculate the fraction AT (pardon my capitalization) x = delta. In replying to your comment, if we consider the double limiting process (of x and y), then the first limit is x -> delta, and the second is y -> infinity. The purpose, however, is to calculate the fraction at x = delta, and to do so, the first limit MUST be x -> delta. 2. In the revised version, it is briefly shown that the fraction is continuous on some interval (delta - epsilon, delta + epsilon) epsilon > 0. Well, if we know that a function is continuous on an interval (a, b) except possibly for c in (a, b) and is equal to a finite value at c, then the whole argument will proceed smoothly. Hisanobu === === Subject: : Re: The proof of the Riemann hypothesis > Why would you have to say things like Do you have a > reason or should I start thinking that perhaps this > is why the paper is ignored? > The last sentence is particularly hurting. But, of > course, if the idea is proved to be wrong, I will > recommend you to ignore it. Sorry if I hurt you. As I told you, I really hope there are no flaws in the argument. But I don't think you should be so sensitive to critics other than the one from a qualified referee. What matters what I or anyone has to say if the paper was revised by a referee? That's why for me it's more important that type of recognition than collecting nice comments in a newsgroup. > Well, I have revised the paper over the past week. In > the revised version, you will see more clearly that > we 1. first calculate the fraction AT (pardon my > capitalization) x = delta. In replying to your > comment, if we consider the double limiting process > (of x and y), then the first limit is x -> delta, and > the second is y -> infinity. The purpose, however, is > to calculate the fraction at x = delta, and to do so, > the first limit MUST be x -> delta. 2. In the revised version, it is briefly shown that > the fraction is continuous on some interval (delta - > epsilon, delta + epsilon) epsilon > 0. Well, if we > know that a function is continuous on an interval (a, > b) except possibly for c in (a, b) and is equal to a > finite value at c, then the whole argument will > proceed smoothly. Not so fast...the continuity of the function is not enough to justify the interchange of limits, because the second limit is taken when y-->oo in the integral from 0 to y. So, in order to prove that you can really interchange limits there, you need something stronger than continuity in (a,b), you will need uniform continuity in the interval 0 delta is useful. Hisanobu === === Subject: : Re: The proof of the Riemann hypothesis > Have you tried using the following theorem in your > verification? Let f_{n}(x) be uniformly convergent on a set E in a > metric space. Suppose that lim f_{n}(x) > x to A exists for each n, where A is a limit point of E. > Then we have lim lim f_{n}(x) > xto A n to infinity = lim lim f_{n}(x). > n to infinity x to A Now my idea in the new version of the paper is as > follows. I consider the integrals f_{n}(x) > = int_{0}^{n}[t^{1/2 + x}sin(Tln(t))]dt/[(e^{t} + > 1)t] and g_{n}(x) > = int_{0}^{n}[t^{1/2 - x}sin(-Tln(t))]dt/[(e^{t} + > 1)t]. It is plain that both functions are uniform > convergent on (0, 1/2). Heyyy...why is it plain that both are uniformly convergent? This is exactly what I was trying to prove and I couldn't. Do you have a proof of this? If this is true, of course, then the limits interchange is justifiable. > But I need to ask one more question on your last > reply. Did you agree on my comment about the > fractions which are not L'Hospital-applicable? I > really do not know why the interchanging matters for > those cases. It doesn't. The thing is that I had modified somewhat the justification for making l'Hopital rule applicable even when g'(y)=0 since in the form it was stated it didn't work. For instance, we can get to the following form of mean value theorem: (f(b)-f(a))g'(c)=(g(b)-g(a))f'(c) as long as g(b)is different from g(a). Your argument is simplifying both f' and g' by sin(Tlny) in order to apply limits and come to l'Hopital rule conclusion, but this may not be possible since then you would be simplifying by zero, which is not allowed. So what I did to bypass this is consider the limit of the quotient R_T(x)/I_T(x) when x-->delta instead of using the quotient R_-T(-x)/R_T(x). This way there would be a cos implied there, which guarantees that f' and g' cannot be BOTH zero. This then implies that both are non zero, provided we chose a and b accordingly (and we can may that). The rest of the proof can be achieved easily using your ideas for getting the contradiction, so I was not too much worried about l'Hopital rule application. What needs still to be settled is the uniform convergence of the integrals, since this is crucial for justifying the double limit process. So if you have any proof of this, it would be very interesting to take a look at it as I was not able to find any. Hisanobu === === Subject: : Re: The proof of the Riemann hypothesis I think the uniform convergence of f_{n}(x) and g_{n}(x) can be obtained easily by the following theorem, which is based on the Cauchy criterion: f_{n}(x) defined on E is uniformly convergent if and only if for every u > 0, there exists an integer N such that m, n geq N, and x in E implies that |f_{n}(x) - f_{m}(x)| leq u. I'm sorry that I can not give a reference to book other than Rudin's, but this theorem is also in Principles of Mathematical Analysis. Now, however, the criterion is trivially satified, for if n geq m, f_{n}(x) - f_{m}(x) = int_{m}^{n}[t^{1/2 + x}sin(Tln(t))]dt/[(e^{t} + 1)t]. The integrand tends to zero as m approaches infinity. I would like to hear more about your idea on making L'Hospital's rule applicable. What would you mean by the following ? > This way there would be a cos implied there... In addition, how could any argument on R(x)/I(x) prove the limit R(-x)/R(x) ? In addition, on the following: >since then you would be simplifying by zero, which is >not allowed My original thought was that just as, say, (x - 1)/(x - 1) = 1 for all x, the above thought may not be necessary. Hisanobu === === Subject: : Re: The proof of the Riemann hypothesis >1. first calculate the fraction AT (pardon my >capitalization) x = delta. In replying to your comment, >if we consider the double limiting process (of x and y), >then the first limit is x -> delta, and the second is > y -> infinity. The purpose, however, is to calculate >the fraction at x = delta, Here, the fraction is the one to which L'Hospital's rule is applicable. >and to do so, the first >limit MUST be x -> delta. It seems that you are concerned with the order of the so-called double limit on the fractions which are not L'Hospital-applicable. For those, we do not need to think about the double limiting process; just honestly let x in the fraction R(x)/I(x) etc. approaches delta. Information on those terms as x to delta is shown in the paper. Hisanobu === === Subject: : Re: Nilpotent, Idempotent days. My association with the Department is that of an alumnus. >Some questions about nilpotenc/idempotence (hope those are words). >>[A denotes nxn square matrix] >>1) Can A^k = A for k > 2, while not for k=2, for some A ? >> Yes. Consider the standard matrix representation of a rotation of the >> plane by 360/(k-1) degrees. To get one in which A^3 = A but A^2 is not >> equal to A, take a rotation by 180 degrees. To get one for k=4, take >> the rotation by 120 degrees. Etc. Excellent. So the example helps? I should point out, then, that your phrasing was ambiguous. I answered the question I thought you had asked, others answered a different question, and both are reasonable interpretations of what you that was sheer chance. (i) Does there exists a matrix A and a positive integer k>2 such that A^k=A, and A^2 is not equal to A? [That's the question I answered, yes, and I gave you an example for each separate choice of k>2; note that the matrix A changes depending on what k is.] (ii) Does there exist a matrix A such that, for ALL positive integers k>2, A^k=A, but such that A^2 is not equal to A? [That's a question that was answered elsewhere; the answer there is NO, because if A^3=A, and A^6 = A, then A = A^6 = (A^3)^2 = A^2]. So you really need to be careful about how you pose the question. > How you remember enough related math to come up with >examples amazes me. This is fairly basic, if you understand that, in a sense, matrices are linear transformations. Finding a linear transformation of the plane that cycles (that is, such that T^k is eventually the same as T) is easy, and finding one that cycles exactly as long as you want it also easy. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Nilpotent, Idempotent Originator: pouya@localhost [Stephen Montgomery-Smith ] [...] Suppose A and B are symmetric real matrices, B non-singular, such that x^tBx=0 implies that x^tAx>=0. (x^t denotes the transpose of the column vector x.) Show that there is a real number s such that A+sB is positive semi definite. >(I'm doing this result from memory so it is not impossible I skipped a hypothesis, but essentially this result is correct.) [...] [Stephen] >Yes, you are correct. If B is non-singular, then I don't need that >extra hypothesis on B. So my original statement still stands uncorrected. I will tell you that it took me several months to find the proof. The >result is not new, as I found out from a post I made to >sci.math.research several years ago, but I don't currently remember the >reference I was given (it was quite old). I don't think it is a well >known result. think I should be able to find the s.m.r. post you referred to. -- Pouya D. Tafti p dot d dot tafti at ieee dot org === === Subject: : How to approximate zeroes (nodes) of a multidimensional function? I have a continuous differentiable real function f(R) defined on R^N (not a polynomial). I am interested in the hypersurface defined by the equation f(R)=0. I know that such an hypersurface exist (we call it a node), and it has dimension N-1, meaning it divides R^N in two disjoint sets (there are other more complicated cases where I have more than one node, but let's keep things simple for now). I would like to know if there exist a way to approximate this node building a sequence of polynomials p1, p2, ... whose node defined by p1(R)=0, p2(R)=0 .... converge to the node I am interested. === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? I have a continuous differentiable real function f(R) defined on R^N > (not a polynomial). continuous differentiable is redundant. Did you mean continuously differentiable? (I might add that using R for a variable in R^N isn't a good idea.) > I am interested in the hypersurface defined by the equation f(R)=0. I > know that such an hypersurface exist (we call it a node), and it has > dimension N-1, meaning it divides R^N in two disjoint sets (there are > other more complicated cases where I have more than one node, but > let's keep things simple for now). I would like to know if there exist a way to approximate this node > building a sequence of polynomials p1, p2, ... whose node defined by > p1(R)=0, p2(R)=0 .... converge to the node I am interested. Assuming f is C^1 and grad(f) never vanishes on the node, the node is locally a graph. By the Weierstrass Approximation theorem, the node of f can then be locally approximated by nodes of polynomials p_n - in the sense that d(node(p_n),node(f)) -> 0. === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? node is locally a graph. By the Weierstrass Approximation > theorem, the node of f can then be locally approximated by nodes > of polynomials p_n - in the sense that d(node(p_n),node(f)) -> 0. so, is there a way to construct those polynomials? Or is the Weierstrass theorem an existence only theorem? === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? >> Assuming f is C^1 and grad(f) never vanishes on the node, the >> node is locally a graph. By the Weierstrass Approximation >> theorem, the node of f can then be locally approximated by nodes >> of polynomials p_n - in the sense that d(node(p_n),node(f)) -> 0. so, is there a way to construct those polynomials? Or is the Weierstrass theorem an existence only theorem? Suppose you want to approximate f on the rectangle G = {x: a_j <= x_j <= a_j + r_j, j=1..n}. For each positive integer k take g_k(x) = M_k^(-1) product_{j=1}^n (4 r_j^2 - x_j^2)^k where M_k = product_{j=1}^n int_{-2 r_j}^{2 r_j} (4 r_j^2 - x_j^2)^k dx_j Let P_k(x) = int_G f(y) g_k(x - y) dy_1 ... dy_n. Then P_k are polynomials, and P_k -> f uniformly on G. Moreover, estimates are available of how closely P_k approximates f on G. Suppose 1) |grad f| > c > 0 on the intersection of the node with G 2) |grad f(x) - grad f(y)| < c/2 for x, y in G with |x-y| < d 3) |f(x)| > e for x in G with distance(x, node) >= d 4) P_k is chosen so |P_k(x) - f(x)| < min(e, cd/2) for x in G. Let G_d = {x in G: dist(x, boundary(G)) > d}. Then a) for every x in G intersect node(P_k), distance(x, node(f)) < d. b) for every x in G_d intersect node(f), distance(x, node(P_k)) < d. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? node is locally a graph. By the Weierstrass Approximation > theorem, the node of f can then be locally approximated by nodes > of polynomials p_n - in the sense that d(node(p_n),node(f)) -> 0. How can I construct those polinomials, given f ? === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? I have a continuous differentiable real function f(R) defined on R^N > (not a polynomial). > I am interested in the hypersurface defined by the equation f(R)=0. I > know that such an hypersurface exist (we call it a node), and it has > dimension N-1, meaning it divides R^N in two disjoint sets (there are > other more complicated cases where I have more than one node, but > let's keep things simple for now). f(v) = 0 for v in R^n will not always define a hypersurface dividing the space R^n into disjoint sets. For example, if v = (x,y) in R^2 then f(v) = |x| + |y| does not do so. I would like to know if there exist a way to approximate this node > building a sequence of polynomials p1, p2, ... whose node defined by > p1(R)=0, p2(R)=0 .... converge to the node I am interested. > There is a theorem due mostly to Weierstrasse to the effect that given a real valued function continuous on a compact region in R^n there is a sequence of polynomial functions which converges uniformly to the given function on that region. === === Subject: : Re: How to approximate zeroes (nodes) of a multidimensional function? the space R^n into disjoint sets. I know, that's why I explicitely said that I know (mainly from physical arguments) that such a hypersurface exists === === Subject: : Re: Mapping of integer functions into reals > Z is the set of all integers. > S is the set of all Z -> Z functions. Is it possible to exist injective S -> R function W such that for every > x, y in Range(W): W(W^-1(x) + W^-1(y)) = u(x, y) > W(W^-1(x) o W^-1(y)) = v(x, y) where u and v are R^2 -> R functions such that: case 1. u is continuous > case 1.1. u(x, y) = x + y > case 2. v is continuous > case 3. u and v are continuous If for some of the cases is possible W to exist, can you give an > example of W? case 1.1: Consider the Q-vector space Q^Z of functions Z->Q. Note that for infinite dimensions, a basis of a Q-vector space has the same cardinality as the vector space itself. Hence Q^Z and R are isomorphic as Q-vector spaces. An isomorphism between Q^Z and R (or even any monomorphism from Q^Z to R) induces an injective mapping S->R such that u is addition. Since finding a basis of either Q^Z or R involves the axiom of choice, we would be very lucky to have v continuous at the same time... -hagman === === Subject: : Re: Mapping of integer functions into reals > Z is the set of all integers. > S is the set of all Z -> Z functions. > Is it possible to exist injective S -> R function W such that for every > x, y in Range(W): > W(W^-1(x) + W^-1(y)) = u(x, y) > W(W^-1(x) o W^-1(y)) = v(x, y) > where u and v are R^2 -> R functions such that: > case 1. u is continuous > case 1.1. u(x, y) = x + y > case 2. v is continuous > case 3. u and v are continuous > If for some of the cases is possible W to exist, can you give an > example of W? case 1.1: > Consider the Q-vector space Q^Z of functions Z->Q. > Note that for infinite dimensions, a basis of a Q-vector space has the > same cardinality as the vector space itself. Hence Q^Z and R are > isomorphic as Q-vector spaces. > An isomorphism between Q^Z and R (or even any monomorphism from Q^Z to > R) induces an injective mapping S->R such that u is addition. Since finding a basis of either Q^Z or R involves the axiom of choice, > we would be very lucky to have v continuous at the same time... So in short: is W constructible for case 1.1 or case 2? If so, can you give an algorithm for W (preferably bijective)? Theron === === Subject: : Re: Mapping of integer functions into reals <011220062048397633%edgar@math.ohio-state.edu.invalid > Z is the set of all integers. > S is the set of all Z -> Z functions. > Is it possible to exist injective S -> R function W such that for every > x, y in Range(W): > W(W^-1(x) + W^-1(y)) = u(x, y) > W(W^-1(x) o W^-1(y)) = v(x, y) > where u and v are R^2 -> R functions such that: > case 1. u is continuous > case 1.1. u(x, y) = x + y > case 2. v is continuous > case 3. u and v are continuous > If for some of the cases is possible W to exist, can you give an > example of W? > Theron > PS. Is there something unclear here? There is a natural topology on S. > Check that the operations + and o are continuous in that topology. > And S is homeomorphic to a subset of R, for example the irrationals. > So let W be such a homeomorphism. Can you give an explicit example of W? In other words: If F = {..., f(-2), f(-1), f(0), f(1), f(2), ...} is a bi-infinite integer sequence corresponding to Z -> Z function f, how to map F into a unique real number in such a way that u and v as described above are continuous (preferably expressible in closed form)? Here is my idea for injective mapping of an arbitrary Z -> Z function f into real number x: 1. the integer part of x = f(0). 2. encode the sequence {f(1), f(-1), f(2), f(-2), ...} as follows: for every f(i) (i <> 0), d = the number of binary digits of f(i). Then place in x: d ones and 0 (the stop bit) if f(i) >= 0 d zeros and 1 otherwise After this place in x the binary digits of f(i). For example if f(n) = n^2 then (in binary): x = 0.101 101 1110100 1110100 111101001 111101001 ... f(0).f(1) f(-1) f(2) f(-2) f(3) f(-3) Does S^2 -> S functions + and o correspond to continuous R^2 -> R functions u and v with this mapping? Theron === === Subject: : Re: Mapping of integer functions into reals <011220062048397633%edgar@math.ohio-state.edu.invalid > Z is the set of all integers. > S is the set of all Z -> Z functions. > Is it possible to exist injective S -> R function W such that for every > x, y in Range(W): > W(W^-1(x) + W^-1(y)) = u(x, y) > W(W^-1(x) o W^-1(y)) = v(x, y) > where u and v are R^2 -> R functions such that: > case 1. u is continuous > case 1.1. u(x, y) = x + y > case 2. v is continuous > case 3. u and v are continuous > If for some of the cases is possible W to exist, can you give an > example of W? > Theron > PS. Is there something unclear here? > There is a natural topology on S. > Check that the operations + and o are continuous in that topology. > And S is homeomorphic to a subset of R, for example the irrationals. > So let W be such a homeomorphism. Can you give an explicit example of W? In other words: If F = {..., f(-2), f(-1), f(0), f(1), f(2), ...} is a > bi-infinite integer sequence corresponding to Z -> Z function f, how to > map F into a unique real number in such a way that u and v as described > above are continuous (preferably expressible in closed form)? Here is my idea for injective mapping of an arbitrary Z -> Z function f > into real number x: 1. the integer part of x = f(0). > 2. encode the sequence {f(1), f(-1), f(2), f(-2), ...} as follows: for every f(i) (i <> 0), d = the number of binary digits of f(i). Then > place in x: > d ones and 0 (the stop bit) if f(i) >= 0 > d zeros and 1 otherwise After this place in x the binary digits of f(i). For example if f(n) = n^2 then (in binary): > x = 0.101 101 1110100 1110100 111101001 111101001 ... > f(0).f(1) f(-1) f(2) f(-2) f(3) f(-3) Does S^2 -> S functions + and o correspond to continuous R^2 -> R > functions u and v with this mapping? Theron I'd do it similar, but avoiding the binary digit stuff: Select a homeomorhism g: R->]0,1[, e.g. x->exp(x)/(1+exp(x)). Then x->a+(b-a)*g(x) is a homeomorphism R->]a,b[. Note that g(x+1)-g(x) < c for some c<1 (namely c=g(1/2)-g(-1/2)). Given f, I will construct an irrational number W(f) in such a manner that [W(f)] = f(0), i.e. I start with interval ]f(0),f(0)+1[. Then in sequence for f(1), f(-1), f(2), f(-2), ..., I shall shorten this interval according to the function value in question, thus resulting in a limit: Assume we have restricted W(f) to ]a,b[ and the next function value is f(n). If we had all of R at our disposal, I would take ]f(n),f(n)+1[ as interval. Since we don't, I use our homeomorphism g and take ]a+(b-a)g(f(n),a+(b-a)g(f(n)+1)[ , which is smaller at least by a factor of c. Therefore, the nested intervals converge to a real number. Since different functions f will lead to disjoint intervals at some stage, the mapping is injective. As G. A. Edgar said, addition and composition induce continuous maps u,v essentially because W is continuous with respect to the product topology on S and +,? are continuous. More explicitly, if f and f' agree on {-n,...,+n} for some n>0 then |W(f)-W(f')| < c^(2n). And vice versa, if |W(f)-W(f')| is small, then the sequences of intervals for f and f' coincide on many terms, i.e. f and f' agree on {-n,...,+n} for some large n. Therefore, W is in fact a homeomorphism between S and Im(W). Addition in S is continuous because sums of functions agreeing on {-n,..,n} agree on the same set. Composition in S is continuous because if f,f' agree on {-n,..,n} and h,h' agree on {-m,..,m} where m = max{|f(k)|:-n<=k<=n}, then h?f and h'?f' agree on {-n,n}. Therefore the induced functions u,v: Im(W) x Im(W) ->Im(W) are continuous. That doesn't imply that they can be extended to continuous functions u,v: R^2->R. At least v cannot be extended. The difficulties come from the fact that for example W(f)->1 if f(0)=0 and f(1)->+oo, hence with h(x)=x+1 we should have v(1,W(h))=lim(W(f?h)) taking the limit over such f, but [W(f?h)]=f(1)->oo. As done here, Im(W) is dense in R and RIm(W) has measure 0. We could just as well have produced a different W such that Im(W) has measure 0. This would make it easier to get continuous extensions. === === Subject: : Re: Mapping of integer functions into reals <011220062048397633%edgar@math.ohio-state.edu.invalid I'd do it similar, but avoiding the binary digit stuff: > Select a homeomorhism g: R->]0,1[, e.g. x->exp(x)/(1+exp(x)). > Then x->a+(b-a)*g(x) is a homeomorphism R->]a,b[. > Note that g(x+1)-g(x) < c for some c<1 (namely c=g(1/2)-g(-1/2)). > Given f, I will construct an irrational number W(f) in such a manner > that [W(f)] = f(0), i.e. I start with interval ]f(0),f(0)+1[. > Then in sequence for f(1), f(-1), f(2), f(-2), ..., I shall shorten > this interval according to the function value in question, thus > resulting in a limit: > Assume we have restricted W(f) to ]a,b[ and the next function value is > f(n). > If we had all of R at our disposal, I would take ]f(n),f(n)+1[ as > interval. > Since we don't, I use our homeomorphism g and take > ]a+(b-a)g(f(n),a+(b-a)g(f(n)+1)[ , which is smaller at least by a > factor of c. Therefore, the nested intervals converge to a real number. > Since different functions f will lead to disjoint intervals at some > stage, the mapping is injective. I'm trying to implement it in a program. I implemented W as follows (in pseudo code): x = abs(f(0)) bound = 1 n = 1 repeat until n < maxn { bound = bound * (g(f(n) + 1) - g(f(n))) x = x + g(f(n)) * bound bound = bound * (g(f(-n) + 1) - g(f(-n))) x = x + g(f(-n)) * bound n = n + 1 } if f(0) < 0 then x = -x If this program is correct, how to implement W^-1? Theron === Content-Length: 2221 Originator: rusin@vesuvius Interesting lecture if you live within travelling distance from Toronto, Ontario, Canada ++++++++++++++++++++++++++++++++++++++++++ Title to risk management of physical options http://toronto.ieee.ca/events/dec0406.htm Speaker Puica Nitu Senior Advisor in Energy Markets Ontario Power Generation Inc. 700 University Avenue Toronto, Ontario Day & Time Monday, December 4, 2006, 6:00 p.m. - 8:30 p.m Location Room BA 2165 Bahen Centre for Information Technology University of Toronto - St. George Campus 40 St. George Street MAP ==> http://oracle.osm.utoronto.ca/map/index2.html (code BA) Organizer The Signals & Computational Intelligence Joint Chapter (http://toronto.ieee.ca/chapters/s_ci.htm) Contact Bruno Di Stefano, b DOT distefano AT ieee DOT org +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Feel free to bring along any colleague who may be interested in this talk. The Signals & Computational Intelligence Joint Chapter is an IEEE Toronto chapter of: - IEEE Aerospace and Electronic Systems Society - IEEE Computational Intelligence Society - IEEE Geoscience and Remote Sensing Society - IEEE Oceanic Engineering Society - IEEE Ultrasound, Ferroelectrics, and Frequency Control Society - IEEE Vehicular Technology Society If you are a member of one or more of these societies and wish to be active in our chapter, please, write to me an short e-mail message. Bruno Di Stefano +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Relevant web pages http://toronto.ieee.ca/index.html http://toronto.ieee.ca/chapters/s_ci.htm http://toronto.ieee.ca/events/dec0406.htm http://ieee-cis.org/ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ -- -- Bruno Di Stefano ---------------------------------------------------------------------------- ----------- http://www3.sympatico.ca/nuptek ---------------------------------------------------------------------------- ----------- === === Subject: : Re: Is Ed Conrad a PSYCHOPATH, sadist, pervert, pedophile and a murderer ? My dear good friend Ed Conrad, Do you work for FBI or NSA as being accused in this thread. How many innocent americans did you murder and how many kids did you rape ? Why are you not replying and denying that you work or worked for FBI or NSA as being accused by this AGCE or AGPE ? >< > FEDERAL LAW: You must be 18 to view these obscene pix. > < > http://mysite.verizon.net/edconrad/FOSSILS/TightFit.jpg > http://mysite.verizon.net/edconrad/FOSSILS/SaveTheWhale.jpg > < > < > LOOK WHAT'S BEEN FOUND BETWEEN ANTHRACITE VEINS > < (Physical Evidence Evolution is a Bunch of Crap) > < > http://mysite.verizon.net/edconrad/FOSSILS/OldestHumanSkull.JPG > http://mysite.verizon.net/edconrad/FOSSILS/HumanBrain.jpg > http://mysite.verizon.net/edconrad/FOSSILS/ManasOldasCoal.jpg > http://mysite.verizon.net/edconrad/FOSSILS/FirstDiscovery.jpg > http://mysite.verizon.net/edconrad/FOSSILS/MoreFossils.jpg > http://mysite.verizon.net/edconrad/FOSSILS/ManasOldasCoal.jpg > http://mysite.verizon.net/edconrad/FOSSILS/TestResults.jpg > http://mysite.verizon.net/edconrad/FOSSILS/OldestTool.jpg > < > < > Or, in other words: > < > < CRIME IN RHYME > The anthros, their interests SO vested > And Poor Truth for too long molested > Their deceit and deception > Near the point of perfection > If it were my call, they'd ALL be arrested > < > Ed Conrad > http://www.psu.edu/ > http://mysite.verizon.net/edconrad/FOSSILS/edyournot.gif > < > < > (cc) New York Times talk.origins David Iain Greig > sci.bio.evolution alt.california MSNBC CBS NBC > Wall Street Journal U.S. News and World Report > Tony Blair London Prehistoric ABC CBS C-span > White House UFOs Intelligent Design Evolution > Hubble Space Pittsburgh Post-Gazette Stock > Market Penn State University 911 Las Vegas > Review-Journal Montreal Gazette Fidel Castro > -- === === Subject: : Charts on projective space. Hello I have a question on the standard way of constructing charts on RP^n. Define RP^n as the quotient space under the map taking a non-zero vector in R^n+1 to the 1-dimensional subspace it spans. Define U_j={(x_1, ... , x_n+1) in R^n+1| x_j is different from 0)} j = 1, ... , n+1 Denote the map taking a non-zero vector to its span for p, then the sets V_j=p(U_j) are charts on RP^n via the maps f_j: V_j ---> R^n [(x_1, ... , x_n+1)] maps to (x_1/x_j, ..., omit j'th coord, ..., x_n+1/ x_j) How do I see that the inverse to f_j is continuous ?? Using the properties of quotient topology I can see f_j is cont. but not the inverse. Are there any propositions about continuity of maps *into* a quotient space ( or about openness of quotient maps) ?? === === Subject: : Re: Charts on projective space. Observe that the quotient map p : R^(n+1) - {0} ---> RP^n is a composition of the retraction r : R^(n+1) - {0} ---> S^n, defined by r(x) = x / ||x||, and the quotient map q : S^n --- > RP^n, which is the restriction of p to S^n and which identifies antipodal point on S^n. To simplifying our reasoning suppose that in your definition of U_j, the j-coordinate x_j = 1. Then the restriction r_j of the map r to U_j is the stereographic projection and therefore r_j : U_j --- > W_j is a homeomorphism, where W_j = r(U_j) is in fact a hemisphere of S^n tangent to U_j in the point (0,...,1,...,0) with 1 on j-th place. Now, consider the restriction q_j of the map q to W_j, which is a continuous one-to-one map and sending W_j onto V_j = q(W_j) = p( U_j). It is easy to prove that such restriction q_j is open (small open disks in W_j are transformed onto small open disks in V_j), therefore q_j : W_j --- > V_j is homeomorphism. (Thm. If f : X ---> Y is a one-to-one continuous map of a space X onto a space Y, then f is open iff f is a homeomorphism.) To complete, suppose u_j : R^n ---> U_j is a linear/affine homeomorphism. Then the composition (q_j r_j u_j) : R^n --- > V_j, as well as (q_j r_j u_j)^{-1} : V_j --- > R^n, is a homeomorphism. Is that clear and obvious ? Zbig === === Subject: : Re: Charts on projective space. Am I missing something obvious here ? :o) M > Hello I have a question on the standard way of constructing charts on > RP^n. Define RP^n as the quotient space under the map taking a non-zero > vector in R^n+1 to the > 1-dimensional subspace it spans. Define U_j={(x_1, ... , x_n+1) in R^n+1| x_j is different from 0)} j = > 1, ... , n+1 Denote the map taking a non-zero vector to its span for p, > then the sets V_j=p(U_j) are charts on RP^n via the maps f_j: V_j ---> R^n > [(x_1, ... , x_n+1)] maps to (x_1/x_j, ..., omit j'th coord, ..., > x_n+1/ x_j) How do I see that the inverse to f_j is continuous ?? Using the properties of quotient topology I can see f_j is cont. but > not the inverse. Are there any propositions about continuity of maps *into* a quotient > space ( or about openness of quotient maps) ?? === === Subject: : Topologist's sine curve. I am reading topology book where it is shown that topoologists sine curve S ([-1,1] on y-axis union the graph of sin(1/x) x in (0,1]) is connected but not path connected. It is said that S is not locally connected at (0,0) since any ball around (0,0) will contain dijoint arcs from the graph. How to see this properly and in nice way ? === === Subject: : Re: Topologist's sine curve. In the space S arbitrary small open neighborhood V of (0,0) is not connected (in fact, it is not path connected) because any such V is the union of infinitely many disjoint arcs without end points and V can be decomposed into two disjoint and open sets V_1 and V_2 of the same form as V . Therefore, S is not locally connected at (0,0). === === Subject: : Re: Topologist's sine curve. > I am reading topology book where it is shown that topoologists sine > curve S ([-1,1] on y-axis union the graph of sin(1/x) x in (0,1]) is > connected but not path connected. It is said that S is not locally > connected at (0,0) since any ball around (0,0) will contain dijoint > arcs from the graph. How to see this properly and in nice way ? > Let r < 1, and set E = B(0,r) intersect S. Then B(0,r) contains (a,0) and (b,0), 0 < a < b, such that sin(1/a) = 1 and sin(1/b) = 0. The sets {x in E : x > a} and {x in E : x < a} form a separation of E. === === Subject: : Re: Topologist's sine curve. >I am reading topology book where it is shown that topoologists sine >curve S ([-1,1] on y-axis union the graph of sin(1/x) x in (0,1]) is >connected but not path connected. It is said that S is not locally >connected at (0,0) since any ball around (0,0) will contain dijoint >arcs from the graph. How to see this properly and in nice way ? > Letting D(e) be the open disc of radius e <= 1, what is the intersection of D(e) and S? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Who wants to write a thesis about singaporean math?? I PAY!! I need to write a thesis about singaporean math until April but I don`t have time. I pay to someone who wants to do it. === === Subject: : Re: Who wants to write a thesis about singaporean math?? I PAY!! > I need to write a thesis about singaporean math until April but I don`t > have time. I pay to someone who wants to do it. > I went to college with someone who is now a professor there. === === Subject: : Re: Who wants to write a thesis about singaporean math?? I PAY!! > I need to write a thesis about singaporean math until April but I don`t > have time. I pay to someone who wants to do it. > I went to college with someone who is now a professor there. Did he pay too? === === Subject: : Re: Who wants to write a thesis about singaporean math?? I PAY!! > I need to write a thesis about singaporean math > until April but I don`t > have time. I pay to someone who wants to do it. > I went to college with someone who is now a professor > there. > I expect that even at that, writing a thesis in four months in his spare time, would be a bit of a strain. :-) Tom === === Subject: : Re: Who wants to write a thesis about singaporean math?? I PAY!! > I need to write a thesis about singaporean math until April but I don`t > have time. I pay to someone who wants to do it. Singapoeran math, like, totally rocks. Five cents, please. === === Subject: : Is this solution correct? Is this solution correct: y'+2y=x^2 ((e^2x)y)'=(e^2x)(x^2) (e^2x)y=(e^2x(2(x-1)x+1))/4 y=(2x^2-2x+1)/4 x^2=x^2 === === Subject: : Re: Is this solution correct? On 3 Dec 2006 15:13:57 -0800, Protoman ((e^2x)y)'=(e^2x)(x^2) >(e^2x)y=(e^2x(2(x-1)x+1))/4 >y=(2x^2-2x+1)/4 x-(1/2)+x^2-x+(1/2)=x^2 >x^2=x^2 > Troll alert, ref: === === Subject: : Re: Is this solution correct? > Is this solution correct: > Idon'treadstufflikethatcausetoomucheyestrain. SoI'llletsomebodyelsecheckyourwork. > y'+2y=x^2 > ((e^2x)y)'=(e^2x)(x^2) > (e^2x)y=(e^2x(2(x-1)x+1))/4 > y=(2x^2-2x+1)/4 x-(1/2)+x^2-x+(1/2)=x^2 > x^2=x^2 > They're ease to read and don't cause eye strain. For example, a = b + xy < c is usually read, a=b+xy Is this solution correct: > Idon'treadstufflikethatcausetoomucheyestrain. > SoI'llletsomebodyelsecheckyourwork. > y'+2y=x^2 > ((e^2x)y)'=(e^2x)(x^2) > (e^2x)y=(e^2x(2(x-1)x+1))/4 > y=(2x^2-2x+1)/4 > x-(1/2)+x^2-x+(1/2)=x^2 > x^2=x^2 > They're ease to read and don't cause eye strain. For example, > a = b + xy < c is usually read, a=b+xy I can solve this one, up to step three: y' + (x^2)y = 2x > ((e^(x^2))y)' = 2x(e^(x^2)) > (e^(x^2))y = S(2x(e^(x^2)) dx //How do you integrate that? integral 2x.e^(x^2) dx = integral e^(x^2) dx^2 It doesn't hurt, ie helps a lot, when limited to ascii, to use extra letters for clarity instead of personal renditions of math notation. > y = (S(2x(e^(x^2)) dx) / (e^(x^2)) What sort of physical system could those two diffy-qs model? > I'm a mathematician. === === Subject: : Re: Is this solution correct? Resulting expression satisfies the starting equation, and it probably is correct. > Is this solution correct: y'+2y=x^2 > ((e^2x)y)'=(e^2x)(x^2) > (e^2x)y=(e^2x(2(x-1)x+1))/4 > y=(2x^2-2x+1)/4 x-(1/2)+x^2-x+(1/2)=x^2 > x^2=x^2 === === Subject: : Re: Continuity of spacetime I try to read ancient text, by characterset so old > I have trouble with > the caligraphy. I contact my old master teacher for > some help to read > this one. > OK, I think I translate one example from ancient > text. It say, consider unit length [ 0, 1 ], and to each > point on [ 0, 1 ] we > assign probability .9 that the point exists. Then, > length is > probabilistic and most likely distance = .9. Reely > wierd. [ 0, 1 ] is both continuous and discrete. I not making this stuff up ! Huang > You are, however, taking this stuff out of context. How do I know that? Tom === === Subject: : Re: Continuity of spacetime <22852323.1165189607770.JavaMail.jakarta@nitrogen.mathforum.org > I try to read ancient text, by characterset so old > I have trouble with > the caligraphy. I contact my old master teacher for > some help to read > this one. > OK, I think I translate one example from ancient > text. > It say, consider unit length [ 0, 1 ], and to each > point on [ 0, 1 ] we > assign probability .9 that the point exists. Then, > length is > probabilistic and most likely distance = .9. Reely > wierd. > [ 0, 1 ] is both continuous and discrete. > I not making this stuff up ! > Huang You are, however, taking this stuff out of context. How do I know that? Tom Einstein talk about some spacetime continuum. How he know it is continuum ? He have no proof. Just call it continuum. If we consider unit interval [ 0, 1 ], and to each point x_n in [ 0, 1 ] we assign probability P(x_n) that the point x_n exists, then we must have that [ 0, 1 ] is connected and also disjoint, or [ 0, 1 ] is neither connected nor disjoint, or maybe somebody going crazy around here, this stuff make my head spinning. I try to get a photo of famous woodcarving for you guy. Huang === === Subject: : Re: Continuity of spacetime > I try to read ancient text, by characterset so > old > I have trouble with > the caligraphy. I contact my old master teacher > for > some help to read > this one. > OK, I think I translate one example from ancient > text. > It say, consider unit length [ 0, 1 ], and to > each > point on [ 0, 1 ] we > assign probability .9 that the point exists. > Then, > length is > probabilistic and most likely distance = .9. > Reely > wierd. > [ 0, 1 ] is both continuous and discrete. > I not making this stuff up ! > Huang > You are, however, taking this stuff out of context. > How do I know that? > Tom > Einstein talk about some spacetime continuum. How he > know it is > continuum ? He have no proof. Just call it > continuum.<< Nonsense. Einstein didn't just make this stuff up either. Space is continuous with time within the Riemannian geometry supported by Minkowski space-time and analyzed in the continuous function matrix of the tensor calculus, in which time has the role of an added coordinate point along with the spatial coordinate points. All of this is most certainly proven mathematically, though the physical problem of the actual shape of space may still be an open question. If we consider unit interval [ 0, 1 ], and to each > point x_n in [ 0, 1 > ] we assign probability P(x_n) that the point x_n > exists, then we must > have that [ 0, 1 ] is connected and also disjoint, or > [ 0, 1 ] is neither connected nor disjoint, or maybe somebody going crazy around here, this stuff > make my head > spinning. Maybe some references will help settle your cranium. Try http://mathworld.wolfram.com/ClosedInterval.html, and perhaps we can pick it up from there? Tom I try to get a photo of famous woodcarving for you > guy. Huang > === === Subject: : Re: Continuity of spacetime <27394564.1165232560336.JavaMail.jakarta@nitrogen.mathforum.org > I try to read ancient text, by characterset so > old > I have trouble with > the caligraphy. I contact my old master teacher > for > some help to read > this one. > OK, I think I translate one example from ancient > text. > It say, consider unit length [ 0, 1 ], and to > each > point on [ 0, 1 ] we > assign probability .9 that the point exists. > Then, > length is > probabilistic and most likely distance = .9. > Reely > wierd. > [ 0, 1 ] is both continuous and discrete. > I not making this stuff up ! > Huang > You are, however, taking this stuff out of context. > How do I know that? > Tom > Einstein talk about some spacetime continuum. How he > know it is > continuum ? He have no proof. Just call it > continuum.<< Nonsense. Einstein didn't just make this stuff up > either. Space is continuous with time within the > Riemannian geometry supported by Minkowski space-time > and analyzed in the continuous function matrix of the > tensor calculus, in which time has the role of an > added coordinate point along with the spatial coordinate > points. All of this is most certainly proven > mathematically, though the physical problem of the > actual shape of space may still be an open question. > If we consider unit interval [ 0, 1 ], and to each > point x_n in [ 0, 1 > ] we assign probability P(x_n) that the point x_n > exists, then we must > have that > [ 0, 1 ] is connected and also disjoint, or > [ 0, 1 ] is neither connected nor disjoint, or > maybe somebody going crazy around here, this stuff > make my head > spinning. Maybe some references will help settle your cranium. Try > http://mathworld.wolfram.com/ClosedInterval.html, and > perhaps we can pick it up from there? Tom Closed interval very interesting. If existence of x_n is probabilistic for all x_n, then there is realvalued probability that interval is closed, but nonzero probability that it is open. Also, same same for limit point. Maybe it contain all limit point, maybe not, this a probabilistic question. Closed interval only well behave if probability gradient is everywhere equal to unity. You can use any metric you want, but actual distance between two points can vary depending on probability gradient. Einstein bent space just a result of this, silly Einstein fooling everybody. Probably drinking Saki and laughing in Heaven right now. Huang === === Subject: : Re: Continuity of spacetime > I try to read ancient text, by characterset > so > old > I have trouble with > the caligraphy. I contact my old master > teacher > for > some help to read > this one. > OK, I think I translate one example from > ancient > text. > It say, consider unit length [ 0, 1 ], and to > each > point on [ 0, 1 ] we > assign probability .9 that the point exists. > Then, > length is > probabilistic and most likely distance = .9. > Reely > wierd. > [ 0, 1 ] is both continuous and discrete. > I not making this stuff up ! > Huang > You are, however, taking this stuff out of > context. > How do I know that? > Tom > Einstein talk about some spacetime continuum. How > he > know it is > continuum ? He have no proof. Just call it > continuum.<< > Nonsense. Einstein didn't just make this stuff up > either. Space is continuous with time within the > Riemannian geometry supported by Minkowski > space-time > and analyzed in the continuous function matrix of > the > tensor calculus, in which time has the role of an > added coordinate point along with the spatial > coordinate > points. All of this is most certainly proven > mathematically, though the physical problem of the > actual shape of space may still be an open > question. > If we consider unit interval [ 0, 1 ], and to > each > point x_n in [ 0, 1 > ] we assign probability P(x_n) that the point x_n > exists, then we must > have that > [ 0, 1 ] is connected and also disjoint, or > [ 0, 1 ] is neither connected nor disjoint, or > maybe somebody going crazy around here, this > stuff > make my head > spinning. > Maybe some references will help settle your > cranium. Try > http://mathworld.wolfram.com/ClosedInterval.html, > and > perhaps we can pick it up from there? > Tom > Closed interval very interesting. If existence of x_n > is probabilistic > for all x_n, then there is realvalued probability > that interval is > closed, but nonzero probability that it is open. > Also, same same for > limit point. Maybe it contain all limit point, maybe > not, this a > probabilistic question. Closed interval only well > behave if probability > gradient is everywhere equal to unity. You can use any metric you want, but actual distance > between two points > can vary depending on probability gradient. Einstein bent space just a result of this, silly > Einstein fooling > everybody. Probably drinking Saki and laughing in > Heaven right now. Huang > Okay, I guess I am a little slow, but I can eventually tell when I'm being put on. Nice troll. Tom === === Subject: : Re: Continuity of spacetime <33418229.1165268129182.JavaMail.jakarta@nitrogen.mathforum.org > I try to read ancient text, by characterset > so > old > I have trouble with > the caligraphy. I contact my old master > teacher > for > some help to read > this one. > OK, I think I translate one example from > ancient > text. > It say, consider unit length [ 0, 1 ], and to > each > point on [ 0, 1 ] we > assign probability .9 that the point exists. > Then, > length is > probabilistic and most likely distance = .9. > Reely > wierd. > [ 0, 1 ] is both continuous and discrete. > I not making this stuff up ! > Huang > You are, however, taking this stuff out of > context. > How do I know that? > Tom > Einstein talk about some spacetime continuum. How > he > know it is > continuum ? He have no proof. Just call it > continuum.<< > Nonsense. Einstein didn't just make this stuff up > either. Space is continuous with time within the > Riemannian geometry supported by Minkowski > space-time > and analyzed in the continuous function matrix of > the > tensor calculus, in which time has the role of an > added coordinate point along with the spatial > coordinate > points. All of this is most certainly proven > mathematically, though the physical problem of the > actual shape of space may still be an open > question. > If we consider unit interval [ 0, 1 ], and to > each > point x_n in [ 0, 1 > ] we assign probability P(x_n) that the point x_n > exists, then we must > have that > [ 0, 1 ] is connected and also disjoint, or > [ 0, 1 ] is neither connected nor disjoint, or > maybe somebody going crazy around here, this > stuff > make my head > spinning. > Maybe some references will help settle your > cranium. Try > http://mathworld.wolfram.com/ClosedInterval.html, > and > perhaps we can pick it up from there? > Tom > Closed interval very interesting. If existence of x_n > is probabilistic > for all x_n, then there is realvalued probability > that interval is > closed, but nonzero probability that it is open. > Also, same same for > limit point. Maybe it contain all limit point, maybe > not, this a > probabilistic question. Closed interval only well > behave if probability > gradient is everywhere equal to unity. > You can use any metric you want, but actual distance > between two points > can vary depending on probability gradient. > Einstein bent space just a result of this, silly > Einstein fooling > everybody. Probably drinking Saki and laughing in > Heaven right now. > Huang Okay, I guess I am a little slow, but I can > eventually tell when I'm being put on. Nice troll. Tom Not really trolling, it's just that I'm such an a-hole that the only way to get any conversation around here is by assuming an alternate identity. Seriously though, this approach does lead to a straightforward explanation of the double slit, and which way inforation. The tradeoff is that it is indeterminately valid. The results which are derived from this approach are quite interesting, but their validity is inherently indeterminate. There's no way around it. Anyway, nobody's really putting up much of a fight around here. You'd think that someone would notice that for example, If the distance from 0 to 1 on [ 0, 1 ] is really equal to say .5, because the probability of the exitence of any given x_n is 50:50, then you might as well be talking about the interval [ 0, .5 ] where the probability of existence of any given x_n is %100. These things seem interchangeable, indeterminately so, and so the whole approach might rightly be characterized as being trivial. I was hoping that someone would notice this. Such arguments do not destroy the approach, however, because my methods embrace existentialism, triviality, and genuine indeterminacy, and so the indeterminacy is actually anticipated and utilized. The Janus faced math you find floating around in here wants the best of both worlds. They want to deny the existence of the trivial, but they dont want to gvie up the randomness. Silly people. You cannot deny the existence of the trivial becase it's existence is indeterminate. To re-toss the word sald, mathematics today deals with two things, that which exists and that which does not. There is no in between. Then they come along and trot out the concept of randomness and use it like one of James Harris's object oriented mathematical modules, just cutting and pasting it wherever they see fit, never really justifying how the hell such a thing could possibly make sense. By virtue of that, mathematics has embraced a subtle inconsistency. And it's better to embrace indeterminacy than inconsistency, in my opinion. So, we have that the existence of a trivial is ideterminate and this solves all of the remaining problems of math, physics, and humankind. Razor blades are sharpened mysteriously, spoons bend all by themselves, plants grow faster and larger, objects are floating in midair. Huang === === Subject: : Re: Continuity of spacetime > I try to read ancient text, by > characterset > so > old > I have trouble with > the caligraphy. I contact my old master > teacher > for > some help to read > this one. > OK, I think I translate one example from > ancient > text. > It say, consider unit length [ 0, 1 ], > and to > each > point on [ 0, 1 ] we > assign probability .9 that the point > exists. > Then, > length is > probabilistic and most likely distance = > .9. > Reely > wierd. > [ 0, 1 ] is both continuous and > discrete. > I not making this stuff up ! > Huang > You are, however, taking this stuff out of > context. > How do I know that? > Tom > Einstein talk about some spacetime continuum. > How > he > know it is > continuum ? He have no proof. Just call it > continuum.<< > Nonsense. Einstein didn't just make this stuff > up > either. Space is continuous with time within > the > Riemannian geometry supported by Minkowski > space-time > and analyzed in the continuous function matrix > of > the > tensor calculus, in which time has the role of > an > added coordinate point along with the spatial > coordinate > points. All of this is most certainly proven > mathematically, though the physical problem of > the > actual shape of space may still be an open > question. > If we consider unit interval [ 0, 1 ], and to > each > point x_n in [ 0, 1 > ] we assign probability P(x_n) that the point > x_n > exists, then we must > have that > [ 0, 1 ] is connected and also disjoint, or > [ 0, 1 ] is neither connected nor disjoint, > or > maybe somebody going crazy around here, this > stuff > make my head > spinning. > Maybe some references will help settle your > cranium. Try > http://mathworld.wolfram.com/ClosedInterval.html, > and > perhaps we can pick it up from there? > Tom > Closed interval very interesting. If existence of > x_n > is probabilistic > for all x_n, then there is realvalued probability > that interval is > closed, but nonzero probability that it is open. > Also, same same for > limit point. Maybe it contain all limit point, > maybe > not, this a > probabilistic question. Closed interval only well > behave if probability > gradient is everywhere equal to unity. > You can use any metric you want, but actual > distance > between two points > can vary depending on probability gradient. > Einstein bent space just a result of this, silly > Einstein fooling > everybody. Probably drinking Saki and laughing in > Heaven right now. > Huang > Okay, I guess I am a little slow, but I can > eventually tell when I'm being put on. > Nice troll. > Tom Not really trolling, it's just that I'm such an > a-hole that the only > way to get any conversation around here is by > assuming an alternate > identity. Seriously though, this approach does lead to a > straightforward > explanation of the double slit, and which way > inforation. The tradeoff > is that it is indeterminately valid. The results > which are derived from > this approach are quite interesting, but their > validity is inherently > indeterminate. There's no way around it. Anyway, nobody's really putting up much of a fight > around here. You'd > think that someone would notice that for example, If the distance from 0 to 1 on [ 0, 1 ] is really > equal to say .5, > because the probability of the exitence of any given > x_n is 50:50, then > you might as well be talking about the interval [ 0, > .5 ] where the > probability of existence of any given x_n is %100. > These things seem > interchangeable, indeterminately so, and so the whole > approach might > rightly be characterized as being trivial. The closed interval [a,b] as I wrongly expected you to understand from the link I gave, contains all of its limit points. If you know what that means, then you should see that the existence question you pose is not an issue. If it is the dichotomy between the continuous line and discrete numbers that troubles you -- Dedekind handled it with a cut principle that straightforwardly instructs us in how to differentiate real numbers without sacrificing continuity. Other than that, as a couple of us have pointed out, you are confusing physical measurability with mathematical models. Two different things. Tom I was hoping that someone would notice this. Such > arguments do not > destroy the approach, however, because my methods > embrace > existentialism, triviality, and genuine > indeterminacy, and so the > indeterminacy is actually anticipated and utilized. The Janus faced math you find floating around in here > wants the best of > both worlds. They want to deny the existence of the > trivial, but they > dont want to gvie up the randomness. Silly people. > You cannot deny the > existence of the trivial becase it's existence is > indeterminate. To re-toss the word sald, mathematics today deals > with two things, that > which exists and that which does not. There is no in > between. Then they > come along and trot out the concept of randomness and > use it like one > of James Harris's object oriented mathematical > modules, just cutting > and pasting it wherever they see fit, never really > justifying how the > hell such a thing could possibly make sense. By virtue of that, mathematics has embraced a subtle > inconsistency. And > it's better to embrace indeterminacy than > inconsistency, in my opinion. > So, we have that the existence of a trivial is > ideterminate and this > solves all of the remaining problems of math, > physics, and humankind. > Razor blades are sharpened mysteriously, spoons bend > all by themselves, > plants grow faster and larger, objects are floating > in midair. Huang > === === Subject: : Re: Continuity of spacetime <734778.1165358411951.JavaMail.jakarta@nitrogen.mathforum.org > If the distance from 0 to 1 on [ 0, 1 ] is really > equal to say .5, > because the probability of the exitence of any given > x_n is 50:50, then > you might as well be talking about the interval [ 0, > .5 ] where the > probability of existence of any given x_n is %100. > These things seem > interchangeable, indeterminately so, and so the whole > approach might > rightly be characterized as being trivial. The closed interval [a,b] as I wrongly expected you > to understand from the link I gave, contains all of > its limit points. Correct. > If you know what that means, then > you should see that the existence question you pose > is not an issue. And what I am saying is that energy is nothing more than an existential probability gradient, and that this is the correct wave-based approach to QM. The word energy has been obsoleted. Everything can be described in terms of probailities. > If it is the dichotomy between > the continuous line and discrete numbers that troubles > you -- Nope - I dont even think that what I said could be read that way. > Dedekind handled it with a cut principle that > straightforwardly instructs us in how to differentiate > real numbers without sacrificing continuity. Other than that, as a couple of us have pointed out, you > are confusing physical measurability with mathematical > models. Two different things. Tom No. I'm giving the reals a new property, just to see what happens. The property which you assign to each real is a real valued probability on [0,1] that the point in question exists. An existential potential. Length is then derived with a combination of metric and an associated PD. In empty flat space, the PD would be everywhere near 1. On quantum scale you can have that the PD is closer to being everywhere near 0, and so you get paths of zero distance. The measurement problem is entirely different. I'm saying that length whatsoever. Again, the problem with an approach based on triviality is that the validity of the approach is always indeterminate, regardless of how well experiment agrees with theory, even if it matches up perfectly, validity will always be indeterminate. === === Subject: : Re: Continuity of spacetime > If the distance from 0 to 1 on [ 0, 1 ] is really > equal to say .5, > because the probability of the exitence of any > given > x_n is 50:50, then > you might as well be talking about the interval [ > 0, > .5 ] where the > probability of existence of any given x_n is > %100. > These things seem > interchangeable, indeterminately so, and so the > whole > approach might > rightly be characterized as being trivial. > The closed interval [a,b] as I wrongly expected you > to understand from the link I gave, contains all of > its limit points. Correct. If you know what that means, then > you should see that the existence question you pose > is not an issue. And what I am saying is that energy is nothing more > than an existential > probability gradient, and that this is the correct > wave-based approach > to QM. The word energy has been obsoleted. Everything can be > described in > terms of probailities. > If it is the dichotomy between > the continuous line and discrete numbers that > troubles > you -- Nope - I dont even think that what I said could be > read that way. > Dedekind handled it with a cut principle that > straightforwardly instructs us in how to > differentiate > real numbers without sacrificing continuity. > Other than that, as a couple of us have pointed > out, you > are confusing physical measurability with > mathematical > models. Two different things. > Tom > No. I'm giving the reals a new property, just to see > what happens. The > property which you assign to each real is a real > valued probability on > [0,1] that the point in question exists. An > existential potential. > Length is then derived with a combination of metric > and an associated > PD. In empty flat space, the PD would be everywhere near > 1. On quantum > scale you can have that the PD is closer to being > everywhere near 0, > and so you get paths of zero distance. The measurement problem is entirely different. I'm > saying that length > this approach > whatsoever. > Again, the problem with an approach based on > triviality is that the > validity of the approach is always indeterminate, > regardless of how > well experiment agrees with theory, even if it > matches up perfectly, > validity will always be indeterminate. > Yes, I suppose it would if one could speak of nonsensical premise, however, the rest of your program fails. Tom === === Subject: : Re: Continuity of spacetime <11732858.1165402530601.JavaMail.jakarta@nitrogen.mathforum.org > If the distance from 0 to 1 on [ 0, 1 ] is really > equal to say .5, > because the probability of the exitence of any > given > x_n is 50:50, then > you might as well be talking about the interval [ > 0, > .5 ] where the > probability of existence of any given x_n is > %100. > These things seem > interchangeable, indeterminately so, and so the > whole > approach might > rightly be characterized as being trivial. > The closed interval [a,b] as I wrongly expected you > to understand from the link I gave, contains all of > its limit points. > Correct. > If you know what that means, then > you should see that the existence question you pose > is not an issue. > And what I am saying is that energy is nothing more > than an existential > probability gradient, and that this is the correct > wave-based approach > to QM. > The word energy has been obsoleted. Everything can be > described in > terms of probailities. > If it is the dichotomy between > the continuous line and discrete numbers that > troubles > you -- > Nope - I dont even think that what I said could be > read that way. > Dedekind handled it with a cut principle that > straightforwardly instructs us in how to > differentiate > real numbers without sacrificing continuity. > Other than that, as a couple of us have pointed > out, you > are confusing physical measurability with > mathematical > models. Two different things. > Tom > No. I'm giving the reals a new property, just to see > what happens. The > property which you assign to each real is a real > valued probability on > [0,1] that the point in question exists. An > existential potential. > Length is then derived with a combination of metric > and an associated > PD. > In empty flat space, the PD would be everywhere near > 1. On quantum > scale you can have that the PD is closer to being > everywhere near 0, > and so you get paths of zero distance. > The measurement problem is entirely different. I'm > saying that length > this approach > whatsoever. > Again, the problem with an approach based on > triviality is that the > validity of the approach is always indeterminate, > regardless of how > well experiment agrees with theory, even if it > matches up perfectly, > validity will always be indeterminate. Yes, I suppose it would if one could speak of > nonsensical premise, however, the rest of your > program fails. Tom Proving that I'm right is not the hard part. Proving it in 25 words or less - that is the hard part. Aquisition of whichway information constitutes destruction of To explain how I view a probabilistic wave - 25 words or less. For each point in space we assign a real valued probability that the point in question exists. The wave is modelled as propagating across this gradient of probabilities. The reason that any length can exist was invented by James Harris, at the Harris Institute for Mathematical Excellence, and can be found here > http://sciphysicsopenmanuscript.blogspot.com/ This photo explains it all http://photos1.blogger.com/blogger/3139/3591/1600/discrete4.jpg === === Subject: : Re: Continuity of spacetime <11732858.1165402530601.JavaMail.jakarta@nitrogen.mathforum.org> Attempted proof in 25 words or less. If sub-Plancklength is nonexistent then length is nonsensical. If sub-Plancklength is trivial then length is probabilistic. If sub-Plancklength is nontrivial then Plancklength does not satisfy it's own definition. ------------------------------------------------- Arguing against this is hopeless. And yet, any geometry, numbers, logic, etc, based on triviality must be inherently trivial or better indeterminately valid. It is inherently indeterminate. That's why Heisenberg Uncertainty principle is true. Nothing else is really possible. I dont expect anyone to actually believe this, even though it happens to be true. === === Subject: : Re: Continuity of spacetime > By virtue of that, mathematics has embraced a subtle inconsistency. And > it's better to embrace indeterminacy than inconsistency, in my opinion. > So, we have that the existence of a trivial is ideterminate and this > solves all of the remaining problems of math, physics, and humankind. > Razor blades are sharpened mysteriously, spoons bend all by themselves, > plants grow faster and larger, objects are floating in midair. Have no idea what your objectives are, but _something_ can be said about the physical meaning of continuity and mathematical exactness. Look up: http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm http://hdebruijn.soo.dto.tudelft.nl/QED/index.htm Han de Bruijn === === Subject: : Re: Continuity of spacetime <33418229.1165268129182.JavaMail.jakarta@nitrogen.mathforum.org> By virtue of that, mathematics has embraced a subtle inconsistency. And > it's better to embrace indeterminacy than inconsistency, in my opinion. > So, we have that the existence of a trivial is ideterminate and this > solves all of the remaining problems of math, physics, and humankind. > Razor blades are sharpened mysteriously, spoons bend all by themselves, > plants grow faster and larger, objects are floating in midair. Have no idea what your objectives are, but _something_ can be said about > the physical meaning of continuity and mathematical exactness. Look up: http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm > http://hdebruijn.soo.dto.tudelft.nl/QED/index.htm Han de Bruijn I hope that you are kidding. Well, even if you're not kidding I suppose that there might be some truth in it, I will concede grudgingly. But the whole materialization thing seems a little bit like the Monty Hall problem if you ask me. That is to say, it sure sounds like you're pulling my leg. For example. I can contemplate the round square. Does this mean that it is materialized ? Something which could never be materialized in the first place ? What really interests me is the double slit, whichway information, and how a probability can be manifested experimentally as a tangible object. How information can be manifested as a tangible item. Got any leads on that one ? I do. === === Subject: : Re: Continuity of spacetime >>Have no idea what your objectives are, but _something_ can be said about >>the physical meaning of continuity and mathematical exactness. Look up: >>http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm >>http://hdebruijn.soo.dto.tudelft.nl/QED/index.htm I hope that you are kidding. I'm not kidding. Han de Bruijn === === Subject: : Re: Continuity of spacetime huangxienchen@yahoo.com a .8ecrit : > Hello. > I read somewhere that physical length is > probabilistic, according to > some guy theory out there. > If length is probabilistic, then continuity is > also probablistic. > It seems to me that the real according to some > guy is that length is > hypothezised as being relativistic, not > probabilistic, and it is > a premise of Special Relativity. > To my knowledge, continuity is not dependant on > length. > abstract. Someone > know something about this one ? > Look up Lorentz - Einstein. > Andr.8e Michaud > I dont think that Lorentz or Einstein, or even > Feynman did anything > like this. Take for example Heine definition of continuity. A real function f is continuous if for any sequence > (xn) such that > lim x_n = x_o, as n->inf > it holds that > lim f(x_n) = f(x_o), as n->inf > (We assume that all points xn, x0 belong to the > domain of f.) > If length is taken to be probabilistic, then for all > x_n there is a > real valued probability that x_n exists, or not. We > simply assign a > real value to each of the x_n and call it a > probability. Heine's definition is changed, but still valid as > long as the > probability is everywhere nonzero. If some of the x_n have probability 0, then f(x_n) is > simply undefined > for those x_n. > But this is continuity of a function, I think we need > probabilistic > compactness. Huang > All real functions are continuous. Something that you wish to call probabilistic compactness, for whatever that may mean, doesn't do anything to change the way that continuous functions work. Once again, you have to say what you mean by length if it has something to do with other than measurement. There is no such thing as a probable measurement. Probabilities are calculated with confidence intervals that depend on the choice of theoretical or calculated endpoints. You run into severe semantic difficulty when you want to call a compact set -- fintely bounded by definition -- probabilistic. If all real functions are continuous, then length exists everywhere with a non- zero probability, and your concluisions above are falsified. Tom === === Subject: : Re: Need help calculating overall network reliability > .....................|-C-|....|-E-| > .....................|....|....|....| > INPUT -- A -- B--|....|----|....|------- OUTPUT > ......|..............|....|....|....|.....| > ......|..............|-D-|....|-F-|.....| > ......|...................................| > ......|--------------G-------------------| > Pretend that the periods (.) are spaces, it was the only way the whitespace > would be respected so the network would display correctly. I need to calculate the overall network reliability if the individual node reliabilities are: A=0.95, B=0.97, C=0.92, D=0.94, E=0.90, F=0.88, G=0.98 I believe that I done that with the following steps (please verify): The reliability of the redundant network including subsystems C and D is: > RCD = RC + RD - (RC)*( RD) = 0.92 + 0.94 - (0.92)(0.94) = 0.9952 The reliability of the redundant network including subsystems E and F is: > REF = RE + RF - (RE)*( RF) = 0.90 + 0.88 - (0.90)(0.88) = 0.988 The reliability of the path is: > RABCDEF = (RA)*(RB)*(RCD)*(REF) = (0.95)(0.97)(0.9952)(0.988) = 0.906072 The reliability of the combined network is: > RSYSTEM = RABCDEF + RG - (RABCDEF)(RG) > = 0.906072 + 0.98 - (0.906072)(0.98) = 0.998121 > The part that I am having trouble understanding is how the overall network > reliability changes if node G is standby-redundant, how do I calculate that? > After searching the web for awhile I found a website that gave me this formula to > use: 1-(P(A)*P(B | A')). Is that correct, and how to I apply it? What do A and B represent in that formula? Presumably A has to do with whether the system ABCDEF succeeds or fails, and B whether G succeeds or fails. If both cases represent failures of their systems, then the answer would be 1 - (1-reliability ABCDEF)*(1 - relability of G) = 1 - (1 - 0.906072)*(1 - 0.98) = 0.99812144, since G is independent of the other system. (This means the conditional probability P(B | A') equals P(B).) --- Christopher Heckman === === Subject: : Re: Need help calculating overall network reliability Ok this diagram was composed with a monospaced font, I am not sure that this will display correctly, because HTML eats spaces for lunch: |-C-| |-E-| | | | | INPUT--A--B------| |----| |------- OUTPUT | | | | | | | |-D-| |-F-| | | | |-------------G-------------------| .....................|-C-|....|-E-| > .....................|....|....|....| > INPUT -- A -- B--|....|----|....|------- OUTPUT > ......|..............|....|....|....|.....| > ......|..............|-D-|....|-F-|.....| > ......|...................................| > ......|--------------G-------------------| > Pretend that the periods (.) are spaces, it was the > only way the > whitespace would be respected so the network would > display correctly. > It's still messes up depending upon what font the > reader is using. > To do ascii art and ascii diagrams, compose with a > monospace font > and advise readers to read with a monospace font. > Any less fails. > I need to calculate the overall network reliability > if the individual node reliabilities are: > A=0.95, B=0.97, C=0.92, D=0.94, E=0.90, F=0.88, > G=0.98 > I believe that I done that with the following steps > (please verify): > The reliability of the redundant network including > subsystems C and D is: > RCD = RC + RD - (RC)*( RD) = 0.92 + 0.94 - > (0.92)(0.94) = 0.9952 > The reliability of the redundant network including > subsystems E and F is: > REF = RE + RF - (RE)*( RF) = 0.90 + 0.88 - > (0.90)(0.88) = 0.988 > The reliability of the path is: > RABCDEF = (RA)*(RB)*(RCD)*(REF) = > (0.95)(0.97)(0.9952)(0.988) = 0.906072 > The reliability of the combined network is: > RSYSTEM = RABCDEF + RG - (RABCDEF)(RG) = 0.906072 > + 0.98 - (0.906072)(0.98) = 0.998121 > The part that I am having trouble understanding is > how the overall network reliability changes if node G > is standby-redundant, how do I calculate that? After > searching the web for awhile I found a website that > gave me this formula to use: 1-(P(A)*P(B | A')). Is > that correct, and how to I apply it? > === === Subject: : Re: Need help calculating overall network reliability <7456259.1165122574996.JavaMail.jakarta@nitrogen.mathforum.org Ok this diagram was composed with a monospaced font, I am not sure that > this will display correctly, because HTML eats spaces for lunch: |-C-| |-E-| > | | | | > INPUT--A--B------| |----| |------- OUTPUT > | | | | | | > | |-D-| |-F-| | > | | > |-------------G-------------------| It appeared correctly in monospace font. A workaround for html is to shut it off. Web based forums don't allow that, thus need for ... instead of spaces. Newsgroups, and email that is not written with html, allow clustering of more than one space. Sorry, I can be of no further help on your network problem other than to assist you with posting ascii graphics and art. > I need to calculate the overall network reliability > if the individual node reliabilities are: > A=0.95, B=0.97, C=0.92, D=0.94, E=0.90, F=0.88, > G=0.98 > I believe that I done that with the following steps > The reliability of the redundant network including > subsystems C and D is: > RCD = RC + RD - (RC)*( RD) = 0.92 + 0.94 - > (0.92)(0.94) = 0.9952 > The reliability of the redundant network including > subsystems E and F is: > REF = RE + RF - (RE)*( RF) = 0.90 + 0.88 - > (0.90)(0.88) = 0.988 > The reliability of the path is: > RABCDEF = (RA)*(RB)*(RCD)*(REF) = > (0.95)(0.97)(0.9952)(0.988) = 0.906072 > The reliability of the combined network is: > RSYSTEM = RABCDEF + RG - (RABCDEF)(RG) = 0.906072 > + 0.98 - (0.906072)(0.98) = 0.998121 > The part that I am having trouble understanding is > how the overall network reliability changes if node G > is standby-redundant, how do I calculate that? After > searching the web for awhile I found a website that > gave me this formula to use: 1-(P(A)*P(B | A')). Is > that correct, and how to I apply it? > === === Subject: : Is the value of e, 2.71828.. because (3-2) = (2-1)? Warning: this is a foolish question. Can we have a number system were (3-2) is NOT equal to (2-1), (4-3) NOT equal to (3-2)... and so on.. so that the value of e = 1? Try imagining a number line where the distance between the numbers is different like a logarithmic graph paper. What I want to ask is: Just like how the value of 'pi' is dependent on the properties of space, is the value of e dependent on how we usually count? .. i.e. When we count, we abstract 3 apples as number 3 although the 3rd apple may not be of the same size as that of the other two. In other words, we like to think of a quantity as the sum of portions that are of equal size. === === Subject: : Re: Is the value of e, 2.71828.. because (3-2) = (2-1)? In sci.math, cybermonk on 3 Dec 2006 19:22:18 -0800 > Warning: this is a foolish question. Can we have a number system were (3-2) is NOT equal to (2-1), (4-3) NOT > equal to (3-2)... and so on.. so that the value of e = 1? Try imagining > a number line where the distance between the numbers is different like > a logarithmic graph paper. What I want to ask is: Just like how the value of 'pi' is dependent on > the properties of space, is the value of e dependent on how we usually > count? .. i.e. When we count, we abstract 3 apples as number 3 although the > 3rd apple may not be of the same size as that of the other two. In > other words, we like to think of a quantity as the sum of portions that > are of equal size. > I'm frankly not sure precisely what you're asking, but it is easily enough done; define 'a-b' as a/b and 'a+b' as a*b and one should get close to what you might want. -- #191, ewill3@earthlink.net Useless C++ Programming Idea #992398129: void f(unsigned u) { if(u < 0) ... } -- === === Subject: : Is the value of e, 2.71828.. because (3-2) = (2-1)? > Warning: this is a foolish question. Yes, it really is!! > Can we have a number system were (3-2) is NOT equal to (2-1), ... Of course. We can have a number system where openparens three hyphen two closeparens has any meaning whatsoever, likewise the other expression, and the word equal can mean anything we want, and the word NOT can mean anything we want. Such a system could bear no resemblance to the kind of number system we usually use. For example, that notation can represent odds, three-to-two odds etc. Or that notation can represent chapter and section in some book. But what do you mean by were in the question? Is that a typo?? As for the question in the === Subject: field: No, that's not why. === === Subject: : Re: Is the value of e, 2.71828.. because (3-2) = (2-1)? > Warning: this is a foolish question. So expect a bunch of foolish answers. Can we have a number system were (3-2) is NOT equal to (2-1), (4-3) NOT > equal to (3-2)... and so on.. so that the value of e = 1? Try imagining > a number line where the distance between the numbers is different like > a logarithmic graph paper. Been there, done that. Haven't you kids ever seen a slide rule? What I want to ask is: Just like how the value of 'pi' is dependent on > the properties of space, is the value of e dependent on how we usually > count? .. i.e. When we count, we abstract 3 apples as number 3 although the > 3rd apple may not be of the same size as that of the other two. In > other words, we like to think of a quantity as the sum of portions that > are of equal size. A parking lot with spaces for 100 cars holds 100 cars regardless of whether they are Beetles or Hummers. === === Subject: : Re: Is the value of e, 2.71828.. because (3-2) = (2-1)? > Warning: this is a foolish question. So expect a bunch of foolish answers. > Can we have a number system were (3-2) is NOT equal to (2-1), (4-3) NOT >> equal to (3-2)... and so on.. so that the value of e = 1? Try imagining >> a number line where the distance between the numbers is different like >> a logarithmic graph paper. Been there, done that. Haven't you kids ever seen a slide rule? > What I want to ask is: Just like how the value of 'pi' is dependent on >> the properties of space, is the value of e dependent on how we usually >> count? >> .. i.e. When we count, we abstract 3 apples as number 3 although the >> 3rd apple may not be of the same size as that of the other two. In >> other words, we like to think of a quantity as the sum of portions that >> are of equal size. A parking lot with spaces for 100 cars holds 100 cars >regardless of whether they are Beetles or Hummers. ... but not 100 railway cars. Does that mean we need to change our system of numbers? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: Is the value of e, 2.71828.. because (3-2) = (2-1)? > Warning: this is a foolish question. >So expect a bunch of foolish answers. >> Can we have a number system were (3-2) is NOT equal to (2-1), (4-3) NOT >> equal to (3-2)... and so on.. so that the value of e = 1? Try imagining >> a number line where the distance between the numbers is different like >> a logarithmic graph paper. >Been there, done that. Haven't you kids ever seen a slide rule? >> What I want to ask is: Just like how the value of 'pi' is dependent on >> the properties of space, is the value of e dependent on how we usually >> count? >> .. i.e. When we count, we abstract 3 apples as number 3 although the >> 3rd apple may not be of the same size as that of the other two. In >> other words, we like to think of a quantity as the sum of portions that >> are of equal size. >A parking lot with spaces for 100 cars holds 100 cars >regardless of whether they are Beetles or Hummers. ... but not 100 railway cars. Does that mean we need to change our system > of numbers? Hmmm...tricky. But then, I'm not the one saying (3-2) doesn't equal (2-1). Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === === Subject: : odds of 7-10 split in bowling I'm interested to know the approximate odds of converting a 7-10 split for a spare in bowling. I've seen different odds: 1/1000, 1/5000, 1/700000. It has occurred 3 times on television in PBA games. That's all I know. Has anyone gone thru the math before in this group. === === Subject: : Re: odds of 7-10 split in bowling For televised PBA games since 1961: 7-10 conversions: 3 number of televised PBA games: 4000-8000 (guess) 7-10 occurences per game both players: 1 (guess) So my order of magnitude guess is between 1:1000 and 1:10000 but closer to 1:1000. === === Subject: : Re: odds of 7-10 split in bowling number of televised PBA games: 4000-8000 (guess) > 7-10 occurences per game both players: 1 (guess) Bad guess. > So my order of magnitude guess is between > 1:1000 and 1:10000 but closer to 1:1000. === === Subject: : Re: odds of 7-10 split in bowling <4i67n2lg10kpjsm353vlg5fbk94rkpadat@4ax.com>, > I'm interested to know the approximate odds of converting a 7-10 split > for a spare in bowling. I've seen different odds: 1/1000, 1/5000, > 1/700000. It has occurred 3 times on television in PBA games. That's all I > know. Has anyone gone thru the math before in this group. Professional bowlers convert this split all the time. The probability it has happened on television fewer than four times in PBA games is less than 1/10000. -- Michael Press === === Subject: : How about the 5-7-10 split? > <4i67n2lg10kpjsm353vlg5fbk94rkpadat@4ax.com>, > I'm interested to know the approximate odds of converting a 7-10 split >> for a spare in bowling. I've seen different odds: 1/1000, 1/5000, >> 1/700000. >> It has occurred 3 times on television in PBA games. That's all I >> know. Has anyone gone thru the math before in this group. Professional bowlers convert this split all the time. > The probability it has happened on television fewer > than four times in PBA games is less than 1/10000. -- > Michael Press How about the 5-7-10 split, also known as the Lilly or Sour Apple? Easier to convert, but I'll bet you'll NEVER see one in a PBA game. http://www.utahbowling.com/Masters/week22.htm === === Subject: : Re: odds of 7-10 split in bowling To make this question more specific, how about for an average professional bowler or average league bowler. I'm aware we can't get an exact odds without specific stats, but we just have to estimate a number of stats for the rough odds. I suppose a start would be to come up with the average number of 7-10 split opportunities that occur in a game for one person. I'm not a bowling expert so I have no idea. === === Subject: : Re: odds of 7-10 split in bowling Mark Miller of USBC gave me an estimate of one 7-10 split converted for every 150,000 games in the 2005/2006 season according. 1:150,000 1:300,000 (for 2 7/10 split opps per game both players) === === Subject: : Re: odds of 7-10 split in bowling > I'm interested to know the approximate odds of converting a 7-10 split > for a spare in bowling. I've seen different odds: 1/1000, 1/5000, > 1/700000. It has occurred 3 times on television in PBA games. That's all I > know. Has anyone gone thru the math before in this group. Unlike Blackjack or [fair] dice, bowling involves skill combined with probability rather than probability alone. Asking the probability of making this split isn't the same type of question as asking for Blackjack what the probability of busting by taking a hit on 16 with a fresh set of decks is. It is player-dependent. The other poster correctly identified that math: (# of successes) / (# of attempts) You could probably get more accuracy by knowing more about the person throwing the ball and the specific circumstances: age, sex, success at this type of split, whether making the split is important (which might affect the level of risk the bowler is willing to carry), etc. You state that this has occurred 3 times on television in PBA games. Could this be because the bowler doesn't always try to make the split, or tries half-heartedly to make it? My understanding is that if you _really_ try to make this split, you risk striking neither pin. Could it be that the bowlers aren't _really_ trying?--they are throwing balls guaranteed to hit one pin or the other, and they're striking the pin off-center and they _might_ get the other pin but they've positioned the probability distribution to avoid a gutter ball? Of course, I don't know how skilled these folks are. My high game bowling is about 120 ... and in a 7-10 split I'd just aim for one pin or the other ... I'm not sure how much control the pros have. === === Subject: : Re: odds of 7-10 split in bowling Here's a video of one. http://www.youtube.com/watch?v=XtDDG5MrxAs I've read you just hit the pin hard and hope for some lucky bounces. Maybe similar to making a full-court shot in basketball. If anyone has estimated the odds I'd be interested to see the math, not necessarily for PBA televised games. === === Subject: : Re: odds of 7-10 split in bowling > Here's a video of one. > http://www.youtube.com/watch?v=XtDDG5MrxAs I've read you just hit the pin hard and hope for some lucky bounces. > Maybe similar to making a full-court shot in basketball. If anyone > has estimated the odds I'd be interested to see the math, not > necessarily for PBA televised games. It looks to me like there isn't enough room on the outside of the pin to hit the pin at an angle to propel it into the other pin (not enough room between the pin and the gutter). So, the strategies for reliably hitting one pin and actually trying for the split ... are not mutually exclusive. A professional bowler would always get one pin or the other, even when trying hard for the split. Shows how much I bowl. === === Subject: : Re: odds of 7-10 split in bowling > I'm interested to know the approximate odds of converting a 7-10 split > for a spare in bowling. I've seen different odds: 1/1000, 1/5000, > 1/700000. It has occurred 3 times on television in PBA games. That's all I > know. Too bad. You also need to know out of how many attempts. > Has anyone gone thru the math before in this group. The math is easy. If you hit the pin in exactly the right spot, the odds of making it are 100%. === === Subject: : Re: odds of 7-10 split in bowling > I'm interested to know the approximate odds of converting a 7-10 split > for a spare in bowling. I've seen different odds: 1/1000, 1/5000, > 1/700000. > It has occurred 3 times on television in PBA games. That's all I > know. Too bad. You also need to know out of how many attempts. > Has anyone gone thru the math before in this group. The math is easy. If you hit the pin in exactly the right spot, > the odds of making it are 100%. I bet you were influenced by my contribution to Atom Totality called Golf Theory which tells how to get a hole in one every time: THE TOTAL ATOM THEORY GUIDE TO WINNING GOLF Total Atom Theory allows you to attain a perfect score in golf (18 on an 18-hole course)! Here's how. (1) Find a golf club which will launch the ball at a 45 degree angle. (2) When you get to the first tee, measure the distance to the hole and call it D. (3) Hit the ball so that its initial velocity is sqrt(D*k), where k is the magic number. (Send me $10,000 for it.) (4) The ball will land in the hole without even bouncing. (5) Pick up the ball and go to the next hole. Clearly golfers have ignored this procedure, which will guarantee a perfect score. Tiger Woods and other players clearly have wasted their time on their technique. Archimedes Plutnoium http://www.amishrakefight.org/gfy/ There is a Plutonium atom in my brain, which makes me a genius! Send me your paycheck, Tiger Woods! === === Subject: : Re: odds of 7-10 split in bowling THE TOTAL ATOM THEORY GUIDE TO WINNING GOLF Total Atom Theory allows you to attain a perfect score in golf (18 on > an 18-hole course)! Here's how. (1) Find a golf club which will launch the ball at a 45 degree angle. (2) When you get to the first tee, measure the distance to the hole and > call it D. (3) Hit the ball so that its initial velocity is sqrt(D*k), where k is > the magic number. (Send me $10,000 for it.) (4) The ball will land in the hole without even bouncing. What about the Coriolis Effect? One has to figure in the direction of the hole. Bob Kolker === === Subject: : Re: odds of 7-10 split in bowling (4) The ball will land in the hole without even bouncing. Wrong. The pin is in the hole. And for the long holes one would have to do Kentucky Windage, i.e. figure the effect of wind. Bob Kolker === === Subject: : Re: wff's definition > In almost every book, including Mendelson's, the definition of wff > doesn't seem satisfactory. First, it should have been defined as class > of members of a known well defined set, say T, and by axiom of subset, > it should have been CONCRETELY settled whether a particular member of T > is a wff or not. > The recursion used in the definition in any other book, like Palyutin's > or Manaster's, does not go in tune with the principle of transfinite > recursion in its strict form. The wffs of a language are not defined by transfinite recursion at all, > the definition is rather a very simple inductive definition. If one > wishes, for some reason, to take a set theoretic approach, the wffs of > a language L are defined as a subset of the set of strings over the > alphabet of L (i.e. finite sequences of symbols of L) as {P | P is a string over the alphabet of L & for all A( if A is > wff-closed P is in A)} where a set A is wff-closed iff - every atomic formula is in A > - if P is in A then ~P is in A > - if P and Q are in A then so are P & Q, P <--> Q, P --> Q, P / Q > - if P is in A and x is a variable then for all x P is in A and so > is exists x P > Well, can you prove the unique readability from your definition? === === Subject: : Re: wff's definition >> If one wishes, for some reason, to take a set theoretic approach, the >> wffs of a language L are defined as a subset of the set of strings over >> the alphabet of L (i.e. finite sequences of symbols of L) as >> {P | P is a string over the alphabet of L & for all A( if A is >> wff-closed P is in A)} >> where a set A is wff-closed iff >> - every atomic formula is in A >> - if P is in A then ~P is in A >> - if P and Q are in A then so are P & Q, P <--> Q, P --> Q, P / Q >> - if P is in A and x is a variable then for all x P is in A and so >> is exists x P Well, can you prove the unique readability from your definition? Adding parentheses in the definition of wff-closed, yes. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : derivative of strictly increasing continuous function Let f be a strictly increasing continous function. Is it true that if f is not differentiable at x0, it must hold that f'(x0) = infinity? (that is, is it true that lim (f(x+h)-f(x))/h exists or be infinity?) === === Subject: : Re: derivative of strictly increasing continuous function > Let f be a strictly increasing continous function. Is it true > that if f is not differentiable at x0, it must hold that > f'(x0) = infinity? (that is, is it true that lim (f(x+h)-f(x))/h > exists or be infinity?) Let d- & d+ be the lower left & lower right Dini derivates and D- & D+ be the upper left & upper right Dini derivates (of some specified function). You appear to be asking if either d- = d+ = D- = D+ = finite or d- = d+ = D- = D+ = infinity must hold at each point of a strictly increasing continuous function. In fact, given any two extended real numbers r and s such that 0 <= r < s <= oo, there exists a strictly increasing continuous function such that we have d- = d+ = r < s = D- = D+ for a co-meager (but necessarily of measure zero, of course) set of real numbers. In particular, this can hold at c-many points in every open interval (c = cardinality of the reals). I'm also pretty sure (but not certain; I'd have to look over some constructions/proofs that I don't have with me right now) that, for a strictly increasing continuous function, we can have 0 <= d- < d+ = r < s = D- < D+ <= oo on a smaller type of set (a certain subclass of the sigma-porous sets) that allows for examples having Hausdorff dimension 1 in every open interval (in particular, it can hold at c-many points in every open interval). ESSAY ON NON-DIFFERENTIABILITY POINTS OF MONOTONE FUNCTIONS Dave L. Renfro === === Subject: : Re: derivative of strictly increasing continuous function How about f(x) given by x for x<0, and 2x for x>=0? It's continuous everywhere, but non-differentiable at x=0 (and not in the sense of infinity). Dr. Michael W. Ecker Associate Professor of Mathematics Pennsylvania State University Wilkes-Barre Campus Lehman, PA 18627 === === Subject: : Re: derivative of strictly increasing continuous function >Let f be a strictly increasing continous function. Is it >true that if f >is not differentiable at x0, it must hold that f'(x0) = >infinity? (that >is, is it true that lim (f(x+h)-f(x))/h exists or be infinity?) No. Given any sequence m_n in (0,infty), it's easy to construct an example where there is a sequence x_n -> x_0 with (f(x_n) - f(x_0))/(x_n - x_0) = m_n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: Is it possible for a probability to be unknown? > if you were forced to make a bet on whether a marble taken out of > the bag at random would be red, what odds would you offer? Being forced to make a bet presumes you're in a contest with an adversary. (Nobody except an adversary would force you to make a bet when you'd rather not make any bet.) In that situation, you need to know the payoff matrix of the contest, the complete set of individual strategies available to your adversary and to yourself, and the sequence of playing. - If you move first, then your adversary can manipulate the situation based on knowledge of what you did, then from your point of view the game is totally hopeless: -- If you guess high probability of red, opponent manipulates the situation so it's something non-red instead. -- If you guess low probability of red, opponent manipulates the situation so it's red. -- If you guess exactly 50%, and if payoff rules are that you win or lose equal amounts depending on statistical estimate of winning based on your probability, then opponent delierately biasses the results long enough to trick you into betting differently, at which point one of the earlier manipulations happens. But you really must read the fine print as to how much you win or lose in different circumstances. Now if your opponent must establish the situation before you make your move, but you can't see the situation until after you make your move, then the minmax per game theory applies. Your best strategy is usually a mixed strategy that maximizes the expected value of the worst that can happen to you per all possilble mixed opponent strategies. So if you are playing against a supernatural being who can change reality out from under, you can't win, so don't worry about it, there's no point worrying about something unfixable. Likewise if you're playing against a sleight-of-hand artist who can change reality out from under you by distracting you from seeing the change, you can't win. But if you are really sure your opponent can't possibly cheat, and the rules say he must set up things before you make your move, and he can't change anything after you've made your move, then minmax game theory applies. So whoever postulated this contest, please tell me what the payoff matrix is for you, given the fact of what probability you guessed, and the fact of what color ball was selected. === === Subject: : Re: Is it possible for a probability to be unknown? >> if you were forced to make a bet on whether a marble taken out of >> the bag at random would be red, what odds would you offer? >Being forced to make a bet presumes you're in a contest with an >adversary. (Nobody except an adversary would force you to make a >bet when you'd rather not make any bet.) In that situation, you >need to know the payoff matrix of the contest, the complete set of >individual strategies available to your adversary and to yourself, >and the sequence of playing. Not necessarily; nature is not an adversary. If there is an adversary, game theory needs to be invoked, not merely probability. Statistical inference is based on the idea that the laws of nature are set, and are not going to change either for or against you. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === === Subject: : Re: Is it possible for a probability to be unknown? Of course it is possible for a probability to be unknown. This is the reason that statistical inference is needed; USUALLY the probability distribution is unknown in practice. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === === Subject: : ???? Suppose you have a 4th degree polynomial. Now consider the lines that intersect this polynomials in 4 points. I was told that the sum of these 4 values are constant regardless of the lines used. How would you show this?? Mkajuma === === Subject: : Re: ???? > Suppose you have a 4th degree polynomial. Now consider the lines that intersect this polynomials in 4 points. I was told that the sum of these 4 values are constant regardless of the lines used. How would you show this?? > Mkajuma Since (x-x1)*(x-x2)*(x-x3)*(x-x4) = x^4-(x1+x2+x3+x4)*x^3+..., the sum of th 4 x-values of the zeroes of the polynomial is just the negative coefficient of x^3. Intersecting with any line instead of the x-axis is like changing the constant and the linear term of the polynomial before computing the roots - this doesn't affect the coefficient of x^3. The same trick works with 3rd degree polynomials and with higher degrees, but not with quadratics. -hagman === === Subject: : Re: ???? > Suppose you have a 4th degree polynomial. Now consider the lines that intersect this polynomials in 4 points. I was told that the sum of these 4 values are constant regardless of the lines used. How would you show this?? > Mkajuma If f(x) = a * x ^ 4 + b * x ^ 3 + ... + e is a fourth-degree polynomial, then the sum of the four roots of f is -b / a (that is, the opposite of the ratio of the x ^ 3 coefficient to the x ^ 4 coefficient). Given a line y = p * x + q that intersects f in four points, the points where this line intersects f are precisely the roots of f(x) - (p * x + q) = 0. But, the coefficients of x ^ 4 and x ^ 3 in f(x) - (p * x + q) are the same as the corresponding coefficients in f(x), so the sum of the roots are the same in both cases. This holds regardless of what line we chose. === === Subject: : Re: ???? Ie the sum of the 4 x values is constant === === Subject: : Re: ???? Solve for the intersection points: a4*x^4+a3*x^3+a2*x^2+a1*x+a0 = b1*x+b0 <=> a4*x^4+a3*x^3+a2*x^2+(a1-b1)*x+(a0-b0) = 0 So, the x-coordinate of the intersection points are the roots of this polynomial. But the sum of the roots of a 4th order polynamial is -a3/a4 because: a4*(x-x1)(x-x2)(x-x3)(x-x4) = a4*x^4 - a4*(x1+x2+x3+x4)*x^3 + a4*(x1*x2+x1*x3+...)*x^2- a4*(x2*x3*x4+x1*x3*x4+...)*x + a4*x1*x2*x3*x4 and by indentifying the two expressions, we get - a4*(x1+x2+x3+x4) = a3 It should work for any polynomial of order n>=3. The sum is -a_{n-1}/a_n It also works if the roots are complex numbers (but in this case the geometrical interpretation is no longer possible). Thomas === === Subject: : The use of probability; was: Re: Baby stat question > > I'm not sure what the exact opposite of a statistian is, but > I'd bet that I'm pretty close. Despite that, I'm teaching the > stats for poets class this semester, and I put the following > problem on my last test: >> An SRS of 10 Iowans is chosen and their weights recorded. The > standard deviation of this sample is 12 lbs. A second SRS of > 10 Iowans is chosen and their weights are recorded. The standard > deviation of this sample is 14 lbs. What is the probability that > difference of the means of these two samples is at least 11 lbs.? >> [...] > Strictly speaking, the difference between of the two sample >> means is either 11 or more, or it isn't. So the correct answer >> is, 0 or 1. Nonsense. Let D be the difference between the sample means. > Assuming a normally distributed population, D, S1, and S2 have a > well-defined joint distribution. We are looking for the conditional > probability, P(|D| > 11 | S1, S2). Or are you sayng that D is random before we draw the sample, but > not after we have drawn it, even though we haven't observed it? Even > before we draw the sample, either |D| will exceed 11 or it will not. Marcus and I continued a discussion of this subject via e-mail. With his permission, I share my latest letter to him. But first his response to the last question, which he addressed in an earlier letter: > The latter. Goddard's question said: what is the probability that > the means of these two samples is at least 11 pounds? The setting > was: the two samples had already been drawn. He was not asking > a question about the means of some hypothetical background > population, nor was he asking about the expected difference between > the means in repeated sampling. He described the results of one > experiment and asked what the probability was of a certain finding. Now my latest letter: >I hope we're not just talking about semantics > No, we are not just talking semantics. >I think there is >at least a meaningful distinction here >between what Goddard asked and what I think >he intended to ask ... > > I think BG asked *exactly* what he meant. >But I thought the question Goddard was >asking was the following. I draw >two random samples of people. I >weigh the people and compute the mean >weight for each random sample. I >KNOW what the difference in the means is. >[After all, he says he had recorded all >the weights and computed the standard deviations. >I assumed, I think reasonably, that he had >computed the means also. Most people would.] > He could have entered the data in a file and requested only the standard deviations from some software. But that is not important: the students do not know what the means are. > >Then he says: What is the probability >that the difference in the means ***for >these samples*** is at least 11? To which I >say, Well, {dummy}, was the difference bigger >than 11, or not? To which Goddard might >say, OK, if your going to be literal and >pedantic about it, I'll rephrase the question. >Which is what he should have done in the >first place. > BG was asking a *probability* question about sampling distributions. Given S1 and S2, what is the the probability that |D| > 11? (BTW, if you have looked in sci.math lately, you will see that I changed my answer to the question.) >The usual context for the kind of >question Goddard was asking is: I have >observed a difference of 11 in the two >means. If the sampling was random, >what is the probability that in >repetitions of the sampling, I would >observe a difference as large as >that or larger? The answer then is >the usual t-statistic p-value, which is >what I think he intended. But he didn't ask >that question. If a student answered >his question with 0 or 1 I would feel >obligated to give full credit. > > You appear to be imposing what you expected on what BG actually asked. Similarly, before I posted the first time, I thought he was asking about the probabliity of a difference in population means. I typed out a tirade about how the posing of such a question indicated that he has no business teaching statistics, since such a probability is not defined in the frequentist paradigm. Fortunately, I reread the question before I posted. Again, BG asked what he meant: a probability question about the difference between sample means from two samples in the same population. A p value would arise in the context of hypothesis testing about the difference between *population* means. >[example about 8 heads in ten flips] > > from an earlier e-mail: I flip a fair coin 10 times without showing you. I write down the number of heads on the card. What is the probability that the number I have written down is an 8? Are you telling me it is either 0 or 1? ***** I guess that shows clearly that >probability is subjective - you know >whether the probability is 0 or 1; I >know that it is 0 or 1, but I don't >know which one. So if forced to bet, >I behave as if the coins had not yet >been flipped. To which you might say, >There's no difference! You're making >an artificial distinction! The coin >flips are going to produce 8 heads or >not If they do, you could say the >probability is 1 and if they do not >then the probability is 0. > > Respectfully, I point out that your position on probabilities as long-term frequencies is quite extreme. (I wonder if you are confusing probabilities and p values right now because of the context of the discussion.) Here is another example. I shuffle a standard deck of playing cards. I will deal you the first card only, but I haven't done so yet. a) What is the probability that the card on top is the ace of spades? b) What is the probability you will receive the ace of spades? (Probability was built on just these kind of questions.) These two events are exactly the same! Saying that the answer to the first question is either 0 or 1 is impractical, to say the least. Even without getting into the whole Bayesian-frequentist debate, I think few would agree that the answers to questions to (a) and (b) above are different. On an exam, I would mark 0 or 1 as an incorrect answer. More examples: - Probability of rain? Well, it either will or not: 0 or 1 (by your reasoning). - Age, weight, height, and cholesterol level amongst men in a certain population have roughly a multivariate normal distribution with known parameters. We picked John Smith at random from this population. He is 42 yrs., 3 mos. old, weighs 180 lbs., and is 5' 9.5 tall. What is the probability his cholesterol level is above 200? Well, it either is or not: 0 or 1 (even more so by your reasoning). - True story: My friends have two children. I visited them the other day, and their daughter Sophia answered the door. What is the probability that their other child is a son? Well, the child is either a son or not, so the answer is 0 or 1. If a student did not answer 1/3 to that question on an exam, I would mark it wrong. Etc. Randomness is not really a property of events, regardless of when they occur (past, present, or future). The event either has happened/is happening/will happen or not - there is no sense is limitiing that argument to the future. Probability provides a numerical description of our *state of knowledge.* We may use long-term frequencies as the basis for choosing these numbers, but we are often not interested in the long-term average (e.g., as in repeated sampling) itself. We are interested in the event at hand. It is perfectly fine to assign probabilities to events that have already happened, as temporality does not necessarily have anything to do with uncertainty. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: The use of probability; was: Re: Baby stat question - True story: My friends have two children. I visited them the > other day, and their daughter Sophia answered the door. What is the > probability that their other child is a son? Well, the child is > either a son or not, so the answer is 0 or 1. If a student did not > answer 1/3 to that question on an exam, I would mark it wrong. D'oh! 1/3 is wrong! The correct answer is 2/3. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: The use of probability; was: Re: Baby stat question > I think BG asked *exactly* what he meant. I think I did too. But let me say (again) that I think it's possible to be overly exact and pedantic when writing questions for students. Marcus(?) suggested a different wording, which, while more exact and pedantic, my students would have had a lot of trouble deciphering. There's something artificial about such wordings, and I came to the conclusion a few years ago that it's part of the student's responsibility to be in context. By which I mean that I can be a little casual in my wording and the student should be familiar enough with the subject that he can put the problem in our context and solve it. It part of the test, actually, as to whether the student can recognize the problem as one he should be able to solve, even though it's not in textbook language. > he has no business teaching statistics, Even though you didn't say it, it's true. The difficulty is in convincing the administration of this fact. Unfortunately, the students adore me (probably because I sympathize with their distaste for stats) and my stellar evaluations keep me at it. I've found y'all's discussion interesting, and I hope you continue to make it public. In case anyone cares, the text (which I didn't choose) is Moore Basic Practice of Statistics 4th ed. It has this Option 1 Option 2 stuff which I also found in Johnson's elementary text. > BG was asking a *probability* question about sampling distributions. > Given S1 and S2, what is the the probability that |D| > 11? (BTW, > if you have looked in sci.math lately, you will see that I changed my > answer to the question.) The main part of your change was in the DF. It is the case that Option 1 gives n1-1+n2-1 when n1=n2, and the text briefly mentions this. But, for the sake of expediency, I made the fiat that we'd always use Option 2 for such problems. We're trying to make these particular students think statistically, not train them for jobs. B. -- The man without a .sig === === Subject: : Re: The use of probability; was: Re: Baby stat question >BG was asking a *probability* question about sampling distributions. >>Given S1 and S2, what is the the probability that |D| > 11? (BTW, >>if you have looked in sci.math lately, you will see that I changed my >>answer to the question.) >> The main part of your change was in the DF. > No! The very important part of my change was that your whole approach was fundamentally wrong - and this a much more serious problem than the wrong degrees of freedom. To wit: Just because P{X / Y > a} = p and you observe that Y = y, you may *not* conclude that P{X > a y} = p. Or to put it another way, in general, P{X / Y > a | Y = y} != P{X > a y}. (This is true if X and Y are independent, but your X and Y are not.) In the case at hand X is the absolute difference in means, Y is the pooled sampled variance, a = 11, and p comes from the t distribution with the appropriate degrees of freedom.. >>he has no business teaching statistics, >> Even though you didn't say it, it's true. The difficulty >is in convincing the administration of this fact. Unfortunately, >the students adore me (probably because I sympathize with their >distaste for stats) and my stellar evaluations keep me at it.>The main part of your change was in the DF. No! Yeah, I realized that just after I posted. > To wit: Just because P{X / Y > a} = p and > you observe that Y = y, you may *not* conclude that P{X > a y} = p. > Or to put it another way, in general, P{X / Y > a | Y = y} != > P{X > ay}. > (This is true if X and Y are independent, but your X and Y > are not.) In the case at hand X is the absolute difference in means, > Y is the pooled sampled variance, a = 11, and p comes from the t > distribution with the appropriate degrees of freedom.. I'm not sure I follow. Aren't there objections to the X and Y are dependent assertions on this thread? In particular, I'm not sure I believe the statement: P{X / Y > a | Y = y} != P{X > ay} Maybe after I think about it a while.... > It is not uncommon for a faculty member to teach one class when an more > informed one is unavailable. As I well know. I've taught Abstract Algebra, Numerical Analysis, ODE's, PDE's, Number Theory (my actual field), Difference Equations, Real Analysis, Complex Analysis, etc. etc. And, yep, I sure have learned a lot of math that way. > What does sfsf mean? stupid fkn smiley face. It's a phrase for those of us who hate emoticons but need one anyway. B. -- The man without a .sig === === Subject: : Re: The use of probability; was: Re: Baby stat question > >>The main part of your change was in the DF. >> >No! >> Yeah, I realized that just after I posted. > >To wit: Just because P{X / Y > a} = p and >>you observe that Y = y, you may *not* conclude that P{X > a y} = p. >>Or to put it another way, in general, P{X / Y > a | Y = y} != >>P{X > a y}. >>(This is true if X and Y are independent, but your X and Y >>are not.) In the case at hand X is the absolute difference in means, >>Y is the pooled sampled variance, a = 11, and p comes from the t >>distribution with the appropriate degrees of freedom.. >> I'm not sure I follow. Aren't there objections to the X and Y >are dependent assertions on this thread? > No. D and Sp (the pooled sample standard deviation) are independent. T and Sp are *not* independent, where T = D / (Sp / sqrt(5)). These are facts which need not be debated. >In particular, I'm not >sure I believe the statement: P{X / Y > a | Y = y} != P{X > a y} Maybe after I think about it a while.... > Suppose P{X = Y = 1} = 1/10, P{X = 2, Y = 1} = 1/5, P{X = 1, Y = 2} = 3/10, and P{X = Y = 2} = 2/5. P(X / Y > 1 | Y = 1) = 2/3; P{X > 1} = 3/5. The correct equality is P(X / Y > a | Y = y} = P(X > a y | Y = y). That is, you must use the *conditional* distribution of X. I hate to admit it, but now I do wonder about your teaching this course. Sorry. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: The use of probability; was: Re: Baby stat question > I hate to admit it, but now I do wonder about your teaching this > course. Sorry. When I said I needed time to think about it, I meant it. I really only skimmed your post before replying, which maybe wasn't the most polite thing, but, heck, who's being polite nowadays? You can relax. I've taught Stats a dozen times in several different venues (for poets and for engineers.) I know what I'm doing, even if I, too, make a mistake once in a while. > No. D and Sp (the pooled sample standard deviation) are independent. > T and Sp are *not* independent, where T = D / (Sp / sqrt(5)). This is what I indicated in my original post. Namely, my question was whether my students might be (accidently) correct in using the z-distribution since there was less variability in the problem than we normally have when using t-distribution. Bart -- The man without a .sig === === Subject: : Re: Reverse (newbie question) Sorry for not being clear in my question. >What i have here is i am given a fixed 8 digit number let say 80385444 >(80385444 + A) * B = 444458308 How to achieve that with minimum calculation? >Is this much clearer? B = 1 A = 44458308-80385444 In the general case, given a number n y = 0 x = n while (x > 0) {at most 8 times to reverse the digits} y = 10*y + x mod 10 x = [x/10] A = n - y B = 1 Remove del for email === === Subject: : analysis solution Can someone show me how to prove this: Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). === === Subject: : Re: analysis solution > Can someone show me how to prove this: Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Can g(x) = exp(-k*x)*f(x) help? === === Subject: : Re: analysis solution , Can someone show me how to prove this: > Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Can g(x) = exp(-k*x)*f(x) help? > Nice. === === Subject: : Re: analysis solution , The World Wide Wade > , > Can someone show me how to prove this: > Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Can g(x) = exp(-k*x)*f(x) help? > Nice. Amen. -- Michael Press === === Subject: : Re: analysis solution > Can someone show me how to prove this: Assume f ' < oo in (a,b) Could f' = -oo somewhere in (a,b)? >, and f continuous continuous where? >, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Hint: Show that f(x) - k*int_[a,x] f(t) dt has a local extremum somewhere in (a,b). === === Subject: : Re: analysis solution On 2006-12-04 03:55:02 -0500, The World Wide Wade said: > >> Can someone show me how to prove this: >> >> Assume f ' < oo in (a,b) Could f' = -oo somewhere in (a,b)? > >> , and f continuous continuous where? > >> , and f(a) = f(b) = 0. Show >> for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Hint: Show that f(x) - k*int_[a,x] f(t) dt has a local extremum > somewhere in (a,b). I'm sorry, but I'm not seeing the obvious here: how do you see f(x) - k int_[a,x] f(t) dt has an extremum inside (a,b)? -- -kira === === Subject: : Re: analysis solution On 2006-12-04 02:03:48 -0500, jraul said: > Can someone show me how to prove this: Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Consider the function g(x) = f'(x) - k f(x) over (a,b). -- -kira === === Subject: : Re: analysis solution > On 2006-12-04 02:03:48 -0500, jraul said: Can someone show me how to prove this: Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). Consider the function > g(x) = f'(x) - k f(x) > over (a,b). After I've considered it, what am I supposed to do with it? === === Subject: : Re: analysis solution On 2006-12-04 15:42:37 -0500, The World Wide Wade said: > >> On 2006-12-04 02:03:48 -0500, jraul said: >> > Can someone show me how to prove this: > > Assume f ' < oo in (a,b), and f continuous, and f(a) = f(b) = 0. Show > for all real k there is a c in (a,b) s.t. f '(c) = k*f(c). >> >> Consider the function >> g(x) = f'(x) - k f(x) >> over (a,b). After I've considered it, what am I supposed to do with it? Um... then you can multiply it by e^{-kx} and realize that e^{-kx} g(x) = (e^{-kx} f(x))' and then use Kovarik's trick in the other post. Sorry, but the truth is I didn't think things through... -- -kira === === Subject: : reflecting back Re: Optimal Strategy for playing stockmarket via VonNeumann; birdflu impact on economics > Portfolio of PAF as of 27 NOV 06 BCE 13,610 > T 8,510 > SGP 1,010 > total share-wealth-units last reported which was 1NOV06 23,120 > total share-wealth-units today 23,130 > realestate land 3APR03 of 3 lots $19,000. > science-art of pictures,porcelain etc starting JAN03 for $17,556. > realestate land 30JUL03 another lot $11,500. > realestate land Sept05 another lot $75,000. Today I sold 1,000 BCE at 24.44 and with the proceeds bought 1,010 SGP > at 21.22 and keeping the cash remainder. I did it for two reasons: I > needed some cash and I wanted to not lose in number of total share > wealth units and I want to diversify into the upcoming birdflu > pandemic. I have a reasonable guess that H5N1 is going pandemic and the drug > companies will be the main beneficiaries of a pandemic. Regardless that > the Democrats control Congress for that is mostly gridlock. I am going to post more to this thread explaining my mistakes for the > past year. The biggest mistake was that I had 23,000 T a year ago and > ended up with 8,510 T and 14,610 BCE. When I should have ended up with > 8,510 T and 14,610 BellSouth. Another mistake this year was the window > of opportunity with BMY, Bristol Myers when it fell to 21 on fears of > losing patent over plavix and now is 24. But I shall hash over these > mistakes later. And I have a nice excuse for making those mistakes and > I will accept that excuse--- I was too deep and involved in physics and > other projects for this past year that I simply let my portfolio > languish. To me, science progress is more important than my own > personal financial progress. Back in August of 2005 the PAF portfolio was this: 15AUG05 BCE 10 BLS 10 BMY 110 T (formerly SBC) 23,270 VZ 10 realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 for $17,331. realestate land 30JUL03 another lot $11,500. I kind of wish I had those shares back again now because T is worth $34 a share and BCE only $24 with a spread of $10 so there is no near future switch campaign going to take place. And because T gives a heftier dividend. But that is old time and none of us is perfect at this game. I should be thankful that the present portfolio is ahead in shares, and money value and in real-estate. But we can learn from past mistakes. These are a few of the mistakes which I made from August 2005 to now, December 2006. (a) should have just kept the 23,270 T and stayed pat and stayed put (b) had a chance sometime in that period to switch out of T and into BLS where T is now worth $34 but BLS is worth $44. I should have seen it coming where T would make a buyout of BLS (c) had the chance to buy some MRK for about $25 with T or BCE worth slightly more than $25, made the switch, and today MRK is worth about $44 with a hefty dividend (d) had the chance to switch some BCE for BMY recently when BMY was depressed at $21 over a scare on plavix patent My excuse was I was too busy with science and other projects this past year. But this year should keep a eye on the market. And because T is so far gone from BCE, that I need a new switching partner for T and that should be VZ. So if VZ goes below T, new opportunities open up. And I should have new switching partners for BCE, and they appear to be both BMY and SGP. I am going back into the pharmaceuticals because of the extent to which I anticipate birdflu, H5N1 will go. I believe it will go pandemic which will be a big boost to big pharma. Also, I think the drug companies will find a drug that will reduce obesity and that works and has almost zero sideeffects. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === === Subject: : Re: reflecting back Re: Optimal Strategy for playing stockmarket via VonNeumann; birdflu impact on economics > [...] > I am going back into the pharmaceuticals because of the extent to which > I anticipate birdflu, H5N1 will go. I believe it will go pandemic which > will be a big boost to big pharma. [...] But it hasn't yet. OTOH, this is a safe bet. If I made a bet with AP, him taking the side of an eventual pandemic, I could never collect from him, since he could always claim that it could happen a few months later. --- Christopher Heckman === === Subject: : quaziblock system of linear equations: Benders decomposition question Hallo! Let us assume we have quaziblock system of linear equations which have quaziblocks with n(j) inner variables & m(j) border variables example A= 1 2 3 0 0 0 0 2 3 4 0 0 0 0 5 2 3 5 0 0 0 0 0 7 8 0 0 0 0 0 0 0 6 7 0 0 0 0 0 5 9 9 0 0 0 0 0 4 8 - 4 quaziblocks, n(1)=2,m(1)=1,n(2)=1,m(2)=1,n(3)=1,m(3)=1,n(4)=1,m (4)=1 let us try to solve it by Benders decomposition using 1 computer for every quaziblock & 1 as server for their interconnecting. So, how many data (how many real numbers) must I transmit from subsystems to server? And what will be the size of linear system in server & number of non-zeros there? It's hard to understand Benders algorithm, because in all internet sourses that I looked throw it operates with restrictions, with dual tasks, with matrix operations that are hard to understand For simplifying, let all n(j)=n, m(j)=m Is that 2*n*m+m^2? or n*m+m^2? or other value? Than less or greater? It's very important for my post-graduate work. Please, provide internet sourse of your answer if possible (that could supply my arguments in future). === === Subject: : Fast Chebyshev Polynomials The well-known recursion formula for Chebyshev polynomials can be written in matrix form: | T_(n+1)(x) | | 2.x -1 || T_(n) (x) | | | = | || | | T_n (x) | | 1 0 || T_(n-1)(x) | When doing the recursion to the end, we find: | T_(n+1)(x) | | 2.x -1 |^n | x | | | = | | | | | T_n (x) | | 1 0 | | 1 | The power of a matrix can be calculated with an amount of work which is of the same order as the amount of work to calculate the power of a real number: order ln(n). Here comes: program Pafnuty; { Tchebyshev Polynomials with Matrix Method type matrix = array[0..1,0..1] of double; function een : matrix; { Unity } var E : matrix; begin E[0,0] := 1; E[1,1] := 1; E[1,0] := 0; E[0,1] := 0; een := E; end; function maal(A,B : matrix) : matrix; { Multiplication } var C : matrix; begin C[0,0] := A[0,0]*B[0,0] + A[0,1]*B[1,0]; C[0,1] := A[0,0]*B[0,1] + A[0,1]*B[1,1]; C[1,0] := A[1,0]*B[0,0] + A[1,1]*B[1,0]; C[1,1] := A[1,0]*B[0,1] + A[1,1]*B[1,1]; maal := C; end; function macht(x : matrix; n : integer) : matrix; { Power x^n } var m : integer; p,y : matrix; begin m := n; y := x; p := een; while m > 0 do begin if (m and 1) > 0 then p := maal(p,y); m := m shr 1; { divide by 2 } y := maal(y,y); end; macht := p; end; function T(n : integer; x : double) : double; { Chebyshev polynomial of the first kind } var M,P : matrix; u : double; begin u := 1; if n = 1 then u := x; if n > 1 then begin M[0,0] := 2*x; M[0,1] := -1; M[1,0] := 1 ; M[1,1] := 0 ; P := macht(M,n-1); u := P[0,0]*x + P[0,1]; end; T := u; end; function hand(n : integer; x : double) : double; { By hand } var u : double; begin u := 0; case n of 0 : u := 1; 1 : u := x; 2 : u := 2*sqr(x) - 1; 3 : u := 4*sqr(x)*x - 3*x; 4 : u := 8*sqr(x)*sqr(x) - 8*sqr(x) + 1; end; hand := u; end; procedure test; { Just a Test } var n,k : integer; begin for n := 0 to 4 do begin Writeln(n); for k := 0 to 10 do begin Writeln(T(n,k/10),' = ',hand(n,k/10)); end; Writeln; end; end; begin test; end. Han de Bruijn === === Subject: : Re: Fast Chebyshev Polynomials And, oh yeah, everything you always wanted to know about Chebyshev and stuff is in: http://hdebruijn.soo.dto.tudelft.nl/jaar2006/pafnuty.pdf http://hdebruijn.soo.dto.tudelft.nl/jaar2006/PAFNUTY.ZIP Han de Bruijn === === Subject: : Re: Fast Chebyshev Polynomials And, oh yeah, everything you always wanted to know about Chebyshev and > stuff is in: http://hdebruijn.soo.dto.tudelft.nl/jaar2006/pafnuty.pdf > http://hdebruijn.soo.dto.tudelft.nl/jaar2006/PAFNUTY.ZIP it is suggested (near the end) that there exists a relationship between Chebyshev Polynomials and the Fibonacci Numbers. This has triggered me to do some research. And indeed found a quite nice reference: http://www.emis.de/journals/PM/52f3/pm52f311.pdf The following theorem is in there: U_n(i/2) = i^n F_(n+1) where i is the imaginary unit: i^2 = -1 . And U = Chebyshev polynomial of the second kind, F = Fibonacci number. Don't know if the following is a well known proof. We employ the matrix method from a previous poster in this thread. The content is adapted to accomodate for Chebyshev polynomials of the second kind (instead of the first). Then the well-known recursion formula can be written as: | U_(n+1)(x) | | 2.x -1 || U_(n) (x) | | | = | || | | U_n (x) | | 1 0 || U_(n-1)(x) | When doing the recursion to the end, we find: | U_(n+1)(x) | | 2.x -1 |^n | 2.x | | | = | | | | | U_n (x) | | 1 0 | | 1 | On the other hand, the Fibonacci numbers are defined as follows: F_0 = 0 , F_1 = 1 , ... , F_(n+1) = F_n + F_(n-1) We can also employ a matrix method for expressing this definition: | F_(n+1) | | 1 1 || F_n | | | = | || | | F_n | | 1 0 || F_(n-1) | When doing the recursion to the end (well, almost to the end) we find: | F_(n+1) | | 1 1 |^(n-1) | 1 | | | = | | | | | F_n | | 1 0 | | 1 | Mind the similarity between this and the Chebyshev matrix formulation! What we want is to construct a mapping between the two. After a bit of puzzling, we find the following: | F_(n+1) | | i -1 || F_n | | | = -i | || | | -i F_n | | 1 0 || -i F_(n-1) | And, in the end: | F_(n+1) | | i -1 | | i | | | = ( -i | |)^(n-1) (-i) | | | -i F_n | | 1 0 | | 1 | Or: | F_(n+1) | | i -1 |^(n-1) | i | i^n | | = | | | | | -i F_n | | 1 0 | | 1 | On the other hand, if we specify for x such that 2.x = i in the matrix method for Chebyshev polynomials of the second kind, then: | U_n(i/2) | | i -1 |^(n-1) | i | | | = | | | | | U_(n-1)(i/2) | | 1 0 | | 1 | Herewith the theorem is proved: i^n F_(n+1) = U_n(i/2) . Han de Bruijn === === Subject: : Calculator for H-index of Scientist Impact and Influence H-index Calculator of Scientist Impact and Influence http://www.epidemiologic.org/2006/12/h-index-calculator-of-scientist-impact. html The H-index, developed by physicist Jorge Hirsch, is an index is designed to go beyond simple statistics such as the total number of citations or publications, to distinguish influential scientists from those who simply publish many papers. The H-index, which accounts for both is also not highly sensitive to single papers that have many citations. The H-index is relatively effective in comparing researchers working in the same scientific field, and not comparable across disciplines due to different publishing and citation patterns. published in Nature, Index aims for fair ranking of scientists Nature. 2005 Aug 18;436:900, which highlighted the H-index's potential use in informing decisions regarding a scientist's tenure, promotion, election to scientific bodies such as the National Academy of Science, and the Royal Society, as well as informative in determining a scientist's impact relative to Nobel Prize laureates. Online calculator to determine an individual scientist's H-index value: http://www.epidemiologic.org/2006/12/h-index-calculator-of-scientist-impact. html === === Subject: : New mathematics/physical sciences positions at http://jobs.phds.org, December 04, 2006 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Statistical Arbitrage Trader: Recruit Associates, NY. Global, multi-strategy hedge fund seeks high calibre stat arb traders. This hedge fund wants the best, and as such is prepared to pay, both in basic and in percentage of upside. You, the ideal candidate will possess an extremely scalable strategy; around 15% annualised... http://jobs.phds.org/jobs/rfrost/statistical * Quant Analyst: Recruit Associates, NY. A US investment bank seeks a mid-level quantitative analyst from cross asset front office team. You will have had experience as a quantitative analyst in an investment bank. 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Must be proficient in... http://jobs.phds.org/jobs/deannarm/strategist-quant === === Subject: : Re: Turing vs. Godel (Newbie Question) >> So, just what is the status of your claim that no-one before CB >> produced a system with axioms, rules of inference, formal queries to >> the system, proofs in the system and corresponding >> executable programs? >> None with any general capability. Your name is Smaill but your brain > is Small. :) >> I'll take this as CB's less than gracious way of admitting that yes, >> No, I said None. >> Actually, you said None with any general capability. Your question does not specify the generality. Quite right. So if you want to say that no system of any sort satisfies the condition, then you would simply say None. Since you failed to make the stronger claim, I conclude that you think there is at least one system before CB with these properties. Now, am I right in thinking so? > It all begins with being honest about the question. Can you be honest in answering my question? > C-B > Do you claim that no-one before CB ever produced a >> system with axioms, rules of inference, formal queries to the system, >> proofs in the system and corresponding executable programs? >> >> >> -- >> Alan Smaill > -- Alan Smaill === === Subject: : Re: Turing vs. Godel (Newbie Question) > produced a system with axioms, rules of inference, formal queries to >> the system, proofs in the system and corresponding >> executable programs? None with any general capability. Your name is Smaill but your brain > is Small. :) >> I'll take this as CB's less than gracious way of admitting that yes, > No, I said None. >> Actually, you said None with any general capability. > Your question does not specify the generality. Quite right. > So if you want to say that no system of any sort satisfies > the condition, then you would simply say None. > Since you failed to make the stronger claim, I conclude > that you think there is at least one system before CB with > these properties. Now, am I right in thinking so? Nobody else has an actual Program Synthesis system. I will state that outright - and stake my reputation on it. The problem is being precise about what the requirement really is. If one has an authentic Automatic Programming System, it is obvious. Once you capture the general principles of programming formally, you will see plenty of examples, variations, and applications for the system. You have no problem coming up with examples from Number Theory - programs to list and decide primes, factors, proper factors, common factors, etc. For example, you can see in the ARXIV paper that CBL is applicable to a dizzying array of problems, from Mathematical programming to Database Query Processing, from the axiomatization of the Theory of Computation to Recursion Theory. And of course, none of this has ever been accomplished in the published literature. > It all begins with being honest about the question. Can you be honest in answering my question? Just for you I'll give it a try. Now, do you have an example of a useful, efficient program being formally derived based on only its specifications? How about listing the numbers from A to B, for given A and B? That's a piece of code that every programmer needs from time to time, but is simple enough to give a complete, self-contained solution. Trade solutions? Or is there a problem with Curry-Howard et. al. coming up with a clear, reasonable, well-explained example? I will certainly supply all the details anyone may ask for in the use of CBL to solve this Program Synthesis problem. Or is Curry-Howard just warm and squishy? C-B > -- > Alan Smaill === === Subject: : algebra with isomorphic- Hello sir~ F and E are field. f : F -> E is a isomorphism. If g : F -> E is a isomorphism, I want to show that f = g. ------------------------------------------------ is this possible thinking ? If possible, How do you show it ? === === Subject: : Re: algebra with isomorphic- > Hello sir~ F and E are field. f : F -> E is a isomorphism. If g : F -> E is a isomorphism, > I want to show that f = g. ------------------------------------------------ > is this possible thinking ? If possible, How do you show it ? The complex numbers form a field with two non-identical automorphisms. (1) the identity map : x + i*y |--> x + i*y and (2) the conjugate map: x + i*y |--> x - i*y === === Subject: : Re: algebra with isomorphic- > F and E are field. f : F -> E is a isomorphism. If g : F -> E is a isomorphism, > I want to show that f = g. ------------------------------------------------ > is this possible thinking ? As you already know by now, the answer is negative. However, it is true when E = F = real field. Don't you wan to try to prove it? Jose Carlos Santos === === Subject: : Re: algebra with isomorphic- > F and E are field. > f : F -> E is a isomorphism. > If g : F -> E is a isomorphism, > I want to show that f = g. > ------------------------------------------------ > is this possible thinking ? > As you already know by now, the answer is negative. However, > it is true when E = F = real field. Don't you wan to try to > prove it? um...interesting... I think....if f : R -> R is a isomorphism, f(1) = f(1*1) = f(1)*f(1). since f(0) = 0, f(1) =1. and f(1) = f(n/n) = n*f(1/n). so, f(1/n) = 1/n. so, f(m/n) = m/n. and 0 = f(1-1) = f(1) + f(-1). so, f(1) = -f(1). so, f(x) = x, (x is rational). If x is irrational, 2 = f(sqrt(2)*sqrt(2)) = f(sqrt(2))*f(sqrt(2)). so, f(sqrt(2)) = sqrt(2). f(pi) = ???..............Hm.............. In this case, I can't progress any more. so, I need your advice. === === Subject: : Re: algebra with isomorphic- > F and E are field. >> f : F -> E is a isomorphism. >> If g : F -> E is a isomorphism, > I want to show that f = g. >> ------------------------------------------------ > is this possible thinking ? >> As you already know by now, the answer is negative. However, >> it is true when E = F = real field. Don't you wan to try to >> prove it? um...interesting... > I think....if f : R -> R is a isomorphism, > f(1) = f(1*1) = f(1)*f(1). > since f(0) = 0, f(1) =1. > and > f(1) = f(n/n) = n*f(1/n). > so, f(1/n) = 1/n. > so, f(m/n) = m/n. > and > 0 = f(1-1) = f(1) + f(-1). > so, f(1) = -f(1). so, f(x) = x, (x is rational). So far, so good. > If x is irrational, > 2 = f(sqrt(2)*sqrt(2)) = f(sqrt(2))*f(sqrt(2)). > so, f(sqrt(2)) = sqrt(2). f(sqrt(2))^2 = 2 and therefore that f(sqrt(2)) = sqrt(2) *or* f(sqrt(2)) = -sqrt(2). > f(pi) = ???..............Hm.............. > In this case, I can't progress any more. > so, I need your advice. In R, x <= y if and only if there is some real number _z_ such that y - x = z^2. Use this fact to prove that _f_ must preserve the order. This fact, together with the fact (already proved by you) that f(x) = x if _x_ is rational allows you to deduce that f = identity. BTW, the same argument also works with the fieal of real algebraic numbers. Jose Carlos Santos === === Subject: : Re: algebra with isomorphic- > F and E are field. >> f : F -> E is a isomorphism. >> If g : F -> E is a isomorphism, > I want to show that f = g. >> ------------------------------------------------ > is this possible thinking ? >> As you already know by now, the answer is negative. However, >> it is true when E = F = real field. Don't you wan to try to >> prove it? > um...interesting... > I think....if f : R -> R is a isomorphism, > f(1) = f(1*1) = f(1)*f(1). > since f(0) = 0, f(1) =1. > and > f(1) = f(n/n) = n*f(1/n). > so, f(1/n) = 1/n. > so, f(m/n) = m/n. > and > 0 = f(1-1) = f(1) + f(-1). > so, f(1) = -f(1). > so, f(x) = x, (x is rational). So far, so good. > If x is irrational, > 2 = f(sqrt(2)*sqrt(2)) = f(sqrt(2))*f(sqrt(2)). > so, f(sqrt(2)) = sqrt(2). f(sqrt(2))^2 = 2 and therefore that f(sqrt(2)) = sqrt(2) *or* > f(sqrt(2)) = -sqrt(2). > f(pi) = ???..............Hm.............. > In this case, I can't progress any more. > so, I need your advice. In R, x <= y if and only if there is some real number _z_ such that > y - x = z^2. Use this fact to prove that _f_ must preserve the order. > This fact, together with the fact (already proved by you) that f(x) = x > if _x_ is rational allows you to deduce that f = identity. Yes, good idea. If y-x >=0, f(y-x) = f(z^2) = f(z)*f(z) ==> f(y) - f(x) >= 0. since sqrt(2) = 1.414...., f(1.4) < f(sqrt(2)) < f(1.5) f(1.41) < f(sqrt(2)) < f(1.42) f(1.414) < f(sqrt(2)) < f(1.415) ....... so, f(sqrt(2)) = sqrt(2). === === Subject: : Re: algebra with isomorphic- > > F and E are field. >> f : F -> E is a isomorphism. >> If g : F -> E is a isomorphism, > I want to show that f = g. >> ------------------------------------------------ > is this possible thinking ? >> As you already know by now, the answer is negative. However, >> it is true when E = F = real field. Don't you wan to try to >> prove it? > um...interesting... > I think....if f : R -> R is a isomorphism, > f(1) = f(1*1) = f(1)*f(1). > since f(0) = 0, f(1) =1. > and > f(1) = f(n/n) = n*f(1/n). > so, f(1/n) = 1/n. > so, f(m/n) = m/n. > and > 0 = f(1-1) = f(1) + f(-1). > so, f(1) = -f(1). > so, f(x) = x, (x is rational). > So far, so good. > If x is irrational, > 2 = f(sqrt(2)*sqrt(2)) = f(sqrt(2))*f(sqrt(2)). > so, f(sqrt(2)) = sqrt(2). > f(sqrt(2))^2 = 2 and therefore that f(sqrt(2)) = sqrt(2) *or* > f(sqrt(2)) = -sqrt(2). > f(pi) = ???..............Hm.............. > In this case, I can't progress any more. > so, I need your advice. > In R, x <= y if and only if there is some real number _z_ such that > y - x = z^2. Use this fact to prove that _f_ must preserve the order. > This fact, together with the fact (already proved by you) that f(x) = x > if _x_ is rational allows you to deduce that f = identity. Yes, good idea. > If y-x >=0, > f(y-x) = f(z^2) = f(z)*f(z) > ==> f(y) - f(x) >= 0. O.K. since sqrt(2) = 1.414...., > f(1.4) < f(sqrt(2)) < f(1.5) > f(1.41) < f(sqrt(2)) < f(1.42) > f(1.414) < f(sqrt(2)) < f(1.415) > ....... > so, f(sqrt(2)) = sqrt(2). Not quite (perhaps you already assumed continuity here?). Instead, suppose that for some irrational number c you had c < f(c). Then there is some rational number q with c < q < f(c). Marc === === Subject: : Re: algebra with isomorphic- <4tio8jF13va9pU1@mid.individual.net> <4tisk5F145mimU1@mid.individual.net> Marc Olschok Ëó.bc.bc: > F and E are field. >> f : F -> E is a isomorphism. >> If g : F -> E is a isomorphism, > I want to show that f = g. >> ------------------------------------------------ > is this possible thinking ? >> As you already know by now, the answer is negative. However, >> it is true when E = F = real field. Don't you wan to try to >> prove it? > um...interesting... > I think....if f : R -> R is a isomorphism, > f(1) = f(1*1) = f(1)*f(1). > since f(0) = 0, f(1) =1. > and > f(1) = f(n/n) = n*f(1/n). > so, f(1/n) = 1/n. > so, f(m/n) = m/n. > and > 0 = f(1-1) = f(1) + f(-1). > so, f(1) = -f(1). > so, f(x) = x, (x is rational). > So far, so good. > If x is irrational, > 2 = f(sqrt(2)*sqrt(2)) = f(sqrt(2))*f(sqrt(2)). > so, f(sqrt(2)) = sqrt(2). > f(sqrt(2))^2 = 2 and therefore that f(sqrt(2)) = sqrt(2) *or* > f(sqrt(2)) = -sqrt(2). > f(pi) = ???..............Hm.............. > In this case, I can't progress any more. > so, I need your advice. > In R, x <= y if and only if there is some real number z such that > y - x = z^2. Use this fact to prove that f must preserve the order. > This fact, together with the fact (already proved by you) that f(x) = x > if x is rational allows you to deduce that f = identity. > Yes, good idea. > If y-x >=0, > f(y-x) = f(z^2) = f(z)*f(z) > ==> f(y) - f(x) >= 0. O.K. > since sqrt(2) = 1.414...., > f(1.4) < f(sqrt(2)) < f(1.5) > f(1.41) < f(sqrt(2)) < f(1.42) > f(1.414) < f(sqrt(2)) < f(1.415) > ....... > so, f(sqrt(2)) = sqrt(2). Not quite (perhaps you already assumed continuity here?). Instead, suppose that for some irrational number c you had c < f(c). > Then there is some rational number q with c < q < f(c). um...I did not assume continuity about f. That is simple contraction method. Anyway, If c < q < f(c), f(q) < f(c). so, f(q-c) < 0 so, q-c < 0 so, q < c contradiction. thank you very much. === === Subject: : Re: algebra with isomorphic- days. My association with the Department is that of an alumnus. >Hello sir~ F and E are field. f : F -> E is a isomorphism. If g : F -> E is a isomorphism, >I want to show that f = g. ------------------------------------------------ >is this possible thinking ? If possible, How do you show it ? It's possible thinking, but it is wrong thinking. Take F = E = C the complex numbers, f as the identity, and g as complex conjugation. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: algebra with isomorphic- >Hello sir~ >F and E are field. >f : F -> E is a isomorphism. >If g : F -> E is a isomorphism, >I want to show that f = g. >------------------------------------------------ >is this possible thinking ? >If possible, How do you show it ? It's possible thinking, but it is wrong thinking. Take F = E = C the complex numbers, f as the identity, and g as > complex conjugation. Oh, I see. I know the complex conjugation isomorphisms. namely, g : R(i) -> R(-i) , g(a+bi) = a-bi. === === Subject: : Re: algebra with isomorphic- On Mon, 4 Dec 2006 20:44:15 +0900, mina_world >Hello sir~ F and E are field. f : F -> E is a isomorphism. If g : F -> E is a isomorphism, >I want to show that f = g. ------------------------------------------------ >is this possible thinking ? Hint: Show that this result implies that a field cannot have a non-trivial automorphism. >If possible, How do you show it ? > ************************ David C. Ullrich === === Subject: : Colimit and top element The following problem comes from Joseph Rotman's book Advanced Modern Algebra Prentice-Hall 2002. I was wondering if someone could kindly provide a proof as I'm going mad trying to formulate one. A partially ordered set $I$ has a *top element* if there exists $infty in I$ with $i leq infty$ for all $i in I$. If ${M_{i}, phi^{i}_{j}}$ is a direct system over $I$, prove that $$varinjlim_{i} M_{i}cong M_{infty}.$$ === === Subject: : Re: Colimit and top element days. My association with the Department is that of an alumnus. >The following problem comes from Joseph Rotman's book Advanced Modern >Algebra Prentice-Hall 2002. I was wondering if someone could kindly >provide a proof as I'm going mad trying to formulate one. Why? >A partially ordered set $I$ has a *top element* if there exists >$infty in I$ with $i leq infty$ for all $i in I$. If ${M_{i}, >phi^{i}_{j}}$ is a direct system over $I$, prove that >$$varinjlim_{i} M_{i}cong M_{infty}.$$ (Avoid exact TeX, especially non-standard macros like varinjlim, which is not part of TeX or LaTeX. Use pseudo-TeX code instead). The elements of the direct limit are equivalence classes of pairs, (a,i), where i is in I and a is in M_i, modulo the equivalence relation that makes (a,i) equivalent to (b,j) if there exists k in I such that k>=i, k>=j, and phi^i_k(a) = phi^j_k(b) (I hope I am following the same convention you are on super and sub indices in the phi; if not, reverse them). Since infty >= i for all i, there is an ->obvius<- way to map [(a,i)] into M_infty: simply map it to phi^i_infty(a). Show this map is well-defined to get a map from the direct limit to M_infty. Now, you also have a natural map from each M_i into the direct limit: M_i maps into the direct limit by mapping a to [(a,i)]. So you have a map from M_infty to the direct limit by sending any x in M_infty to [(x,infty)]. Show the two maps are inverses of each other, and that will give you the fact that they are isomorphic. Why are you having trouble with it? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Colimit and top element fed this fish to the penguins: >The following problem comes from Joseph Rotman's book Advanced Modern >Algebra Prentice-Hall 2002. I was wondering if someone could kindly >provide a proof as I'm going mad trying to formulate one. A partially ordered set $I$ has a *top element* if there exists >$infty in I$ with $i leq infty$ for all $i in I$. If ${M_{i}, >phi^{i}_{j}}$ is a direct system over $I$, prove that >$$varinjlim_{i} M_{i}cong M_{infty}.$$ > A top element in a poset is usualy called a (the) maximum. Well, you have an obvious cone phi_{i, infty}:M_{i} -> M_{infty} and you have to prove the universal property for it. Hint: use the fact that infty is a maximum in I. G. Rodrigues === === Subject: : cancel Control: cancel This message was cancelled from within Mozilla. === === Subject: : growth rate problem I am working on a homework problem for compuer complexity. I have studied my calculus book on limits as my calc class was 15 years ago, but still am not sure how to reduce the problem, though I did learn quite a bit from reading up on calc. I have convinced myself of the big int class, but I seem to be having a hard time figuring out a good proof. If anyone can give me a hint or what reduction to use it would be appreciated. there are 3 functions below, if for any 2 function f(), g(): limit (as n approaches infinity) f(n)/g(n) = 0 then g() has a faster growth rate than f(). this is du and ko book, page 283, and prob 6.1.3 on page 285. for the 3 problems 1. 2 ** (n ** (log2 n)) 2. n ** ((log2 n) ** 2) 3. log2 n ** (2 ** n) and the more obvious case was that 1 grows faster than 2. I emailed the instructor the following argument for the case of 3 & 1, but he said it is not a mathmatical proof. It's been hard for me to get help on this and I live far from the university. I have been doing ok in the class thus far, but the material has been getting harder and my schedule has been getting busy. Any suggestions or help would be appreciated: We now look at the exponents for problems 1 and 3 for n = 1..50 and then n = 1016..1025. We see that initially for n < 30, the difference between problem 1 and 3 are not that great. However by the time n=1020, we see that the exponent value of problem 3 is a number with 308 digits, whereas problem 1 has 31 for n=1020. For n=1020: problem 1 is of the form 2 ** (1020**log2(1020)) =(approx.) 2 ** (1020**10) problem 3 is of the form: log2(1020) ** (2 ** 1020) =(approx.) 10 ** (2**1020) Thus we can see that since problem varies more greatly because it's higher level exponent is n, whereas problem 1 higher level exponent is only log2(n), problem 3 grows much faster as n increases twards infinity. thus limit(problem 1(n)/problem 3(n)) = 0 since problem 3 grows faster twards infinity. thus we say: problem 1 << problem 3 === === Subject: : Re: growth rate problem > for the 3 problems > 1. 2 ** (n ** (log2 n)) > 2. n ** ((log2 n) ** 2) > 3. log2 n ** (2 ** n) Let k = lg n. Then, 1) 2^[(2^k)^k] = 2^[2^(k^2)] 2) (2^k)^k? = 2^(k^3) 3) k^[2^(2^k)] Just from eyeballing the equations, I can see that 3 grows faster than 1 which grows faster than 2. I'll leave the formal proof up to you. === === Subject: : Re: growth rate problem > for the 3 problems > 1. 2 ** (n ** (log2 n)) > 2. n ** ((log2 n) ** 2) > 3. log2 n ** (2 ** n) Let k = lg n. Then, 1) 2^[(2^k)^k] = 2^[2^(k^2)] > 2) (2^k)^kî = 2^(k^3) > 3) k^[2^(2^k)] Just from eyeballing the equations, I can see that 3 grows faster than > 1 which grows faster than 2. I'll leave the formal proof up to you. figured out which grows faster ... === === Subject: : Re: growth rate problem On 4 Dec 2006 13:17:51 -0800, wbsurfver@yahoo.com > for the 3 problems >> 1. 2 ** (n ** (log2 n)) >> 2. n ** ((log2 n) ** 2) >> 3. log2 n ** (2 ** n) >> Let k = lg n. Then, >> 1) 2^[(2^k)^k] = 2^[2^(k^2)] >> 2) (2^k)^kî = 2^(k^3) >> 3) k^[2^(2^k)] >> Just from eyeballing the equations, I can see that 3 grows faster than >> 1 which grows faster than 2. I'll leave the formal proof up to you. figured out which grows faster ... How about taking log2 of each expression twice. Then the limits involving those expressions should be obvious. It might make matters even more obvious to consider the suggestion made by eKo1 to let k = log2 n, 2^k = n. Example (log := log2): log log #1 = [log n]^2 With k = log n, log log #1 = k^2 === === Subject: : Re: growth rate problem #1: (2^n)^log(2,n) #2: (n^log(2,n))^2 #3: (log(2,n)^2)^n For a relationship between #1 and #2, say x(n)=n^log(2,n) (2^n)^log(2,n)=2^x (n^log(2,n))^2=x^2 Since #1 comes out on top here, and we suspect #3 has a higher growth rate, we will compare them. Call y(n)=2^n. (2^n)^log(2,n)=y^log(2,n) (log(2,n)^2)^n=log(2,n)^y y(n)=2^n implies log(2,y)=n, so... y^log(2,n)=y^log(2,log(2,y)) log(2,n)^y=log(2,log(2,y))^y === === Subject: : Re: growth rate problem > Let k = lg n. Then, > 1) 2^[(2^k)^k] = 2^[2^(k^2)] > 2) (2^k)^kî = 2^(k^3) > 3) k^[2^(2^k)] > figured out which grows faster ... Hint: Take lg of both sides and apply some logarithmic identities. Then consider what happens when k = 2 and when k > 2. === === Subject: : Re: growth rate problem Hi. I'm going to run with the assumption that (**) is the same as (^), or exponentiation. If such is not the case, disregard this post. Try making a new function for two of the three parts of each exponent. For example, if you say g(n)=n^log(2,n), then the first equation becomes 2^g(n), and the second becomes g(x)^2. Making a similar definition for the first and third should just about do it. === === Subject: : Group on arbitrary ordinal While imposing the group structure on $omega$, we used the fact that in it there are no limit ordinals, i.e., each of its elements is a successor of some other ordinal; given an arbitry ordinal $alpha$, in which there might also be some limit ordinals, can there be a group structure on it? If there is at least one, consider the set G=G(alpha) :={* :alpha x alpha ->alpha | (alpha,*) is a group}; what is the cardinality of G? The problem is simplified in the following way : A is any set, with cardinality beta; how many distinct(non isomorphic) group structures are there on A? === === Subject: : Re: Group on arbitrary ordinal >While imposing the group structure on $omega$, we used the fact that >in it there are no limit ordinals, i.e., each of its elements is a >successor of some other ordinal; given an arbitry ordinal $alpha$, in >which there might also be some limit ordinals, can there be a group >structure on it? If there is at least one, consider the set G=G(alpha) >:={* :alpha x alpha ->alpha | (alpha,*) is a group}; what is the >cardinality of G? >The problem is simplified in the following way : A is any set, with >cardinality beta; how many distinct(non isomorphic) group structures >are there on A? Without choice, if beta = beta^2, and there is a group structure, there are exactly 2^beta. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === === Subject: : Re: Group on arbitrary ordinal The problem is simplified in the following way : A is any set, with >cardinality beta; how many distinct(non isomorphic) group structures >are there on A? Without choice, if beta = beta^2, and there is a > group structure, there are exactly 2^beta. Do you need to assume that there is a group structure on beta? Isn't beta^2 = beta enough to imply that the free group with beta generators has cardinality beta? (Assuming beta > 1 of course.) === === Subject: : Re: Group on arbitrary ordinal days. My association with the Department is that of an alumnus. >While imposing the group structure on $omega$, we used the fact that >in it there are no limit ordinals, i.e., each of its elements is a >successor of some other ordinal; given an arbitry ordinal $alpha$, in >which there might also be some limit ordinals, can there be a group >structure on it? If there is at least one, consider the set G=G(alpha) >:={* :alpha x alpha ->alpha | (alpha,*) is a group}; what is the >cardinality of G? >The problem is simplified in the following way : A is any set, with >cardinality beta; how many distinct(non isomorphic) group structures >are there on A? Later: >First you construct a group on an arbitrary nonempty set. I think it >will make you understand my question. I'm going to have to agree with David Ullrich. First, this question is much different from the one which finishes your original post. Second, the question that finishes the original post seems to be completely unrelated to all the discussion that precedes it. In fact, as far as I can tell, connecting a group structure on omega with the fact that its elements are successors is nonsense. And note that each of its elements is a successor of some ordinal is a FALSE statement anyway; 0 is not a successor. Now, ANY set can be given a group structure. Finite sets are easy. For a set of cardinality kappa, you just consider the direct sum of kappa copies Z_2 (the cyclic group of order 2); this is equivalent to the group of finite subsets of kappa under boolean addition (symmetric difference). It is not hard to verify the set of finite subsets of an infinite ordinal is bijectable with the ordinal itself, and so it gives a group structure on that ordinal via the bijection. As to how many non-isomorphic group structures are there on A, I doubt the answer is known for any but the simplest values of beta. It is not even known for beta = p^n for p a prime and n sufficiently large. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Group on arbitrary ordinal And note that each of its elements > is a successor of some ordinal is a FALSE statement anyway; 0 is not > a successor. OK, sorry for the mistake! The successor issue was for the following reason: Constructing N, a monoid, on omega, we used finite induction: P_1 is true; P_n istrue=>P_(n+1) is true : togather, P_n is true for all n; this is true for omega doesn't have any limit ordinals. Why I have raised the question. However, I had another question, which you didn't favour with your kind glance : how many groups are there on S, if S is of cardinality beta? === === Subject: : Re: Group on arbitrary ordinal days. My association with the Department is that of an alumnus. And note that each of its elements >> is a successor of some ordinal is a FALSE statement anyway; 0 is not >> a successor. OK, sorry for the mistake! >The successor issue was for the following reason: >Constructing N, a monoid, on omega, we used finite induction: P_1 is >true; P_n istrue=>P_(n+1) is true : togather, P_n is true for all n; >this is true for omega doesn't have any limit ordinals. Why I have >raised the question. But this is an issue that has to do with ->INDUCTION<-, not with monoids, groups, or anything else. Yes. Mathematical induction of the form P(1) holds; If P(x) holds then P(x+1) holds; will establish P(a) for all a in omega, and the inductive step is insufficient for higher ordinals. Thus, that proof will not show that omega itself has property P. But, as noted, that's an issue about induction, not about anything else. You focused on the wrong thing. >However, I had another question, which you didn't favour with your kind >glance : how many groups are there on S, if S is of cardinality beta? I ->did<- comment. I noted that it is a hard question. The answer is not even generally known for the FINITE cardinalities, let alone the infinite ones. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Group on arbitrary ordinal And note that each of its elements >> is a successor of some ordinal is a FALSE statement anyway; 0 is not >> a successor. >OK, sorry for the mistake! >The successor issue was for the following reason: >Constructing N, a monoid, on omega, we used finite induction: P_1 is >true; P_n istrue=>P_(n+1) is true : togather, P_n is true for all n; >this is true for omega doesn't have any limit ordinals. Why I have >raised the question. But this is an issue that has to do with ->INDUCTION<-, not with > monoids, groups, or anything else. Yes. Mathematical induction of the form P(1) holds; > If P(x) holds then P(x+1) holds; will establish P(a) for all a in omega, and the inductive step is > insufficient for higher ordinals. Thus, that proof will not show that > omega itself has property P. But, as noted, that's an issue about induction, not about anything > else. You focused on the wrong thing. > Well, Sir, I wanted to focus on the construction of Z from N. Kindly call it up, and see if there was at all an application of finite induction. To construct a commutative group on higher ordinals, the same trick might not be applicable. And again, such other questions can be raised. For example, given a field F of cardinality alpha, and a vector space V (of card beta) over F, how many distinct bases do V possess? Well, you can use AC. === === Subject: : Re: Group on arbitrary ordinal However, I had another question, which you didn't favour with your kind >glance : how many groups are there on S, if S is of cardinality beta? I ->did<- comment. I noted that it is a hard question. The answer is > not even generally known for the FINITE cardinalities, let alone the > infinite ones. But such questions are usually easier for infinite cardinalities than for finite ones, at least if the axiom of choice is available. === === Subject: : Re: Group on arbitrary ordinal days. My association with the Department is that of an alumnus. >>However, I had another question, which you didn't favour with your kind >>glance : how many groups are there on S, if S is of cardinality beta? >> I ->did<- comment. I noted that it is a hard question. The answer is >> not even generally known for the FINITE cardinalities, let alone the >> infinite ones. But such questions are usually easier for infinite cardinalities than >for finite ones, at least if the axiom of choice is available. Fair enough; the answer for infinite cardinality may well be known. I would expect the answer to be 2^{kappa} for each set of cardinality kappe; it is certainly no more than that, since each group structure is given by a table with is kappa x kappa = kappe entries, each any of kappa possibilities, giving kappa^kappa = 2^kappa possible tables at most. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Group on arbitrary ordinal On 5 Dec 2006 22:27:57 -0800, Saurav And note that each of its elements >> is a successor of some ordinal is a FALSE statement anyway; 0 is not >> a successor. OK, sorry for the mistake! >The successor issue was for the following reason: >Constructing N, a monoid, on omega, we used finite induction: P_1 is >true; P_n istrue=>P_(n+1) is true : togather, P_n is true for all n; >this is true for omega doesn't have any limit ordinals. Why I have >raised the question. >However, I had another question, which you didn't favour with your kind >glance : how many groups are there on S, if S is of cardinality beta? ************************ David C. Ullrich === === Subject: : Re: Group on arbitrary ordinal On 4 Dec 2006 05:48:35 -0800, Saurav in it there are no limit ordinals, i.e., each of its elements is a >successor of some other ordinal; given an arbitry ordinal $alpha$, in >which there might also be some limit ordinals, can there be a group >structure on it? I don't understand the question. If n and m are in omega then n + m is in omega, and I suppose that one might say that when we defined addition for elements of omega we used the fact that there are no limit ordinals in omega. Note the word might - I really don't quite see what this has to do with that. But: That notion of addition does _not_ make omega into a group. Of course there do exist many group structures on omega, but I don't see at _all_ what that has to do with the fact that every element of omega is a successor. >If there is at least one, consider the set G=G(alpha) >:={* :alpha x alpha ->alpha | (alpha,*) is a group}; what is the >cardinality of G? Now I'm really confused. Because you do seem to be talking about arbitrary group structures, not defined by actual addition. So what could that possibly have to do with the fact that omage has no limit ordinals for elements? It's incredibly clear that the number of group structures on omega is the same as on any infinite countable ordinal. >The problem is simplified in the following way : A is any set, with >cardinality beta; how many distinct(non isomorphic) group structures >are there on A? ************************ David C. Ullrich === === Subject: : Re: Group on arbitrary ordinal Now I'm really confused. Because you do seem to be talking about > arbitrary group structures, not defined by actual addition. So > what could that possibly have to do with the fact that omage > has no limit ordinals for elements? It's incredibly clear that > the number of group structures on omega is the same as on > any infinite countable ordinal. > First you construct a group on an arbitrary nonempty set. I think it will make you understand my question. === === Subject: : Re: Group on arbitrary ordinal On 4 Dec 2006 20:09:40 -0800, Saurav > Now I'm really confused. Because you do seem to be talking about >> arbitrary group structures, not defined by actual addition. So >> what could that possibly have to do with the fact that omage >> has no limit ordinals for elements? It's incredibly clear that >> the number of group structures on omega is the same as on >> any infinite countable ordinal. First you construct a group on an arbitrary nonempty set. Ok. Say S is nonempty. If S is finite then we can make S into a cyclic group. Suppose that S is infinite. Say k is the cardinality of S. Recall that k^n = k for any finite n. Also k * Aleph_0 = k. That means that S has the same cardinality as the class of all finite subsets of S. That is, S has the same cardinality as G, where G is the set of all mappings of S into {0,1} which equal 0 except at finitely many points. Now G is a group in an obvious way, and since there is a bijection between G and S we have a group structure on S. >I think it >will make you understand my question. Nope. I'll ask again: What does the question of a group structure on an ordinal have to do with the question of whether the original's elements are all successors? You might try _answering_ the question this time, instead of just giving me another assignment... ************************ David C. Ullrich === === Subject: : Modeling Interval Data Supposing I have interval data like from scenario. Risk Count 20k-100K 51 100K-1MM 20 1MM-10MM 2 10MM - 100MM 0.03 How can I model this data i.e. I want to fit a distribution like Transformed beta distribution through this data. How do I estimate the parameters of the function? Since one of the count is a fraction can I use the interval likelihood function i.e the cumulative product of the difference of the cumulative distribution values to the power of their frequencies? Would appreciate your thoughts Sujan === === Subject: : Proving lim int | f_n - f | = 0. Hello I'd like some hints to prove the following assertion: Suppose the functions (f_n) are integrable, with lim f_n = f. If lim int | f_n | = int | f |, then lim int | f_n - f | = 0. The problem statement is not very clear, but I think the integrals are Lebesgue integrals over a measure space. Amanda === === Subject: : Re: Proving lim int | f_n - f | = 0. sorry, I didn't mean to top post. I'm confused with Google. Let g_n | f_n | + | f | - |f_n - f|. Then g_n >=0 and converges to 2 | f | By fatou's Lemma, 2 int | f | < = lim inf Int g_n = 2 int | f | - lim sup Int |f_n - f| => lim sup Int | f_n - f | <=0. Since Int | f_n - f | >=0 for every n, it follows lim sup Int | f_n - f | = lim Int | f_n - f | = 0. Amanda === === Subject: : Re: Proving lim int | f_n - f | = 0. On 2006-12-04 14:32:37 -0500, Amanda said: > Let g_n | f_n | + | f | - |f_n - f|. Then g_n >=0 and converges to 2 | > f | How did you conclude so fast that g_n -> 2 |f|? |f_n - f| -> 0. > By fatou's Lemma, 2 int | f | < = lim inf Int g_n = 2 int | f | - > lim sup Int |f_n - f| => lim sup Int | f_n - f | <=0. Since Int | f_n > - f | >=0 for every n, it follows lim sup Int | f_n - f | = lim Int | > f_n - f | = 0. > Amanda -- -kira === === Subject: : Re: Proving lim int | f_n - f | = 0. > On 2006-12-04 14:32:37 -0500, Amanda said: Let g_n | f_n | + | f | - |f_n - f|. Then g_n >=0 and converges to 2 | > f | How did you conclude so fast that > g_n -> 2 |f|? |f_n - f| -> 0. ???? > By fatou's Lemma, 2 int | f | < = lim inf Int g_n = 2 int | f | - > lim sup Int |f_n - f| => lim sup Int | f_n - f | <=0. Since Int | f_n > - f | >=0 for every n, it follows lim sup Int | f_n - f | = lim Int | > f_n - f | = 0. > Amanda === === Subject: : Re: Proving lim int | f_n - f | = 0. On 2006-12-04 15:22:34 -0500, The World Wide Wade said: > >> On 2006-12-04 14:32:37 -0500, Amanda said: >> > Let g_n | f_n | + | f | - |f_n - f|. Then g_n >=0 and converges to 2 | > f | >> >> How did you conclude so fast that >> g_n -> 2 |f|? >> >> |f_n - f| -> 0. ???? Oooo... I was thinking L^p norm. But here, the |.| is just the absolute value. I got cross-wired with another result that says something like || f_n || -> || f || implies f_n -> f, but I can't recall what kind of convergence the second expression takes. Sorry... I need sleep. :) > > By fatou's Lemma, 2 int | f | < = lim inf Int g_n = 2 int | f | - > lim sup Int |f_n - f| => lim sup Int | f_n - f | <=0. Since Int | f_n > - f | >=0 for every n, it follows lim sup Int | f_n - f | = lim Int | > f_n - f | = 0. > Amanda -- -kira === === Subject: : Re: Proving lim int | f_n - f | = 0. On 2006-12-04 15:50:24 -0500, Kira Yamato said: I got cross-wired with another result that says something like > || f_n || -> || f || > implies > f_n -> f, > but I can't recall what kind of convergence the second expression takes. Ok... I found it. It says for 1 < p < oo if f_n ->w f weakly in L^p and || f_n || -> || f ||, then f_n -> f strongly in L^p. -- -kira === === Subject: : Re: Proving lim int | f_n - f | = 0. > Hello I'd like some hints to prove the following assertion: Suppose the functions (f_n) are integrable, with lim f_n = f. If lim > int | f_n | = int | f |, then lim int | f_n - f | = 0. Consider | f_n | + | f | - | f_n - f |. === === Subject: : Re: Proving lim int | f_n - f | = 0. The World Wide Wade escreveu: > Hello > I'd like some hints to prove the following assertion: > Suppose the functions (f_n) are integrable, with lim f_n = f. If lim > int | f_n | = int | f |, then lim int | f_n - f | = 0. Consider | f_n | + | f | - | f_n - f |. Ok, thank you. But we can't apply Dominated Convergence Theorem, because we can't ensure | f_n | + | f | - | f_n - f | is dominated by a integrable function. Aren't you implicitly assuming uniform converge of f_n -> f? Amanda === === Subject: : Re: Proving lim int | f_n - f | = 0. > The World Wide Wade escreveu: Hello > I'd like some hints to prove the following assertion: > Suppose the functions (f_n) are integrable, with lim f_n = f. If lim > int | f_n | = int | f |, then lim int | f_n - f | = 0. > Consider | f_n | + | f | - | f_n - f |. Ok, thank you. But we can't apply Dominated Convergence Theorem, > because we can't ensure | f_n | + | f | - | f_n - f | is dominated by a > integrable function. Aren't you implicitly assuming uniform converge of > f_n -> f? Let's see, the big 3 are FL, MCT and DCT. Find an acronym that works for you. === === Subject: : Re: Proving lim int | f_n - f | = 0. I'm confused. DCT and MCT don't apply here, I think. And Fatou's Lemma doesn't prove the assertion.. Amanda The World Wide Wade escreveu: > The World Wide Wade escreveu: > Hello > I'd like some hints to prove the following assertion: > Suppose the functions (f_n) are integrable, with lim f_n = f. If lim > int | f_n | = int | f |, then lim int | f_n - f | = 0. > Consider | f_n | + | f | - | f_n - f |. > Ok, thank you. But we can't apply Dominated Convergence Theorem, > because we can't ensure | f_n | + | f | - | f_n - f | is dominated by a > integrable function. Aren't you implicitly assuming uniform converge of > f_n -> f? Let's see, the big 3 are FL, MCT and DCT. Find an acronym that > works for you. === === Subject: : Re: Proving lim int | f_n - f | = 0. > I'm confused. DCT and MCT don't apply here, I think. And Fatou's Lemma > doesn't prove the assertion.. Try again. And please don't top-post. > Amanda > The World Wide Wade escreveu: The World Wide Wade escreveu: > Hello > I'd like some hints to prove the following assertion: > Suppose the functions (f_n) are integrable, with lim f_n = f. If lim > int | f_n | = int | f |, then lim int | f_n - f | = 0. > Consider | f_n | + | f | - | f_n - f |. > Ok, thank you. But we can't apply Dominated Convergence Theorem, > because we can't ensure | f_n | + | f | - | f_n - f | is dominated by a > integrable function. Aren't you implicitly assuming uniform converge of > f_n -> f? > Let's see, the big 3 are FL, MCT and DCT. Find an acronym that > works for you. === === Subject: : Re: interpolation/approximation functions question, and I don't really know of any books that > discuss it. In general, I would try plotting the data points and > seeing what they look like. Try transforming the dependent > and independent variables. > When you say transform, i'm assuming that you mean Fourier transform and then look at the spectral components of the data? > Also, your data come from some application. Often, from > your knowledge of the application you can say something about > the function hidden in the data. Even if you don't have > an exact theory, an approximate theory will often suggest > a functional form to try. Another consideration is how the function f(t) should behave > at large or small t. If you know that f(t) should approach > zero as t tends to +infinity, then this restricts the functions > you should use. A polynomial would definitely be out. > I'm sorry, but can you explain why a polynomial would not be optimal here? I think that this is the crux of my problem, i don't have an intuitive feeling for when/where to use the various interpolation functions....can you expand on this a little? > In the example you gave, of a periodic function, you would know > from your knowledge of the physical system to expect a periodic > function, and periodicity should also be evident from a plot > of the data. Olin Perry Norton === === Subject: : Re: interpolation/approximation functions I don't know of a general procedure that answers your > question, and I don't really know of any books that > discuss it. > In general, I would try plotting the data points and > seeing what they look like. Try transforming the dependent > and independent variables. When you say transform, i'm assuming that you mean Fourier transform > and then look at the spectral components of the data? > Also, your data come from some application. Often, from > your knowledge of the application you can say something about > the function hidden in the data. Even if you don't have > an exact theory, an approximate theory will often suggest > a functional form to try. > Another consideration is how the function f(t) should behave > at large or small t. If you know that f(t) should approach > zero as t tends to +infinity, then this restricts the functions > you should use. A polynomial would definitely be out. I'm sorry, but can you explain why a polynomial would not be optimal > here? I think that this is the crux of my problem, i don't have an > intuitive feeling for when/where to use the various interpolation > functions....can you expand on this a little? > If you have samples of mortality rates in a certain country per year, e.g. 2% in 2000 and 1% in 2005 (for example due to improved medical services), it may look wise to conclude that the rate was probably that the mortality is 0% in 2010 (and negative after that). So maybe it would be better to look for an interpolation of /logarithms/ of but would never go below zero) If you have observed a comet and thus have approximate positions for severalpoints in time, you should use your knowledge of Kepler's law to narrow down the set of approximation functions. If it turns out that this restriction doesn't allow a good fit, contact the SETI program. === === Subject: : Re: interpolation/approximation functions I agree completely with everything that hagman and R.G. Vickson said. Olin Perry Norton === === Subject: : Re: interpolation/approximation functions I don't know of a general procedure that answers your > question, and I don't really know of any books that > discuss it. > In general, I would try plotting the data points and > seeing what they look like. Try transforming the dependent > and independent variables. When you say transform, i'm assuming that you mean Fourier transform > and then look at the spectral components of the data? No, I doubt that is what he means. He likely means thay you should try looking at the square root, or the square, or the log, or the exponential of x an/or y instead of the original variables x and y themselves. In statistics, for example, this is a common technique used when performing data exploration. A Google search on transformation of variables + data exploration turns up some hits: http://documents.wolfram.com/applications/eda/ExplorationAndTransformationOf Data.html http://www.research.ibm.com/visualanalysis/papers/interex.pdf http://www.brodgar.com/Chapter4BM.pdf http://csdl2.computer.org/persagen/DLAbsToc.jsp http://www.math.wisc.edu/~craciun/PAPERS/Raghu_David_Jaya_Ming_GCraciun_Data Mining.pdf You can also try searching under the topic of data mining. > Also, your data come from some application. Often, from > your knowledge of the application you can say something about > the function hidden in the data. Even if you don't have > an exact theory, an approximate theory will often suggest > a functional form to try. > Another consideration is how the function f(t) should behave > at large or small t. If you know that f(t) should approach > zero as t tends to +infinity, then this restricts the functions > you should use. A polynomial would definitely be out. I'm sorry, but can you explain why a polynomial would not be optimal > here? Well, think about it. Terms like x, x^2, x^3, etc. go to + or - infinity for x going to infinity. Therefore, using them directly cannot possibly lead to a bounded value for y. R.G. Vickson Adjunct Professor, University of Waterloo > I think that this is the crux of my problem, i don't have an > intuitive feeling for when/where to use the various interpolation > functions....can you expand on this a little? > In the example you gave, of a periodic function, you would know > from your knowledge of the physical system to expect a periodic > function, and periodicity should also be evident from a plot > of the data. Olin Perry Norton === === Subject: : Re: interpolation/approximation functions [snipped] > No, I doubt that is what he means. He likely means thay you should try > looking at the square root, or the square, or the log, or the > exponential of x an/or y instead of the original variables x and y > themselves. In statistics, for example, this is a common technique used > when performing data exploration. A Google search on transformation of > variables + data exploration turns up some hits: > http://documents.wolfram.com/applications/eda/ExplorationAndTransformationOf D ata.html > http://www.research.ibm.com/visualanalysis/papers/interex.pdf > http://www.brodgar.com/Chapter4BM.pdf > http://csdl2.computer.org/persagen/DLAbsToc.jsp > http://www.math.wisc.edu/~craciun/PAPERS/Raghu_David_Jaya_Ming_GCraciun_Data M ining.pdf > And this one: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Visualisation/Visual isation.html I think your question is basically one of what is called data exploration or exploratory data analysis, of which plotting the data and looking at it is a very important part. There are some good books on this topic. Of course, it would also be helpful if you had an intuitive feel for how different functions behave. Consider a polynomial, for instance, the function y(x) given by: y = c0 + c1*x + c2*x^2 + c3+x^3 + c4*x^4 Here, c0, c1, c2, c3, and c4 are constants. Possibly you may determine the values of these constants by curve-fitting. I'm using the ^ to indicate an exponent, so x^2 is the square of x. The point is that, over some limited region, all these terms may be of roughly equal magnitude, and may even cancel each other out to some degree. The function y(x) may go up and down. But, if we consider the behavior of this function as x becomes large, as long as the coefficient C4 is not zero, then the last term, c4*x^4 will totally dominate the others. So, this polynomial doesn't go to zero as x becomes large, it's not even bounded. === === Subject: : Re: interpolation/approximation functions And this one: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Visualisation/Visual i sation.html > I think your question is basically one of what is called > data exploration or exploratory data analysis, > of which plotting the data and looking at it is a > very important part. There are some good books on this topic. Of course, it would also be helpful if you had an intuitive > feel for how different functions behave. Consider a polynomial, for instance, the function y(x) > given by: y = c0 + c1*x + c2*x^2 + c3+x^3 + c4*x^4 Here, c0, c1, c2, c3, and c4 are constants. Possibly you may > determine the values of these constants by curve-fitting. > I'm using the ^ to indicate an exponent, so x^2 is the square > of x. The point is that, over some limited region, all these terms may be > of roughly equal magnitude, and may even cancel each other out to > some degree. The function y(x) may go up and down. But, if we > consider the behavior of this function as x becomes large, as long > as the coefficient C4 is not zero, then the last term, c4*x^4 > will totally dominate the others. So, this polynomial doesn't > go to zero as x becomes large, it's not even bounded. - and others - have posted. cheers, col === === Subject: : Counting Colours in a Graph Suppose I am given a graph G (for simplicity, we may take it as regular with degree k). If I am allowed to colour the edges of G so that each vertex will have at most d colours on the edges incident to it. There could be more that d colours in the system as two vertices do not need to share the same set of colours. We can count the number of times that a particular colour appears in the resulting graph. Suppose the colour which occurs most appears M times. Suppose we want to minimize M. (1) Prove that if M is minimal, each vertex should use all d colours on its incident edge. (2) If G can be decomposed into complete subgraphs of degree k/d, by colouring each of these subgraph with the same colour will give a minimum M. I found this problem peculiar. Could anyone let me know whether there === === Subject: : Re: Counting Colours in a Graph > Suppose I am given a graph G (for simplicity, we may take it as regular > with degree k). If I am allowed to colour the edges of G so that each > vertex will have at most d colours on the edges incident to it. There > could be more that d colours in the system as two vertices do not need > to share the same set of colours. It seems to me that d is always equal to k - since G is k-regular so that each vertex is incident to exactly k-different-color-edges. We can count the number of times that a particular colour appears in > the resulting graph. Suppose the colour which occurs most appears M > times. Suppose we want to minimize M. (1) Prove that if M is minimal, each vertex should use all d colours on > its incident edge. Trivially true. Since G is k-regular, d = k, and we are considering a proper edge coloring, each vertex will have d incident edges, all colored differently. > (2) If G can be decomposed into complete subgraphs of degree k/d, by > colouring each of these subgraph with the same colour will give a > minimum M. Can you explain what you mean by this with an example? Maybe you can consider a clique of size 3 or 4, or a cycle of length 4. Pradip Caulagi === === Subject: : Re: Counting Colours in a Graph > Suppose I am given a graph G (for simplicity, we may take it as regular > with degree k). If I am allowed to colour the edges of G so that each > vertex will have at most d colours on the edges incident to it. There > could be more that d colours in the system as two vertices do not need > to share the same set of colours. It seems to me that d is always equal to k - since G is k-regular so > that each vertex is incident to exactly k-different-color-edges. Note that d< k is a given restriction. > We can count the number of times that a particular colour appears in > the resulting graph. Suppose the colour which occurs most appears M > times. Suppose we want to minimize M. > (1) Prove that if M is minimal, each vertex should use all d colours on > its incident edge. Trivially true. Since G is k-regular, d = k, and we are considering a > proper edge coloring, each vertex will have d incident edges, all > colored differently. It might or might not hold for d < k. > (2) If G can be decomposed into complete subgraphs of degree k/d, by > colouring each of these subgraph with the same colour will give a > minimum M. Can you explain what you mean by this with an example? Maybe you can > consider a clique of size 3 or 4, or a cycle of length 4. > I meant partitioning the edge set of the given graph so that each partition forms a complete subgraph. === === Subject: : Re: Counting Colours in a Graph > Suppose I am given a graph G (for simplicity, we may take it as regular > with degree k). If I am allowed to colour the edges of G so that each > vertex will have at most d colours on the edges incident to it. There > could be more that d colours in the system as two vertices do not need > to share the same set of colours. We can count the number of times that a particular colour appears in > the resulting graph. Suppose the colour which occurs most appears M > times. Suppose we want to minimize M. (1) Prove that if M is minimal, each vertex should use all d colours on > its incident edge. > (2) If G can be decomposed into complete subgraphs of degree k/d, by > colouring each of these subgraph with the same colour will give a > minimum M. I found this problem peculiar. Could anyone let me know whether there I think you need to clarify how many different colors are allowed. For example, if the largest degree of any vertex is less than or equal to d, then the minimum M is 1 (color every edge a different color!) Similar reasoning implies to me that if N is the smallest integer larger than D/d, where D is the maximum degree of a vertex, it seems that M=N should work (that is definitely the minimum possible value for M, but I haven't proven that it is attainable.) === === Subject: : Re: Counting Colours in a Graph > I think you need to clarify how many different colors are allowed. For > example, if the largest degree of any vertex is less than or equal to > d, then the minimum M is 1 (color every edge a different color!) > Similar reasoning implies to me that if N is the smallest integer > larger than D/d, where D is the maximum degree of a vertex, it seems > that M=N should work (that is definitely the minimum possible value for > M, but I haven't proven that it is attainable.) I forget to say I am considering regular graph. === === Subject: : Re: Counting Colours in a Graph > I think you need to clarify how many different colors are allowed. For > example, if the largest degree of any vertex is less than or equal to > d, then the minimum M is 1 (color every edge a different color!) Let's neglect the trivial case d > D. We assume d Similar reasoning implies to me that if N is the smallest integer > larger than D/d, where D is the maximum degree of a vertex, it seems > that M=N should work (that is definitely the minimum possible value for > M, but I haven't proven that it is attainable.) Why N is minimum? M is the number of times the colour used most appears. If one colour is used N times at a particular vertex, keeping the constraint on d, some other colours must be used more that N times at other vertices. === === Subject: : boolean formula for maximum Suppose a and b are n-bit integers. How to express the max(a,b) as a boolean equation concisely? Say a recursive formula. === === Subject: : Re: boolean formula for maximum > Suppose a and b are n-bit integers. How to express the max(a,b) as a > boolean equation concisely? Say a recursive formula. > Go from high bits to low bits and find the first which are different. If you find none, they are equal. Otherwise, the smaller has 0, the larger has 1 at the bit location described above. For your convenience, pick the larger. === === Subject: : Groups and commutators Hi ! I found the following problem in Lang's Algebra : Let G be a group and x, y, z in G such that : [x,y]=xyx(-1)y(-1)=y [y,z]=z [z,x]=x show that these relations imply x=y=z=e (neutral element)... Can anyone help me prove this result ? kalcifire === === Subject: : Re: Groups and commutators > Hi ! I found the following problem in Lang's Algebra : Let G be a group and x, y, z in G such that : > [x,y]=xyx(-1)y(-1)=y > [y,z]=z > [z,x]=x > show that these relations imply x=y=z=e (neutral element)... Can anyone help me prove this result ? Well you can type it into GAP or Magma and it will tell you instantly that G is trivial, but I guess that will not satisfy you? Rather than just manipulating equations, let's try to use a few more ideas. First we show: Lemma. The group X = < x,y | x y x^-1 = y^2 > is solvable. Proof. Since x^i y x^-i = y^(2^i), we have [y, x^i y x^-i] = 1 for all i, and hence (conjugating by x^-i) [y, x^-i y x^i] = 1 for all in, so y commutes with x^i y x^-i for all integers i, and hence conjugating by powers of x, all such conjugates commute with each other, so the subgroup N of X generated by all x^i y x^-i is abelian. Also N is normal in X with X/N cyclic (since = X), so X is solvable. Now let G = with x y x^-1 y^-1 = y (1) y z y^-1 z^-1 = z (2) z x z^-1 x^-1 = x (3) Since x,y, and z are commutators, G is perfect. The subgroup H = of G is a quotient group of the group X above, so is solvable. So if we can prove that H = G, then G will be perfect and solvable and hence trivial. We can prove H = G using a bit of coset enumeration. Since by (3) z x = x^2 z, we have Hzx = Hz (4) (3) implies z^2 x = x^4 z^2, so Hz^2 x = Kz^2 (5) By (2), Hz^2 y = Hyz = Hz (6) and by (1) Hz^2 = Hz^2 x y x^-1 y^-2 =(5) Hz^2 y x^-1 y^-2 =(6) Hz x^-1 y^-2 =(4) Hz y^-2 =(6) Hz^2 y^-1, so Hz^2 y = Hz^2, and hence by (6) Hz = Hz^2, so Hz = H and H = G. Derek Holt. === === Subject: : Re: Groups and commutators Hi Derek ! In fact when trying to solve the problem I was manipulating the 3 relations heavily but failed to get anywhere ! (I did see that was perfect but was not able to use it !) Kalcifire Derek Holt a ?crit : > Hi ! > I found the following problem in Lang's Algebra : > Let G be a group and x, y, z in G such that : > [x,y]=xyx(-1)y(-1)=y > [y,z]=z > [z,x]=x > show that these relations imply x=y=z=e (neutral element)... > Can anyone help me prove this result ? Well you can type it into GAP or Magma and it will tell you instantly > that G is trivial, but I guess that will not satisfy you? Rather than just manipulating equations, let's try to use a few more > ideas. First we show: Lemma. The group X = < x,y | x y x^-1 = y^2 > is solvable. Proof. Since x^i y x^-i = y^(2^i), we have [y, x^i y x^-i] = 1 for all > i, and hence (conjugating by x^-i) [y, x^-i y x^i] = 1 for all in, so > y commutes with x^i y x^-i for all integers i, and hence conjugating > by powers of x, all such conjugates commute with each other, so the > subgroup N of X generated by all x^i y x^-i is abelian. Also N is > normal in X with X/N cyclic (since = X), so X is solvable. Now let G = with x y x^-1 y^-1 = y (1) > y z y^-1 z^-1 = z (2) > z x z^-1 x^-1 = x (3) Since x,y, and z are commutators, G is perfect. The subgroup H = we can prove that H = G, then G will be perfect and solvable and hence > trivial. We can prove H = G using a bit of coset enumeration. Since by (3) z x = x^2 z, we have Hzx = Hz (4) (3) implies z^2 x = x^4 z^2, so Hz^2 x = Kz^2 (5) By (2), Hz^2 y = Hyz = Hz (6) and by (1) Hz^2 = Hz^2 x y x^-1 y^-2 =(5) Hz^2 y x^-1 y^-2 =(6) Hz x^-1 y^-2 =(4) > Hz y^-2 =(6) Hz^2 y^-1, so Hz^2 y = Hz^2, and hence by (6) Hz = Hz^2, so Hz = H and H = G. Derek Holt. === === Subject: : Re: Groups and commutators > I found the following problem in Lang's Algebra : Let G be a group and x, y, z in G such that : > [x,y]=xyx(-1)y(-1)=y > [y,z]=z > [z,x]=x > show that these relations imply x=y=z=e (neutral element)... Can anyone help me prove this result ? x^{-1} = y^2xy^{-1} (from relation #1) = y^2xzy^{-1}z^{-2} (from relation #2) = y^2x^{-1}x^2zy^{-1}z^{-2} (because x = x^{-1}x^2) = y^2x^{-1}zxy^{-1}z^{-2} (from relation #3) = y^2x^{-1}zxz^{-1}y^{-1} (from relation #2) = y^2x^{-1}x^2y^{-1} (from relation #3) = y^2xy^{-1} (because x = x^{-1}x^2) = xyy^{-1} (from relation #1) = x, and therefore x = e. Jose Carlos Santos === === Subject: : Re: Groups and commutators <4tit7uF13no5pU1@mid.individual.net > I found the following problem in Lang's Algebra : > Let G be a group and x, y, z in G such that : > [x,y]=xyx(-1)y(-1)=y > [y,z]=z > [z,x]=x > show that these relations imply x=y=z=e (neutral element)... > Can anyone help me prove this result ? x^{-1} = y^2xy^{-1} (from relation #1) = y^2xzy^{-1}z^{-2} (from relation #2) = y^2x^{-1}x^2zy^{-1}z^{-2} (because x = x^{-1}x^2) = y^2x^{-1}zxy^{-1}z^{-2} (from relation #3) = y^2x^{-1}zxz^{-1}y^{-1} (from relation #2) = y^2x^{-1}x^2y^{-1} (from relation #3) = y^2xy^{-1} (because x = x^{-1}x^2) = xyy^{-1} (from relation #1) = x, and therefore x = e. x^{-1} = x doesn't imply x=e. This only shows xx=e (and by symmetry, yy=zz=e). But then, from the original relations x = z{-1}zx (because z{-1}z=e) =z{-1}[x,z]xz (by definition of [,]) =z{-1}xxz (given relation) =z{-1}ez (Jos? Carlos Santos) =z{-1}z = e (and by symmetry y=z=e). -hagman === === Subject: : Re: Groups and commutators > But then, from the original relations > x = z{-1}zx (because z{-1}z=e) > =z{-1}[x,z]xz (by definition of [,]) Well, no. By definition, [x,z] = xzx^{-1}z^{-1}. But, obviously, what you meant here was: = z^{-1}[z,x]xz, which is correct. > =z{-1}xxz (given relation) > =z{-1}ez (Jos.8e Carlos Santos) I'm glad I was helpful. :-) > =z{-1}z = e > (and by symmetry y=z=e). Jose Carlos Santos === === Subject: : Re: Groups and commutators <4tit7uF13no5pU1@mid.individual.net> Jos? Carlos Santos a ?crit : > I found the following problem in Lang's Algebra : > Let G be a group and x, y, z in G such that : > [x,y]=xyx(-1)y(-1)=y > [y,z]=z > [z,x]=x > show that these relations imply x=y=z=e (neutral element)... > Can anyone help me prove this result ? x^{-1} = y^2xy^{-1} (from relation #1) = y^2xzy^{-1}z^{-2} (from relation #2) = y^2x^{-1}x^2zy^{-1}z^{-2} (because x = x^{-1}x^2) = y^2x^{-1}zxy^{-1}z^{-2} (from relation #3) = y^2x^{-1}zxz^{-1}y^{-1} (from relation #2) = y^2x^{-1}x^2y^{-1} (from relation #3) = y^2xy^{-1} (because x = x^{-1}x^2) = xyy^{-1} (from relation #1) = x, and therefore x = e. > Jose Carlos Santos I still have a problem with the very first line : x^{-1} = y^2xy^{-1} since relation #1 implies x = y^2xy^{-1} Am i missing something ? === === Subject: : Re: Groups and commutators Quick and silly! :-( > I still have a problem with the very first line : > x^{-1} = y^2xy^{-1} > since relation #1 implies > x = y^2xy^{-1} > Am i missing something ? Nothing. On the other hand, I am missing a brain. :-( Did you see one? Just one question: *which* exercise is this (chapter + number), if you don't mind? Jose Carlos Santos === === Subject: : Re: Groups and commutators <4tit7uF13no5pU1@mid.individual.net> <4tj369F14a3htU1@mid.individual.net> Hi Jos? ! In my edition of Lang's Algebra (addison wesley third ed.), the exercise appear as an example on page 69 under the chapter I,?12 about Free Groups...Hope that helps ! I do not even know if this example is supposed to be easy or difficult... Jos? Carlos Santos a ?crit : > Quick and silly! :-( > I still have a problem with the very first line : > x^{-1} = y^2xy^{-1} > since relation #1 implies > x = y^2xy^{-1} > Am i missing something ? Nothing. On the other hand, I am missing a brain. :-( Did you see > one? Just one question: *which* exercise is this (chapter + number), if > you don't mind? > Jose Carlos Santos === === Subject: : Re: mean value theorem - historical irony Reaction to all posts in this thread The French mathematician Michel Rolle (1652 - 1719) formulated a not-too-rigourous version of the mean value theorem in the course of his attempts to =debunk= differential and integral calculus. Source: http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Rolle.html >My book gives the theorem on an open set (a,b): Assume that f has a derivative (finite or infinte) at each point of an >open interval (a,b), and assume also that f is continuous at both >endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = >f'(c)(b-a). Another book I have defines it on [a,b]. In the above, is the >hypothesis f continuous at a,b is so that we can extend f to [a,b], and >thereby evaluate f(a), f(b)? Otherwise, if f is just defined on (a,b), >how can we write f(a), f(b)? > === === Subject: : Re: mean value theorem My book gives the theorem on an open set (a,b): > Assume that f has a derivative (finite or infinte) at each point of an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). I doubt very much your books says this. Can you give the > statement exactly? It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 (Mean-Value Theorem). === === Subject: : Re: mean value theorem > My book gives the theorem on an open set (a,b): > Assume that f has a derivative (finite or infinte) at each point of an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). > I doubt very much your books says this. Can you give the > statement exactly? It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 > (Mean-Value Theorem). The theorem as you state it is false: Define f(x) = 0 if -1 <= x < 0, f(0) = 1/2, f(x) = 1 if 0 < x <= 1. Then on [-1,1], f has a derivative (f'(x) = 0 except at 0, where f'(0) = +oo), f is continuous at the end points, but there is no c in (-1,1) such that f(1) - f(0) = f'(c)*2. === === Subject: : Re: mean value theorem My book gives the theorem on an open set (a,b): > Assume that f has a derivative (finite or infinte) at each point of an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). > I doubt very much your books says this. Can you give the > statement exactly? It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 > (Mean-Value Theorem). > Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1). f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. However you have f'(x) = -1/x^2 < 0. Tilt! === === Subject: : Re: mean value theorem , > My book gives the theorem on an open set (a,b): > Assume that f has a derivative (finite or infinte) at each point of an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). > I doubt very much your books says this. Can you give the > statement exactly? > It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 > (Mean-Value Theorem). > Let f(x) = 1/x. Then f'(0) = -oo. No, there is no way to define f(0) so that f'(0) exists in [-oo,oo]. > Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! === === Subject: : Re: mean value theorem Assume that f has a derivative (finite or infinte) at each point of an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). > I doubt very much your books says this. Can you give the > statement exactly? > It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 > (Mean-Value Theorem). > Let f(x) = 1/x. Then f'(0) = -oo. No, there is no way to define f(0) so that f'(0) exists in > [-oo,oo]. > f(0) = 0 f'(0) = lim(h->0) (f(0 + h) - f(0)/h = lim(h->0) 1/h^2 = oo. > Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 > f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! > f'(x) < 0 for x /= 0 and f'(0) = oo. === === Subject: : Re: mean value theorem > Assume that f has a derivative (finite or infinte) at each point of > an > open interval (a,b), and assume also that f is continuous at both > endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = > f'(c)(b-a). > I doubt very much your books says this. Can you give the > statement exactly? > It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 > (Mean-Value Theorem). > Let f(x) = 1/x. Then f'(0) = -oo. > No, there is no way to define f(0) so that f'(0) exists in > [-oo,oo]. > f(0) = 0 > f'(0) = lim(h->0) (f(0 + h) - f(0)/h = lim(h->0) 1/h^2 = oo. My mistake, you are absolutely correct. Not only that, any way you define f(0) in R leads to f'(0) = oo. > Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 > f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! > f'(x) < 0 for x /= 0 and f'(0) = oo. === === Subject: : Re: mean value theorem On 2006-12-04 00:00:57 -0500, William Elliot said: > >> My book gives the theorem on an open set (a,b): >> >> Assume that f has a derivative (finite or infinte) at each point of an >> open interval (a,b), and assume also that f is continuous at both >> endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = >> f'(c)(b-a). > > I doubt very much your books says this. Can you give the > statement exactly? >> >> It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 >> (Mean-Value Theorem). >Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! However, 1/x is not continuous at x=0. If we add the assumption that f be continuous on [a,b], the situation may be different. Suppose f' is defined on (a,b) except at c where f'(c) = +infty. Then, the Mean-Value Theorem over (a,c) gives a point p in (a,c) such that (c-a)f'(p) = f(c) - f(a) and over (c,b) a point q in (c,b) such that (b-c)f'(q) = f(b) - f(c). Then, f(b)-f(a) = (b-c)f'(q) - (c-a)f'(p). So, [f(b)-f(a)]/(b-a) = (b-c)/(b-a) f'(q) - (c-a)/(b-a) f'(p). Denote the left side M := [f(b)-f(a)]/(b-a). Then, the above equation says M is in the convex hull of { f'(q), f'(p) }. So, M is greater than or equal to one of the two values f'(p) or f'(q). Let's say M >= f'(p). Now, we can apply the Intermediate-Value Theorem for Derivatives to (a,c) to see that there exists a point d in (a,c) such that f'(d) = M. So we've found a point d in (a,b) such that f'(d) = [f(b)-f(a)]/(b-a). I think this idea can be extended to cover the more general case where f be 1) a *continuous* function on [a,b] and 2) has derivative on (a,b) that takes values from the extended set Rv{+infty, -infty} and 3) has at most a countable number of points {p_i} contained in (a,b) at which the derivative does not take a finite value and 4) the set {p_i} is bounded. Then there is a point c in (a,b) such that f(b) - f(a) = f'(c)(b-a). -- -kira === === Subject: : Re: mean value theorem > On 2006-12-04 00:00:57 -0500, William Elliot said: >> My book gives the theorem on an open set (a,b): >> >> Assume that f has a derivative (finite or infinte) at each point of an >> open interval (a,b), and assume also that f is continuous at both >> endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = >> f'(c)(b-a). > > I doubt very much your books says this. Can you give the > statement exactly? >> >> It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 >> (Mean-Value Theorem). >Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! However, 1/x is not continuous at x=0. If we add the assumption that f be continuous on [a,b], the situation > may be different. Suppose f' is defined on (a,b) except at c where f'(c) = +infty. Then, > the Mean-Value Theorem over (a,c) gives a point p in (a,c) such that > (c-a)f'(p) = f(c) - f(a) > and over (c,b) a point q in (c,b) such that > (b-c)f'(q) = f(b) - f(c). Then, > f(b)-f(a) = (b-c)f'(q) - (c-a)f'(p). > So, > [f(b)-f(a)]/(b-a) = (b-c)/(b-a) f'(q) - (c-a)/(b-a) f'(p). > Denote the left side M := [f(b)-f(a)]/(b-a). Then, the above equation > says M is in the convex hull of { f'(q), f'(p) }. So, M is greater > than or equal to one of the two values f'(p) or f'(q). Let's say > M >= f'(p). > Now, we can apply the Intermediate-Value Theorem for Derivatives to > (a,c) to see that there exists a point d in (a,c) such that > f'(d) = M. > So we've found a point d in (a,b) such that > f'(d) = [f(b)-f(a)]/(b-a). I think this idea can be extended to cover the more general case where f be > 1) a *continuous* function on [a,b] and > 2) has derivative on (a,b) that takes values from the extended set > Rv{+infty, -infty} and > 3) has at most a countable number of points {p_i} contained in (a,b) at > which the derivative does not take a finite value and > 4) the set {p_i} is bounded. How could {p_i} not be bounded? It's a subset of (a,b). > Then there is a point c in (a,b) such that > f(b) - f(a) = f'(c)(b-a). There's a simpler proof: Just use the ordinary proof of the MVT. Rolle goes through just fine (note f can't have an extreme value where f' is infinite), and Rolle => MVT. === === Subject: : Re: mean value theorem <2006120402144816807-kirakun@earthlinknet On 2006-12-04 00:00:57 -0500, William Elliot said: >> My book gives the theorem on an open set (a,b): >> Assume that f has a derivative (finite or infinte) at each point of an >> open interval (a,b), and assume also that f is continuous at both >> endpoints a, b. Then there is a c in (a,b) such that f(b)-f(a) = >> f'(c)(b-a). >> I doubt very much your books says this. Can you give the > statement exactly? >> It does; the book is Apostol's Mathematical Analysis., Theorem 5.11 >> (Mean-Value Theorem). > Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1). > f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 > f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. > However you have f'(x) = -1/x^2 < 0. Tilt! However, 1/x is not continuous at x=0. > It does not require continuity except at the end points. > If we add the assumption that f be continuous on [a,b], the situation > may be different. > The question was to state exactly the problem. It was not stated. Only a mystical number was conjured. > Suppose f' is defined on (a,b) except at c where f'(c) = +infty. Then, Astounding. Now produce an example of a continuous function over (a,b) that has some c in (a,b) with f'(c) = oo. > the Mean-Value Theorem over (a,c) gives a point p in (a,c) such that > (c-a)f'(p) = f(c) - f(a) > and over (c,b) a point q in (c,b) such that > (b-c)f'(q) = f(b) - f(c). Then, > f(b)-f(a) = (b-c)f'(q) - (c-a)f'(p). > So, > [f(b)-f(a)]/(b-a) = (b-c)/(b-a) f'(q) - (c-a)/(b-a) f'(p). > Denote the left side M := [f(b)-f(a)]/(b-a). Then, the above equation > says M is in the convex hull of { f'(q), f'(p) }. So, M is greater > than or equal to one of the two values f'(p) or f'(q). Let's say > M >= f'(p). > Now, we can apply the Intermediate-Value Theorem for Derivatives to > (a,c) to see that there exists a point d in (a,c) such that > f'(d) = M. > So we've found a point d in (a,b) such that > f'(d) = [f(b)-f(a)]/(b-a). I think this idea can be extended to cover the more general case where f be > 1) a *continuous* function on [a,b] and > 2) has derivative on (a,b) that takes values from the extended set > Rv{+infty, -infty} and > 3) has at most a countable number of points {p_i} contained in (a,b) at > which the derivative does not take a finite value and > 4) the set {p_i} is bounded. Then there is a point c in (a,b) such that > f(b) - f(a) = f'(c)(b-a). -- -kira === === Subject: : Re: mean value theorem Mail-To-News-Contact: abuse@dizum.com >> On 2006-12-04 00:00:57 -0500, William Elliot said: Assume that f has a derivative (finite or infinte) at each point of an open interval (a,b), and assume also that f is continuous at both endpoints a, b. >> Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1). >> f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1 >> f(b) - f(a) = 2 = b - a. Thus f'(c) = 1. >> However you have f'(x) = -1/x^2 < 0. Tilt! >> However, 1/x is not continuous at x=0. >It does not require continuity except at the end points. 1/x doesn't have a derivative at x=0, either, and that's specifically required above. -- Michael F. Stemper #include Visualize whirled peas! === === Subject: : Re: mean value theorem On 2006-12-04 04:37:33 -0500, William Elliot said: Astounding. Now produce an example of a continuous function over (a,b) > that has some c in (a,b) with f'(c) = oo. How about f(x) = x^{1/3}? -- -kira === === Subject: : Re: mean value theorem <2006120402144816807-kirakun@earthlinknet> <2006120404442616807-kirakun@earthlinknet On 2006-12-04 04:37:33 -0500, William Elliot said: > Astounding. Now produce an example of a continuous function over (a,b) > that has some c in (a,b) with f'(c) = oo. How about f(x) = x^{1/3}? > Hm... f'(x) = (1/3) x^(-2/3) === === Subject: : geometry question: spring volume and surface Hi all, Can someone provide me with some formulas to calculate the volume and surface of a spring (or coil, spiral?) given the radius of the wire, the winding radius and the z-distance between to similar points of adjacent windings? If you can provide a formula + derivation how to get there, it would save me a lot of time and worries (no homework problem, I assure you). gert === === Subject: : Re: geometry question: spring volume and surface On Mon, 04 Dec 2006 15:57:16 +0100, Gert Van den Eynde >Hi all, Can someone provide me with some formulas to calculate the volume and >surface of a spring (or coil, spiral?) given the radius of the wire, the >winding radius and the z-distance between to similar points of >adjacent windings? If you can provide a formula + derivation how to get >there, it would save me a lot of time and worries (no homework problem, >I assure you). gert I made a Maple worksheet giving a development of a parameterization for the surface of the spring and its surface area. See: http://math.asu.edu/~kurtz/WireSurface.pdf You can check it over. I will leave the volume for someone else.. --Lynn === === Subject: : Re: geometry question: spring volume and surface ETAuAhUAkj4FVymD2dpwwgMnyh1m87Jjfr4CFQCC5x8xIOcyIuIfnRr+vdEoMSCOlQ== If you cut a short length of spring along planes perpendicular to the length along the wire, you get a cylindrivcal volume. So the volume is the diameter of the wire squared, times pi/4, times the total length of wire. --OL === === Subject: : Re: geometry question: spring volume and surface <26133-4576432C-1112@storefull-3254.bay.webtv.net If you cut a short length of spring along planes perpendicular to the > length along the wire, you get a cylindrivcal volume. So the volume is > the diameter of the wire squared, times pi/4, times the total length of > wire. --OL Trying to recall from a textbook I used some 50 years ago while teaching mechanical engineering, the following might help you: Consider a wire of circular cross-section with radius r , which is tightly wrapped around a cylindrical mandrel of radius R, with a pitch of p. Then a = cross-sectional area of the wire = pi*r^2 Rbar= R + r c = circumference of the wire = 2*pi*r C = circumference of the circle of radius Rbar = 2*pi*Rbar P = pitch angle = arctan(p/C) L = length of one wrap of the wire = C/cos(P) A = surface area of wire per single coil (wrap) = L*c V = volume of wire per single coil (wrap) = L*a Hope I remembered it correctly-- anyway, maybe this'll help. Best wishes, Grover Hughes === === Subject: : Re: geometry question: spring volume and surface Hi all, >Can someone provide me with some formulas to calculate the volume and >surface of a spring (or coil, spiral?) given the radius of the wire, the >winding radius and the z-distance between to similar points of >adjacent windings? If you can provide a formula + derivation how to get >there, it would save me a lot of time and worries (no homework problem, >I assure you). >gert I made a Maple worksheet giving a development of a parameterization > for the surface of the spring and its surface area. See: http://math.asu.edu/~kurtz/WireSurface.pdf You can check it over. I will leave the volume for someone else.. Well, if it's really a physical object made of wire, I'd weigh the spring and divide the mass by the density of the material to arrive at the volume ;) === === Subject: : Jech's Set Theory Now that Springer has reprinted Jech's Set Theory and sells it for the astronomical price of $160 (almost enough to purchase a used copy of Woodin's book on large cardinals), what can one read instead that will cover all the same material at a reasonable price? -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === === Subject: : Re: Jech's Set Theory > Now that Springer has reprinted Jech's Set Theory and sells it for the > astronomical price of $160 (almost enough to purchase a used copy of > Woodin's book on large cardinals), what can one read instead that will > cover all the same material at a reasonable price? Nothing. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: Jech's Set Theory Now that Springer has reprinted Jech's Set Theory and sells it for the > astronomical price of $160 (almost enough to purchase a used copy of > Woodin's book on large cardinals), what can one read instead that will > cover all the same material at a reasonable price? Nothing. As far as explanations given and ease of learing from are concerned, how do you compare Kunen with Jech? MoeBlee === === Subject: : Re: Jech's Set Theory how do you compare Kunen with Jech? I wouldn't recommend Jech as a primary source for anyone who isn't already well acquainted with set theory. The primary strength of Jech's book is its encyploaedic scope, covering as it does, in varying level of detail, pretty much all important areas of set theory where one can speak of an established theory in the sense of a relatively stable body of results, techniques ideas. Kunen's _An Introduction to Independence Proofs_ has another aim, however, and despite certain technical flaws - there are non-trivial mathematical errors in the book - it's a fine introduction to the subject. If one is not already pretty much up to the speed with independence results, Kunen is probable a better book as far as ease of learning is concerned. If one is interested in just the basics of forcing, though, instead of Kunen or Shoenfield, [1971], Unramified Forcing in Axiomatic Set Theory, V. 1, -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: Jech's Set Theory As far as explanations given and ease of learing from are concerned, > how do you compare Kunen with Jech? I wouldn't recommend Jech as a primary source for anyone who isn't > already well acquainted with set theory. The primary strength of Jech's > book is its encyploaedic scope, covering as it does, in varying level > of detail, pretty much all important areas of set theory where one can > speak of an established theory in the sense of a relatively stable > body of results, techniques ideas. Kunen's _An Introduction to > Independence Proofs_ has another aim, however, and despite certain > technical flaws - there are non-trivial mathematical errors in the book > - it's a fine introduction to the subject. If one is not already pretty > much up to the speed with independence results, Kunen is probable a > better book as far as ease of learning is concerned. If one is > interested in just the basics of forcing, though, instead of Kunen or Shoenfield, [1971], Unramified Forcing in Axiomatic Set Theory, V. > 1, MoeBlee === === Subject: : Re: Jech's Set Theory > Now that Springer has reprinted Jech's Set Theory and sells it for the > astronomical price of $160 (almost enough to purchase a used copy of > Woodin's book on large cardinals), what can one read instead that will > cover all the same material at a reasonable price? Kunen is not as many pages as Jech, but it covers basically the same ground as Jech, even if not everything in Jech. Though I don't know how much Kunen sells for. MoeBlee === === Subject: : Re: Jech's Set Theory > Now that Springer has reprinted Jech's Set Theory and sells it for the > astronomical price of $160 (almost enough to purchase a used copy of > Woodin's book on large cardinals), what can one read instead that will > cover all the same material at a reasonable price? Kunen is not as many pages as Jech, but it covers basically the same > ground as Jech, even if not everything in Jech. Though I don't know how > much Kunen sells for. P.S. Don't forget that Shoenfield, which is very reasonably priced, also covers the independence proofs, though that is only part of the subject matter of Jech and Kunen. MoeBlee === === Subject: : Re: Jech's Set Theory > P.S. Don't forget that Shoenfield, which is very reasonably priced, > also covers the independence proofs, though that is only part of the > subject matter of Jech and Kunen. Shoenfield's _Mathematical Logic_, splendid as it is, is not even remotely in the same ballpark as Jech's _Set Theory_ in scope or depth. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: Jech's Set Theory P.S. Don't forget that Shoenfield, which is very reasonably priced, > also covers the independence proofs, though that is only part of the > subject matter of Jech and Kunen. Shoenfield's _Mathematical Logic_, splendid as it is, is not even > remotely in the same ballpark as Jech's _Set Theory_ in scope or depth. Of course I agree MoeBlee === === Subject: : Re: Jech's Set Theory > Now that Springer has reprinted Jech's Set Theory and sells it for > the > astronomical price of $160 (almost enough to purchase a used copy of > Woodin's book on large cardinals), what can one read instead that > will > cover all the same material at a reasonable price? Amazon has it for $134.40 -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === === Subject: : Re: Jech's Set Theory > Now that Springer has reprinted Jech's Set Theory and sells it for >> the >> astronomical price of $160 (almost enough to purchase a used copy >> of >> Woodin's book on large cardinals), what can one read instead that >> will >> cover all the same material at a reasonable price? Amazon has it for $134.40 I have just noticed that when I bought my copy from Amazon, back in August, it was six cents cheaper than now! So, best to get it there quickly. -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === === Subject: : Right cosets intersection Ok, this is just a little bit harder than previous exercises I did, but I guess I found a quite clear image of how to solve it. The problem is that I find hard to transpose those tips in a formal way. The proposition is the following. --------------------------------------- Let {S_i : i in I} a family of subgroups of a group G, let {S_i t_i : i in I} a family of right cosets, and let _ D = | | S_i. Prove that either _ 1) {} = | | S_i t_i or _ 2) Dg = | | S_i t_i for some g in G. --------------------------------------- (1) is quite trivial, let's focus on (2). I guess see that the partition of G induced by the coset equivalence relation could be the key to the solution. In fact _ _ (*) _ x in | | S_i t_i => x in T_i in G/S_i for all i => x in | |T_i in G/| |S_i. So x is in Dg for some g in G. Maybe one should prove (*), I am not certain but I assume that it is so... it seems clear to me. If previous things aren't good, let me know about a possible way (hint) to solve the problem. === === Subject: : Re: Right cosets intersection > Ok, this is just a little bit harder than previous exercises I did, but I > guess I found a quite clear image of how to solve it. The problem is that I > find hard to transpose those tips in a formal way. The proposition is the > following. > --------------------------------------- > Let {S_i : i in I} a family of subgroups of a group G, let {S_i t_i : i in > I} a family of right cosets, and let > _ > D = | | S_i. Prove that either > _ > 1) {} = | | S_i t_i or > _ > 2) Dg = | | S_i t_i for some g in G. > --------------------------------------- (1) is quite trivial, let's focus on (2). I guess see that the partition of > G induced by the coset equivalence relation could be the key to the > solution. In fact _ _ (*) _ > x in | | S_i t_i => x in T_i in G/S_i for all i => x in | |T_i in G/| > |S_i. > So x is in Dg for some g in G. > Maybe one should prove (*), I am not certain but I assume that it is so... > it seems clear to me. > If previous things aren't good, let me know about a possible way (hint) to > solve the problem. If the intersection is not empty,let g be an element of it. It can be written as g=s_i t_i for all i. If g' is another element of the intersection, then g'=s'_i t_i for all i, g'g^{-1}=s'_i s_i^{-1} is in all S_i, i.e. g' in Dg. If d is in D, then d g = d s_i t_i is in S_i t_i for all i === === Subject: : Re: Right cosets intersection > Let {S_i : i in I} a family of subgroups of a group G, let {S_i t_i : i in > I} a family of right cosets, and let > _ > D = | | S_i. > Prove that either > _ > 1) {} = | | S_i t_i or > 2) Dg = | | S_i t_i for some g in G. If not 1), then some g in /_j Sj tj. for all j, g.tj^-1 in Sj h in Dg iff hg^-1 in D = /_j Sj iff for all j, hg^-1 in Sj iff for all j, hg^-1 g.tj^-1 in Sj (check this step) iff for all j, h.tj^-1 in Sj iff for all j, h in Sj tj iff h in /_j Sj tj === === Subject: : Re: Right cosets intersection > Prove that either >> _ >> 1) {} = | | S_i t_i or >> 2) Dg = | | S_i t_i for some g in G. If not 1), then some g in /_j Sj tj. > for all j, g.tj^-1 in Sj h in Dg > iff hg^-1 in D = /_j Sj > iff for all j, hg^-1 in Sj > iff for all j, hg^-1 g.tj^-1 in Sj (check this step) > iff for all j, h.tj^-1 in Sj > iff for all j, h in Sj tj > iff h in /_j Sj tj === === Subject: : Re: Right cosets intersection > If the intersection is not empty,let g be an element of it. > It can be written as g=s_i t_i for all i. > If g' is another element of the intersection, then > g'=s'_i t_i for all i, g'g^{-1}=s'_i s_i^{-1} is in all S_i, i.e. g' in > Dg. > If d is in D, then d g = d s_i t_i is in S_i t_i for all i I think your proof is good, but (maybe I misunderstood) it supposes the existence of at least 2 elements in the intersection, right? === === Subject: : Let programming contest kids do the research! I propose a people's approach to solving math/cs research problems: researchers publish their current subproblems (not as grand as famous open problems) on a blog or a centralized place (e.g. a dedicated and invite worldwide students (especially those fighting for IMO, IOI, ACMICPC or other kinds of math/cs contests) to have a ponder at them in their spare time. This gives the students a chance to work on real research problems and share credits in case they solve one. There is already an example of this in biochemistry R&D: www.innocentive.com. It involves cash rewards, but I think math/cs can be free. Yao Ziyuan === === Subject: : Re: dini derivative Let f be a continuous function on [0,1]. The upper right >> Dini derivative is bounded below on (0,1). Show that f >> is differentiable almost everywhere. > If anyone is interested, perhaps googling Denjoy Young Saks > (for the theorem with this name) might work. In fact, I think > it may be a consequence of the DYS theorem, which would mean > continuous can be replaced with measurable, but using the > DYS theorem would be like my observation in another thread that > the Lebesgue density theorem implies that the graph of a monotone > function has zero planar measure. I was busy with other things this weekend, but I did manage to glance at the Denjoy Young Saks Theorem, and it doesn't imply this result. DYS provides an almost everywhere Dini derivate decomposition behavior, but one of the four classifications besides having a finite derivative includes the upper right Dini derivate being finite, so DYS doesn't help. Anyway, Robert Israel has already posted a more reasonable proof. For what it's worth, these notions (bounded upper unilateral derivate(s) and/or bounded lower unilateral derivate(s)) come up in various pointwise monotonic type conditions. This is where the function may not be increasing or decreasing at a point, but after the addition of some linear function, it will be. http://www.ams.org/proc/1999-127-01/S0002-9939-99-04571-2/ Dave L. Renfro === === Subject: : Re: dini derivative WOW...this is elegant. >Let f be a continuous function on [0,1]. The upper right Dini >derivative is bounded below on (0,1). Show that f is differentiable >almost everywhere.Suppose the upper right Dini derivative D^+ f is bounded below > by c. By changing f to f + (1-c) t, we may assume c > 0. Now I claim f is increasing on (0,1). Suppose not: then > there are a, b with 0 < a < b < 1 and f(a) > f(b). > Let m = (f(b) - f(a))/(b - a) < 0. The > continuous function g(x) = f(x) - m/2 (x-a) has a maximum > at some point c in [a,b]. Since > g(b) = (f(a)+f(b))/2 < f(a) = g(a), c < b. And then > D^+ f(c) = m/2 + D^+ g(c) <= m/2 < 0, contradiction. So now we can apply the theorem that a nondecreasing > function is differentiable almost everywhere. Robert Israel isr...@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === === Subject: : Re: dini derivative <2006120123472816807-kirakun@earthlinknet> Unfortunately it is not. It is a closed book exam... > On 2006-12-01 23:11:17 -0500, Li Yi said: >> This is something I could probably find a dozen references on >> at home, but I'm not there right now, it's not something I've >> ever worked out myself > Oh dear. > It is a problem from my real analysis exam of previous year.That was a take-home exam, I hope. -- -kira === === Subject: : Re: dini derivative >>Let f be a continuous function on [0,1]. The upper right Dini >>derivative is bounded below on (0,1). Show that f is differentiable >>almost everywhere. Suppose the upper right Dini derivative D^+ f is bounded below >by c. By changing f to f + (1-c) t, we may assume c > 0. Yeah, I got that far, and decided that that might imply f was nodecreasing... oh well. >Now I claim f is increasing on (0,1). Suppose not: then >there are a, b with 0 < a < b < 1 and f(a) > f(b). >Let m = (f(b) - f(a))/(b - a) < 0. The >continuous function g(x) = f(x) - m/2 (x-a) has a maximum >at some point c in [a,b]. Since >g(b) = (f(a)+f(b))/2 < f(a) = g(a), c < b. And then >D^+ f(c) = m/2 + D^+ g(c) <= m/2 < 0, contradiction. So now we can apply the theorem that a nondecreasing >function is differentiable almost everywhere. Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich === === Subject: : Re: dini derivative >Let f be a continuous function on [0,1]. The upper right Dini >derivative is bounded below on (0,1). Show that f is differentiable >almost everywhere. Suppose the upper right Dini derivative D^+ f is bounded below > by c. By changing f to f + (1-c) t, we may assume c > 0. Now I claim f is increasing on (0,1). Suppose not: then > there are a, b with 0 < a < b < 1 and f(a) > f(b). > Let m = (f(b) - f(a))/(b - a) < 0. The > continuous function g(x) = f(x) - m/2 (x-a) has a maximum > at some point c in [a,b]. Since > g(b) = (f(a)+f(b))/2 < f(a) = g(a), c < b. And then > D^+ f(c) = m/2 + D^+ g(c) <= m/2 < 0, contradiction. So now we can apply the theorem that a nondecreasing > function is differentiable almost everywhere. If I am following this, then you used `c' for two different parameters. -- Michael Press === === Subject: : Re: dini derivative Let f be a continuous function on [0,1]. The upper right Dini >derivative is bounded below on (0,1). Show that f is differentiable >almost everywhere. > Suppose the upper right Dini derivative D^+ f is bounded below > by c. By changing f to f + (1-c) t, we may assume c > 0. > Now I claim f is increasing on (0,1). Suppose not: then > there are a, b with 0 < a < b < 1 and f(a) > f(b). > Let m = (f(b) - f(a))/(b - a) < 0. The > continuous function g(x) = f(x) - m/2 (x-a) has a maximum > at some point c in [a,b]. Since > g(b) = (f(a)+f(b))/2 < f(a) = g(a), c < b. And then > D^+ f(c) = m/2 + D^+ g(c) <= m/2 < 0, contradiction. > So now we can apply the theorem that a nondecreasing > function is differentiable almost everywhere. If I am following this, then you used `c' for two > different parameters. Yes, I did. Sorry. Replace the c's after some point by d's. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : IS THERE NICE GUYS AND GIRLS:))) IF YOU WANNA SEE ME COME HERE: click here: http://adultfriendfinder.com/go/g823382 NICK:perfect_body85 === === Subject: : Re: Baby stat question [Correction] <45724AD3.90403@netscape.net We assume the underlying population is normally distributed. Let D be > the difference in sample means, and let S1 and S2 be the sample > standard deviations. These three random variables are independent. I'm not sure if they are really independant. I see the following correlation: If sample deviations S1 or S2 are bigger/smaller than the population deviation, it is likely that there were unusually many/few samples far from the population mean, then the sample means are less/more accurate and the difference D can be expected to be bigger/smaller === === Subject: : Re: Baby stat question [Correction] <45724AD3.90403@netscape.net > We assume the underlying population is normally distributed. Let D be > the difference in sample means, and let S1 and S2 be the sample > standard deviations. These three random variables are independent. I'm not sure if they are really independant. > I see the following correlation: > If sample deviations S1 or S2 are bigger/smaller than the population > deviation, it is likely that there were unusually many/few samples far > from the population mean, then the sample means are less/more accurate > and the difference D can be expected to be bigger/smaller You may be thinking of the scenario where M and V are the sample mean and variance of a sample of size N. Then M and V are independent. If the sample is split into subsamples of size N1 and N2, then M1 and V1 are independent, and M2 and V2 are independent. However, given M, V, N1, and N2: 1. M1 and M2 are linearly dependent because M = (N1*M1+N2*M2)/(N1+N2) 2. V1, V2 and D^2 = (M1- M2)^2 are linearly dependent because (N-1)V = SUM(i=1,N1){ (x1-M)^2 } + SUM(i=1,N2){ (x2-M)^2 } = N1*(M-M1)^2 + SUM(i=1,N1){ (x1-M1)^2 } + { 1-->2} = (N1-1)*V1+(N2-1)*V2 +2*N1*N2*D^2/N^2 Hope this helps. Greg === === Subject: : Re: Baby stat question [Correction] > > >We assume the underlying population is normally distributed. Let D be >the difference in sample means, and let S1 and S2 be the sample >standard deviations. These three random variables are independent. > >I'm not sure if they are really independant. >>I see the following correlation: >>If sample deviations S1 or S2 are bigger/smaller than the population >>deviation, it is likely that there were unusually many/few samples far >>from the population mean, then the sample means are less/more accurate >>and the difference D can be expected to be bigger/smaller >> You may be thinking of the scenario where M and V are the sample mean >and variance of a sample of size N. Then M and V are independent. If >the >sample is split into subsamples of size N1 and N2, then M1 and V1 are >independent, and M2 and V2 are independent. However, given M, V, N1, >and N2: 1. M1 and M2 are linearly dependent because M = (N1*M1+N2*M2)/(N1+N2) 2. V1, V2 and D^2 = (M1- M2)^2 are linearly dependent because (N-1)V = SUM(i=1,N1){ (x1-M)^2 } + SUM(i=1,N2){ (x2-M)^2 } > = N1*(M-M1)^2 + SUM(i=1,N1){ (x1-M1)^2 } + { 1-->2} > = (N1-1)*V1+(N2-1)*V2 +2*N1*N2*D^2/N^2 Hope this helps. > Only to confuse matters. M1 and M2 are determined by completely different, independent samples, hence they are independent. Same for V1 and V2. As I said, M1, M2, V1, V2 are jointly independent. Greg, I wish you would stop piping in when you clearly don't know what you are talking about. Sorry, I don't mean to reef fish here, but it really is becoming a repeated problem. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: Baby stat question [Correction] <45724AD3.90403@netscape.net> <45766EE0.9010105@netscape.net >We assume the underlying population is normally distributed. Let D be >the difference in sample means, and let S1 and S2 be the sample >standard deviations. These three random variables are independent. >I'm not sure if they are really independant. >>I see the following correlation: >>If sample deviations S1 or S2 are bigger/smaller than the population >>deviation, it is likely that there were unusually many/few samples far >>from the population mean, then the sample means are less/more accurate >>and the difference D can be expected to be bigger/smaller >You may be thinking of the scenario where M and V are the sample mean >and variance of a sample of size N. Then M and V are independent. If >the >sample is split into subsamples of size N1 and N2, then M1 and V1 are >independent, and M2 and V2 are independent. However, given M, V, N1, >and N2: >1. M1 and M2 are linearly dependent because >M = (N1*M1+N2*M2)/(N1+N2) >2. V1, V2 and D^2 = (M1- M2)^2 are linearly dependent because >(N-1)V = SUM(i=1,N1){ (x1-M)^2 } + SUM(i=1,N2){ (x2-M)^2 } > = N1*(M-M1)^2 + SUM(i=1,N1){ (x1-M1)^2 } + { 1-->2} > = (N1-1)*V1+(N2-1)*V2 +2*N1*N2*D^2/N^2 >Hope this helps. Only to confuse matters. M1 and M2 are determined by completely > different, independent samples, hence they are independent. Same for > V1 and V2. As I said, M1, M2, V1, V2 are jointly independent. Greg, I wish you would stop piping in when you clearly don't know what > you are talking about. I may have not made my point clearly; however I know exactly what I am talking about. hagman: >> I'm not sure if they are really independant Heath thought: Why would he think that they are dependent? Heath reply: > You may be thinking of the scenario where ... random draws from the normal population yielded independent Ms and Vs. I failed to *emphasize* that the only way (that I can see) he could possibly think that they were dependent was if he thought that the random draws were made from a finite sample, *without replacement*, until all cases were chosen. My apologies for not making that explicitly clear. Question 1: What about the bootstrap case where N1+N2 (N1,N2 arbitrary) random draws are made *with replacement*, from a finite sample of size N? I think the sample stats can be considered independent as long as N >=2 However, I'm not sure. > Sorry, I don't mean to reef fish here, but it > really is becoming a repeated problem. carefully, some of the misunderstandings would not have happened. Hope this helps. Greg === === Subject: : Re: Baby stat question [Correction] >We assume the underlying population is normally distributed. Let D be >>the difference in sample means, and let S1 and S2 be the sample >>standard deviations. These three random variables are independent. >> I'm not sure if they are really independant. >I see the following correlation: >If sample deviations S1 or S2 are bigger/smaller than the population >deviation, it is likely that there were unusually many/few samples far >from the population mean, then the sample means are less/more accurate >and the difference D can be expected to be bigger/smaller > It is a standard result that the sample mean and sample variance from a normally distributed population are independent Consult any basic text on statistics; some specific references are Mood and Graybill or either of SM Ross's probabilty texts. Let M and V denote respectively sample mean and variance. M1, M2, V1, and V2 are jointly independent, so M1 - M2 and (V1 + V2) / 2 are independent. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === === Subject: : Re: Baby stat question [Correction] >> We assume the underlying population is normally distributed. Let D be >> the difference in sample means, and let S1 and S2 be the sample >> standard deviations. These three random variables are independent. I'm not sure if they are really independant. >I see the following correlation: >If sample deviations S1 or S2 are bigger/smaller than the population >deviation, it is likely that there were unusually many/few samples far >from the population mean, then the sample means are less/more accurate >and the difference D can be expected to be bigger/smaller It is a theorem that the sample mean and sample standard deviation for iid normal random variables are independent. See e.g. Evans and Rosenthal, Probability and Statistics: The Science of Uncertainty, Theorem 4.6.6. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === Subject: : Re: How to solve a stiff ODE system? I want to solve a system of first order ODEs with a given initial > value: > d Y > ---- = f(t, Y) Y(t_0) = Y_0 > d t > Where, f(t, Y) is nonlinear. > I'm currently modeling the space charge flow of the high voltage direct > current power lines. This is a ODE system with four state variables. What are the equations? I have a Calculus-level compiler (FC) that may solve this. Would you be willing to put it up on a web site for others to view and learn from? (see http://www.digitalCalculus.com/page/593562 for an example write-up.) Phil === === Subject: : Re: How to solve a stiff ODE system? > d Y > ---- = f(t, Y), Y(t_0) = Y_0 > d t > Warning: Failure at t=1.533227e-003. Unable to meet integration > tolerances without reducing the step size below the smallest value > allowed (5.447117e-018) at time t. > Suspect stiffness is not a problem,cannot say further as you do not > indicate f(t,Y).For eps =10^-8, if initial condition Y(t_eps) = Y_0 is > used and ode behaves well, then problem was with usage of first order > equation.To fix such cases switch over to a second order differential > equation providing two initial conditions. > The ode system of mine is as follows: dy(1) / dt = y(2) > dy(2) / dt = - (y(3) - y(4)) / epsilon0 / E^2 > dy(3) / dt = 1 / epsilon0 / y(2) / E^2 * (y(3)^2 - y(3) * y(4) * (1 - > epsilon0 * Ri / mu_+ / e)) > dy(4) / dt = 1 / epsilon0 / y(2) / E^2 * (y(3) * y(4) * (1 - epsilon0 * > Ri / mu_- / e) - y(4)^2) Where, > epsilon0: permittivity of vacuum 8.85e-12 > E = E(t): the electric field. it is a function of t. After integrating > one step of the ODE, the electric field should be recalculated from > y(1), y(2), y(3), y(4), and this formula doesn't belong to the ODE > system, so it is not given here. > Ri: the recombination coefficient of bipolar charge. 2.2e-12 > mu_+, mu_-: the mobility of positive and negative charge, 1.6e-4 > e: elementary charge 1.6e-19 The above system is used to model the space charge flow of the high > voltage direct current power line. I still don't quite follow you: how should I switch to a second order > differential equation? What is the reason of doing this? And how to > apply this procedure to my ODE system? > Before going to second order, at first redefine terms so that you compute quantities in absolute value range (.01-10).Tried to solve the ODE numerically.It crashes at start of any chosen time interval, even as an initial value problem,found no errors. Perhaps you need to at first check the physics modeling even as an IVP.In a simulation program start with the simplest electrostatic situations of equipotentials or force lines and equations depicting them.When it works,gradually increase its complexity in steps.That traps the error at a particular stage. Narasimham === === Subject: : Re: How to solve a stiff ODE system? > d Y > ---- = f(t, Y), Y(t_0) = Y_0 > d t > Warning: Failure at t=1.533227e-003. Unable to meet integration > tolerances without reducing the step size below the smallest value > allowed (5.447117e-018) at time t. > Suspect stiffness is not a problem,cannot say further as you do not > indicate f(t,Y).For eps =10^-8, if initial condition Y(t_eps) = Y_0 is > used and ode behaves well, then problem was with usage of first order > equation.To fix such cases switch over to a second order differential > equation providing two initial conditions. > The ode system of mine is as follows: dy(1) / dt = y(2) > dy(2) / dt = - (y(3) - y(4)) / epsilon0 / E^2 > dy(3) / dt = 1 / epsilon0 / y(2) / E^2 * (y(3)^2 - y(3) * y(4) * (1 - > epsilon0 * Ri / mu_+ / e)) > dy(4) / dt = 1 / epsilon0 / y(2) / E^2 * (y(3) * y(4) * (1 - epsilon0 * > Ri / mu_- / e) - y(4)^2) Where, > epsilon0: permittivity of vacuum 8.85e-12 > E = E(t): the electric field. it is a function of t. After integrating > one step of the ODE, the electric field should be recalculated from > y(1), y(2), y(3), y(4), and this formula doesn't belong to the ODE > system, so it is not given here. > Ri: the recombination coefficient of bipolar charge. 2.2e-12 > mu_+, mu_-: the mobility of positive and negative charge, 1.6e-4 > e: elementary charge 1.6e-19 The above system is used to model the space charge flow of the high > voltage direct current power line. I still don't quite follow you: how should I switch to a second order > differential equation? What is the reason of doing this? And how to > apply this procedure to my ODE system? > Tried to solve the ODE numerically. It crashes at start of any chosen time interval, even as an initial value problem,found no errors. Perhaps you need to at first check the physics modeling even as an IVP.In a simulation program start with the simplest electrostatic situations like equipotentials or force lines and equations depicting them.When it works,gradually increase its complexity in steps.That traps the error at a particular stage. Narasimham === === Subject: : Re: How to solve a stiff ODE system? > Hello everybody! > I want to solve a system of first order ODEs with a given initial > value: > d Y > ---- = f(t, Y) Y(t_0) = Y_0 > d t > Where, f(t, Y) is nonlinear. > Up till now, I've tried to deal with this IVP using a series of ode** > solvers (stiff or nonstiff) in Matlab. However, when integrating to > some certain time points, a warning message is thrown out from the > ode** solver as follows: > Warning: Failure at t=1.533227e-003. Unable to meet integration > tolerances without reducing the step size below the smallest value > allowed (5.447117e-018) at time t. > So I examined the Jacobian matrix (frac{partial f}{partial Y}) of > the nonlinear mapping at the current time point and obtained its > eigenvalues as follows: > lambda1 = -1.72864992567911e9 > lambda2 = 0.29658911109064e9 > lambda3 = -0.17145123351451e9 > lambda4 = 0 > We can see that this Jacobian matrix has a negative eigenvalue with > very large absolute value, and also a very large positive eigenvalue. > According to the definition of the stiffness of the ODEs, which is > given by Shampine and Gear: a ODE system is stiff, when: > 1) the real parts of all the eigenvalues of the Jacobian matrix should > not be very large positive real numbers; > 2) there should be at least one eigenvalue of the Jacobian matrix, > which has a negative real part with very large absolute value; > 3) ... > My present situation complies with second requirement of the stiffness > definition, but not the first one, due to the existence of the > eigenvalue: 0.29658911109064e9. > Then my questions are: > 1) Is my ode system stiff? > 2) if so, why do the stiff ode solvers of Matlab failed at my problem? > What will be the possible reasons? > 3) if my ode system is not stiff, then is the method for solving stiff > ode also suitable for my problem? And if the method of solving stiff > ode does not work, are there any other methods that can be effective? > your help! > > I was hoping someone who knows what they are doing would answer, and > save me from possibly embarrassing myself. > Anyway, my guess is that stiffness is not your problem. Maybe your > system of equations really is blowing up, something like dy/dt = y^2. > That might account for your super large lambda2. > What are you modeling? If it is something like motion of planets, it > would indicate something like two planets getting very close to each > other (close in the sense that their centers are close to each other, so > in effect the planets have crashed into each other, and your model is no > longer appropriate). > Stephen I'm currently modeling the space charge flow of the high voltage direct > current power lines. This is a ODE system with four state variables. > And I'm now using shooting methods to convert a BVP to IVP. Because I > need to provide initial guess to ignite the optimization process in > order to shoot the boundary condition, when the initial guess is not > good enough or the optimizer steps on a not so good feasible point, I > met wit the situation that the integration of the system blew up > halfway. I examined the Jacobian matrix, before the blowup, when things go > normal, there is no large positive or negative eigenvalue. But when > the error jumps out, something that may be like dy/dt = y^2 that you've > given, come into play. At the present, I can't make it clear what will happen in the reality > when ODE solver blows up. However, it must indicate a initial guess > that won't appear in real world anyway. So the cause of the error may be the shooting method? Is there any way > to avoid integration blowup by using other methods for BVP? (Currently, > I only know multishoot and relaxation methods, but I'm not sure if they > will solve my problem) There are good collocation methods for this. What is the advantage of a shooting method? Can your method do y'' = sinh (50 * y), y(0) = 0, y(1) = 1? -- Michael Press === === Subject: : Re: How to solve a stiff ODE system? There are good collocation methods for this. What is > the advantage of a shooting method? Can your method do y'' = sinh (50 * y), y(0) = 0, y(1) = 1? -- > Michael Press Hello Michael! I have tried the bvp4c function in Matlab, which I think to be using collocation method. But as soon as it starts, error occurs, which indicates that jacobian matrix is singular. How should I do with this phenomenon? === === Subject: : Re: How to solve a stiff ODE system? , > There are good collocation methods for this. What is > the advantage of a shooting method? Can your method do > y'' = sinh (50 * y), y(0) = 0, y(1) = 1? > -- > Michael Press Hello Michael! I have tried the bvp4c function in Matlab, which I think to be using > collocation method. But as soon as it starts, error occurs, which > indicates that jacobian matrix is singular. How should I do with this > phenomenon? > COLNEW is a robust solver in FORTRAN I once ported it to C. Best ask the experts at sci.math.num-analysis for help, rather than me. I only commented because you mentioned using a shooting method. I can only guess. Maybe the representation in ODE's of the problem allows non-physical solutions, or the posing of the problem misses a bit. Or the ODE's are every bit as awkward as you have found. Maybe you could impose some constraint that gets a numerical solution, then relax the constraint and watch what happens. -- Michael Press === === Subject: : Re: How to solve a stiff ODE system? There are good collocation methods for this. What is > the advantage of a shooting method? Can your method do > y'' = sinh (50 * y), y(0) = 0, y(1) = 1? > -- > Michael Press > Hello Michael! > I have tried the bvp4c function in Matlab, which I think to be using > collocation method. But as soon as it starts, error occurs, which > indicates that jacobian matrix is singular. How should I do with this > phenomenon? COLNEW is a robust solver in FORTRAN > rather than me. I only commented because you mentioned > using a shooting method. I can only guess. Maybe the representation in ODE's of > the problem allows non-physical solutions, or the > posing of the problem misses a bit. Or the ODE's are > every bit as awkward as you have found. Maybe you could > impose some constraint that gets a numerical solution, > then relax the constraint and watch what happens. -- > Michael Press they are good experiences and valuable for reference. I'll try COLNEW === === Subject: : Re: How to solve a stiff ODE system? Hello everybody! >I want to solve a system of first order ODEs with a given initial >value: >d Y >---- = f(t, Y) Y(t_0) = Y_0 >d t >Where, f(t, Y) is nonlinear. >>Up till now, I've tried to deal with this IVP using a series of ode** >solvers (stiff or nonstiff) in Matlab. However, when integrating to >some certain time points, a warning message is thrown out from the >ode** solver as follows: >>Warning: Failure at t=1.533227e-003. Unable to meet integration >tolerances without reducing the step size below the smallest value >allowed (5.447117e-018) at time t. >>So I examined the Jacobian matrix (frac{partial f}{partial Y}) of >the nonlinear mapping at the current time point and obtained its >eigenvalues as follows: >>lambda1 = -1.72864992567911e9 >lambda2 = 0.29658911109064e9 >lambda3 = -0.17145123351451e9 >lambda4 = 0 >>We can see that this Jacobian matrix has a negative eigenvalue with >very large absolute value, and also a very large positive eigenvalue. >According to the definition of the stiffness of the ODEs, which is >given by Shampine and Gear: a ODE system is stiff, when: >>1) the real parts of all the eigenvalues of the Jacobian matrix should >not be very large positive real numbers; >2) there should be at least one eigenvalue of the Jacobian matrix, >which has a negative real part with very large absolute value; >3) ... >>My present situation complies with second requirement of the stiffness >definition, but not the first one, due to the existence of the >eigenvalue: 0.29658911109064e9. >>Then my questions are: >>1) Is my ode system stiff? >2) if so, why do the stiff ode solvers of Matlab failed at my problem? >What will be the possible reasons? >3) if my ode system is not stiff, then is the method for solving stiff >ode also suitable for my problem? And if the method of solving stiff >ode does not work, are there any other methods that can be effective? >>your help! >I was hoping someone who knows what they are doing would answer, and >>save me from possibly embarrassing myself. >>Anyway, my guess is that stiffness is not your problem. Maybe your >>system of equations really is blowing up, something like dy/dt = y^2. >>That might account for your super large lambda2. >>What are you modeling? If it is something like motion of planets, it >>would indicate something like two planets getting very close to each >>other (close in the sense that their centers are close to each other, so >>in effect the planets have crashed into each other, and your model is no >>longer appropriate). >>Stephen > I'm currently modeling the space charge flow of the high voltage direct > current power lines. This is a ODE system with four state variables. > And I'm now using shooting methods to convert a BVP to IVP. Because I > need to provide initial guess to ignite the optimization process in > order to shoot the boundary condition, when the initial guess is not > good enough or the optimizer steps on a not so good feasible point, I > met wit the situation that the integration of the system blew up > halfway. > I examined the Jacobian matrix, before the blowup, when things go > normal, there is no large positive or negative eigenvalue. But when > the error jumps out, something that may be like dy/dt = y^2 that you've > given, come into play. > At the present, I can't make it clear what will happen in the reality > when ODE solver blows up. However, it must indicate a initial guess > that won't appear in real world anyway. > So the cause of the error may be the shooting method? Is there any way > to avoid integration blowup by using other methods for BVP? (Currently, > I only know multishoot and relaxation methods, but I'm not sure if they > will solve my problem) > Looking at your ODE, I can see it might blow up if y(2) gets close to 0 > (i.e. it is a lot more straightforward than the dy/dy=y^2 example I gave > - more akin to two planets getting too close to each other). As a matter of interest, what are your boundary conditions in your BVP? > Is it on y(1) and y(2) for t taking two different values? And then > you are guessing (i.e. shooting with) y(3) and y(4)? I can see a real > possibility that you might be hitting the y(2)=0 issue. One way you could possibly fix this problem is maybe to replace the y(2) > in the 3rd and 4th equation with f(y(2)), where f(x) is defined to be x > if |x| is greater than some threshhold value x0, and sign(x) x0 > otherwise. In this way you artificially avoid the stumbling block. It > won't affect your final answer, because you happen to know that you > never get close to this, but it will allow your intermediate shots to > compute. My boundary conditions are on y(1) and y(2), and I make a initial guess for y(3) and y(4), then shoot to the other end and compare the results with y(1), y(2) on that boundary. Excuse me for forgetting to make it clear that the quantities of different state variables differ a lot: y(1): its dimension is usually of 1e5, y(1) increases monotonously when t is increasing, so y(1) can never stay constant. y(2): it usually belongs to domain (0,10), and it can never be zero y(3), y(4): their dimension is usually of 1e-8 I think you are right: according to your suggestion, when there are error thrown out from ODE solver, I found out y(2) is actually small: about 1e-7 (although it can never be zero). I'll try the method you've given to see if the ODE solver and optimizer will work. By the way, will the remarkable dimension difference between the different variables matter a lot? I'm very grateful for your patient and enlightening suggestions and instructions, thank you ! === === Subject: : Re: How to solve a stiff ODE system? > >>Hello everybody! I want to solve a system of first order ODEs with a given initial value: d Y ---- = f(t, Y) Y(t_0) = Y_0 d t Where, f(t, Y) is nonlinear. >Up till now, I've tried to deal with this IVP using a series of ode** solvers (stiff or nonstiff) in Matlab. However, when integrating to some certain time points, a warning message is thrown out from the ode** solver as follows: >Warning: Failure at t=1.533227e-003. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.447117e-018) at time t. >So I examined the Jacobian matrix (frac{partial f}{partial Y}) of the nonlinear mapping at the current time point and obtained its eigenvalues as follows: >lambda1 = -1.72864992567911e9 lambda2 = 0.29658911109064e9 lambda3 = -0.17145123351451e9 lambda4 = 0 >We can see that this Jacobian matrix has a negative eigenvalue with very large absolute value, and also a very large positive eigenvalue. According to the definition of the stiffness of the ODEs, which is given by Shampine and Gear: a ODE system is stiff, when: >1) the real parts of all the eigenvalues of the Jacobian matrix should not be very large positive real numbers; 2) there should be at least one eigenvalue of the Jacobian matrix, which has a negative real part with very large absolute value; 3) ... >My present situation complies with second requirement of the stiffness definition, but not the first one, due to the existence of the eigenvalue: 0.29658911109064e9. >Then my questions are: >1) Is my ode system stiff? 2) if so, why do the stiff ode solvers of Matlab failed at my problem? What will be the possible reasons? 3) if my ode system is not stiff, then is the method for solving stiff ode also suitable for my problem? And if the method of solving stiff ode does not work, are there any other methods that can be effective? >your help! >>I was hoping someone who knows what they are doing would answer, and >>save me from possibly embarrassing myself. >>Anyway, my guess is that stiffness is not your problem. Maybe your >>system of equations really is blowing up, something like dy/dt = y^2. >>That might account for your super large lambda2. >>What are you modeling? If it is something like motion of planets, it >>would indicate something like two planets getting very close to each >>other (close in the sense that their centers are close to each other, so >>in effect the planets have crashed into each other, and your model is no >>longer appropriate). >>Stephen I'm currently modeling the space charge flow of the high voltage direct >current power lines. This is a ODE system with four state variables. >And I'm now using shooting methods to convert a BVP to IVP. Because I >need to provide initial guess to ignite the optimization process in >order to shoot the boundary condition, when the initial guess is not >good enough or the optimizer steps on a not so good feasible point, I >met wit the situation that the integration of the system blew up >halfway. >>I examined the Jacobian matrix, before the blowup, when things go >normal, there is no large positive or negative eigenvalue. But when >the error jumps out, something that may be like dy/dt = y^2 that you've >given, come into play. >>At the present, I can't make it clear what will happen in the reality >when ODE solver blows up. However, it must indicate a initial guess >that won't appear in real world anyway. >>So the cause of the error may be the shooting method? Is there any way >to avoid integration blowup by using other methods for BVP? (Currently, >I only know multishoot and relaxation methods, but I'm not sure if they >will solve my problem) >>Looking at your ODE, I can see it might blow up if y(2) gets close to 0 >>(i.e. it is a lot more straightforward than the dy/dy=y^2 example I gave >>- more akin to two planets getting too close to each other). >>As a matter of interest, what are your boundary conditions in your BVP? >> Is it on y(1) and y(2) for t taking two different values? And then >>you are guessing (i.e. shooting with) y(3) and y(4)? I can see a real >>possibility that you might be hitting the y(2)=0 issue. >>One way you could possibly fix this problem is maybe to replace the y(2) >>in the 3rd and 4th equation with f(y(2)), where f(x) is defined to be x >>if |x| is greater than some threshhold value x0, and sign(x) x0 >>otherwise. In this way you artificially avoid the stumbling block. It >>won't affect your final answer, because you happen to know that you >>never get close to this, but it will allow your intermediate shots to >>compute. My boundary conditions are on y(1) and y(2), and I make a initial guess > for y(3) and y(4), then shoot to the other end and compare the results > with y(1), y(2) on that boundary. Excuse me for forgetting to make it > clear that the quantities of different state variables differ a lot: > y(1): its dimension is usually of 1e5, y(1) increases monotonously > when t is increasing, so y(1) can never stay constant. > y(2): it usually belongs to domain (0,10), and it can never be zero > y(3), y(4): their dimension is usually of 1e-8 I think you are right: according to your suggestion, when there are > error thrown out from ODE solver, I found out y(2) is actually small: > about 1e-7 (although it can never be zero). I'll try the method you've > given to see if the ODE solver and optimizer will work. By the way, will the remarkable dimension difference between the > different variables matter a lot? I'm very grateful for your patient and enlightening suggestions and > instructions, thank you ! > First, whoever said you should try a numerical analysis group probably is right. I really am not an expert - I hadn't even heard of the collocation method until I read this thread. Nevertheless I'm glad you posted here, because it does look like an interesting question. Now that it looks like the problem is when y(2) gets close to 0, it seems to me that a good method might be to put in some kind of artificial barrier in the ODE so that y(2) can never become 0. E.g. dy(2) / dt = - (y(3) - y(4)) / epsilon0 / E2 + f(y(2)) where f(x) is 0 if |x|>x0 (x0 is some small threshhold), and f(x) blows up quite quickly to infinity or minus infinity (depending on whether x is positive or negative) - in effect, if y(2) gets close to 0, it bounces away from zero. Then you try any of your favorite methods (shooting or collocation). If the final solution hits this barrier (i.e. |y(2)| gets smaller than x_0), first try smaller and smaller x0, and if that doesn't work then you know your ODE is physically wrong. In other words, I really think the problem is with the ODE itself - not with any method you are using. And the solution to your problem will be to play with the ODE itself, not to play with the method. Stephen === === Subject: : Re: How to solve a stiff ODE system? >Hello everybody! I want to solve a system of first order ODEs with a given initial value: d Y ---- = f(t, Y) Y(t_0) = Y_0 d t Where, f(t, Y) is nonlinear. >Up till now, I've tried to deal with this IVP using a series of ode** solvers (stiff or nonstiff) in Matlab. However, when integrating to some certain time points, a warning message is thrown out from the ode** solver as follows: >Warning: Failure at t=1.533227e-003. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.447117e-018) at time t. >So I examined the Jacobian matrix (frac{partial f}{partial Y}) of the nonlinear mapping at the current time point and obtained its eigenvalues as follows: >lambda1 = -1.72864992567911e9 lambda2 = 0.29658911109064e9 lambda3 = -0.17145123351451e9 lambda4 = 0 >We can see that this Jacobian matrix has a negative eigenvalue with very large absolute value, and also a very large positive eigenvalue. According to the definition of the stiffness of the ODEs, which is given by Shampine and Gear: a ODE system is stiff, when: >1) the real parts of all the eigenvalues of the Jacobian matrix should not be very large positive real numbers; 2) there should be at least one eigenvalue of the Jacobian matrix, which has a negative real part with very large absolute value; 3) ... >My present situation complies with second requirement of the stiffness definition, but not the first one, due to the existence of the eigenvalue: 0.29658911109064e9. >Then my questions are: >1) Is my ode system stiff? 2) if so, why do the stiff ode solvers of Matlab failed at my problem? What will be the possible reasons? 3) if my ode system is not stiff, then is the method for solving stiff ode also suitable for my problem? And if the method of solving stiff ode does not work, are there any other methods that can be effective? >your help! >>I was hoping someone who knows what they are doing would answer, and >>save me from possibly embarrassing myself. >>Anyway, my guess is that stiffness is not your problem. Maybe your >>system of equations really is blowing up, something like dy/dt = y^2. >>That might account for your super large lambda2. >>What are you modeling? If it is something like motion of planets, it >>would indicate something like two planets getting very close to each >>other (close in the sense that their centers are close to each other, so >>in effect the planets have crashed into each other, and your model is no >>longer appropriate). >>Stephen I'm currently modeling the space charge flow of the high voltage direct >current power lines. This is a ODE system with four state variables. >And I'm now using shooting methods to convert a BVP to IVP. Because I >need to provide initial guess to ignite the optimization process in >order to shoot the boundary condition, when the initial guess is not >good enough or the optimizer steps on a not so good feasible point, I >met wit the situation that the integration of the system blew up >halfway. >>I examined the Jacobian matrix, before the blowup, when things go >normal, there is no large positive or negative eigenvalue. But when >the error jumps out, something that may be like dy/dt = y^2 that you've >given, come into play. >>At the present, I can't make it clear what will happen in the reality >when ODE solver blows up. However, it must indicate a initial guess >that won't appear in real world anyway. >>So the cause of the error may be the shooting method? Is there any way >to avoid integration blowup by using other methods for BVP? (Currently, >I only know multishoot and relaxation methods, but I'm not sure if they >will solve my problem) >>Looking at your ODE, I can see it might blow up if y(2) gets close to 0 >>(i.e. it is a lot more straightforward than the dy/dy=y^2 example I gave >>- more akin to two planets getting too close to each other). >>As a matter of interest, what are your boundary conditions in your BVP? >> Is it on y(1) and y(2) for t taking two different values? And then >>you are guessing (i.e. shooting with) y(3) and y(4)? I can see a real >>possibility that you might be hitting the y(2)=0 issue. >>One way you could possibly fix this problem is maybe to replace the y(2) >>in the 3rd and 4th equation with f(y(2)), where f(x) is defined to be x >>if |x| is greater than some threshhold value x0, and sign(x) x0 >>otherwise. In this way you artificially avoid the stumbling block. It >>won't affect your final answer, because you happen to know that you >>never get close to this, but it will allow your intermediate shots to >>compute. > > My boundary conditions are on y(1) and y(2), and I make a initial guess > for y(3) and y(4), then shoot to the other end and compare the results > with y(1), y(2) on that boundary. Excuse me for forgetting to make it > clear that the quantities of different state variables differ a lot: > y(1): its dimension is usually of 1e5, y(1) increases monotonously > when t is increasing, so y(1) can never stay constant. > y(2): it usually belongs to domain (0,10), and it can never be zero > y(3), y(4): their dimension is usually of 1e-8 > I think you are right: according to your suggestion, when there are > error thrown out from ODE solver, I found out y(2) is actually small: > about 1e-7 (although it can never be zero). I'll try the method you've > given to see if the ODE solver and optimizer will work. > By the way, will the remarkable dimension difference between the > different variables matter a lot? > I'm very grateful for your patient and enlightening suggestions and > instructions, thank you ! First, whoever said you should try a numerical analysis group probably > is right. I really am not an expert - I hadn't even heard of the > collocation method until I read this thread. Nevertheless I'm glad you > posted here, because it does look like an interesting question. Now that it looks like the problem is when y(2) gets close to 0, it > seems to me that a good method might be to put in some kind of > artificial barrier in the ODE so that y(2) can never become 0. E.g. dy(2) / dt = - (y(3) - y(4)) / epsilon0 / E2 + f(y(2)) where f(x) is 0 if |x|>x0 (x0 is some small threshhold), and f(x) blows > up quite quickly to infinity or minus infinity (depending on whether x > is positive or negative) - in effect, if y(2) gets close to 0, it > bounces away from zero. Then you try any of your favorite methods (shooting or collocation). If > the final solution hits this barrier (i.e. |y(2)| gets smaller than > x_0), first try smaller and smaller x0, and if that doesn't work then > you know your ODE is physically wrong. In other words, I really think the problem is with the ODE itself - not > with any method you are using. And the solution to your problem will be > to play with the ODE itself, not to play with the method. Stephen Hello Stephen! I've begun to gradually understand your ideas. You are right, the first thing I should do is to control the value of y(2) so that it won't be too small. No matter what new and sound methods there will be, as long as y(2) is zero, none of them could help a little. Previously, I had the illusion that when there was an error, I should resort to a newer and more complicated mathematical method. But up to now, I've learned from you that a good and effective way of doing things is firstly to reexamine the model carefully and check if there will be some clues and something that can be reduced or controlled. In addition, try to grasp the physical meaning and background is important. Just as you said, take planet movement as an example, when ode solver is wrong, the model has already become unreasonable. This has given me a hint, and I indeed find out the serious deviation of the model from the real world. The derivation of my model is based on some assumptions. When bad initial guess of the IVP is given, these assumptions are not valid any more. This fact tells me that I must prevent the value of y(2) under some threshhold. this I've really begun to know more than before! === === Subject: : Re: How to solve a stiff ODE system? > Hello everybody! I want to solve a system of first order ODEs with a given initial value: d Y ---- = f(t, Y) Y(t_0) = Y_0 d t Where, f(t, Y) is nonlinear. >Up till now, I've tried to deal with this IVP using a series of ode** solvers (stiff or nonstiff) in Matlab. However, when integrating to some certain time points, a warning message is thrown out from the ode** solver as follows: >Warning: Failure at t=1.533227e-003. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.447117e-018) at time t. >So I examined the Jacobian matrix (frac{partial f}{partial Y}) of the nonlinear mapping at the current time point and obtained its eigenvalues as follows: >lambda1 = -1.72864992567911e9 lambda2 = 0.29658911109064e9 lambda3 = -0.17145123351451e9 lambda4 = 0 >We can see that this Jacobian matrix has a negative eigenvalue with very large absolute value, and also a very large positive eigenvalue. According to the definition of the stiffness of the ODEs, which is given by Shampine and Gear: a ODE system is stiff, when: >1) the real parts of all the eigenvalues of the Jacobian matrix should not be very large positive real numbers; 2) there should be at least one eigenvalue of the Jacobian matrix, which has a negative real part with very large absolute value; 3) ... >My present situation complies with second requirement of the stiffness definition, but not the first one, due to the existence of the eigenvalue: 0.29658911109064e9. >Then my questions are: >1) Is my ode system stiff? 2) if so, why do the stiff ode solvers of Matlab failed at my problem? What will be the possible reasons? 3) if my ode system is not stiff, then is the method for solving stiff ode also suitable for my problem? And if the method of solving stiff ode does not work, are there any other methods that can be effective? >your help! >I was hoping someone who knows what they are doing would answer, and >>save me from possibly embarrassing myself. >Anyway, my guess is that stiffness is not your problem. Maybe your >>system of equations really is blowing up, something like dy/dt = y^2. >>That might account for your super large lambda2. >What are you modeling? If it is something like motion of planets, it >>would indicate something like two planets getting very close to each >>other (close in the sense that their centers are close to each other, so >>in effect the planets have crashed into each other, and your model is no >>longer appropriate). >Stephen >>I'm currently modeling the space charge flow of the high voltage direct current power lines. This is a ODE system with four state variables. And I'm now using shooting methods to convert a BVP to IVP. Because I need to provide initial guess to ignite the optimization process in order to shoot the boundary condition, when the initial guess is not good enough or the optimizer steps on a not so good feasible point, I met wit the situation that the integration of the system blew up halfway. >I examined the Jacobian matrix, before the blowup, when things go normal, there is no large positive or negative eigenvalue. But when the error jumps out, something that may be like dy/dt = y^2 that you've given, come into play. >At the present, I can't make it clear what will happen in the reality when ODE solver blows up. However, it must indicate a initial guess that won't appear in real world anyway. >So the cause of the error may be the shooting method? Is there any way to avoid integration blowup by using other methods for BVP? (Currently, I only know multishoot and relaxation methods, but I'm not sure if they will solve my problem) Looking at your ODE, I can see it might blow up if y(2) gets close to 0 >>(i.e. it is a lot more straightforward than the dy/dy=y^2 example I gave >>- more akin to two planets getting too close to each other). >>As a matter of interest, what are your boundary conditions in your BVP? >> Is it on y(1) and y(2) for t taking two different values? And then >>you are guessing (i.e. shooting with) y(3) and y(4)? I can see a real >>possibility that you might be hitting the y(2)=0 issue. >>One way you could possibly fix this problem is maybe to replace the y(2) >>in the 3rd and 4th equation with f(y(2)), where f(x) is defined to be x >>if |x| is greater than some threshhold value x0, and sign(x) x0 >>otherwise. In this way you artificially avoid the stumbling block. It >>won't affect your final answer, because you happen to know that you >>never get close to this, but it will allow your intermediate shots to >>compute. >>My boundary conditions are on y(1) and y(2), and I make a initial guess >for y(3) and y(4), then shoot to the other end and compare the results >with y(1), y(2) on that boundary. Excuse me for forgetting to make it >clear that the quantities of different state variables differ a lot: >y(1): its dimension is usually of 1e5, y(1) increases monotonously >when t is increasing, so y(1) can never stay constant. >y(2): it usually belongs to domain (0,10), and it can never be zero >y(3), y(4): their dimension is usually of 1e-8 >>I think you are right: according to your suggestion, when there are >error thrown out from ODE solver, I found out y(2) is actually small: >about 1e-7 (although it can never be zero). I'll try the method you've >given to see if the ODE solver and optimizer will work. >>By the way, will the remarkable dimension difference between the >different variables matter a lot? >>I'm very grateful for your patient and enlightening suggestions and >instructions, thank you ! >First, whoever said you should try a numerical analysis group probably >>is right. I really am not an expert - I hadn't even heard of the >>collocation method until I read this thread. Nevertheless I'm glad you >>posted here, because it does look like an interesting question. >>Now that it looks like the problem is when y(2) gets close to 0, it >>seems to me that a good method might be to put in some kind of >>artificial barrier in the ODE so that y(2) can never become 0. E.g. >>dy(2) / dt = - (y(3) - y(4)) / epsilon0 / E2 + f(y(2)) >>where f(x) is 0 if |x|>x0 (x0 is some small threshhold), and f(x) blows >>up quite quickly to infinity or minus infinity (depending on whether x >>is positive or negative) - in effect, if y(2) gets close to 0, it >>bounces away from zero. >>Then you try any of your favorite methods (shooting or collocation). If >>the final solution hits this barrier (i.e. |y(2)| gets smaller than >>x_0), first try smaller and smaller x0, and if that doesn't work then >>you know your ODE is physically wrong. >>In other words, I really think the problem is with the ODE itself - not >>with any method you are using. And the solution to your problem will be >>to play with the ODE itself, not to play with the method. >>Stephen > Hello Stephen! I've begun to gradually understand your ideas. You are right, the first > thing I should do is to control the value of y(2) so that it won't be > too small. No matter what new and sound methods there will be, as long > as y(2) is zero, none of them could help a little. Previously, I had the illusion that when there was an error, I should > resort to a newer and more complicated mathematical method. But up to > now, I've learned from you that a good and effective way of doing > things is firstly to reexamine the model carefully and check if there > will be some clues and something that can be reduced or controlled. In addition, try to grasp the physical meaning and background is > important. Just as you said, take planet movement as an example, when > ode solver is wrong, the model has already become unreasonable. This > has given me a hint, and I indeed find out the serious deviation of the > model from the real world. The derivation of my model is based on some > assumptions. When bad initial guess of the IVP is given, these > assumptions are not valid any more. This fact tells me that I must > prevent the value of y(2) under some threshhold. this I've really begun to know more than before! > I was very glad to be of help. === === Subject: : Re: How to solve a stiff ODE system? > By the way, will the remarkable dimension difference between the > different variables matter a lot? It depends on how the ODE solver works. I guess that if it is designed well, it won't. But I guess you could check by renormalizing the dimensions of the equation (e.g. replace y(1) by z(1)=y(1)/(typical value of y(1)), same for y(2), etc). Stephen === === Subject: : Re: How to solve a stiff ODE system? By the way, will the remarkable dimension difference between the > different variables matter a lot? It depends on how the ODE solver works. I guess that if it is designed > well, it won't. But I guess you could check by renormalizing the > dimensions of the equation (e.g. replace y(1) by z(1)=y(1)/(typical > value of y(1)), same for y(2), etc). Stephen I've tried your idea to do variable scaling. Seems it doesn't do anything good. :( === === Subject: : Given a closed curve, how to determine if a piont is inside the curve? I don't know how to determine if a point is inside a curve. The curve is simple and closed (like Jordan Curve). The curve can be convex or concave. If the curve is convex, I can draw a line passing through the point and test if the two cross points are on the same side or on both sides of the point. But it does not work when the curve is concave. Any comments and suggestions are welcome. Ben === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? Ben I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. Here at the Sortov Institute we worked out a method involving _three_ rays, which avoids any hassle when the curve meets a ray without crossing it, or meets a ray infinitely many times. I'll sketch the idea in a finite discrete setting, such as we see in computer graphics. Our curve is a mapping f: Z to ZxZ-{(0,0)} which is periodic with period p, say, such that |f(x+1) - f(x)|^2 is 0,1, or 2 for any x (continuity). We want the winding number, with respect to the origin, of the restriction of f to [0,p]. Define three rays from the origin: A = {(x,0) | x > 0} B = {(0,x) | x > 0} C = {(x,0) | x < 0} If f misses all three rays, the winding number is zero. Suppose otherwise. We will define a mapping k of Z into the group {0,1,2} of integers mod 3. For any z in Z such that f(z) is on one of the rays, write k(z)={ 0 if f(z) in A 1 if f(z) in B 2 if f(z) in C. For any z in Z such that f(z) is not in any of the rays, let y(z) be the greatest integer Ben >> I don't know how to determine if a point is inside a curve. The curve >> is simple and closed (like Jordan Curve). The curve can be convex or >> concave. > Here at the Sortov Institute we worked out a method involving _three_ > rays, which avoids any hassle when the curve meets a ray without crossing > it, or meets a ray infinitely many times. I'll sketch the idea in a finite > discrete setting, such as we see in computer graphics. > Our curve is a mapping > f: Z to ZxZ-{(0,0)} > which is periodic with period p, say, such that > |f(x+1) - f(x)|^2 is 0,1, or 2 for any x (continuity). > We want the winding number, with respect to the origin, of the restriction > of f to [0,p]. > Define three rays from the origin: > A = {(x,0) | x > 0} > B = {(0,x) | x > 0} > C = {(x,0) | x < 0} > If f misses all three rays, the winding number is zero. Suppose otherwise. > We will define a mapping k of Z into the group {0,1,2} of integers mod 3. > For any z in Z such that f(z) is on one of the rays, write > k(z)={ > 0 if f(z) in A > 1 if f(z) in B > 2 if f(z) in C. > For any z in Z such that f(z) is not in any of the rays, let y(z) be the > greatest integer one of the rays, and define k(z)=k(y(z)). > k is now defined everywhere on Z. Write > h(z) = k(z)-k(z-1) (an element of the group {0,1,2} of 3 elements). I slipped here. That should say an element of Z. > Add up the sum h(1)+h(2)+...+h(p) _without_ reducing mod 3, and the result > is 3W where W is the winding number desired. W may be postive, negative, > or zero; no problem. With some effort, this method can be adapted for the more theoretical > problem of an arbitrary continuous curve in the plane (continuous in the > usual sense). As usual, the notion of compactness serves as a substitute > for the notion of finiteness. LH The idea is pretty simple. You travel once around the curve, and when you meet a ray different from the most recent ray you met, you increment or decrement the winding number by one-third; increment for A then B or B then C or C then A, and decrement for the reverse cases. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. If the curve is convex, I can draw a line passing through the point and > test if the two cross points are on the same side or on both sides of > the point. But it does not work when the curve is concave. Any comments and suggestions are welcome. Everything is just here, including free software and source code: http://hdebruijn.soo.dto.tudelft.nl/www/grondig/crossing.htm Han de Bruijn === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. Let the curve C be f(x,y) = 0 and the given point be P(x1,y1).Then f(x1,y1) = 0, > 0,< 0 depending on whether P is on, outside or inside of C.Implementable by a function switch in computer code. HTH === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. Let the curve C be f(x,y) = 0 and the given point be P(x1,y1).Then > f(x1,y1) = 0, > 0,< 0 depending on whether P is on, outside or inside > of C.Implementable by a function switch in computer code. HTH This doesn't necessarily work. For example, suppose f(x,y) = (x^2 + y^2 - 1)^2. David === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? >> I don't know how to determine if a point is inside a curve. The curve >> is simple and closed (like Jordan Curve). The curve can be convex or >> concave. >> Let the curve C be f(x,y) = 0 and the given point be P(x1,y1).Then >> f(x1,y1) = 0, > 0,< 0 depending on whether P is on, outside or inside >> of C.Implementable by a function switch in computer code. HTH This doesn't necessarily work. >For example, suppose f(x,y) = (x^2 + y^2 - 1)^2. Well, there's a (reasonable) sense in which for that f, f^{-1}(0) is not simple: it's double, in fact! Lee Rudolph === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. If the curve is convex, I can draw a line passing through the point and > test if the two cross points are on the same side or on both sides of > the point. But it does not work when the curve is concave. Any comments and suggestions are welcome. > Ben Let the curve C be f(x,y) = 0 and the given point be P(x1,y1).Then f(x1,y1) = 0, > 0,< 0 depending on whether P is on, outside or inside of C. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? days. My association with the Department is that of an alumnus. I don't know how to determine if a point is inside a curve. The curve >is simple and closed (like Jordan Curve). The curve can be convex or >concave. If the curve is convex, I can draw a line passing through the point and >test if the two cross points are on the same side or on both sides of >the point. But it does not work when the curve is concave. Any comments and suggestions are welcome. Let x be the point you are wondering about. Pick any point which is definitely outside the curve, and call it p. Draw a line from p to x. If the line crosses the curve an odd number of times, then the point x is inside the curve. If it crosses the curve an even number of times, then x is outside the curve. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? I don't know how to determine if a point is inside a curve. The curve >is simple and closed (like Jordan Curve). The curve can be convex or >concave. >If the curve is convex, I can draw a line passing through the point and >test if the two cross points are on the same side or on both sides of >the point. But it does not work when the curve is concave. This is a well-known result, but the original poster wanted a method that could be implamanted in a computer program. There might be difficulties in findind a point which is certain to be outside the curve algorithmically. >Any comments and suggestions are welcome.Let x be the point you are wondering about. Pick any point which is definitely outside the curve, and call it > p. Draw a line from p to x. If the line crosses the curve an odd > number of times, then the point x is inside the curve. If it crosses > the curve an even number of times, then x is outside the curve. -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin > magidin-at-member-ams-org === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > ...the original poster wanted a method > that could be implamanted in a computer program. There might be > difficulties in findind a point which is certain to be outside the > curve algorithmically. If the closed curve is given algorithmically, its perpendicular projection on some line, say on a line parallel to one of the coordinate axes, should be an interval. The endpoints of that interval should then be algorithmically determinable, at least roughly. This would allow finding of points on that line are reasonably certain to be outside the closed curve. This assumes that the curve is, saY, a continuous image of the real interval [0,1] with the points for 0 and 1 coinciding. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > Let x be the point you are wondering about. Pick any point which is definitely outside the curve, and call it > p. Draw a line from p to x. If the line crosses the curve an odd > number of times, then the point x is inside the curve. If it crosses > the curve an even number of times, then x is outside the curve. Hehe! This has got to be the cutest proof I've seen the past month in sci.math! Still growing and learning, I guess.. :-) > Arturo Magidin -- Ioannis ------- The best way to predict reality, is to know exactly what you DON'T want. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? days. My association with the Department is that of an alumnus. > Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it >> p. Draw a line from p to x. If the line crosses the curve an odd >> number of times, then the point x is inside the curve. If it crosses >> the curve an even number of times, then x is outside the curve. Hehe! This has got to be the cutest proof I've seen the past month in >sci.math! Well, it's not really a proof. It's an assertion. (-: But then, the original poster only asked for a way to figure it out, not for a proof that the method works. I confess I do not know if I would be able to write out a formal proof of the assertion; I believe it follows from Jordan's Curve theorem, though. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? >> Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it >> p. Draw a line from p to x. If the line crosses the curve an odd >> number of times, then the point x is inside the curve. If it crosses >> the curve an even number of times, then x is outside the curve. >Hehe! This has got to be the cutest proof I've seen the past month in >sci.math! Well, it's not really a proof. It's an assertion. (-: wrong word from acquired speed. See .sig :-) > Arturo Magidin -- Ioannis ------- The best way to predict reality, is to know exactly what you DON'T want. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? <1165252705.325483@athprx04> > Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it > p. Draw a line from p to x. If the line crosses the curve an odd > number of times, then the point x is inside the curve. If it crosses > the curve an even number of times, then x is outside the curve. >>Hehe! This has got to be the cutest proof I've seen the past month in >>sci.math! Well, it's not really a proof. It's an assertion. (-: But then, the > original poster only asked for a way to figure it out, not for a proof > that the method works. I confess I do not know if I would be able to > write out a formal proof of the assertion; I believe it follows from > Jordan's Curve theorem, though. Also, aside from it not being a proof, there are other issues which can arise (of which I expect Arturo was aware). For example, how do you decide whether the line crosses the curve? You have to be able to distinguish crossing from intersecting tangentially without crossing. (Consider the x-axis and the curves y=x^2 vs. y=x^3.) What if the boundary curve has a straight line segment for part of it, and what if your line coincides with that line segment? How do you count that? And what if your line crosses the curve infinitely many times (as it might if you had something like the topologists' sine curve)? For a reasonable boundary curve and a generic point, these issues shouldn't arise, so it's a good method. Just like Arturo, though, I don't want to offer an actual proof. If I were looking for an elementary argument, I would try methods like those in Henle's proof of the Jordan curve theorem, in his book A combinatorial introduction to topology. -- J. H. Palmieri Associate Professor of Mathematics University of Washington Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu http://www.math.washington.edu/~palmieri/ === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? <1165252705.325483@athprx04> I am going to implement Arturo's method which is close to the method described in http://www.acm.org/tog/editors/erich/ptinpoly/ Could it be easier in implementation to count total number of intersections. For example I can draw a horizontal line passing through the point and count the total intersections. If the number is even, the point is inside. Is it equivelent to Arturo's method? -------------------------- What if the boundary curve has a straight line segment for part of it, and what if your line coincides with that line segment? --------------------------- This could cause problems because my curve is basically a polygon. I will add some code to avoid this. Ben > Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it > p. Draw a line from p to x. If the line crosses the curve an odd > number of times, then the point x is inside the curve. If it crosses > the curve an even number of times, then x is outside the curve. >>Hehe! This has got to be the cutest proof I've seen the past month in >>sci.math! > Well, it's not really a proof. It's an assertion. (-: But then, the > original poster only asked for a way to figure it out, not for a proof > that the method works. I confess I do not know if I would be able to > write out a formal proof of the assertion; I believe it follows from > Jordan's Curve theorem, though. Also, aside from it not being a proof, there are other issues which > can arise (of which I expect Arturo was aware). For example, how do > you decide whether the line crosses the curve? You have to be able to > distinguish crossing from intersecting tangentially without crossing. > (Consider the x-axis and the curves y=x^2 vs. y=x^3.) What if the > boundary curve has a straight line segment for part of it, and what if > your line coincides with that line segment? How do you count that? > And what if your line crosses the curve infinitely many times (as it > might if you had something like the topologists' sine curve)? For a reasonable boundary curve and a generic point, these issues > shouldn't arise, so it's a good method. Just like Arturo, though, I > don't want to offer an actual proof. If I were looking for an > elementary argument, I would try methods like those in Henle's proof > of the Jordan curve theorem, in his book A combinatorial introduction > to topology. -- > J. H. Palmieri > Associate Professor of Mathematics University of Washington > Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu > http://www.math.washington.edu/~palmieri/ === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I am going to implement Arturo's method which is close to the method > described in > http://www.acm.org/tog/editors/erich/ptinpoly/ Could it be easier in implementation to count total number of > intersections. For example I can draw a horizontal line passing through > the point and count the total intersections. If the number is even, the > point is inside. Is it equivelent to Arturo's method? -------------------------- > What if the boundary curve has a straight line segment for part of it, > and what if your line coincides with that line segment? > --------------------------- This could cause problems because my curve is basically a polygon. I > will add some code to avoid this. This _will_ cause problems. And you must not add code, you must design your code from the very start in such a way that such problems shall not occur: http://hdebruijn.soo.dto.tudelft.nl/www/grondig/crossing.htm Han de Bruijn === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? <1165252705.325483@athprx04> described in > http://www.acm.org/tog/editors/erich/ptinpoly/ Could it be easier in implementation to count total number of > intersections. For example I can draw a horizontal line passing through > the point and count the total intersections. If the number is even, the > point is inside. Is it equivelent to Arturo's method? -------------------------- > What if the boundary curve has a straight line segment for part of it, > and what if your line coincides with that line segment? > --------------------------- This could cause problems because my curve is basically a polygon. I > will add some code to avoid this. If your curve is a polygon, the winding number method proposed by Ben Rudiak-Gould is better and quite easy to implement: The given point p=(x0,y0) splits the plane (without the point) into four disjoint sets: TR={(x,y): x>x0 & y>=y0} TL={(x,y): x<=x0 & y>y0} BL={(x,y): x=x0 & y <1165252705.325483@athprx04> described in > http://www.acm.org/tog/editors/erich/ptinpoly/ Could it be easier in implementation to count total number of > intersections. For example I can draw a horizontal line passing through > the point and count the total intersections. If the number is even, the > point is inside. Is it equivelent to Arturo's method? You need to use a horizontal ray starting at the point, not a horizontal line passing through the point. Otherwise, consider a circle centered at the origin, and consider the point (2,0), which is outside the circle. The x-axis goes through the point and intersects the curve in two points, which is even. On the other hand, if you consider the point (0,0), inside the circle, the x-axis goes through the point and intersects the curve in two points, which is even. With a ray starting at the point, an odd number of intersections means it's inside. > -------------------------- > What if the boundary curve has a straight line segment for part of it, > and what if your line coincides with that line segment? > --------------------------- This could cause problems because my curve is basically a polygon. I > will add some code to avoid this. For example, just compute the slopes of all of your boundary edges, and then test using lines (well, rays) with slopes not equal to any of those edges. Ben Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it >> p. Draw a line from p to x. If the line crosses the curve an odd >> number of times, then the point x is inside the curve. If it crosses >> the curve an even number of times, then x is outside the curve. >>Hehe! This has got to be the cutest proof I've seen the past month in >sci.math! >> Well, it's not really a proof. It's an assertion. (-: But then, the >> original poster only asked for a way to figure it out, not for a proof >> that the method works. I confess I do not know if I would be able to >> write out a formal proof of the assertion; I believe it follows from >> Jordan's Curve theorem, though. >> Also, aside from it not being a proof, there are other issues which >> can arise (of which I expect Arturo was aware). For example, how do >> you decide whether the line crosses the curve? You have to be able to >> distinguish crossing from intersecting tangentially without crossing. >> (Consider the x-axis and the curves y=x^2 vs. y=x^3.) What if the >> boundary curve has a straight line segment for part of it, and what if >> your line coincides with that line segment? How do you count that? >> And what if your line crosses the curve infinitely many times (as it >> might if you had something like the topologists' sine curve)? >> For a reasonable boundary curve and a generic point, these issues >> shouldn't arise, so it's a good method. Just like Arturo, though, I >> don't want to offer an actual proof. If I were looking for an >> elementary argument, I would try methods like those in Henle's proof >> of the Jordan curve theorem, in his book A combinatorial introduction >> to topology. -- J. H. Palmieri Associate Professor of Mathematics University of Washington Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu http://www.math.washington.edu/~palmieri/ === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? >> Let x be the point you are wondering about. >> Pick any point which is definitely outside the curve, and call it >> p. Draw a line from p to x. If the line crosses the curve an odd >> number of times, then the point x is inside the curve. If it crosses >> the curve an even number of times, then x is outside the curve. >Hehe! This has got to be the cutest proof I've seen the past month in >sci.math! Well, it's not really a proof. It's an assertion. (-: But then, the > original poster only asked for a way to figure it out, not for a proof > that the method works. I confess I do not know if I would be able to > write out a formal proof of the assertion; I believe it follows from > Jordan's Curve theorem, though. One must be concerned about the possibility that the line is tangent to the curve someplace. Of course, the probability of that is low. But this is intended for computer application (using floating-point arithmetic, I assume) and so, due to inherent inaccuracies, there will occur, although infrequently, instances when the point is judged to be inside the curve when it's really outside, and vice versa. David === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. One way to do it is to add up the differential changes in the angle the curve makes with the point and some reference line through the point. The sum will be 0 if the point is outside the curve and 360 degrees if it's inside. More specifically, if the curve is given by x(t) and y(t), then integral d/dt arctan (y(t)/x(t)) dt will be zero if the origin is outside the curve and +/- 2pi if it's inside. The above is equivalent to integral (x(t) y'(t) - y(t) x'(t)) / (x(t)^2 + y(t)^2) dt -- Ben === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? understanding is wrong. Starting from a reference point on the curve, an ant walks along the curve in one direction. It adds up the angles between the reference line and the line passing through itself and the testing point. The angle should be 360 degree if the point is inside the curve when the ant walks back to the starting reference point. I need to implement this on computer. How to deal with the numerical errors. I am afraid the errors will be accumulative. Polygon - Ray Method' in this group, but I lost in the middle of the discussion when they talked about the winding number and infinite crossing. Ben > I don't know how to determine if a point is inside a curve. The curve > is simple and closed (like Jordan Curve). The curve can be convex or > concave. One way to do it is to add up the differential changes in the angle the > curve makes with the point and some reference line through the point. The > sum will be 0 if the point is outside the curve and 360 degrees if it's > inside. More specifically, if the curve is given by x(t) and y(t), then integral d/dt arctan (y(t)/x(t)) dt will be zero if the origin is outside the curve and +/- 2pi if it's inside. > The above is equivalent to integral (x(t) y'(t) - y(t) x'(t)) / (x(t)^2 + y(t)^2) dt -- Ben === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > understanding is wrong. Starting from a reference point on the curve, an ant walks along the > curve in one direction. It adds up the angles between the reference > line and the line passing through itself and the testing point. The > angle should be 360 degree if the point is inside the curve when the > ant walks back to the starting reference point. I need to implement this on computer. How to deal with the numerical > errors. I am afraid the errors will be accumulative. Polygon - Ray Method' in this group, but I lost in the middle of the > discussion when they talked about the winding number and infinite > crossing. The winding number method and the crossing number method are equivalent, but the winding number method is far more _expensive_ than the crossing number method. Han de Bruijn === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > I need to implement this on computer. How to deal with the > numerical errors. I am afraid the errors will be accumulative. This is sci.math, for pure-math theoretical responses. If you want advice on algorithms for computer, you should post to comp.programming instead. When you do, please tell what representation you are (already) using for the closed curve. For example, if it's the zero-set of a function of two variables, then the answer is trivial. If it's expressed parametrically, then I can tell you how to avoid accumulated errors using the bug-crawl angle method somebody else described mathematically. If your representation for the Jordan curve is something else, I'd like to know what! === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? I need to implement this on computer. How to deal with the > numerical errors. I am afraid the errors will be accumulative. This is sci.math, for pure-math theoretical responses. If you want > advice on algorithms for computer, you should post to > comp.programming instead. When you do, please tell what > representation you are (already) using for the closed curve. For > example, if it's the zero-set of a function of two variables, then > the answer is trivial. If it's expressed parametrically, then I can > tell you how to avoid accumulated errors using the bug-crawl angle > method somebody else described mathematically. If your > representation for the Jordan curve is something else, I'd like to > know what! Mathematically, the inquiry is about the index of a point with respect to a closed curve (on which the point does not lie); intuitively, how many times the curve ran around the point. For a simple closed curve oriented counterclockwise, the index will be +1 for points inside, and 0 outside. Formula: let (x(t), y(t)) be a parametrization of the piecewise smooth curve C for y in [t1,t2], then the index of the point (a,b) is 1/(2*pi) * int[over C] ((x-a)*dy-(y-b)*dx)/((x-a)^2+(y-b)^2) (It is the real part of (1/2*pi*i)*int[over C] dz/(z-(a+i*b))) For C a polygon, there is a MATLAB program I put together years ago: (No math after this point) % runs.m calculates the index of a point P with respect % to a closed polygonal path given by n points. % ******> Input: XY , an n-by-2 matrix XY (the rows % defining the points), % P, a 2-component vector. % ******> Output: r, the net number of runs of XY around P % It is assumed that the last point is connected % to the first point. % ******> Call: r=runs(XY,P) % By Z.V. Kovarik, May 3, 1996 % function r=runs(XY,P); i=sqrt(-1); P=P(1)+i*P(2); Z=XY(:,1)+i*XY(:,2); n=length(Z); Z=[Z;Z(1)]-P; % closing up the path Z=Z(1+(1:n))./Z(1:n); % complex rotations - unscaled r=round(sum(atan(imag(Z)./(abs(Z)+real(Z))))/pi); === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? > understanding is wrong. Starting from a reference point on the curve, an ant walks along the > curve in one direction. It adds up the angles between the reference > line and the line passing through itself and the testing point. The > angle should be 360 degree if the point is inside the curve when the > ant walks back to the starting reference point. I need to implement this on computer. How to deal with the numerical > errors. I am afraid the errors will be accumulative. If you keep the total error < Pi, then you'll have a definitive answer. === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? curve in one direction. It adds up the angles between the reference > line and the line passing through itself and the testing point. The > angle should be 360 degree if the point is inside the curve when the > ant walks back to the starting reference point. If an ant moves anticlockwise around the closed path, the interior is always on the left and the outside on the right. Now draw a line through your point in question until it crosses the curve ( first time). The ant will notice if this line is coming from the left (interior) or not. Wit friendly greetings Hero === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? understanding is wrong. > Starting from a reference point on the curve, an ant walks along the > curve in one direction. It adds up the angles between the reference > line and the line passing through itself and the testing point. The > angle should be 360 degree if the point is inside the curve when the > ant walks back to the starting reference point. > If an ant moves anticlockwise around the closed path, the interior is > always on the left and the outside on the right. > Now draw a line through your point in question until it crosses the > curve ( first time). The ant will notice if this line is coming from > the left (interior) or not. > It seems, some can not translate this into math. So, transform Your equation of the the closed curve into a parametric one. A line from the point in question to the curve (where it crosses the first time) has a direction in this point of intersection. So has the parametric curve to.These are given by the first derivates. The vector product of these let You determine by it's sign, whether the point in question is to the left ( inside) or not. With friendly greetings Hero === === Subject: : Re: Given a closed curve, how to determine if a piont is inside the curve? understanding is wrong. > Starting from a reference point on the curve, an ant walks along the > curve in one direction. It adds up the angles between the reference > line and the line passing through itself and the testing point. The > angle should be 360 degree if the point is inside the curve when the > ant walks back to the starting reference point. > If an ant moves anticlockwise around the closed path, the interior is > always on the left and the outside on the right. > Now draw a line through your point in question until it crosses the > curve ( first time). The ant will notice if this line is coming from > the left (interior) or not. > It seems, some can not translate this into math. So, > transform Your equation of the the closed curve into a parametric one. > A line from the point in question to the curve (where it crosses the > first time) has a direction in this point of intersection. So has the > parametric curve to.These are given by the first derivates. The vector > product of these let You determine by it's sign, whether the point in > question is to the left ( inside) or not. With friendly greetings > Hero Possibly useful: http://cgafaq.info/wiki/Point_in_polygon A polygon is not a smooth curve, but perhaps approximations are sufficient or perhaps the OP has the curve data already translated into polygons. === === Subject: : Re: Calculate Gibbs phenomenon (how to?) <456daeb2.779893534@news-south.giganews.com> <19ktm2hm1ra14vf7l1u0kj3t4dlkbfin5i@4ax.com> <88mtm2l9ahi1c62rgb156fiqa52ba3k3o7@4ax.com> <9b60n253tp0cmmjjnh03ch5up496tcgev7@4ax.com> I guess I need to take a look in the book (Folland Real Analysis) you are recommending in order to fully understand why the limit is 9 percent. b_n correct. Hopefully this thread will help others trying to solve Gibbs phenomenon too. All the best, Badger skrev: > On Fri, 01 Dec 2006 06:01:33 -0600, David C. Ullrich >On Thu, 30 Nov 2006 08:58:21 -0500, Badger >On Thu, 30 Nov 2006 06:43:42 -0600, David C. Ullrich >>An example of a fourier series that I am trying to prove Gibbs' 9% with >>is: >>-x^2-4x, -4 < x < 0 >>x, 0 < x < 4 >>This for instance gives me e.g. an a0 = 14/3, >>and I can calculate that the it has a max discontinuity close to 4,36 >>(which is the 9%). >>Maybe I finally see what you've been talking about when you >ask about trig identities. For certain simple functions one >can use a trig identity to find a simple form for the n-th >partial sum of the Fourier series. But you shouldn't expect >that to happen for all functions - for most functions >trig identities are no help in simplifying the n-th >partial sum. >>Hopefully the post by David Ullrich will serve to point you in the >>right direction. >>I do not know where you got the function you gave as an example, or >>what you hope to show with it, but here's what I got for the >>coefficients (which I believe are correct, but be warned that I might >>have gone wrong somewhere): > -x^2-4x, -4 < x < 0 > x, 0 < x < 4 >> a_0 = 14/3 >> a_n = -4/(n^2 pi^2) [ 3 cos(n pi) + 5 ] >> b_n = -4/(n^3 pi^3) [ (n^2 pi^2 - 8) cos(n pi) + 8 ] >>Where you go from here I don't know, but good luck! >I suspect that proving a general result, for example for >functions of bounded variation with a jump discontinuity, >will be no harder than trying to work out the details >starting with those coefficients... Yes, and probably like you, I was wondering what the OP hoped to show > by analyzing some particular function(s). One reason I posted the coefficients, JIC this is coursework of some > sort, was so the OP could check his results. Another, more important > reason, was to underscore your point that, in general, they're not > generally nice. Hopefully your post has provided the OP with a more useful question to > consider. > >************************ >David C. Ullrich === === Subject: : Re: Calculate Gibbs phenomenon (how to?) >I guess I need to take a look in the book (Folland Real Analysis) you >are recommending in order to fully understand why the limit is 9 >percent. b_n correct. Hopefully this thread will help others trying to solve Gibbs phenomenon >too. I think there are better ways to explain it than what Folland does. But first you need to know something about the _theory_ of Fourier series. Ask again about the Gibbs phenomenon when the following sentence makes perfect sense to you: The n-th partial sum of the Fourier series of f is given by the convolution of f with the Diriclet kernel D_n. >All the best, Badger skrev: >> On Fri, 01 Dec 2006 06:01:33 -0600, David C. Ullrich >>On Thu, 30 Nov 2006 08:58:21 -0500, Badger >On Thu, 30 Nov 2006 06:43:42 -0600, David C. Ullrich >>An example of a fourier series that I am trying to prove Gibbs' 9% with is: -x^2-4x, -4 < x < 0 x, 0 < x < 4 >This for instance gives me e.g. an a0 = 14/3, and I can calculate that the it has a max discontinuity close to 4,36 (which is the 9%). >>Maybe I finally see what you've been talking about when you >>ask about trig identities. For certain simple functions one >>can use a trig identity to find a simple form for the n-th >>partial sum of the Fourier series. But you shouldn't expect >>that to happen for all functions - for most functions >>trig identities are no help in simplifying the n-th >>partial sum. >>Hopefully the post by David Ullrich will serve to point you in the >right direction. >>I do not know where you got the function you gave as an example, or >what you hope to show with it, but here's what I got for the >coefficients (which I believe are correct, but be warned that I might >have gone wrong somewhere): -x^2-4x, -4 < x < 0 >> x, 0 < x < 4 >> a_0 = 14/3 >> a_n = -4/(n^2 pi^2) [ 3 cos(n pi) + 5 ] >> b_n = -4/(n^3 pi^3) [ (n^2 pi^2 - 8) cos(n pi) + 8 ] >>Where you go from here I don't know, but good luck! >>I suspect that proving a general result, for example for >>functions of bounded variation with a jump discontinuity, >>will be no harder than trying to work out the details >>starting with those coefficients... >> Yes, and probably like you, I was wondering what the OP hoped to show >> by analyzing some particular function(s). >> One reason I posted the coefficients, JIC this is coursework of some >> sort, was so the OP could check his results. Another, more important >> reason, was to underscore your point that, in general, they're not >> generally nice. >> Hopefully your post has provided the OP with a more useful question to >> consider. >> >>************************ >>David C. Ullrich ************************ David C. Ullrich === === Subject: : Re: ZFC in 4 Axioms. When it is said that NBG is a conservative extension of ZF(C) what is > meant is that NBG when formulated using two different kinds of > variables - one for sets and one for classes - is a conservative > extension of ZF(C). > If we formulate NBG without using two different > kinds of variables the result is, as you note, not a conservative > extension of ZF(C) in the technical sense. > theory from a 2-sorted language into a 1-sorted language, then the > consverativism would not technically carry over (or maybe I should have > surmised that for myself)! I've ACTUALLY been losing sleep over this. I > thought that I might have to become a crank, Hmmm..., Moe Blee being a crank, that's interesting. Moe according to > your speach everybody in this forum is a crank, including the people > crank, Arturo Magidin is a crank, etc,etc.... Wrong. > arguing with you and the > ghost of Torkel Franzen for the next fifteen years that mathematicians > have it all wrong since NBG is not a conservative extension of ZF! So > this is a relief to find that I am not disputed in the very particular > sense I presented. > Even then NBG is morally a > conservative extension since we can, as Stephen suggests, define a > predicate Set in the language of NBG so that all theorems of ZF are > theorems of NBG when relativized to the predicate Set and conversely. > Okay, I can picture that. My picture will be sharper as eventually I > investigate more. > MoeBlee === === Subject: : Re: ZFC in 4 Axioms. > hmm..., then I should go read this book you told me to read, but can > you help me get it? > If you are in usa then you can perchace it at the site I showed it once > in this forum, and I can give u my mailing address, and I will pay > back, to your account. I cant' do that. MoeBlee === === Subject: : Re: ZFC in 4 Axioms. > >> Axiom 2) (ExAy yex<->P(y))<->x=/=y The variables x and y are bound on the left hand side and free on the > right hand side. This axiom does not say what you think it says. Just curious: what do you think that he intends Axiom 2 to say? -- Herman Jurjus === === Subject: : Re: ZFC in 4 Axioms. > Axiom 2) (ExAy yex<->P(y))<->x=/=y >> The variables x and y are bound on the left hand side and free on the >> right hand side. This axiom does not say what you think it says. Just curious: what do you think that he intends Axiom 2 to say? I haven't the foggiest idea. But I'm pretty confident he didn't mean to leave x and y free on the right hand side while bound on the left. -- Puts his arm around you, fiddles with your hair. You know, and he says, come on, you know, just because you like a bit of a kiss and a cuddle with another man doesn't make you gay. Which, you know, I've thought a lot about. But I think it does. I think it does. --- The Office (interviews) === === Subject: : Re: ZFC in 4 Axioms. > Just curious: what do you think that he intends Axiom 2 to say? I would guess he wants something like ExAy(y in x iff P(y) & y =/= x). -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === === Subject: : Re: ZFC in 4 Axioms. > I just occured to my mind, that ZFC can be reduced to four axioms as > below: > 1)Extensionality 2) Comprehension 3)Infinity 4)AC. > 1),2),3) are as in the traditional presentation of ZFC. > Axiom 2) (ExAy yex<->P(y))<->x=/=y > I am not sure of the above. > Then > (ExAy yex<->P(y))<->~x=y > should rather be written as either > (ExAy (yex<->P(y)) <-> ~u=v) > with free variables u and v (and whatever is free in P(y) besides x and > y) > or > Ex (Ay (yex<->P(y)) <-> ~x=v) > with free variables u and whatever is free in P(y) besides x and y > or > ExAy ((yex<->P(y)) <-> ~x=y) Yes, this is the one. > with no free variables, except what is free in P(y) apart from x and y. > Of course, all these variants have different meaning. Then let P(y):= y=y (about the simplest choice possible). Then your axiom scheme claims ExAy ((yex<->y=y) <-> ~x=y) ExAy ((yex<->T) <-> ~x=y) ExAy (yex <-> ~x=y) There is a set that contains all sets except itself. Call this set A. Now, since you also claimed that given sets a,b there is a set (denoted {a,b}) having exactly a and b as elements xe{a,b} <-> (x=a v x=b) (the proof was not conclusive, but the fact should be true anyway), and also we should have that, given a set a, the union Ua of all its elements exists, xeUa <-> Ez(xez & zea) we conclude that there is a set V, V := U{{A,A},A} apparently the set of all sets !? Especially, VeV which you thought you had proved to be wrong. === === Subject: : Re: ZFC in 4 Axioms. > I just occured to my mind, that ZFC can be reduced to four axioms as > below: > 1)Extensionality 2) Comprehension 3)Infinity 4)AC. > 1),2),3) are as in the traditional presentation of ZFC. > Axiom 2) (ExAy yex<->P(y))<->x=/=y > I am not sure of the above. > Then > (ExAy yex<->P(y))<->~x=y > should rather be written as either > (ExAy (yex<->P(y)) <-> ~u=v) > with free variables u and v (and whatever is free in P(y) besides x and > y) > or > Ex (Ay (yex<->P(y)) <-> ~x=v) > with free variables u and whatever is free in P(y) besides x and y > or > ExAy ((yex<->P(y)) <-> ~x=y) > Yes, this is the one. > with no free variables, except what is free in P(y) apart from x and y. > Of course, all these variants have different meaning. Then let P(y):= y=y (about the simplest choice possible). > Then your axiom scheme claims > ExAy ((yex<->y=y) <-> ~x=y) > ExAy ((yex<->T) <-> ~x=y) > ExAy (yex <-> ~x=y) > There is a set that contains all sets except itself. Call this set A. > Now, since you also claimed that given sets a,b there is a set (denoted > {a,b}) having exactly a and b as elements > xe{a,b} <-> (x=a v x=b) > (the proof was not conclusive, but the fact should be true anyway), and > also we should have that, given a set a, the union Ua of all its > elements exists, > xeUa <-> Ez(xez & zea) > we conclude that there is a set V, > V := U{{A,A},A} > apparently the set of all sets !? Especially, VeV which you thought you > had proved to be wrong. you mean V=U{{A},A} yea apparentlly this is true. but this should be checked by comprehesion. what do I mean is that union and pairing are theorums in this theory, while comprehension is an axiom. so union and pairing should be designed to avoid this. Zuhair === === Subject: : Re: ZFC in 4 Axioms. > I just occured to my mind, that ZFC can be reduced to four axioms as > below: > 1)Extensionality 2) Comprehension 3)Infinity 4)AC. > 1),2),3) are as in the traditional presentation of ZFC. > Axiom 2) (ExAy yex<->P(y))<->x=/=y > I am not sure of the above. > Then > (ExAy yex<->P(y))<->~x=y > should rather be written as either > (ExAy (yex<->P(y)) <-> ~u=v) > with free variables u and v (and whatever is free in P(y) besides x and > y) > or > Ex (Ay (yex<->P(y)) <-> ~x=v) > with free variables u and whatever is free in P(y) besides x and y > or > ExAy ((yex<->P(y)) <-> ~x=y) > Yes, this is the one. > with no free variables, except what is free in P(y) apart from x and y. > Of course, all these variants have different meaning. > Then let P(y):= y=y (about the simplest choice possible). > Then your axiom scheme claims > ExAy ((yex<->y=y) <-> ~x=y) > ExAy ((yex<->T) <-> ~x=y) > ExAy (yex <-> ~x=y) > There is a set that contains all sets except itself. Call this set A. > Now, since you also claimed that given sets a,b there is a set (denoted > {a,b}) having exactly a and b as elements > xe{a,b} <-> (x=a v x=b) > (the proof was not conclusive, but the fact should be true anyway), and > also we should have that, given a set a, the union Ua of all its > elements exists, > xeUa <-> Ez(xez & zea) > we conclude that there is a set V, > V := U{{A,A},A} > apparently the set of all sets !? Especially, VeV which you thought you > had proved to be wrong. you mean V=U{{A},A} I had not yet defined {A}. But let us define {A} to be the set havingexactly A as element xe{a} <-> x=a By Extension, this is the same as {A,A}. yea apparentlly this is true. but this should be checked by > comprehesion. what do I mean is that union and pairing are theorums in