mm-140 >My research can be dif'cult to understand, so I thought I'd try out>yet another way of explaining it. Some of you may have 'gured out>that I test out explanations on Usenet for use elsewhere, to re'ne my>own understanding, or just in case someone out there might 'nally get>it.Now then, again here's my discovery:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So far, no discovery. We agree on this part and it hasno particular signi'cance.>In that form it's hard to understand what follows next unless you pay>attention to what you have, speci'cally that cubic de'ning the a's.I can get it because of the symmetry of(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where I've gone ahead and substituted a_3(x) back in to replace>b_3(x), and it's important that you focus on that symmetry.It's that symmetry which allows the cubic(*) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)to de'ne ALL the a's, but something happens when I divide by 49.> Only if you divide by 49 in a certain way: factoring it as 7, 7, 1. There is *another* factorization which works asdesired when x <> 0.>Then the symmetry is broken. Physics jargon, used super'cially here to give theimpression that the writer sees and understands a pattern.I doubt anyone is either fooled or impressed by it.>Without that symmetry it's impossible to>'nd a SINGLE cubic to handle what results when you divide both sides>by 49.> False. There is a cubic. But it does not correspondto the 7, 7, 1 factorization. But ironically symmetry IS the key to all this: symmetry in the form of Galois permutations of the roots of irreducible polynomials. That is what tells you that if one of a_i(x) is non-coprime to 7, then they all are. And that, of course, tells you that yourfactorization of 49 as 7, 7, 1 is wrong, wrong, wrong wheneveryour polynomial in the a's is irreducible - which it is for almost all x.>That's important because it's why the functions are NOT algebraic>integer functions!!!> I think you have accepted this fact - that a_1(x)/7 is not an algebraic integer - which of course we pointedout months and months ago, and you fought tooth and nail fora very long time. But I bet you don't really understand the proof of it. As a test, why don't you explain to the folks here in your ownwords why it is true?>Now then, I'll recap. Symmetry allows the a's to be de'ned by a>cubic, which shows them to be algebraic integer functions, but>dividing by 49 *breaks* that symmetry, taking away the ability to 'nd>some cubic to de'ne the results, which proves that the resulting>functions are not algebraic integer functions.(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the b's are roots ofb^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.My point is that the second and third coef'cients are impossible to>de'ne in general.> If by impossible to de'ne in general you mean that they cannot be algebraic integers, I agree. That is because 7, 7, 1 is the wrong factorization.>You may 'nd them for some particular x, but in general, they are>forever hidden from you.Notice that doing that substitution with a_3(x) for b_3(x) gives me(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22but you have broken symmetry since the other constant terms are 1 and>1, so you're still stuck.> Right. b_1(x) and b_2(x) cannot be algebraic integers.We all agree on this. It comes back to your having made thewrong choice in factoring 49: 7, 7, 1 doesn't work. Somethingelse does.>Now by emphasizing what happens *after* 49 is divided from both sides>I'm trying to get at least some of you to face the mathematical>realities here, and I've made other posts pointing it out as well.> Only if you divide by 49 in the wrong way, as you keep insistingon doing. OK, here is another way to think about this. Consider your polynomialin a, a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice that the constant term always has a factor of 49.Basic Galois theory says that EACH of the roots must benon-coprime to 7. This is implies in particular that ifyou factor in the 7, 7, 1 pattern, then the term (5 a_3(x) + 7),if divided by 1, is going to have some bits of 7 left over.This is bad for good old 7, 7, 1 because in general, (2401 x^3 - 147 x^2 + 3x)is coprime to 7. This just cries out to you that 7, 7, 1 is the wrong factorization of 49 for your purposes - but *also* that there has to be a correct factorization, 49 = w1(x) * w2(x) * w3(x),where each of the w's is a nonunit and none of the w's equals 7, and each of the terms (5 a_i(x) + 7)/wi(x)is an algebraic integer. In particular, a_3(x)/w3(x) isan algebraic integer: a_3(x) is divisible by a piece of 7! After all, there IS the Magidin-Mckinnon theorem. Therelevance of that theorem here is that it says that for P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22,there must be a factorization of the form (5*c1(x) + d1(x))*(5*c2(x) + d2(x))*(5*c3(x) + d3(x))where c1, c2, c3, d1, d2, and d3 are algebraic integers.This factorization must be related to your original factorization for P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)of the form (5 a_1(x) + 7)*(5 a_2(x) + 7)*(5 a_3(x) + 7).It implies that there must exist w1(x), w2(x), w3(x)such that c1(x) = a_1(x)/w1(x), d1(x) = 7/w1(x), c2(x) = a_2(x)/w2(x), d2(x) = 7/w2(x), c3(x) = a_3(x)/w3(x), d3(x) = 7/w3(x),and w1(x)*w2(x)*w3(x) = 49. It all hangs together so nicely, so unlike the bad old 7, 7, 1factorization. Doesn't it cry out to you? Perhaps not. It would if you knew a little more math. A little side note here. You might perhaps have caught in theabove that d1(x)*d2(x)*d3(x) = 7. You might say, Hey, waita minute, doesn't d1(x)*d2(x)*d3(x) have to equal 22 ? Ha,Red Baron I mean Nora Baron, I got you this time!!! No such luck. This objection might make sense if you were factoring P(x) and P(x)/49 as polynomials in x. Oddly enough that is not what you are doing. This is going to sound strange because you have to go quite a ways back to understand it. You are factoringP(x) and P(x)/49 not as polynomials in x, but as polynomials in ... 5 !!! After all, in your *own* original factorization, (5 a_1(x) + 7)*(5 a_2(x) + 7)*(5 a_3(x) + 7),you might notice that 7*7*7 is not equal to 49*22, eh? Like I say,it all hangs together. >Courage in mathematics might sound new to you as an idea, but those>who came before you had courage. That's how they learned of sqrt(2),>and sqrt(-1), and much other mathematics that some refused, fought,>and bitterly attacked as they lacked that courage.Of course, the mathematics should win, assuming that humanity survives>long enough, but regardless each of you now faces an individual test>of your own courage. It's not yet in the history books this time,>where you can read and just imagine that you'd never have fought over>sqrt(2) or refused to accept sqrt(-1) as an imaginary number.> If you were right SOMEONE would notice it. It is not the casethat 100% of people who read your posts are either dimwits ordishonest. If some honest person noticed that you were right, we whooppose you would be proven wrong and humiliated. If you wereright it would take both enormous courage and foolhardiness to sayotherwise. That leaves you with two alternative conclusions:either we are ALL enormously courageous but dishonest fools, or you are wrong.>Here it's the present, and your test of courage is now.Oh, and be sure to check my blog archives: OK, checked. Your drivel there is much the same as yourdrivel here. Drivel is drivel. Sorry, I didn't sign up for anycontributions. Say, what is happening with those papers you submitted? About time they got back to you with the referees' reports, eh? I betseeing some of the thoughtful and pungently worded arguments you have posted here would help them decide on publication. Whynot send them a line and some examples? >And hey, if you're moping and miserable because mathematics tests you,>then maybe, if you think you're a mathematician, you might want to try>a different 'eld.> Well, hey, I am neither moping nor miserable. I have complete con'dence that I am right and you are wrong. It is just a matter of time before you see that 7, 7, 1 is the wrong factorization. Then everything goes down the drain. Nora B.James Harris =>My research can be dif'cult to understand, so I thought I'd try out>yet another way of explaining it. Some of you may have 'gured out>that I test out explanations on Usenet for use elsewhere, to re'ne my>own understanding, or just in case someone out there might 'nally get>it.Now then, again here's my discovery:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So far, no discovery. We agree on this part and it has> no particular signi'cance.>In that form it's hard to understand what follows next unless you pay>attention to what you have, speci'cally that cubic de'ning the a's.I can get it because of the symmetry of(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where I've gone ahead and substituted a_3(x) back in to replace>b_3(x), and it's important that you focus on that symmetry.It's that symmetry which allows the cubic(*) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)to de'ne ALL the a's, but something happens when I divide by 49.> Only if you divide by 49 in a certain way: factoring it as> 7, 7, 1. There is *another* factorization which works as> desired when x <> 0.>Then the symmetry is broken.> Physics jargon, used super'cially here to give the> impression that the writer sees and understands a pattern.> I doubt anyone is either fooled or impressed by it.>Without that symmetry it's impossible to>'nd a SINGLE cubic to handle what results when you divide both sides>by 49.> False. There is a cubic. But it does not correspond> to the 7, 7, 1 factorization. But ironically symmetry IS the key to all this: symmetry in the> form of Galois permutations of the roots of irreducible polynomials.> That is what tells you that if one of a_i(x) is non-coprime to 7,> then they all are. And that, of course, tells you that your> factorization of 49 as 7, 7, 1 is wrong, wrong, wrong whenever> your polynomial in the a's is irreducible - which it is for> almost all x.>That's important because it's why the functions are NOT algebraic>integer functions!!!> I think you have accepted this fact - that a_1(x)/7 is> not an algebraic integer - which of course we pointed> out months and months ago, and you fought tooth and nail for> a very long time. But I bet you don't really understand the proof of it. As a> test, why don't you explain to the folks here in your own> words why it is true?Now then, I'll recap. Symmetry allows the a's to be de'ned by a>cubic, which shows them to be algebraic integer functions, but>dividing by 49 *breaks* that symmetry, taking away the ability to 'nd>some cubic to de'ne the results, which proves that the resulting>functions are not algebraic integer functions.(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the b's are roots ofb^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.My point is that the second and third coef'cients are impossible to>de'ne in general.> If by impossible to de'ne in general you mean that they cannot be> algebraic integers, I agree. That is because 7, 7, 1 is the> wrong factorization.>You may 'nd them for some particular x, but in general, they are>forever hidden from you.Notice that doing that substitution with a_3(x) for b_3(x) gives me(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22but you have broken symmetry since the other constant terms are 1 and>1, so you're still stuck.> Right. b_1(x) and b_2(x) cannot be algebraic integers.> We all agree on this. It comes back to your having made the> wrong choice in factoring 49: 7, 7, 1 doesn't work. Something> else does.>Now by emphasizing what happens *after* 49 is divided from both sides>I'm trying to get at least some of you to face the mathematical>realities here, and I've made other posts pointing it out as well. Only if you divide by 49 in the wrong way, as you keep insisting> on doing.> OK, here is another way to think about this. Consider your polynomial> in a, a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice that the constant term always has a factor of 49.Oh Nora, Nora dude, how can you be so mean? This is blowing James' mind!He's going crazy here setting x's to zero trying to 'nd the constant term.No James it's true! The constant term of your polynomial in a is: - 49(2401 x^3 - 147 x^2 + 3x) !!Here is a _constant_ term of a polynomial that is a function of x!!! Itchanges when x changes!!!Your silent admirer,KeithK> Basic Galois theory says that EACH of the roots must be> non-coprime to 7. This is implies in particular that if> you factor in the 7, 7, 1 pattern, then the term (5 a_3(x) + 7), if divided by 1, is going to have some bits of 7 left over.> This is bad for good old 7, 7, 1 because in general, (2401 x^3 - 147 x^2 + 3x) is coprime to 7. This just cries out to you that 7, 7, 1 is the wrong> factorization of 49 for your purposes - but *also* that there> has to be a correct factorization, 49 = w1(x) * w2(x) * w3(x), where each of the w's is a nonunit and none of the w's> equals 7, and each of the terms (5 a_i(x) + 7)/wi(x) is an algebraic integer. In particular, a_3(x)/w3(x) is> an algebraic integer: a_3(x) is divisible by a piece of 7! After all, there IS the Magidin-Mckinnon theorem. The> relevance of that theorem here is that it says that for P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22, there must be a factorization of the form (5*c1(x) + d1(x))*(5*c2(x) + d2(x))*(5*c3(x) + d3(x)) where c1, c2, c3, d1, d2, and d3 are algebraic integers.> This factorization must be related to your original> factorization for P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)> of the form (5 a_1(x) + 7)*(5 a_2(x) + 7)*(5 a_3(x) + 7). It implies that there must exist w1(x), w2(x), w3(x)> such that c1(x) = a_1(x)/w1(x), d1(x) = 7/w1(x),> c2(x) = a_2(x)/w2(x), d2(x) = 7/w2(x),> c3(x) = a_3(x)/w3(x), d3(x) = 7/w3(x), and w1(x)*w2(x)*w3(x) = 49.> It all hangs together so nicely, so unlike the bad old 7, 7, 1> factorization. Doesn't it cry out to you? Perhaps not. It would if> you knew a little more math.> A little side note here. You might perhaps have caught in the> above that d1(x)*d2(x)*d3(x) = 7. You might say, Hey, wait> a minute, doesn't d1(x)*d2(x)*d3(x) have to equal 22 ? Ha,> Red Baron I mean Nora Baron, I got you this time!!! No such luck. This objection might make sense if you were factoring> P(x) and P(x)/49 as polynomials in x. Oddly enough that is not> what you are doing. This is going to sound strange because you> have to go quite a ways back to understand it. You are factoring> P(x) and P(x)/49 not as polynomials in x, but as polynomials in ... 5 !!! After all, in your *own* original factorization, (5 a_1(x) + 7)*(5 a_2(x) + 7)*(5 a_3(x) + 7), you might notice that 7*7*7 is not equal to 49*22, eh? Like I say,> it all hangs together.>Courage in mathematics might sound new to you as an idea, but those>who came before you had courage. That's how they learned of sqrt(2),>and sqrt(-1), and much other mathematics that some refused, fought,>and bitterly attacked as they lacked that courage.Of course, the mathematics should win, assuming that humanity survives>long enough, but regardless each of you now faces an individual test>of your own courage. It's not yet in the history books this time,>where you can read and just imagine that you'd never have fought over>sqrt(2) or refused to accept sqrt(-1) as an imaginary number. If you were right SOMEONE would notice it. It is not the case> that 100% of people who read your posts are either dimwits or> dishonest. If some honest person noticed that you were right, we who> oppose you would be proven wrong and humiliated. If you were> right it would take both enormous courage and foolhardiness to say> otherwise. That leaves you with two alternative conclusions:> either we are ALL enormously courageous but dishonest fools, or you> are wrong.>Here it's the present, and your test of courage is now.Oh, and be sure to check my blog archives: drivel here. Drivel is drivel. Sorry, I didn't sign up for any> contributions. Say, what is happening with those papers you submitted? About> time they got back to you with the referees' reports, eh? I bet> seeing some of the thoughtful and pungently worded arguments you> have posted here would help them decide on publication. Why> not send them a line and some examples?>And hey, if you're moping and miserable because mathematics tests you,>then maybe, if you think you're a mathematician, you might want to try>a different 'eld. Well, hey, I am neither moping nor miserable. I have complete> con'dence that I am right and you are wrong. It is just a matter> of time before you see that 7, 7, 1 is the wrong factorization. Then> everything goes down the drain.> Nora B.>James Harris =This post from James earns my nomination as the WORST math exposition whichhas ever appeared in this newsgroup. Although he titles the postUnderstanding the math he never explains what he is trying to explain orprove, preferring to post a string of polynomial expressions withtransformations and substitutions accompanied by ambiguous and unsupportedassertions. This manner of presenting math would *never* pass a peerreview. (See my comments, embedded in the quoted post below)> My research can be dif'cult to understand, so I thought I'd try out> yet another way of explaining it. Some of you may have 'gured out> that I test out explanations on Usenet for use elsewhere, to re'ne my> own understanding, or just in case someone out there might 'nally get> it.Are you explaining your research? -- or some speci'c result? It is *not*clear where you are going throughout this exposition. Nor is it clear whereyou end up when have 'nished.> Now then, again here's my discovery: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.What, exactly, is the discovery to which you refer? You have given certaincubics and conditions on them, but no hint of why you are dealing withthem. Note that the polynomial in ïa' reduces to: a^3 - a^2 when x = 0.Hence, if the ïroots' you cite are solutions of the cubic equationa^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x) = 0which roots can be designated a_1(x), a_2(x) and a_3(x). Then thepolynomial equation at x = 0 reduces toa^3 - 3a^2 = 0so that the values of the roots are (for x = 0) : a_1(0) = 0,a_2(0) = 0 anda_3(0) = 3}. Since the polynomial in ïa' is monic and the coef'cients areintegers *when ïx' is an integer*, the roots of the original polynomial arealgebraic integers (again, when ïx' is an integer).> In that form it's hard to understand what follows next unless you pay> attention to what you have, speci'cally that cubic de'ning the a's. I can get itIt's hard to follow anything with no hint of where it's going. Get what?The cubic de'ning the ïa's? The expression for the roots? What are you*getting*?> because of the symmetry of (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where I've gone ahead and substituted a_3(x) back in to replace> b_3(x), and it's important that you focus on that symmetry.What symmetry? You have used the terms ïsymmetry' in a completelyunconventional manner without explaining what it means in this context. Areyou referring to the similarity of the expressions involving a_1(x), a_2(x)and a_3(x), or what? If so, that's not a typical use of the term ïsymmetry'and you ought to state what it is you mean by it.> It's that symmetry which allows the cubic a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) to de'ne ALL the a's,Symmetry has nothing whatsoever to do with de'ning the ïa's. They arecompletely de'ned by the polynomial you posted and the fact that thefundamental theorem of algebra guarantees a cubic to have three roots. Inthe case of a monic polynomial with integer coef'cients you further haveroots which are algebraic integers, but you certainly have three roots forevery cubic of the forma^3 + N1 a^2 + N2 a + N4 = 0where the ïN's are computable expressions. There is no symmetryrequirement, unless you have imposed one *without* explanation.> but something happens when I divide by 49.Of course something happens. Especially in the case of equations involvingelements of rings, as you so far appear to be doing. Division is notnecessarily valid in rings, only the operations of addition andmultiplication are speci'ed.> Then the symmetry is broken. Without that symmetry it's impossible to> 'nd a SINGLE cubic to handle what results when you divide both sides> by 49.WRONG! Symmetry has nothing whatsoever to do with the problems associatedwith division in rings. Furthermore the above assertions the impossibilityof 'nding a cubic are completely unsupported by any evidence, proof,citations or even an explanation of what you mean by ïsymmetry' or of what'results' cannot be handled.> That's important because it's why the functions are NOT algebraic> integer functions!!!Stop.What is an ïalgebraic integer function'? To what are you referringhere? This is totally out of the blue and it sounds as if you are trying toestablish that certain ïsymmetries' govern whether a function is an'algebraic integer function'. It's certainly not surprising that divisionmay destroy some properties of expressions involving elements of a ring.Dividing any prime integer by any other integer exceeding 2 produces anumber which is not in the ring of integers. So what?> Now then, I'll recap. Symmetry allows the a's to be de'ned by a> cubic,WRONG! Symmetry has nothing whatsoever to do with it. The cubic was alreadygiven asa^3 + (-1 + 49x)a^2 - 49(2401x^3 -147x^2 + 3x)The ïa's, if you mean the roots ofa^3 + (-1 + 49x)a^2 - 49(2401x^3 -147x^2 + 3x) = 0are completely determined (as functions of ïx') by the fundamental theoremof algebra, using Cardan's formula or some other method. Since thepolynomial is monic and the coef'cients are integers (if ïx' is aninteger) the roots found for any integer value of ïx' will be algebraicintegers. But they will obviously be different for every different(integer) value of ïx'.> which shows them to be algebraic integer functions,So now you suggest that the *roots* of the cubic in ïa' are algebraicinteger functions. If you mean that for every value of ïx' the roots arealgebraic integers, this is false. If you mean that for every *integer*value of ïx' the roots are algebraic integers, this is true -- but, again,this has nothing whatsoever to do with symmetry. It is simply because thecubic in ïa' is monic with integer coef'cients. Plug any integer value of'x' into the expression for each root of ïa' and an algebraic integer willresults. If that's what you mean, then no concept of symmetry is required.> but> dividing by 49 *breaks* that symmetry, taking away the ability to 'nd> some cubic to de'ne the results, which proves that the resulting> functions are not algebraic integer functions.You have completely misunderstood what is going on with the math and ofwhat constitutes a ïproof'. There is absolutely no explanation for yourconclusion that *breaking* some kind of symmetry transforms functions fromalgebraic integer functions to non-algebraic integer functions.On the other hand, *dividing* arbitrary expressions involving integers byother integers may well produce fractional quantities which are neitherintegers nor even algebraic integers. But that is not novel, it isexpected.As I stated at the beginning of this post, you have presented a series ofpolynomial expressions and substitutions with ambiguous and unsupportedassertions -- no explanations, no proofs.> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the b's are roots of b^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x) and when x=0, b_1(0) = b_2(0) = b_3(0) = 0. My point is that the second and third coef'cients are impossible to> de'ne in general.Who cares?. You now have(5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360x + 22where you have de'ned b_1(x) = a_1(x) / 7 and b_2(x) = a_2(x) / 7 so thatthe previous values of a_1(x) and a_2(x), which were algebraic integerswhen ïx' is an integer, are now *divided* by 7. Again, division is not oneof the ring operations for algebraic integers -- only addition andmultiplication. It isn't surprising that you have quantities which are*not* algebraic integers.> You may 'nd them for some particular x, but in general, they are> forever hidden from you.What is hidden? The values of the ïb's have already been de'ned by theirrelation to the ïa's. Your proposed form for a cubic de'ning the ïb's isunnecessary.> Notice that doing that substitution with a_3(x) for b_3(x) gives me (5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22 but you have broken symmetry since the other constant terms are 1 and> 1, so you're still stuck.What symmetry? What has symmetry got to do with this exposition? How do theconstant terms enter into it? Who's stuck? Stuck with what? Your so-calledexplanations are completely incomprehensible.All you have done in this entire exposition is divide two of the algebraicinteger roots of the cubic equation (assuming ïx' is an integer)a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0by 7, and handwaving about ïsymmetry' and the impossibility of 'nd valuesfor some hypothetical polynomial in ïb', etc. What's your point?> Now by emphasizing what happens *after* 49 is divided from both sides> I'm trying to get at least some of you to face the mathematical> realities here, and I've made other posts pointing it out as well.The *mathematical reality* is that division is not generally allowed as anoperation in rings. If you persist in dividing expressions involvingalgebraic integers by arbitrary quantities and then throw up your hands indisgust at the error in core mathematics you have no one to blame butyourself. You're simply stuf'ng a rabbit into a hat and then triumphantlyannouncing its presence.> Courage in mathematics might sound new to you as an idea, but those> who came before you had courage. That's how they learned of sqrt(2),> and sqrt(-1), and much other mathematics that some refused, fought,> and bitterly attacked as they lacked that courage.The guy that fought to have pi de'ned as 3 (or whatever) also had courage.But, like you, he was wrong!> Of course, the mathematics should win, assuming that humanity survives> long enough, but regardless each of you now faces an individual test> of your own courage. It's not yet in the history books this time,> where you can read and just imagine that you'd never have fought over> sqrt(2) or refused to accept sqrt(-1) as an imaginary number.Mathematics will win. Pi will *not* be de'ned as 3. Your research willend up where it belongs.> Here it's the present, and your test of courage is now.Your exposition is test of patience, not courage. All you have succeeded inproving is that you are an idiot.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com = [.snip.]>What, exactly, is the discovery to which you refer? You have given certain>cubics and conditions on them, but no hint of why you are dealing with>them. Note that the polynomial in ïa' reduces to: a^3 - a^2 when x = 0.>Hence, if the ïroots' you cite are solutions of the cubic equationa^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x) = 0which roots can be designated a_1(x), a_2(x) and a_3(x). Then the>polynomial equation at x = 0 reduces toa^3 - 3a^2 = 0so that the values of the roots are (for x = 0) : a_1(0) = 0,a_2(0) = 0 and>a_3(0) = 3}. Since the polynomial in ïa' is monic and the coef'cients are>integers *when ïx' is an integer*, the roots of the original polynomial are>algebraic integers (again, when ïx' is an integer).In fact, the roots are algebraic integers whenever x is an algebraicinteger, since roots of monic polynomials with algebraic integercoef'cients are also algebraic integers. [.snip.]>WRONG! Symmetry has nothing whatsoever to do with the problems associated>with division in rings. Actually, there is some symmetry: if the polynomial de'ning the rootsis irreducible over Q, then the Galois group of the splitting 'eldgives you a group of symmetries among the roots. This symmetry is whatallows you to deduce, for example, that for any rational prime p, ifone root is coprime to p then all roots are coprime to p, if one rootis divisible by p then all roots are divisible by p, etc. Provided thepolynomial is irreducible over Q (and in particular, has integer coef'cients). [.snip.]-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu [.snip.]What, exactly, is the discovery to which you refer? You have given certain>cubics and conditions on them, but no hint of why you are dealing with>them. Note that the polynomial in ïa' reduces to: a^3 - a^2 when x = 0.>Hence, if the ïroots' you cite are solutions of the cubic equationa^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x) = 0which roots can be designated a_1(x), a_2(x) and a_3(x). Then the>polynomial equation at x = 0 reduces toa^3 - 3a^2 = 0so that the values of the roots are (for x = 0) : a_1(0) = 0,a_2(0) = 0 and>a_3(0) = 3}. Since the polynomial in ïa' is monic and the coef'cients are>integers *when ïx' is an integer*, the roots of the original polynomial are>algebraic integers (again, when ïx' is an integer). In fact, the roots are algebraic integers whenever x is an algebraic> integer, since roots of monic polynomials with algebraic integer> coef'cients are also algebraic integers.> [.snip.]WRONG! Symmetry has nothing whatsoever to do with the problems associated>with division in rings. Actually, there is some symmetry: if the polynomial de'ning the roots> is irreducible over Q, then the Galois group of the splitting 'eld> gives you a group of symmetries among the roots. This symmetry is what> allows you to deduce, for example, that for any rational prime p, if> one root is coprime to p then all roots are coprime to p, if one root> is divisible by p then all roots are divisible by p, etc. Provided the> polynomial is irreducible over Q (and in particular, has integer coef'cients).Is the symmetry to which refer the explanation for James' assertion that some'symmetry' in the polynomial, which he alludes to but does not identify, isresponsible for the cubic having a solution at all?> [.snip.] --> => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> = Arturo Magidin> magidin@math.berkeley.edu--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =[cut]> Is the symmetry to which refer the explanation for James' assertion that some> ïsymmetry' in the polynomial, which he alludes to but does not identify, is> responsible for the cubic having a solution at all?I took the symmetry to mean the three 7's that appear as theconstant terms of the three factors. When one divides by 49=7*7*1,the symmetry is lost since two 7's go to 1 and the third 7 remains.-- Bill Hale => [cut]> Is the symmetry to which refer the explanation for James' assertion that some> ïsymmetry' in the polynomial, which he alludes to but does not identify, is> responsible for the cubic having a solution at all? I took the symmetry to mean the three 7's that appear as the> constant terms of the three factors. When one divides by 49=7*7*1,> the symmetry is lost since two 7's go to 1 and the third 7 remains. -- Bill HaleI wasn't sure what James meant, although it is clear he believed that some kind ofsymmetry was of critical importance. I simply noted that the roots of the cubic canbe found regardless of any real or imagined symmetries and that the division by 49(or some other number) is likely to result in terms which are not algebraicintegers.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =>My research can be dif'cult to understand, so I thought I'd try out>yet another way of explaining it. You need to actually refute the explanations of what's wronginstead of just repeating the explanation. Over and over yousay the only way that 49 can split is like so - people say no,that's not the only way, and you never reply with a _proof_that it's the only way, just vigorous assertion.>[...]Courage in mathematics might sound new to you as an idea, but those>who came before you had courage. That's how they learned of sqrt(2),>and sqrt(-1), and much other mathematics that some refused, fought,>and bitterly attacked as they lacked that courage.Of course, the mathematics should win, assuming that humanity survives>long enough, but regardless each of you now faces an individual test>of your own courage. It's not yet in the history books this time,>where you can read and just imagine that you'd never have fought over>sqrt(2) or refused to accept sqrt(-1) as an imaginary number.Here it's the present, and your test of courage is now.Also, as many people have explained many times, this sortof pompous nonsense is a bad idea in terms of strategy -anyone reading the last three paragraphs is going to concludeyou're a crackpot, even if they have no idea what the mathis about.Honest.>Oh, and be sure to check my blog archives:>And hey, if you're moping and miserable because mathematics tests you,>then maybe, if you think you're a mathematician, you might want to try>a different 'eld.Giggle. Still looking for that mirror, eh?>James Harris************************David C. Ullrich => >My research can be dif'cult to understand, so I thought I'd try out>yet another way of explaining it. > > You need to actually refute the explanations of what's wrong> instead of just repeating the explanation. Over and over you> say the only way that 49 can split is like so - people say no,> that's not the only way, and you never reply with a _proof_> that it's the only way, just vigorous assertion.> Well sure, delete out the argument which settles the issue, and thencome back and claim that it's not settled.Now math history is full of people like David Ullrich, a guy with atitle, 'ghting against some new idea. In the past it was sqrt(-1),as mathematicians fought against an idea they thought of as silly.The hallmark of such people is a refusal to follow mathematical logic. You give a detailed mathematical position, and they just delete itall out, and claim it's not what it is.Yet the facts speak for themselves:My work depends on advanced mathematical tools for factoringpolynomials, where I use one polynomial to factor another polynomialinto non-polynomial factors.I'd be curious if anyone can 'nd anything like it in the recordedhistory of mathematics.Along with that work, I have my prime number research, where I have apartial difference equation used to count prime numbers--a 'rst inmath history.People like David Ullrich are the anti-knowledge crowd. Those peoplewho know just enough to feel comfortable in a complex world. The kindof people who say, no more knowledge.Ullrich probably feels very comfortable with whatever mathematicalknowledge he has. He can pay his bills using the money given him bytaxpayers, as he's a professor at Oklahoma State University. Heprobably has lots of feedback from various people in society to makehim feel good about himself.What I represent is the unknowable future. The power that comes nomatter how comfortable people are, no matter how satis'ed they arewith themselves or their positions--the power of change.In that sense I'm a force of Nature, a force of the Universe, a livingemodiment of change itself, so I understand the fear I can induce inpeople like David Ullrich.But the spirit of mathematics is ultimately about what's not known,what's beyond what's known today, as that de'nes the work of thefuture.I stand here part of a great tradition of discoverers, people who gotpast the Ullrich's to push forward knowledge so that today you havesqrt(2), i, e, e^x, planes, trains, automobiles, spacecraft, andcomputers.It's too easy to look at the past as if it were easy. As if new ideaswere just accepted because they were right, and could be provenlogically, mathematically, or practically, so that you stand by today,forcing yet another discovery to break through.But that is your loss, as my strength comes from the knowledge I have,the knowledge of those who came before me, the knowledge of those whowill come after.The knowledge that the truth is beautiful.All of you can resist, 'ght until you die, as you will die someday. I can't look to any of you for permanence as you're in a sense, deadalready. But I can, and you can look at the truth--beautiful,permanent, absolute.The discoverers are the forces of Nature, born to a grand tradition,tasked with the seeming impossible, 'ghting throughout time,throughout history, to make history itself.The discoverers are the living embodiment of the forces of Goodagainst Evil. The people who push the limits of knowledge for thebene't of all, and the love of Truth.We are the greatest warriors for Good, of all time, through all time,through all worlds and realities where sentient beings live.We are the discoverers.James Harris The power that comes no> matter how comfortable people are, no matter how satis'ed they are> with themselves or their positions--the power of change.Good grief! Someone get me a shovel!> In that sense I'm a force of Nature, a force of the Universe, a living> emodiment of change itself, so I understand the fear I can induce in> people like David Ullrich.Stand back! He's gonna blow!> But the spirit of mathematics is ultimately about what's not known,> what's beyond what's known today, as that de'nes the work of the> future. I stand here part of a great tradition of discoverers, people who got> past the Ullrich's to push forward knowledge so that today you have> sqrt(2), i, e, e^x, planes, trains, automobiles, spacecraft, and> computers.one-man craft and pilot it yourself, because you'll never sign up a volunteer things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =>> >>My research can be dif'cult to understand, so I thought I'd try out>>yet another way of explaining it. You need to actually refute the explanations of what's wrong>> instead of just repeating the explanation. Over and over you>> say the only way that 49 can split is like so - people say no,>> that's not the only way, and you never reply with a _proof_>> that it's the only way, just vigorous assertion.>> Well sure, delete out the argument which settles the issue, and then>come back and claim that it's not settled.Now math history is full of people like David Ullrich, a guy with a>title, 'ghting against some new idea. In the past it was sqrt(-1),>as mathematicians fought against an idea they thought of as silly.The hallmark of such people is a refusal to follow mathematical logic.> You give a detailed mathematical position, and they just delete it>all out, and claim it's not what it is.Yet the facts speak for themselves:My work depends on advanced mathematical tools for factoring>polynomials, where I use one polynomial to factor another polynomial>into non-polynomial factors.I'd be curious if anyone can 'nd anything like it in the recorded>history of mathematics.Along with that work, I have my prime number research, where I have a>partial difference equation used to count prime numbers--a 'rst in>math history.People like David Ullrich are the anti-knowledge crowd. Those people>who know just enough to feel comfortable in a complex world. The kind>of people who say, no more knowledge.Ullrich probably feels very comfortable with whatever mathematical>knowledge he has. He can pay his bills using the money given him by>taxpayers, as he's a professor at Oklahoma State University. He>probably has lots of feedback from various people in society to make>him feel good about himself.What I represent is the unknowable future. The power that comes no>matter how comfortable people are, no matter how satis'ed they are>with themselves or their positions--the power of change.In that sense I'm a force of Nature, a force of the Universe, a living>emodiment of change itself, so I understand the fear I can induce in>people like David Ullrich.But the spirit of mathematics is ultimately about what's not known,>what's beyond what's known today, as that de'nes the work of the>future.I stand here part of a great tradition of discoverers, people who got>past the Ullrich's to push forward knowledge so that today you have>sqrt(2), i, e, e^x, planes, trains, automobiles, spacecraft, and>computers.It's too easy to look at the past as if it were easy. As if new ideas>were just accepted because they were right, and could be proven>logically, mathematically, or practically, so that you stand by today,>forcing yet another discovery to break through.But that is your loss, as my strength comes from the knowledge I have,>the knowledge of those who came before me, the knowledge of those who>will come after.The knowledge that the truth is beautiful.All of you can resist, 'ght until you die, as you will die someday. >I can't look to any of you for permanence as you're in a sense, dead>already. But I can, and you can look at the truth--beautiful,>permanent, absolute.The discoverers are the forces of Nature, born to a grand tradition,>tasked with the seeming impossible, 'ghting throughout time,>throughout history, to make history itself.The discoverers are the living embodiment of the forces of Good>against Evil. The people who push the limits of knowledge for the>bene't of all, and the love of Truth.We are the greatest warriors for Good, of all time, through all time,>through all worlds and realities where sentient beings live.We are the discoverers.Sweet Jesus but we discoverers are full of crap today, are we not?>James Harris************************David C. Ullrich =* James Harris> Now math history is full of people like David Ullrich, a guy with a> title, 'ghting against some new idea. In the past it was sqrt(-1),> as mathematicians fought against an idea they thought of as silly.are dead and buried, (the ideas). -- Jon Haugsand => * James Harris> Now math history is full of people like David Ullrich, a guy with a> title, 'ghting against some new idea. In the past it was sqrt(-1),> as mathematicians fought against an idea they thought of as silly.> > are dead and buried, (the ideas).Most people show worth through effort. For instance, if you valueyour job, one would assume that you put a lot of effort into it. Similarly, if you value a relationship, you're willing to work to keepit going.People like Ullrich value trying to attack the new, as evidenced byhis *efforts* in that regard.There are a lot of people out there I probably would suppose arenutty, who have ideas that I don't care at all about, and then againthere are people who are supposedly very competent, or expert, whohave ideas that I don't care about, and you know what?I don't spend time on them.Evil is the pursuit of ignorance, challenging new truths to hold on toan old, and comfortable point of view. It's also stupid as change isinevitable, and the energy of the evil person is wasted.However, in spite of its stupidity evil poses challenges, which thediscoverer is ever tasked with 'ghting through, including handlingthose who 'ght for evil in their attempts to maintain their owncomfort against knowledge.That 'ght is one of the continuing burdens of the Universe's 'rst,greatest, and last 'ghting force.James Harris => However, in spite of its stupidity evil poses challenges, which the> discoverer is ever tasked with 'ghting through, including handling> those who 'ght for evil in their attempts to maintain their own> comfort against knowledge. That 'ght is one of the continuing burdens of the Universe's 'rst,> greatest, and last 'ghting force.Mr. Harris,When were you chosen as the supreme Galactic Minister of Mathematics?I somehow missed the announcement.I suppose that is a great title for you as those who cannot produce go intomanagement in most cases.Congratulations, 'nally a title you can be proud of and you have nomathematical achievements to speak of!Bravo Universo Mathematico! continuing burdens of the Universe's 'rst,> greatest, and last 'ghting force.Keep up the good work. Your application for Galactic Chief Technical Of'cer, W.O.O.O. is underconsideration.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =>dead and buried, (the ideas).Most people show worth through effort. For instance, if you value>your job, one would assume that you put a lot of effort into it. >Similarly, if you value a relationship, you're willing to work to keep>it going.People like Ullrich value trying to attack the new, as evidenced by>his *efforts* in that regard.There are a lot of people out there I probably would suppose are>nutty, who have ideas that I don't care at all about, and then again>there are people who are supposedly very competent, or expert, who>have ideas that I don't care about, and you know what?I don't spend time on them.Evil is the pursuit of ignorance, challenging new truths to hold on to>an old, and comfortable point of view. It's also stupid as change is>inevitable, and the energy of the evil person is wasted.However, in spite of its stupidity evil poses challenges, which the>discoverer is ever tasked with 'ghting through, including handling>those who 'ght for evil in their attempts to maintain their own>comfort against knowledge.That 'ght is one of the continuing burdens of the Universe's 'rst,>greatest, and last 'ghting force.Does anybody else think that James has just seen Lord of the Rings 3 -Return of The King? And is projecting again?-- MinSo where are all the buffaloes? => * James Harris> Now math history is full of people like David Ullrich, a guy with a> title, 'ghting against some new idea. In the past it was sqrt(-1),> as mathematicians fought against an idea they thought of as silly.> > are dead and buried, (the ideas). Furthermore, the example of sqrt(-1) is not quite the way Harrispresents it. It is not really a good example.The original proposer of sqrt(-1) realized that sqrt(-1) had no logicalfoundation. Harris seems to indicate that the original proposer hadto 'ght against the other mathematicians to have his idea accepted.In fact, he himself did not accept it as logically sound.The same goes for sqrt(2), but in the other direction of acceptance.I don't think any of the Pythagoreans thought the proof was invalidwhen they saw it. It did casue a problem with their philosophy onthe nature of the universe, but that is something different.-- Bill Hale =Reading Ch 12 of Lee Smolin's Three Roads to Quantum Gravity shows that Lenny's World Hologram is Virtual Reality Max in sense ofTipler's Physics of Immortality of Omega with a Tech-Gnostic Demented There is an inversion of Plato's Allegory of The Cave with The Screen (a quantum gravity computer) as the reality. There is no Light at the End of the Tunnel in this version of The Thirteenth Floor.I mean the WHP Weak Holographic Principle.It is interesting that in my theory the Planck area Lp^2 is the connection between IT guv and BIT the MACRO-QUANTUM VACUUM COHERENCE FIELD that I replace Hawking's Wave Function of the Universe with.guv(Einstein) = Minkowski + Lp^2(Coherent Phase of LOCAL Vacuum Coherence)(,u,v)Lp^2 is like h/m in a quantized circulation vortex wherev = (h/m)Grad(Coherent Phase of the LOCAL order parameter)i.e. 4D World Crystal Lattice distortion 'eld isdu(x) = Lp^2 (Coherent Phase of LOCAL Vacuum Coherence),uEinstein's Cosmological Constant at FRW scale vanishes for a critical Vacuum Coherence intensity|Vacuum Coherence|^2 = (Planck Volume)^-3 criticalPlanck Volume = (String Bit Link)(Planck Area)Quantizing string links implies quantizing dual Bekenstein areas.Going off critical gives either repulsive dark energy of negative zero point pressure or attractive dark matter of positive zero point pressure, both merely different densities, or intensities, of the w = -1 Vacuum Coherence Field. =Is it possible to evaluate following integral as a function of k ( 0 Is it possible to evaluate following integral as a function of k ( 0 > > int_0^{2 pi}{ ( ( ln( 1+k sin(x) ) )^2 ) dx }> integrand as a power series in k. The coef'cient of k^nis rational * sin(x)^n; integrate sin(x)^n from 0 to 2*pi (using arecursive reduction formula, perhaps). Sum the resulting series. => Is it possible to evaluate following integral as a function of k ( 0> int_0^{2 pi}{ ( ( ln( 1+k sin(x) try:Expand the integrand as a power series in k. The coef'cient of k^n>is rational * sin(x)^n; integrate sin(x)^n from 0 to 2*pi (using a>recursive reduction formula, perhaps). Sum the resulting series.The rational coef'cient of k^n sin^n(x), for n >= 2, is n-1 n 2 --- 1 n (-1) - > - = (-1) H(n-1) [1] n --- j j=1where H(n) is the n^{th} harmonic number.However, the integral from 0 to 2pi of any odd power of sin(x) is 0.So you only need to consider the even n, and for even n, |2pi n | sin (x) dx = 2pi C(n,n/2)/2^n [2] | 0Putting this all together, and substituting n -> 2n, we get that |2pi 2 | ln(1 + k sin(x)) dx | 0 oo --- 2n = 2pi > H(2n-1)/n C(2n,n)/4^n k [3] --- n=1The summation in [3] converges absolutely for |k| <= 1.Rob Johnson take out the trash before replying =let A be an open interval on the real line (R), ie. A = (a, b) with a < b(where a, b in R). Now let x = min(A).x is an interior point of A if there exist an e > 0 such that O_e(x) (whereO_e() is the e-neighbourhood of x) but any neighbourhood of x is going tocontain a point, y, with y < x and since x = min(A) y cannot be in A (unlessthe set {x} is a neighbourhood of x??).I'm having similar dif'culties with other ideas but believe they are allderived from this one simple example. For example is it possible to de'ne avariable with valuee = min(|x - y| : x, y in R and x != y)as the smallest distances possible between two distinct points in the set ofreal numbers. This number would be equivalent toA = (b, c)B = [a, b]e = min(A) - max(B) with a <= b < c(with a = b simply min(A) - b)There is obviously a problem with this reasoning so could someone pleaseexplain it to => let A be an open interval on the real line (R), ie. A = (a, b) with a < b> (where a, b in R). Now let x = min(A).> > x is an interior point of A if there exist an e > 0 such that O_e(x)> (where O_e() is the e-neighbourhood of x) but any neighbourhood of x is> going to contain a point, y, with y < x and since x = min(A) y cannot be> in A (unless the set {x} is a neighbourhood of x??).> > I'm having similar dif'culties with other ideas but believe they are all> derived from this one simple example. For example is it possible to de'ne> a variable with value> > e = min(|x - y| : x, y in R and x != y)> > as the smallest distances possible between two distinct points in the set> of real numbers. This number would be equivalent to> > A = (b, c)> B = [a, b]> > e = min(A) - max(B) with a <= b < c> > (with a = b simply min(A) - b)> > There is obviously a problem with this reasoning so could also learning this material for the 'rst time...By de'nition you have A = {x in R : a < x < b}. Now examining theinequality a < x < b, observe that we can 'nd a number c thatlies between a and x, choose c = (a+x)/2. But this works for *all*x that lie in the interval (a,b). That is, we can take x as closeas we like to a, but we can always 'nd a c such that a < c < x.Therefore A has no minimum element.Hope I haven't lead you, myself or others astray!Bruce. =>let A be an open interval on the real line (R), ie. A = (a, b) with a < b>(where a, b in R). Now let x = min(A).x does not exist. The open interval A has no minimum.>x is an interior point of A if there exist an e > 0 such that O_e(x) (where>O_e() is the e-neighbourhood of x) but any neighbourhood of x is going to>contain a point, y, with y < x and since x = min(A) y cannot be in A (unless>the set {x} is a neighbourhood of x??).No. This is exactly what tells you that A does not have aminimum. Given any element r of A, there always exists an element s ofA such that sI'm having similar dif'culties with other ideas but believe they are all>derived from this one simple example. For example is it possible to de'ne a>variable with valuee = min(|x - y| : x, y in R and x != y)No. There is no minimum for that set either.>as the smallest distances possible between two distinct points in the set of>real numbers. This number would be equivalent toA = (b, c)>B = [a, b]e = min(A) - max(B) with a <= b < cThere is no minimum of A (there is a maximum of B, namely b), so thisexpression makes no sense.>(with a = b simply min(A) - b)There is obviously a problem with this reasoning so could someone please>explain it to me?Just because you write min(A) does not mean such a thingexist. Sets of reals can fall into any of the following categories:(1) Has both a maximum and a minimum.(2) Has a maximum but no minimum.(3) Has no maximum, has a minimum.(4) Has neither a maximum nor a minimum.You can re'ne them further by introducing the notion of in'mumand supremum, but I suspect that may only confuse you at this point,so I will not.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu>let A be an open interval on the real line (R), ie. A = (a, b) with a < b>(where a, b in R). Now let x = min(A). x does not exist. The open interval A has no minimum.x is an interior point of A if there exist an e > 0 such that O_e(x)(where>O_e() is the e-neighbourhood of x) but any neighbourhood of x is going to>contain a point, y, with y < x and since x = min(A) y cannot be in A(unless>the set {x} is a neighbourhood of x??). No. This is exactly what tells you that A does not have a> minimum. Given any element r of A, there always exists an element s of> A such that sderived from this one simple example. For example is it possible tode'ne a>variable with valuee = min(|x - y| : x, y in R and x != y) No. There is no minimum for that set either.as the smallest distances possible between two distinct points in the setof>real numbers. This number would be equivalent toA = (b, c)>B = [a, b]e = min(A) - max(B) with a <= b < c There is no minimum of A (there is a maximum of B, namely b), so this> expression makes no sense.(with a = b simply min(A) - b)There is obviously a problem with this reasoning so could someone please>explain it to me? Just because you write min(A) does not mean such a thing> exist. Sets of reals can fall into any of the following categories: (1) Has both a maximum and a minimum.> (2) Has a maximum but no minimum.> (3) Has no maximum, has a minimum.> (4) Has neither a maximum nor a minimum. You can re'ne them further by introducing the notion of in'mum> and supremum, but I suspect that may only confuse you at this point,> so I will not. --> => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> heaps for this, I think I have it clear now.So in effect min(A) is not de'ned because if you tried to create a Turingmachine capable of computing this value it would not halt (because you canalways 'nd a smaller value for any give x in A). So in computer sciencespeak min(A) is uncomputable?I guess I'm trying to look at it from a computer science point of view. Asin, is it possible to create some operation, min, that applied to some openinterval (on the real line) returned an x s.t. x <= y for all y in A. Andthe answer is no because of the above reasons.!Q. => > So in effect min(A) is not de'ned because if you tried to create a Turing> machine capable of computing this value it would not halt...Actually, you yourself gave an elegant proof that min(A) does notexist: min(A) means the smallest element belonging to A, and youproved that given any x in A, it cannot be min(A). Therefore, nomember of A is min(A), and min(A) does not exist.> I guess I'm trying to look at it from a computer science point of view. As> in, is it possible to create some operation, min, that applied to some open> interval (on the real line) returned an x s.t. x <= y for all y in A. And> the answer is no because of the above reasons.The real question is what you're after. If A is the open interval(a,b), it's clear what you want the min to be: exactly a. But ofcourse a is not a member of the open interval.You are probably 'shing for the useful concept called a lowerbound. The interval (0,1) is bounded below by -10, or -100, or-1...or, notice, by 0. Although 0 is not a member of of (0,1), it hasthe interesting property that any number greater than 0 (but, ofcourse, less than 1) is a member of the interval, and hence, by yourown argument, not a lower bound. Conversely, any number which is alower bound of (0,1) must necessarily be <=0.One might call 0 the greatest lower bound of the interval (0,1). Ithas all the properties you desire in a minimum except one: it is not amember of the set.If you're studying analysis now, you will encounter greatest lowerbounds, and their mirror image, least upper bounds, very soon.HTH,Len. =>let A be an open interval on the real line (R), ie. A = (a, b) with a < b>>(where a, b in R). Now let x = min(A).>> x does not exist. The open interval A has no minimum.>>x is an interior point of A if there exist an e > 0 such that O_e(x)>(where>>O_e() is the e-neighbourhood of x) but any neighbourhood of x is going to>>contain a point, y, with y < x and since x = min(A) y cannot be in A>(unless>>the set {x} is a neighbourhood of x??).>> No. This is exactly what tells you that A does not have a>> minimum. Given any element r of A, there always exists an element s of>> A such that smachine capable of computing this value it would not halt (because you can>always 'nd a smaller value for any give x in A).No. Turing machines don't enter into it. Besides, there are plenty ofthings that exist but are not computable.min(A) does not exist because there is no element of A that satis'esthe conditions necessary to be the minimum of A.Remember: given a subset of real numbers A, a real number x is said tobe the minimum of A, or min(A) if and only if two conditions aremet: (1) For every a in A, x<= a; and (2) x is in A.The reason why min(A) does not exist is because no number satis'esthe conditions: you can prove that any x that satis'es (2) cannotsatisfy (1) (e.g., by your e-neighborhood argument). So no number cansatisfy both conditions, so no number can satisfy both.> So in computer science>speak min(A) is uncomputable?No. min(A) does not exist, period. This has nothing to do with beingcomputable or not computable. min(A) and the real number whose squareis equal to -1 are in the same boat.>I guess I'm trying to look at it from a computer science point of view. As>in, is it possible to create some operation, min, that applied to some open>interval (on the real line) returned an x s.t. x <= y for all y in A. And>the answer is no because of the above reasons.Has nothing to do with computability.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu>>So in effect min(A) is not de'ned because if you tried to create a>>Turing machine capable of computing this value it would not halt (because>>you can always 'nd a smaller value for any give x in A).> > No. Turing machines don't enter into it. Besides, there are plenty of> things that exist but are not computable.This is trivial, but it follows doesn't it that if we take as a de'nitionof computationhttp://mathworld.wolfram.com/Computation.htmlgiven an algorithm to 'nd the smallest element in an open interval, thealgorithm will not terminate because what the algorithm is looking fordoesn't exist. That is, in our case, if the smallest element in theinterval doesn't exist it isn't by de'nition computable. =>>So in effect min(A) is not de'ned because if you tried to create a>Turing machine capable of computing this value it would not halt (because>you can always 'nd a smaller value for any give x in A).No. Turing machines don't enter into it. Besides, there are plenty of>> things that exist but are not computable.This is trivial, but it follows doesn't it that if we take as a de'nition>of computation>http://mathworld.wolfram.com/Computation.html> given an algorithm to 'nd the smallest element in an open interval, the>algorithm will not terminate because what the algorithm is looking for>doesn't exist.This does not necessarily follow. It would be trivial to write analgorithm that, given an interval in the reals, will terminate andgive you its minimum if it exists, or tell you it does notexist. Whether or not a given interval has a minimum is computable.> That is, in our case, if the smallest element in the>interval doesn't exist it isn't by de'nition computable.It is incorrect to con§ate computability with existence.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu>>So in effect min(A) is not de'ned because if you tried to create a>>Turing machine capable of computing this value it would not halt>>(because you can always 'nd a smaller value for any give x in A).> > No. Turing machines don't enter into it. Besides, there are plenty of> things that exist but are not computable.>>This is trivial, but it follows doesn't it that if we take as a>>de'nition of computation>>http://mathworld.wolfram.com/Computation.html given an algorithm to>>'nd the smallest element in an open interval, the algorithm will not>>terminate because what the algorithm is looking for doesn't exist.> > This does not necessarily follow. It would be trivial to write an> algorithm that, given an interval in the reals, will terminate and give> you its minimum if it exists, or tell you it does not exist. Whether or> not a given interval has a minimum is computable.> That is, in our case, if the smallest element in the>>interval doesn't exist it isn't by de'nition computable. It is incorrect to con§ate computability with existence.If we take as an algorithmbegin loop if current element is smallest in A return current element else current element <-- next smallest element end-loopendNow assuming we can always 'nd a next smallest element(which we can) this algorithm does not terminate. The implication(in this case then) is that the non-existence of the smallestelement in the set implies the non-computablity of the smallestelement. As a general statement maybe we could saybegin Loop if solution found return answer else set up next test case, data etc end-loopendin this general case wouldn't non existence of answer imply non-computability of answer? =>If we take as an algorithm>begin> loop> if current element is smallest in A> return current element> else> current element <-- next smallest element> end-loop>end>Now assuming we can always 'nd a next smallest element>(which we can) this algorithm does not terminate.There is no next smallest element, just as there is no smallestelement. Or do you mean some smaller element? In that case,depending on what you use for some smaller element, your algorithmmight not terminate for a closed interval either.But the failure of one attempt at an algorithm does notmean that the quantity you're trying to compute doesn't exist, or is not computable.> The implication>(in this case then) is that the non-existence of the smallest>element in the set implies the non-computablity of the smallest>element.>in this general case wouldn't >non existence of answer imply non-computability of answer?You can't eat unicorns because they don't exist. Does thatmean unicorns are inedible?Again, obviously you can't compute what doesn't exist, but that really has nothing to do with the theory of computability.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => Does that mean unicorns are inedible?Good question, but some people will be confusedagain, since it's true: all unicorns are inedible.:-)Rainer Rosenthalr.rosenthal@web.de--Theorem 0: The set of edible unicorns is an improper subset of the set of all inedible unicorns. =>So in effect min(A) is not de'ned because if you tried to create a>Turing machine capable of computing this value it would not halt>(because you can always 'nd a smaller value for any give x in A).No. Turing machines don't enter into it. Besides, there are plenty of>> things that exist but are not computable.>>This is trivial, but it follows doesn't it that if we take as a>de'nition of computation>http://mathworld.wolfram.com/Computation.html given an algorithm to>'nd the smallest element in an open interval, the algorithm will not>terminate because what the algorithm is looking for doesn't exist.This does not necessarily follow. It would be trivial to write an>> algorithm that, given an interval in the reals, will terminate and give>> you its minimum if it exists, or tell you it does not exist. Whether or>> not a given interval has a minimum is computable.>> > That is, in our case, if the smallest element in the>interval doesn't exist it isn't by de'nition computable.It is incorrect to con§ate computability with existence.If we take as an algorithmbegin> loop> if current element is smallest in A> return current element> else> current element <-- next smallest element> end-loop>endThis is not an algorithm. There is no next smallest element, so thisis not well-de'ned. Even assuming it were an algorithm, providing an algorithm that doesnot stop does not prove that something is not computable. By thatstandard, the smallest positive even integer is not computable, byconsidering the algorithmbegin set current element = 1; loop if current element is even, stop and return current element else current element <--- current element + 2 end-loopend>Now assuming we can always 'nd a next smallest element>(which we can) No, we cannot. Please tell me, if A is the interval [0,1], andcurrent element is .5, what is the next smallest element? [.snip.]>in this general case wouldn't >non existence of answer imply non-computability of answer?It also implies computability of the answer. Because very number whichis an answer to the problem of 'nding the minimum of (0,1) iscomputable.Which is why it is pretty much nonsense to talk about computability ornon-computability of numbers that do not exist in the 'rst place.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu algorithm. There is no next smallest element, so this is> not well-de'ned.Imagine we have a machine that is able to perform §oating point calculations with arbitrary precision, we can tryto 'nd the smallest element in the interval (a,b) by// pre: a a do currentSmallest <-- (currentSmallest+a)/2 end-while return currentSmallestend> > Even assuming it were an algorithm, providing an algorithm that does not> stop does not prove that something is not computable. By that standard,> the smallest positive even integer is not computable, by considering the> algorithmI'm not trying to prove that. I'm saying that non-existence of asolution implies the non-computability of a solution.I don't know how to *prove* that though :)> > begin> set current element = 1;> loop> if current element is even, stop and return current element else> current element <--- current element + 2> end-loop> end> > >>Now assuming we can always 'nd a next smallest element (which we can)> > No, we cannot. Please tell me, if A is the interval [0,1], and current> element is .5, what is the next smallest element?The A that the OP was talking about was an *open* interval... => I'm not trying to prove that. I'm saying that non-existence of a> solution implies the non-computability of a solution.> > I don't know how to *prove* that though :)You are saying roughly that if something doesn't exist, then you can't 'nd it? => >> I'm not trying to prove that. I'm saying that non-existence of a>> solution implies the non-computability of a solution.I don't know how to *prove* that though :)> > You are saying roughly that if something doesn't exist, then you can't> 'nd it?Yes! smallest element, so this is>> not well-de'ned.Imagine we have a machine that is able to perform >§oating point calculations with arbitrary precision, we can try>to 'nd the smallest element in the interval (a,b) by// pre: abegin> currentSmallest <-- (b+a)/2> while currentSmallest > a do> currentSmallest <-- (currentSmallest+a)/2> end-while> return currentSmallest>endThis is a ridiculous way of trying to 'nd the smallest element of aninterval. This algorithm would fail to terminate in looking for thesmallest element of the CLOSED interval [0,1]: it would start at .5,then .25, then .125, etc.>> Even assuming it were an algorithm, providing an algorithm that does not>> stop does not prove that something is not computable. By that standard,>> the smallest positive even integer is not computable, by considering the>> algorithmI'm not trying to prove that. I'm saying that non-existence of a>solution implies the non-computability of a solution.I don't know how to *prove* that though :)Easy: any element which is a solution is non-computable. It is alsogreen, red, plaid, tastes like diet coke, is odorless, and saysOf course, any element which is a solutions also ->is<- computable.You are describing properties of elements of the empty set. Anamusing, but ultimately futile, exercise.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu>So in effect min(A) is not de'ned because if you tried to create a Turing>machine capable of computing this value it would not halt (because you can>always 'nd a smaller value for any give x in A). So in computer science>speak min(A) is uncomputable?You can't compute something that doesn't exist. On the other hand, allmembers of the empty set are computable... But seriously, x does not exist is not the same thing as x is notcomputable.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => >>So in effect min(A) is not de'ned because if you tried to create a Turing>>machine capable of computing this value it would not halt (because you can>>always 'nd a smaller value for any give x in A). So in computer science>>speak min(A) is uncomputable?> > You can't compute something that doesn't exist. On the other hand, all> members of the empty set are computable...> > But seriously, x does not exist is not the same thing as x is not> computable.Hardline constructivists may disagree :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =Robin Chapman> But seriously, x does not exist is not the same thing as x is not> computable. Hardline constructivists may disagree :-)Do they exist?RR =>Robin Chapman> But seriously, x does not exist is not the same thing as x is not>> computable.>> Hardline constructivists may disagree :-)Do they exist?I know someone whose advisor made him a hardline constructivist.Lee Rudolph =[added sci.logic, sci.math]>I got my math masters in 1992 and my specialty was Algebra.But, there has always been something that bothers me. In'nity!You are not alone. There were (in)famous l-o-n-g threads on sci.logic and sci.math by Phil, who claimed that there was a largest integer. In'nity is outside our experience and outside our naive intuitions, and that can be uncomfortable.>I know that the set of integers contains the same number of elements as the>set of even integers. I know the proof - there is a 1-1 functionI think you mean 1-1 correspondence, i.e. a bijection. A 1-1 function usually means an injection.> from Z -2Z, etc... Therefore there are the same number of elements. It just>bothers me!!! I wish mathmaticians would come up with different sizes of>in'nity besides the countably in'nite (eg Z) and uncountably in'nite (eg>R).Well, set theorists have; but you won't like it! There are an (ahem) in'nite number of in'nities. And there can also be large cardinal in'nities that are even bigger than those.>Another one I hate - the set of prime integers is the same number as Z+.>UGH! I hate that!So, I have a question....If I have a Set A with a proper subset B and there is a 1-1 function from>that set B to a set C then does the set A contain more elements than C?>hehe...In this example, A is at least as large as B, and B is the same size as C (assuming you mean 1-1 correspondence), therefore A is at least as large as C; A might be larger, but we can't tell that from the information given so far.Note that one of the standard de'nitions of an in'nite set (Dedekind's) is a set that has some 1-1 correspondance with a proper subset of itself. So your A can be the same size as B only if they are in'nite.>We will make the set B = 100Z and the set C = Z.Now there is a 1-1 function from B->C, f(x) = x/100. So, there are the same>number of elements in 100Z as there are in Z.Yes.>Let A = B U (union) {a,b,c,d,e,....,x,y,z}.>So, does A have more elements than C???In this case, A and C are the same size. To see that, let {a,...,z} <-> {1,...,26}, {0,-1,-2,...} <-> itself, and {1,2,...} <-> {27,28,...}.>haha This drives me crazy! I hate in'nity.Well, there's a philosophy-of-mathaematics view called Finitism (and there's even Ultra'nitism) that you might want to look into.>Totally frustrated!!!I hate in'nity! Give me things I can deal with and conceptulize like>Z/<5>! or S3 (S with subscript 3)...The good news is if you work with in'nite sets for a while your intuitions will become more comfortable with them.-- ---------------------------| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- = : The application of one-to-one mapping as applied to potentially in'nite : sets is intuitively unsatisfactory. : : eg {2n}U {2n-1} = {n} is true for all 'nite n, so why would it not be true : ïas n tends to in'nity'?Well, limits are just different, that's why. : Similarly {n} - {f(n)} = {f ï(n)}, where the latter set is the complement of : {f(n)}, ought to be universally true on intuitive and aesthetic grounds.In the usual set theory around these parts, complements don't even exist --the universal set doesn't exist. : It is clear that ïincomplete' sets like {n} have no greatest memberJust because a set is in'nite does NOT mean that it is in ANY senseincomplete. Indeed, if you want a set that COMPLETELY contains ALLthe natural numbers, it would HAVE to be In'nite. It is the NON-in'nite sets of naturals that DESERVE to be called incomplete, sincethey, unlike the set of all of them, are the sets that leave some of thenaturals out. More to the point, even though N does not have a lastelement, it DOES have a 'rst one, so if you think not having a last elementis so horrible, just use > instead of < as your order relation. Then NWILL have a last element. My point is simply that there are INNUMERABLY MANYDIFFERENT ways in which you can order N; the usual way is just ONE way andit is NOT special. : (in the sense of Dadekind sections) so, intuitively, they can have no meaningful : totals, however de'ned.No, I'm sorry; that is NOT intuitive. : If there were such a total, the supposed process of : one-to-one correspondence would fail anyway, as the function : an + b exceeded it.That is not clear. : Unless such a process can be viewed from the ïtop end' : the way it works ïat in'nity' cannot be sensibly spoken about.That, though, is true. But why is it a problem?When you can view it from the top end, you do, and that's the answer. : The next hurdle is to reject the idea there is a unique entity called : ïin'nity'. In'nite, in common parlance, means ïnot-'nite'. The subject : matter of in'nity pertains to the in'nitely small as well as the : in'nitely large.Not really. The in'nitely large comes 'rst.The in'nitely small, to the extent that you can de'ne it at all,is de'ned in terms of the in'nitely large. : Consider the set of proper fractions {F(n+)} = {1/n}. This in'nite set, : like {n+}, has no greatest memberWrong; it does. Under the usual ordering, its greatest member is (1/1)=1. : but, unlike {n+}, has no least member.Right. Why should this be any more or less problematic than thesame problem in the other direction? : An : interesting property of {F(n+)} is that all of its members are less than any : member of {n+}. : : Suppose, now, we de'ne a set {G(n)}where all of its members exceed those of : {n+}.THAT ISN'T a de'nition. THAT de'nition (if it were one) is satis'edby the empty set: all of ITS members -- ALL ZERO of them -- exceed everyelement of {n}.You 'rst have to expect that its POSSIBLE for there to existsomething bigger than all elements of N. The above is an existenceassertion, NOT a de'nition. What you meant was, Suppose there is somethingbigger than every element of {n}. Suppose further that there is a set whereall of its members have this property (a set of supernatural numbers).The 'rst thing to point out is that this thing, whatever it maybe, is NOT a number in the USUAL intuitive sense. : Formally:(paraphrased)Aabn[ ( a e{F(n+)} & b e{G(n)} ) -> a a* : : Df: a*/a = b*/b = 1* so a*/b* = a/b : : This provides a link with 'nite numbers. : : It follows that 1* x a = a*. : : For any rational or real number r there will be a corresponding in'nite : number r*. : : The desire for a workable in'nite arithmetic is satis'ed. : All that is necessary is to ditch the theory of in'nite cardinals founded : on set theory.Just because the supernatural numbers in non-standard models of Peanoarithmetic are not the traditional ZFC in'nite cardinals does NOT meanthat those cardinals have to be ditched. Peano arithmetic is one thing.Cardinal arithmetic is another.-- --- It's dif'cult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America Thomasalt.algebra.help,sci.logic,sci.math:> The application of one-to-one mapping as applied to potentially in'nite> sets is intuitively unsatisfactory.> eg {2n}U {2n-1} = {n} is true for all 'nite n, so why would it not be true> ïas n tends to in'nity'?It isn't true, if you're using the notation correctly: {2n},{2n-1}, and {n} are all singletons, and {2n} U {2n-1} = {2n, 2n-1}. I can only guess that you're misusing the notationand actually intend ï{2n}' to signify {2n : n in I} for someunspeci'ed index set I. Of course it's not true in general that{2n : n in I} U {2n-1 : n in I} = {n : n in I}, so this doesn'thelp you very much. I can think of at least one other notionthat you might be attempting to convey, but it doesn't helps youeither, since it's true for all ordinals, not just 'nite ones.> Similarly {n} - {f(n)} = {f ï(n)}, where the latter set is the complement of> {f(n)}, ought to be universally true on intuitive and aesthetic grounds.Again, the notation is grossly misused, and ïcomplement' requiresa speci'cation of the set in which the complement is beingtaken. On the most reasonable interpretation of this mishmash,however, it *is* a true statement; intuition and aesthetics don'tcome into it.> It is clear that ïincomplete' sets like {n} have no greatest member (in the> sense of Dadekind sections) so, intuitively, they can have no meaningful> totals, however de'ned. If you want to talk mathematics, learn the damn' language. Theset {n} has a single element and hence obviously a largestelement. If you mean N, the set of natural numbers (non-negativeintegers), or Z+, the set of positive integers, say so. In anycase, totals have nothing to do with the matter.> If there were such a total, the supposed process of> one-to-one correspondence would fail anyway, as the function> an + b exceeded it. Okay, you *are* either an ignoramus or a nut-case. Never mind,then. F'ups set.[...] = If you want to talk mathematics, learn the damn' language. The> set {n} has a single element and hence obviously a largest> element. If you mean N, the set of natural numbers (non-negative> integers), or Z+, the set of positive integers, say so. In any> case, totals have nothing to do with the matter.> Okay, you *are* either an ignoramus or a nut-case. Never mind,> then. F'ups set.>I believe this is called a rush to judgement. True nut-cases--among which Imomentarily include you--treat problems of communication as problems best dealtwith using rude verbiage.I strongly suspect that you are a better person than that. Perhaps you can showeveryone?:-)mitch =I will copy/paste the last half of my post made &prev=and then bottom-post more.> ....> And A(r) is, if I am right, ...> > oo> --- (1+1/2+1/3+...+1/k)> -------------------> / k^(r+1)> ---> k=1> > = sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^(1+r),> > which is an Euler sum.> [ http://mathworld.wolfram.com/EulerSum.html ]> > (For example,> A(1) = 2*zeta(3).)> > (Right?)> > > So, since q(r,m) = > > sum{k|m,k<=sqrt(m)} k^r /m^r> > (note inequality's direction here, as opposed to in q()'s de'nition),> > I wonder naively> if there is some kind of zeta-function-like re§ection formula for the> Euler-sum analytical continuation which relates> > E(r) and E(2-r),> > where E(r) is > sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^r.> > (I know of no study regarding analytical continuating Euler sums at> all, nor anything beyond the consideration of such at r = integers >=> 2.)I must ask again, is there anything commonly known about theanalytical continuation of Euler-sums, such as where its zero's (ifany) are?? Is there any kind of zeta-like continuation-formula (analogous to thatwhich relates zeta(z) with zeta(1-z)) for Euler sums?Are there any known closed forms, in terms of zeta-functions orwhatever, for any E(r)'s where r is *not* a =Let s(r,m) =--- r > k/---k|m1<= k <= sqrt(m)(which is, in linear-mode)sum{k|m,1<= k<= sqrt(m)} k^r.So, we have s(r,m) is the sum of the r-powers taken over the lower half of the positivedivisors of m.For example, s(1,m) is:http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A066839 If r is > 0 (r = any *positive* real), then:limit{m -> oo} m --- 1 ------- > s(2r,k) =m^(r+1) / --- k=1 1---------- (?)2 r (r+1)Linear-mode:limit{m->oo} (1/m^(r+1)) sum{k=1 to m} s(2r,k) = 1/(2 r (r+1)) (?)(I am err-prone today, so I hope I thwarted fate...)Example: If I am right, the sum of the 'rst m terms of the EIS'sA066839 divided by m^(3/2) approaches =1)k in CQ such that A={a+bk; a,b in Z} is a ring with +, x operations of CSuppose A has exactely 4 elements invetibleProve that A= Z[i]---------------------------------------------------------- --------------2)K 'nite 'eld n>=2 an integer f in K[X] irreductible polynomial of degreeng= product of all nonconstant polynomials of K[X] wich have degree 1)>k in CQ such that A={a+bk; a,b in Z} is a ring with +, x operations of C>Suppose A has exactely 4 elements invetible>Prove that A= Z[i]Since A is a ring, k^2 must lie in A, so k^2 = a_0+b_0*k for someintegers a_0, b_0, not both zero. Therefore, k satis'es the quadraticpolynomial x^2 -b_0 x - a_0 = 0.It cannot be that a_0 = 0, for then k=0 or k=b_0, and k was not arational number. So a_0 is nonzero. Let D = b^2+4a_0, thediscriminant. D cannot be the square of a rational, so write D =r^2*d, with d square free. Then A = Z[sqrt(d)] (prove it).Now assume that A=Z[sqrt(d)] has exactly four invertible elements. Wewant to show that d=-1.If a+b*sqrt(d) is invertible, then so is a-b*sqrt(d); since the onlyunits with b=0 are 1 and -1 (prove it), the only other two units area+b*sqrt(d) and a-b*sqrt(d) for some a,b, b nonzero, and they areinverses of each other. So1 = (a+b*sqrt(d))*(a-b*sqrt(d)) = a^2 -db^2.The only possible solutions of a^2-db^2 = 1 with d<0 are a=1,b=0;a=-1,b=0 (any value of d), and d=-1, a=0, b=1, d=-1, a=0, b=-1. Thisshows that A=Z[sqrt(d)] = Z[sqrt(-1)] = Z[i].-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu =Islimit{n-> oo} (1/n) (product{k=1 to n} product{j|k} (1 +1/j)^j )^(1/n)= =*** post for FREE via your newsreader at post.newsfeed.com ***> Is there anyone who can tell me where to 'nd resources on Goedels> theorem gives me 3510 hits. Rene. -- > Ren.8e Meyer> Student of Physics & Mathematics> Zhejiang University, Hangzhou, ChinaIt should be Godel's INcompleteness theorem. Mathworld has an entry on thesubject:http://mathworld.wolfram.com/ GoedelsIncompletenessTheorem.htmlWikipedia also covers the subject: http://en.wikipedia.org/wiki/G%F6del's_incompleteness_ theoremQuite how reliable these sources are on the subject I'm not sure, as I amnot that familiar with it. They provide further links to other sources.http://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- =>> Is there anyone who can tell me where to 'nd resources on Goedels>> completeness theorem gives me 3510 hits.>It should be Godel's INcompleteness theorem. No, that's a different theorem. Goedel proved both completenesson both.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =2 Players.Each player has an equal number of cards from the shuf§ed deck dealtto them.Each player places the cards down face-up, one each turn, in one oftwo rows (one row per player). Each player chooses the card he/she will play simultaneously astheir opponent chooses and chooses secretly on each turn, but thenplaces their card into their row at the same time as their opponent,and places it so the opponent is then aware of it. When two cards are next to each other in a player's row, the cards have a direction (greater-than, less-than, equal-to), whichis the newly placed card's numerical value in relation to thenumerical value of the card immediately previously played.A player gets a point every time the direction of his/her two mostrecently played cards matches the direction of his/her opponent'scard-pair which were played one and two moves previously.(ace = 1, jack = 11, queen = 12, king = 13) Example:(random rows, no stategy, ignoring suits)2 > A < 3 > A < 10 < K > Q > 4 0 1 1 1 0 0 0 = 3 points4 < 6 > A < 3 > 2 < 9 = 9 > 7 0 1 1 1 1 0 1 = 5 pointsAny suggestions for strategy???And we can of course play with the integers 1 through n instead ofusing cards, where each Quet =Rationals are UncountableLet S be the set of all rational numbers [0,1).s is a member of S if 0.000... <= s < 1.000...and s is rational.Assume s is represented in base factorial (!).Base ! is used because every rational numberhas a 'nite representation in base !.In base ! the allowable digits forposition k are (0,1,...,k).(k starts at 1)Every position, k, represents 1 / (k+1)!.k1 1/2! = 1/22 1/3! = 1/63 1/4! = 1/24.123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10).Every rational number has an unique 'nite base !representation. Any 'nite base ! number is rational.We can create the set S de'ned above by taking theset produced by counting in base !..0.1.01.11.02.12.001.101...There exists a rational number, x, not in S.If S(i) is of the form .111...1 andits length is equal to or greater than xthen set x to a string of 1's one longer than S(i).x differs from every member of S.x is a rational number because it has a 'nite numberof digits. The length of x is exactly one greaterthan some member of S.0.0 <= x < 1x = 1/2! + 1/3! + ... + 1/k!and equals the largest rational numberless than the fractional part of e.Actually, it is the smallest rational approximation of e that is not in setSwhere a rational approximation of e is any number of the form .111...111.Russell- Zeno is right. Motion is impossible. =There are some really interesting results in base !, and you seem toreally be progressing through them, but there are a couple of issueswith what you have said (not the least of which being that rationals/are/ countable).> If S(i) is of the form .111...1 and> its length is equal to or greater than x> then set x to a string of 1's one longer than S(i).But the set as you de'ned it contains .1, .11, .111, .1111, .11111,...So actually your x has to have in'nity+1 digits! Every 'nite numberof that form already is in S.> x = 1/2! + 1/3! + ... + 1/k!> and equals the largest rational number> less than the fractional part of e.Are you serious about this? The rationals are dense in the reals, sobetween every two reals there is another rational. You can see thatyourself, since as k goes to in'nity you get rational numbers thatget arbitrarily close to e.It also seems that you attempted to demonstrate that for this setupthere is a rational not in the set you de'ned, but that does not meenthat there is a /different/ countable set that does contain all reals(and if you check your replies you will see one).I have found time and time again that it is a lot easier to provesomething that is true (and to read the question, but that is adifferent story altogether). =>Rationals are UncountableLet S be the set of all rational numbers [0,1).>s is a member of S if 0.000... <= s < 1.000...>and s is rational.Assume s is represented in base factorial (!).>Base ! is used because every rational number>has a 'nite representation in base !.In base ! the allowable digits for>position k are (0,1,...,k).>(k starts at 1)Every position, k, represents 1 / (k+1)!.k>1 1/2! = 1/2>2 1/3! = 1/6>3 1/4! = 1/24.123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10).Every rational number has an unique 'nite base !>representation. Any 'nite base ! number is rational.We can create the set S de'ned above by taking the>set produced by counting in base !..0>.1>.01>.11>.02>.12>.001>.101>...This list contains all allowable 'nite base ! fractions. By yourstatement above that every rational number has a unique, 'nite base !representation, this list contains all rational numbers in [0,1).If you are assuming that this list is 'nite, then all you are provingbelow is that there are in'nitely many rationals, not uncountably many.Thus, I will assume that this list is in'nite and contains all possiblebase ! fractions.>There exists a rational number, x, not in S.If S(i) is of the form .111...1 and>its length is equal to or greater than x>then set x to a string of 1's one longer than S(i).Wait a minute; if x has been de'ned, then why are we changing x? Evenif this were part of a method to 'nd x, for all x, there will always bean S(i) of the form .111...1 whose length will be greater than x; thus,this process will never end and x will equal e-2, which is not rational.>x differs from every member of S.As mentioned above, x = e - 2, which is not rational.>x is a rational number because it has a 'nite number>of digits. The length of x is exactly one greater>than some member of S.Assuming that you are thinking of the preceding method as de'ning x,then x has an in'nite number of digits.>0.0 <= x < 1x = 1/2! + 1/3! + ... + 1/k!>and equals the largest rational number>less than the fractional part of e.There is no largest rational number less than e-2. This should tell youthat something is amiss.>Actually, it is the smallest rational approximation of e that is not in set>S>where a rational approximation of e is any number of the form .111...111.Since S is the set of all rational numbers in [0,1), any rationalapproximation of e-2 in [0,1) is in S.Rob Johnson take out the trash before replying =On UncountableYou've certainly given this some thought, but might I suggest nexttime try to prove something not trivially false.>Let S be the set of all rational numbers [0,1).>s is a member of S if 0.000... <= s < 1.000...>and s is rational.Fine.>Assume s is represented in base factorial (!).>Base ! is used because every rational number>has a 'nite representation in base !.In base ! the allowable digits for>position k are (0,1,...,k).>(k starts at 1)>Every position, k, represents 1 / (k+1)!.k>1 1/2! = 1/2>2 1/3! = 1/6>3 1/4! = 1/24.123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10).Every rational number has an unique 'nite base !>representation. Any 'nite base ! number is rational.Yes but you also assume a special form for the base-! representationthat you really should prove to be i) unique ii) cover all rationalsbetween 0 and 1. I'll take it as given for now.>We can create the set S de'ned above by taking the>set produced by counting in base !..0>.1>.01>.11>.02>.12>.001>.101>...There exists a rational number, x, not in S.I suppose between 0 and 1, otherwise this is not relevant.>If S(i) is of the form .111...1 and>its length is equal to or greater than x>then set x to a string of 1's one longer than S(i).What is S(i)? The largest number whose base-! representation consistsof all ones? It doesn't exist! Your de'nition creates a set thatincludes among other things all numbers of the form:S(i) = Sum(k:2->i, 1/k!)where i is in N. The only number *not* in the list is the limit ofsaid sum as i tends towards in'nity, which converges to theirrational number e - 2.In fact, the unique representation of rationals in base-! is one wayof showing there exists a bijection from N to Q.>Actually, it is the smallest rational approximation of e that is not in set>S where a rational approximation of e is any number of the form .111...111.In other words, the maximum of an open set. It does not exist. => Rationals are UncountableThen how do you explain the numerous _injective_ mappings one may construct from Q to N?For example:An arbitary rational may be represented uniquely as p/q, where, WLOG, we may require that: p is an integer, and q is a positive integer, and p and q have no common factors except 1 and if p = 0 then q = 1De'ne f : Q -> N by f(p,q) = 2^p*5^q if p is positive or zero, and f(p/q) = 3^(-p)*5^q if p is negative.It is trivial to show that f is injective, thus the cardinality of the rationals cannot be larger than that of the naturals.And, of course, the trivial injective mapping g:N -> Q : n -> n/1completes the issue. =How about the Chinese Remainder Theorem. At least the name issomewhat suggestive, but why can't we just call it the remaindertheorem. What makes it inherently Chinese. => How about the Chinese Remainder Theorem. At least the name is> somewhat suggestive, but why can't we just call it the remainder> theorem. What makes it inherently Chinese.It is attrbuted to the 13-th century Chinese mathematicianCh'in Chiu-Shao.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) => It is attrbuted to the 13-th century Chinese mathematician> Ch'in Chiu-Shao.Yes, I know that, but my point is that if you look at it one way, itstill seems strange. While more memorable than Chiu-Shao's remaindertheorem, the actual theorem isn't too chinese. Pardon my ramblings. =a simple question:Somebody knows the notation in the following picture:http://www.mycgiserver.com/~devwebcl/notation.gi's it the highest integer less than or equal to (n-j)/2 ?(like §oor function in computer laguages?)thus if those parenthesis are backward will be like Ceil?(Ceil to obtain the lowest integer =* German Gonzalez> a simple question:> > Somebody knows the notation in the following picture:> http://www.mycgiserver.com/~devwebcl/notation.gif> > is it the highest integer less than or equal to (n-j)/2 ?> (like §oor function in computer laguages?)> > thus if those parenthesis are backward will be like Ceil?> (Ceil to obtain the lowest integer greater than or equal to X)> Well, notation is just notation, but this is quite common yes. Lookhere (found on Google):http://www.chilton.com/~jimw/symbols.html(Differences between mathematics and computer science may take placewhen the numbers are negative.)-- Jon Haugsand =I have been struggling with this problem for the last few days. Doesanybody have any advice? This is from Royden I think.Construct (or prove existence of) a closed set K in the unit squaresuch that any closed set in the unit interval is equal to {x: (x,c) much all! =>I have been struggling with this problem for the last few days. Does>anybody have any advice? This is from Royden I think.Construct (or prove existence of) a closed set K in the unit square>such that any closed set in the unit interval is equal to {x: (x,c) >much all!> >Let (I_n: n in N) count the open intervals with rational end points.For x in [0,1] let O_x be the union of the intervals I_n for which x_n=1; wherex=0.x_1x_2x_3... (in binary notation)The set K = {(x,y): y notin O_x} is closed and does exactly what you want.KP => Construct (or prove existence of) a closed set K in the unit square> such that any closed set in the unit interval is equal to {x: (x,c)> much all!>Is the unit square open or closed?Is the unit interval open or closed?K can be any closed set containing C = { (x,c) | x in unit interval }In particular, C will do. =I am looking for some web sites and references that deal withalgorithms of optimization that permit to arrange geometric shapes inorder to uniformly cover a surface (example of problem : How tooptimize the parameters of a sprayer in order to have an uniformthickness coating?). Have you some informations about these sorts ofproblems?Jean => I am looking for some web sites and references that deal with> algorithms of optimization that permit to arrange geometric shapes in> order to uniformly cover a surface (example of problem : How to> optimize the parameters of a sprayer in order to have an uniform> thickness coating?). Have you some informations about these sorts of> problems?> JeanI'll give you a real world problem that I encountered a few days ago thatmight be what you're looking for.I work for a meteorology website. We are currently undergoing some changesand want to promote our website even more. As a result, we consideredlooking at making rain gauges. Being an environmental website, we want touse the least amount of material, but we'd like it to hold 50 cc of liquid.What dimensions do you make the gauge?E-Mail me for the answer.-- David MoranChief MeteorologistOklahoma Storm Team => I am looking for some web sites and references that deal with> algorithms of optimization that permit to arrange geometric shapes in> order to uniformly cover a surface (example of problem : How to> optimize the parameters of a sprayer in order to have an uniform> thickness coating?). Have you some informations about these sorts of> =>I am a high-school student in Greece and I have read in a Geometry>book about Fermat's problem - that is, given a triangle ABC, to 'nd a>point P, so that>PA + PB + PC = minimum . The book mentions that P is so, so that the>angles> ^ ^ ^>APB = BPC = CPA = 120 degrees. However, this doesn't seem right to me.>What happens if the triangle has an obtuse angle greater than 120>degrees?As was mentioned by Sylvain Croussette, the Fermat point of an obtusetriangle is the vertex with the obtuse angle.For a generalization of this theorem and a proof of an algorithm to 'ndthe Fermat point of a 'nite set of points, see http://www.whim.org/nebula/math/fermatpt.htmlThere it is proven that if the points are not in a line, their Fermatpoint is unique. It is also shown that if, for some k, | --- p_k - p_j | | > ----------- | <= 1 [1] | --- |p_k - p_j| | j<>kthen p_k is the Fermat point. This corresponds to the vertex with theobtuse angle in an obtuse triangle. Otherwise, iterating the function --- p_k > --------- --- |p - p_k| k M(p) = ------------- [2] --- 1 > --------- --- |p - p_k| kconverges to the Fermat point except possibly on a countable set. Notethat [2] is a positively weighted mean of the {p_k}. If [1] does nothold, then if p is the Fermat point of {p_k}, we have --- p - p_k 0 = > --------- [3] --- |p - p_k| kEquation [3] is the generalization of the 120 degree condition mentionedabove.Rob Johnson take out the trash before replying =This is a problem that should be very easy to get rid off, yet it's tormenting me.(warning: sorry for my broken english, I am from Italy)suppose you've got a list of integersN1, N2, N3, N4, N5... Nmsorted from the smallest to the biggest one. The distribution is absolutelyrandom, so that the list could be(1,2,2,3,3,4,5,5,5,6,7)but also(5,26,6246,20123400)(3,3,27235235,12347472479679234592646) etc.starting from such a list, I want to multiply its elements until I obtain anotherlist made up just of two integers, whose difference must be as little as possible.Let me explain: for example(2,2,2,3,3,5) may be grouped as (2*2*2*3,3*5) = (18,15)are there known algorithms or this proceeding? Anything will work, heuristicor deterministic... speed isn't important. - andreab.93, http://www.escaped.it = > suppose you've got a list of integers> > N1, N2, N3, N4, N5... Nm> > sorted from the smallest to the biggest one. The distribution is absolutely> random, so that the list could be> > (1,2,2,3,3,4,5,5,5,6,7) but also> > (5,26,6246,20123400)> > (3,3,27235235,12347472479679234592646)> > etc.> > starting from such a list, I want to multiply its elements until I > obtain another list made up just of two integers, whose difference> must be as little as possible.> are there known algorithms or this proceeding? Anything will work,> heuristic or deterministic... speed isn't important.It's not clear if you're asking for the absolute smallest difference,or something guaranteed to be close to it.You might look at bin packing algorithms, where you'd be packingthe logarithms of the integers into two bins. =>suppose you've got a list of integersN1, N2, N3, N4, N5... Nmsorted from the smallest to the biggest one. The distribution is absolutely>random, so that the list could be(1,2,2,3,3,4,5,5,5,6,7)but also(5,26,6246,20123400)(3,3,27235235,12347472479679234592646) etc.starting from such a list, I want to multiply its elements until I>obtain another>list made up just of two integers, whose difference must be as little as>possible.>Let me explain: for example(2,2,2,3,3,5) may be grouped as (2*2*2*3,3*5) = (18,15)are there known algorithms or this proceeding? Anything will work, heuristic>or deterministic... speed isn't important.A very nice approximate method for the problem of splitting a set intotwo subsets of approximately equal sum is due to Karmarkar and Karp. Itcan be adapted to your problem with equal products, as follows:remove the two largest elements N_{m-1} and N_m and replace them by N_m/N_{m-1} (inserting this in the appropriate place in your list). Repeat this until there is only one element in the list. Label thisby A. Then repeat the following: if x is in the list and came fromy/z, give y the same label as x and give z the other label (chosen fromA and B). When all the original elements are labelled, those labelledA go in one subset and those labelled B in the other. Note that if youhave different copies of the same number, these are considered separateand can have different labels.Here's an example: start with2, 3, 6, 7, 8, 12replace 8 and 12 by 3/2 so you have 3/2, 2, 3, 6, 7replace 6 and 7 by 7/6 so you have 7/6, 3/2, 2, 3replace 2 and 3 by 3/2 so you have 7/6, 3/2, 3/2replace 3/2 and 3/2 by 1 so you have 1, 7/6replace 1 and 7/6 by 7/6 so you have 7/6.Label 7/6 by A. This came from 1 and another 7/6: label that 7/6 A, 1 BThat 7/6 came from 6 and 7: label 7 A, 6 BThe 1 came from 3/2 and 3/2: label one 3/2 A and the other BOne 3/2 (say the one we labelled A) came from 3 and 2: label 3 A and 2 BThe other 3/2 came from 8 and 12: label 12 B and 8 A.Now we have the subsets (from the original set)3, 7, 8 labelled A for product 1682, 6, 12 labelled B for product 144Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => > This is a problem that should be very easy to get rid off, yet it's > tormenting me.... > suppose you've got a list of integers> > N1, N2, N3, N4, N5... Nm...> starting from such a list, I want to multiply its elements until I obtain [two integers]> whose difference must be as little as possible.... > are there known algorithms or this proceeding? Anything will work, heuristic> or deterministic... speed isn't important.The easy way to dispose of this problem is to translate it to a knapsack problem, and algorithms, they must be awfully easy to come up with, wouldn't you think? Anyhow, say the total of the logs of your numbers is t. Set up two knapsacks that hold a little more than t/2 each, and proceed in the obvious manner.-jiw => This is a problem that should be very easy to get rid off, yet it'stormenting me.> (warning: sorry for my broken english, I am from Italy) suppose you've got a list of integers N1, N2, N3, N4, N5... Nm sorted from the smallest to the biggest one. The distribution isabsolutely> random, so that the list could be (1,2,2,3,3,4,5,5,5,6,7) but also (5,26,6246,20123400) (3,3,27235235,12347472479679234592646) etc. starting from such a list, I want to multiply its elements until I obtainanother> list made up just of two integers, whose difference must be as little aspossible.> Let me explain: for example (2,2,2,3,3,5) may be grouped as (2*2*2*3,3*5) = (18,15) are there known algorithms or this proceeding? Anything will work,heuristic> or deterministic... speed isn't important.If there are n elements in your list then the list can be split into twoparts in 2^n ways. (Some of these splits will not be distinct if the listhas repeated values.) Just try all 2^n splits and see which is best.Applying this method to your (2,2,3,3,4,5,5,5,6,7) gives a least differenceof 11: 2*2*3*3*4*6=864 and 5*5*5*7=875 .-- Clive Toothhttp://www.clivetooth.dk =The point is that it's not necessary to split the list. I can take elements and multiply them from ANY position.Example:(1,1,1,3,3,5,5,7,7,11,13)can be:a=1*3*5*7*7*11b=3*5*14(a,b) => The point is that it's not necessary to split the list. I can takeelements and multiply them from ANY position.> Example: (1,1,1,3,3,5,5,7,7,11,13) can be: a=1*3*5*7*7*11> b=3*5*14 (a,b)Seems to be some 13/14 confusion.Also, starting your example with 1,1,1 does not seem very helpful.Erm... anyway... I am not sure I understand you.In my example, which you have not quoted, I split the list (2,2,3,3,4,5,5,5,6,7)into (2,2,3,3,4,-,-,-,6,-)and (-,-,-,-,-,5,5,5,-,7)Is there something wrong with that?-- Clive Toothhttp://www.clivetooth.dk => Is there something wrong with that?No but there's something seriously wrong with ME : )Sorry, hours of work on 10,000 digits numbers... I am starting to feel dizzy. - andrea => Is there something wrong with that? No but there's something seriously wrong with ME : )> Sorry, hours of work on 10,000 digits numbers... I am starting to feeldizzy.Ok.Another way would be to compute the logarithm, to any base, of each of thenumbers, to double precision §oating point accuracy (say).Then add those logs together and halve the answer, that is the log of thesquare root of the product.Then compute the logarithm of each of each of the 2^n partial products usingHowever, if several sets give a sum which is very close to the log of thesquare root, you should recompute those products exactly to decide theissue. This should reduce the amount of high precision arithmetic involved.-- Clive Toothhttp://www.clivetooth.dk => > starting from such a list, I want to multiply its elements until I obtain > another list made up just of two integers, whose difference must be as> little as possible.Sounds like a homework problem to me.> are there known algorithms or this proceeding? Anything will work, heuristic> or deterministic... speed isn't important.There's one obvious one: multiply the 'rst K integers and the lastN-K, and compute the difference. Pick K such that the difference isminimized. (I'm assuming that you must multiply successiveintegers--i.e., that you are not allowed to reorder them beforemultiplying.)Unfortunately, if this is a homework problem, your teacher probablydoesn't want the brute-force algorithm. Also, an algorithm question isprobably better directed to a comp.* newsgroup.Len. => There's one obvious one: multiply the 'rst K integers and the last> N-K, and compute the difference. Pick K such that the difference is> minimized. (I'm assuming that you must multiply successive> integers--i.e., that you are not allowed to reorder them before> multiplying.)Right. I thought to this approach already, but.... well, when I saidspeed isn't important, I wasn't counting this one! :)> Unfortunately, if this is a homework problem, your teacher probably> doesn't want the brute-force algorithm. Also, an algorithm question is> probably better directed to a comp.* newsgroup.No, this isn't homework at all... but brute force isn't that good anyway,since I am dealing with numbers which range from 1 to 1,000,000 digits(yes, quite large beasts).You're right about the comp.* thing, but I think that it's more likely thata mathematician will generate the sparkle which will light the 're of the algorithm,rather than I'll 'nd that 're already burning in the comp.* world ;) =I want to know what is the theory behind solving the following equation.a*u^2=b mod p, the 'eld is F_p of char equal 3.And the equation a*u^4=b & u^6*c=c'+ ra+ r^3. I need to 'nd out u,resp r based on a discussion depending on coef'cients.A => I want to know what is the theory behind solving the following equation.> a*u^2=b mod p, the 'eld is F_p of char equal 3.> And the equation a*u^4=b & u^6*c=c'+ ra+ r^3. > I need to 'nd out u,resp r based on a discussion depending on coef'cients.> AF_p is the 'eld with 3 elements, yes?Then, assuming a is not zero, a^2=1, (a zero being trivial), then you areasking for the roots (u) of u^2 = ab (3). Well, as long as ab=1, there aretwo roots u=1 or 2. If ab=0 (ie b=0) there is one root, u=0, and if ab=2there are no roots. Looking at the next line makes me think F_p isn't the 'eld with threeelements, as U^4=u^2=u^6 in this 'eld, and r^3=r. the units in F_3 are acyclic group of order 2. Is there perhaps something else? Like F_q where q is a power of 3? =Excuse my poor english.I want to calculateint(exp(u*x+v/x,x=a..b) with 0I want to calculate>int(exp(u*x+v/x,x=a..b) with 0To simplify, i'm want to calculate int(exp(u*x+v/x,x=1..inf) : it's the >same problem.No it isn't. But I suspect that this isn't quite what you want either,because it will diverge at in'nity if u > 0.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Robert Israel a .8ecrit :> >I want to calculate>>int(exp(u*x+v/x,x=a..b) with 0 > > I assume that's int(exp(u*x+v/x),x=a..b). I doubt that there's> a closed form.yes> > >>To simplify, i'm want to calculate int(exp(u*x+v/x,x=1..inf) : it's the >>same problem.> > > No it isn't. But I suspect that this isn't quite what you want either,> because it will diverge at in'nity if u > 0.yes. converge for u<0 and v<0.y=alpha*x, alpha>0, so int(exp(u*x+v/x),x=1..inf)=int(exp(u*y/alpha+v*alpha/y),y= alpha..inf) same as before.> > Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i021sSx07352; =It is clear that you don't know what Cantor's proof actually says, since >Cantor did not assume a mapping from the naturals ONTO the reals, but >version of the proof is direct, not by contradiction, It is others who >insist on recasting it into an indirect proof.It is obvious that I dont know that direct proof. After reading your post, Ive just got three questions:1)Where can I get this proof?2)Why there are so many books (written by outstanding mathematicians, I think) that invariably show the proof by contradiction?3)May you explain what does exactly mean *arbitrary mapping* from the naturals to the reals? Perhaps does it mean a mapping from the naturals to an in'nite set of reals, but no the set of all the real numbers? Nicolas de la Foz =It is clear that you don't know what Cantor's proof actually says, since >Cantor did not assume a mapping from the naturals ONTO the reals, but >version of the proof is direct, not by contradiction, It is others who >insist on recasting it into an indirect proof.> It is obvious that I dont know that direct proof. After reading your > post, Ive just got three questions:> > 1)Where can I get this proof?> 2)Why there are so many books (written by outstanding > mathematicians, I think) that invariably show the proof by > contradiction?> 3)May you explain what does exactly mean *arbitrary mapping* > from the naturals to the reals? Perhaps does it mean a mapping from > the naturals to an in'nite set of reals, but no the set of all the > real numbers? > > > Nicolas de la Foz> (1) Try Google, though it is probably in German.(2) I don't know. Possibly, the authors think that proofs by contradiction are more dramatic.(3) By arbitrary I mean that all you are allowed to know, or assume, about the function is that it has domain the naturals and codomain the reals: f : N -> R, and that's all you can say about it.You do not know whether it is injective nor whether it is surjective nor its range, nor anything except its domain and its codomain. =>>It is clear that you don't know what Cantor's proof actually says, since >>Cantor did not assume a mapping from the naturals ONTO the reals, but >>version of the proof is direct, not by contradiction, It is others who >>insist on recasting it into an indirect proof.> It is obvious that I dont know that direct proof. After reading your > post, Ive just got three questions:> 1)Where can I get this proof?See below.> 2)Why there are so many books (written by outstanding > mathematicians, I think) that invariably show the proof by > contradiction?Probably because they are used to speaking to mathematicians and it hasnot occurred to them that an indirect proof might be harder to understandthan a direct proof.> 3)May you explain what does exactly mean *arbitrary mapping* > from the naturals to the reals? Perhaps does it mean a mapping from > the naturals to an in'nite set of reals, but no the set of all the > real numbers? Let f: N -> R be a mapping from the naturals to the reals. Note: we donot assume anything at all about f except that it is a mapping. We makeno assumptions about whether f is injective (1-1) or not, whether it issurjective (onto) or not, or whether the range of f is 'nite orin'nite. The mapping f is arbitrary because we assume absolutelynothing about it except that it is a mapping from N to R.This mapping determines a list of reals in the sense that we can writedown all the reals in the range of f in order: f(1) f(2) f(3) . . .Proposition: Let f: N -> R be given. Then f is not a surjection.Proof. We are to show that there exists x in R such that x is not in therange of f. That is, x != f(k) for any k in N.We do this by de'ning, for each k, the k-th digit in the decimalrepresentation of x. Given k > 0, we 'rst look at d_k, the k-th digitfollowing the decimal point in the representation of f(k) from our list.We next de'ne the k-th digit of x, x_k, as follows: If d_k is a 1, set x_k = 2. If d_k is not a 1, set x_k = 1.Then the number x = .(x_1)(x_2)(x_3)... is the required number. It isnot in the list because for each k, x differs from f(k) in the k-thdigit.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. =>>It is clear that you don't know what Cantor's proof actually says, since >>Cantor did not assume a mapping from the naturals ONTO the reals, but >>version of the proof is direct, not by contradiction, It is others who >>insist on recasting it into an indirect proof.> > > It is obvious that I dont know that direct proof. After reading your > post, Ive just got three questions:> > 1)Where can I get this proof?> > See below.> > 2)Why there are so many books (written by outstanding > mathematicians, I think) that invariably show the proof by > contradiction?> > Probably because they are used to speaking to mathematicians and it has> not occurred to them that an indirect proof might be harder to understand> than a direct proof.> > 3)May you explain what does exactly mean *arbitrary mapping* > from the naturals to the reals? Perhaps does it mean a mapping from > the naturals to an in'nite set of reals, but no the set of all the > real numbers? > > Let f: N -> R be a mapping from the naturals to the reals. Note: we do> not assume anything at all about f except that it is a mapping. We make> no assumptions about whether f is injective (1-1) or not, whether it is> surjective (onto) or not, or whether the range of f is 'nite or> in'nite. The mapping f is arbitrary because we assume absolutely> nothing about it except that it is a mapping from N to R.> > This mapping determines a list of reals in the sense that we can write> down all the reals in the range of f in order:> > f(1)> f(2)> f(3)> .> .> .> > Proposition: Let f: N -> R be given. Then f is not a surjection.> > Proof. We are to show that there exists x in R such that x is not in the> range of f. That is, x != f(k) for any k in N.> > We do this by de'ning, for each k, the k-th digit in the decimal> representation of x. Given k > 0, we 'rst look at d_k, the k-th digit> following the decimal point in the representation of f(k) from our list.> We next de'ne the k-th digit of x, x_k, as follows:> > If d_k is a 1, set x_k = 2.> If d_k is not a 1, set x_k = 1.> > Then the number x = .(x_1)(x_2)(x_3)... is the required number. It is> not in the list because for each k, x differs from f(k) in the k-th> digit.And note that none of its digits can be 0 or 9, so that it cannot be any of the numbers having two potential decimal representations, such as 1.000... = support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i021sTC07369; =>>The diagonal argument needs a premise (a list, otherwise it cant >>work). If you know a *neutral* premise, then I believe that the proof >>by contradiction would be valid.Assume a set of reals S and an injection f from the naturals to this>set (effectively making it a list). Then show that for any such f we>can always 'nd a real not in the map of f using the oh-so-wonderful>diagonal argument, so f can't be a surjection onto any S that contains>all reals. It follows that f cannot be a bijection.It doesn't really matter anyway since you've not given any argument>against proof-by-contradiction (in fact you accept it as correct in>the case of sqrt(2)) except it can be used to prove unintuitive>things so it must be wrong.If with the premise *Assume a set of reals S and an injection f from the naturals to this* you are supposing that S is *the set of ALL reals*, then it is a false premise because breaks the rules of naturals, but if S represents an INCOPLETE set of reals, then the premise is worse than false, it is stupid, since in this case you dont need the diagonal argument to 'nd out that at least one real number is not in S.To get a good conclusion, the proof by contradiction needs coherent premises, and this is not the case. To assume that sqrt(2) is rational is not a false and stupid premise because it doesnt break any rule and moreover, the probability of being rational before the proof is one half.Nicolas de la Foz the premise *Assume a set of reals S and an injection f from >the naturals to this* you are supposing that S is *the set of ALL >reals*, then it is a false premise because breaks the rules of >naturalsWhat? f:N->R, f(n) = n is an injection from the naturals to reals. No,it does not get all real values since it's not a surjection - thatwould be impossible. What do mean breaks the rules of naturals?>but if S represents an INCOPLETE set of reals, then the >premise is worse than false, it is stupid, since in this case you >dont need the diagonal argument to 'nd out that at least one real >number is not in S.Of course it's not stupid. It shows that any mapping between naturalsand reals can only be to a countable subset of reals. We can not startfrom an incomplete mapping and complete it by induction (adding realsone-by-one) because there's always some reals that we can add.>To get a good conclusion, the proof by contradiction needs coherent >premises, and this is not the case.De'ne coherent.>To assume that sqrt(2) is rational is not a false and stupid premise>because it doesnt break any rule and moreover, the probability of>being rational before the proof is one half.First you must prove that sqrt(2) is real. Then the probability of anyreal number being rational is 0.Sqrt(2) is of the form n/m where n and m are relatively prime and mis not 0 is a false premise, which leads to a direct contradictionwith itself. There exists a bijection between N and R is a similarlyfalse premise, which also leads to a direct contradiction with itself.I fail to see the difference between your suggested logical values offalse and false and support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i02Dn'20058; =>
My problem is
:>>I would like to minimise a function with N arguments. The
principle is>>to>>change the N args until the function is
equal to zero. My method is to>>add a deplacment to each arg,
and to see witch of them is the best, and>>to itter>>I don't
have a lot of af'nity with maths... Is it a well known
problem,>>with >>typic algorithms ? I don't really know 
where
to search on the web to>>'nd this>>type of informationThis is
unconstrained minimization. The method which you describe>is
slow and will not converge well. If your function is
continuous and>differentiable, a conjugate-directions method
is probably the best way>to go. A conjugate-directions solver
is part of my ANSI C subroutine library WNLIB.>You can
download WNLIB as described below. The conjugate-directions
>package is called
wnconj.>------------------------------------------------------
------------------------FEATURES: WNLIB is an ANSI C
subroutine library that contains numerous tools that> I have
found to be useful in my programming practice. The following
are included: miscellaneous: improved memory allocator and
memory debugger; parsing package data structures and
algorithms: linked list; balanced binary tree; hash table -
with common hash> functions; dd tree; sorting for lists and
arrays;> high-quality pseudo-random number generator
numerical: various matrix and vector operations, including
matrix inverse, > least-squares inverse linear programming:
simplex method; transportation problem; shortest path
problem;> critical path problem non-linear optimization:
conjugate directions and conjugate gradient methods for
constrained> and unconstrained problems; simulated annealing
random numbers: very good (cryptographically-strong)
pseudo-random number generator. > True random number generator
(no kidding!). Normal distribution,> Poisson distribution,
Cauchy distribution, etc.>LEGAL: BY DOWNLOADING OR COMPILING
THIS LIBRARY, YOU ACCEPT AND AGREE TO THE TERMS > AND
CONDITIONS PRINTED BELOW. IF YOU DO NOT AGREE, DO NOT DOWNLOAD
OR > COMPILE THIS LIBRARY. The author and Spike Technologies
provide this C code in the hope> that it will be helpful,
however, we assume no responsibility > for the use of this
code, nor any responsibility for its support.> The software is
distributed on an AS IS basis, without warranty.> Neither the
authors, the software developers, nor Spike Technologies> make
any representation, or warranty, either express or implied,
with> respect to the software programs and subroutines, their
quality, accuracy, > or 'tness for a speci'c 
purpose.
Therefore, neither the authors, the > software developers, nor
Spike Technologies shall have any liability to > you or any
other person or entity with respect to any liability, loss, >
or damage caused or alleged to have been caused directly or
indirectly by > the programs and subroutines contained in this
library. This includes, but > is not limited to, interruption
of service, loss of data, loss of classroom > time, loss of
consulting or anticipatory pro'ts, or consequential damages >
from the use of these programs. >COPYRIGHT NOTICE: The source
code in this directory is provided free of charge to anybody >
who wants it. It is in the public domain and therefore may be
used by > anybody for any purpose. This copyright notice and
the above legal notice> may not be removed.>AUTHOR: Will
Naylor Spike Technologies> 500 E. Calaveras Blvd #206>
Milpitas, CA 95035 Spike Technologies provides algorithm and
CAD software consulting> and contract programming and IC
design services.>DOWNLOADING and INSTALLATION:1) Create a
directory called wnlib wherever you want this code to> reside
and go into it. Type $ cd  $ cd wnlib2) Download using ftp or rftp or a
web browser. > > a) For ftp or rftp type $ ftp ftp.rahul.net
Type ïanonymous' at login prompt.> After you are 
logged in
type ftp> cd pub/spiketech/softlib/wnlib> ftp> get
INSTALL.txt> ftp> get uuwnlib.z> ftp> quit b) For a web
browser, go to > > http://www.spiketech.com/spike> > Click
over ïSpike Freeware' link. Read instructions 
and > click over
 to download. Your browser should throw a >
dialog box; you can select a directory where you want to store
it> and click ok. The browser will download it.3) Decode the
install tar 'le. Type $ uudecode uuwnlib.z> $ uncompress <
tar'le | tar xvf -4) Make the code. Type $ make all5) If this
fails, it may be necessary to modify variables in 
acc/make'le>
to 't the platform you are running on.6) When you believe you
have compiled correctly, type $ make selftest to run the
self-testing diagnostics. If selftest runs successfully> to
completion with no error messages, you have successfully
installed> wnlib.>USE:1) Summary: ARCHIVE is in:
wnlib/acc/text.a> H FILES are in: wnlib/acc/h/*.h> MAN PAGES
are in: wnlib/doc/man/*.man 2) Under wnlib, the directory acc
contains .h 'les to include and archive> 'les to 
link to. The
directory acc/h contains links to all of the > .h 'les. The
directory acc contains an archive 'le called text.a. > Use 
the
-I switch> of cc to include the .h 'les. See the 
make'le's in
acc for examples> of linking text.a 'les and including .h
'les.3) man pages are to be found in doc/man. The script 
wnman
in the> directory command is a convenient way to view these man
pages.> Type wnman  to view a man page.> The command
wnman -k  does a keyword search on all> the man
pages.> > The variable wnlib must point to your wnlib
directory for> wnman to work. A csh command of the form $ set
wnlib = xxx/yyy/wnlib sets this up. Include such a command in
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i02Dngq20080;
=Quoting from the
introduction of the book Rings,Fields, and Groups
(allenby):There seems little point in giving here a list of
contentsmore detailed than that which follows. To the
experienced reader those section headings will reveal all; for
the beginnereven more detail will reveal little...I suppose
that applies to names as well: For an absolute beginner fresh
out of school,abelian group probably won't be much more
tantalizing than the descriptive noncommunative group,where
the student may even feel a little intimidated;after all,
he/she has probably almost exclusivelyencountered variables
such as x, y, z, etc. in connection with real numbers and not
with n*n matrices (usually written capatilized, if they were
discussed as variables in equations at all), permutation
functions, elements of abstract groups, etc. during the last
3-4 years with the exception of the vector cross product; a
student thinking so what they're telling mehere is that
noncommunative group elements are similar tovectors forming a
cross product will probably do more to confuse. At least when
a student reads abelian groupand sees what is being de'ned
she/he can say: xy != yx? that Abel person must have been
onecrazy guy... Not that being descriptive is a bad thing in
general; an advanced reader may well be inclined to make his
own descriptive terms from names, such as GODel (whichwas
mentioned earlier in this thread...). By the way, in German
for example, words are concantinated much more often than in
English, making them naturally moredescriptive: Wertebereich
der Funktion =Werte-bereich der Funktion = value-area of the
function =range of the functionUrbild der Funktion =Ur-bild
der Funktion =Pre-picture of the function =domain of the
function Eine Einfuerung in die Mathematik =Eine Ein-fuerung
in die Mathematik =A In-Leading to Mathematics =An
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i02Dnj520184;
=Still being kind of a newcomer to ('nite)
group theory, I read over the proof of Lagrange's theorem:H
subgroup G -> |G| = |H||G:H|. And of the manysubsequent
examples (and problems). The authorof the introductory text
book I was reading misseda very easy example, in my opinion:If
G has a subgroup H of even order, then G is even.Proof: If G
has a subgroup H of even order, then |G|= |H||G:H|. The index
of H in G, |G:H|, is either even or odd. It doesn't matter;
since odd*even = even and even*even = even, q.e.d.This is not
my question, however. It is:(Idea- since {1,-1} is a
nontrivial group, ...), doesit follow that every group with
nontrivial center has even order?C. Dement 
= [.snip.]> This
is not my question, however. It is:> (Idea- since {1,-1} is a
nontrivial group, ...),What does -1 mean in the context of an
arbitrary group? And what makes you think that in your group,
even if -1 makes sense,-1 is not equal to 1?> does> it follow
that every group with nontrivial center > has even order?No.
Every group whose order is the power of a prime has
nontrivialcenter (you will deduce it as a consequence of the
class formula),in particular, any group of order p^n, p an odd
prime, n>0, hasnontrivial center.Arturo Magidin, sans .sig
=>
If G has a subgroup H of even order, then G is even.>What about
Z/2 x Z ? Are in'nite groups even?> Proof: If G has a 
subgroup
H of even order, then> |G|= |H||G:H|. The index of H in G,
|G:H|, is> either even or odd. It doesn't matter;> since
odd*even = even and even*even = even, q.e.d. This is not my
question, however. It is:> (Idea- since {1,-1} is a nontrivial
group, ...), does> it follow that every group with nontrivial
center> has even order?>1 is the identity? Then -1 = 1.1 
isn't
identity? Then 0 is identity and 1+1 is in that group.
=> >
If G has a subgroup H of even order, then G is even. What
about Z/2 x Z ? Are in'nite groups even?> > Proof: If G has a
subgroup H of even order, then> |G|= |H||G:H|. The index of H
in G, |G:H|, is> either even or odd. It doesn't matter;> 
since
odd*even = even and even*even = even, q.e.d. This is not my
question, however. It is:> (Idea- since {1,-1} is a nontrivial
group, ...), does> it follow that every group with nontrivial
center> has even order? 1 is the identity? Then -1 = 1.> 1
isn't identity? Then 0 is identity and 1+1 is in that
group.It's an ABELIAN group, meaning it's 
multiplication. 1 is
the identityand -1!=1. {-1, 1} is isomorphic to the cyclic
group Z/2Z, (thatisn't Z/2 * Z, but actually Z/(2Z).) Z/2Z 
has
2 members, 0 and 1.
=If G has a subgroup H of even order,
then G is even.>> What about Z/2 x Z ? Are in'nite groups
even?Proof: If G has a subgroup H of even order, then>> |G|=
|H||G:H|. The index of H in G, |G:H|, is>> either even or odd.
It doesn't matter;>> since odd*even = even and even*even =
even, q.e.d.>> This is not my question, however. It is:>>
(Idea- since {1,-1} is a nontrivial group, ...), does>> it
follow that every group with nontrivial center>> has even
order?>> 1 is the identity? Then -1 = 1.>> 1 isn't identity?
Then 0 is identity and 1+1 is in that group.It's an ABELIAN
group, meaning it's multiplication. 1 is the identity>and
-1!=1. -1 has no intrinsic meaning. Take the cyclic
multiplicative group of 3elements, 1, x, x^2, with x^3 = 1.
What is -1 then? > {-1, 1} is isomorphic to the cyclic group
Z/2Z,Not every group has a unique central element of order 2.
What is -1?--
=It's not denial. I'm just very selective 
about what I
accept as reality. --- Calvin (Calvin and
Hobbes)
Arturo Magidinmagidin@math.berkeley.edu>
This is not my question, however. It is:> (Idea- since {1,-1}
is a nontrivial group, ...), does> it follow that every group
with nontrivial center > has even order?> > C. DementNo.
Trivially every abelian group has non-trivial centre , just
pick onethat has odd order. If you were going to consider
non-abelian groups, trya group of order p^n for p odd. The
book should prove that at some pointthis has non-trivial
centre; this is the same as possessing at least twoconjugacy
approve@localhost)
=>
My problem is :>>I would like to
minimise a function with N arguments. The principle
is>>to>>change the N args until the function is equal to zero.
My method is to>>add a deplacment to each arg, and to see witch
of them is the best, and>>to itter>>I don't have a lot of
af'nity with maths... Is it a well known problem,>>with
>>typic algorithms ? I don't really know where to search on
the web to>>'nd this>>type of informationThis is 
unconstrained
minimization. The method which you describe>is slow and will
not converge well. If your function is continuous
and>differentiable, a conjugate-directions method is probably
the best way>to go. A conjugate-directions solver is part of
my ANSI C subroutine library WNLIB.>You can download WNLIB as
described below. The conjugate-directions >package is called
wnconj.>------------------------------------------------------
------------------------FEATURES: WNLIB is an ANSI C
subroutine library that contains numerous tools that> I have
found to be useful in my programming practice. The following
are included: miscellaneous: improved memory allocator and
memory debugger; parsing package data structures and
algorithms: linked list; balanced binary tree; hash table -
with common hash> functions; dd tree; sorting for lists and
arrays;> high-quality pseudo-random number generator
numerical: various matrix and vector operations, including
matrix inverse, > least-squares inverse linear programming:
simplex method; transportation problem; shortest path
problem;> critical path problem non-linear optimization:
conjugate directions and conjugate gradient methods for
constrained> and unconstrained problems; simulated annealing
random numbers: very good (cryptographically-strong)
pseudo-random number generator. > True random number generator
(no kidding!). Normal distribution,> Poisson distribution,
Cauchy distribution, etc.>LEGAL: BY DOWNLOADING OR COMPILING
THIS LIBRARY, YOU ACCEPT AND AGREE TO THE TERMS > AND
CONDITIONS PRINTED BELOW. IF YOU DO NOT AGREE, DO NOT DOWNLOAD
OR > COMPILE THIS LIBRARY. The author and Spike Technologies
provide this C code in the hope> that it will be helpful,
however, we assume no responsibility > for the use of this
code, nor any responsibility for its support.> The software is
distributed on an AS IS basis, without warranty.> Neither the
authors, the software developers, nor Spike Technologies> make
any representation, or warranty, either express or implied,
with> respect to the software programs and subroutines, their
quality, accuracy, > or 'tness for a speci'c 
purpose.
Therefore, neither the authors, the > software developers, nor
Spike Technologies shall have any liability to > you or any
other person or entity with respect to any liability, loss, >
or damage caused or alleged to have been caused directly or
indirectly by > the programs and subroutines contained in this
library. This includes, but > is not limited to, interruption
of service, loss of data, loss of classroom > time, loss of
consulting or anticipatory pro'ts, or consequential damages >
from the use of these programs. >COPYRIGHT NOTICE: The source
code in this directory is provided free of charge to anybody >
who wants it. It is in the public domain and therefore may be
used by > anybody for any purpose. This copyright notice and
the above legal notice> may not be removed.>AUTHOR: Will
Naylor Spike Technologies> 500 E. Calaveras Blvd #206>
Milpitas, CA 95035 Spike Technologies provides algorithm and
CAD software consulting> and contract programming and IC
design services.>DOWNLOADING and INSTALLATION:1) Create a
directory called wnlib wherever you want this code to> reside
and go into it. Type $ cd  $ cd wnlib2) Download using ftp or rftp or a
web browser. > > a) For ftp or rftp type $ ftp ftp.rahul.net
Type ïanonymous' at login prompt.> After you are 
logged in
type ftp> cd pub/spiketech/softlib/wnlib> ftp> get
INSTALL.txt> ftp> get uuwnlib.z> ftp> quit b) For a web
browser, go to > > http://www.spiketech.com/spike> > Click
over ïSpike Freeware' link. Read instructions 
and > click over
 to download. Your browser should throw a >
dialog box; you can select a directory where you want to store
it> and click ok. The browser will download it.3) Decode the
install tar 'le. Type $ uudecode uuwnlib.z> $ uncompress <
tar'le | tar xvf -4) Make the code. Type $ make all5) If this
fails, it may be necessary to modify variables in 
acc/make'le>
to 't the platform you are running on.6) When you believe you
have compiled correctly, type $ make selftest to run the
self-testing diagnostics. If selftest runs successfully> to
completion with no error messages, you have successfully
installed> wnlib.>USE:1) Summary: ARCHIVE is in:
wnlib/acc/text.a> H FILES are in: wnlib/acc/h/*.h> MAN PAGES
are in: wnlib/doc/man/*.man 2) Under wnlib, the directory acc
contains .h 'les to include and archive> 'les to 
link to. The
directory acc/h contains links to all of the > .h 'les. The
directory acc contains an archive 'le called text.a. > Use 
the
-I switch> of cc to include the .h 'les. See the 
make'le's in
acc for examples> of linking text.a 'les and including .h
'les.3) man pages are to be found in doc/man. The script 
wnman
in the> directory command is a convenient way to view these man
pages.> Type wnman  to view a man page.> The command
wnman -k  does a keyword search on all> the man
pages.> > The variable wnlib must point to your wnlib
directory for> wnman to work. A csh command of the form $ set
wnlib = xxx/yyy/wnlib sets this up. Include such a command in
your .cshrc.>
http://www.spiketech.com/spikeGot: ïThe requested URL /spike approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i02Dngm20074; =>numbers in your head in a matter of seconds. Any truth to this? Any>suggestions as to by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i02Dni920157; =>Ultimately, one would have to come up with a means for>ef'ciently refering to quasi-in'nite regressions of operation>numbers embedded inside of other operation numbers...but the present>set of symbols and terminology should suf'ce for now.>Someone has already come up with a notation for thisWhat you are looking for, is Conway chained arrow notation explained herehttp://en.wikipedia.org/wiki/Conway_chained_arrowhttp:// home.earthlink.net/~mrob/pub/math/largenum-3.html =>I need to do optimization to optimize for a matrix that is as othogonal as>>possible, or as diagonal as possible...>>How to I set the objective measure of the orthogonality and diagonality>>of matrices?This comes immediately to mind.Orthogonals: how close is AA^T to the identity matrix (maybe using> square-sums of all elements in the matrix AA^T - I).Diagonals: how close is A - A^T to the zero matrix (again, using> square-sums of A - A^T).This would measure how symmetric A was, not how diagonal it is.Better might be to use the square-sums of off-diagonal elements.>Not knowing your algorithm for creating these matrices it's hard to>say whether these metrics are useful or not.Rob Johnson take out the trash before replying =I made a post about a quadratic example that Rick Decker, a professorat Hamilton (I said University before but it might be College), gavein a recent post. I haven't seen any replies to that yet in GoogleGroups, so I'll leave it and make another thread to consider Decker'sexample in more detail, again here are some headers to allow you to'nd his original post:a_1(x) a_2(x) = 7(x^2 + x)or decomposition, where here 7(x^2 + x) is being broken up into twoalgebraic integer functions which are its factors.Now that can be done with my work and my factorizations, but I thinkthat a quadratic is simple enough for me to explain some more points,so I'll use Decker's example without feeling a need to use morecomplicated work of my own.So there's this second factorization, and from your understanding ofalgebraic integers, you should accept that in general algebraicintegers are in'nitely decomposable, 0 trivially.So from a *common sense* perspective, that is, the commonunderstanding of the ability to decompose algebraic integers, youmight *reasonably* suppose that you can always 'nd decompositions inthe ring of algebraic integers such thata_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x),where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,so that you could havew_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x)in the ring of algebraic integers.However, that common sense view is wrong, as although algebraicintegers *are* in'nitely decomposable into algebraic integers, itturns out that it's rather easy to show that the peculiar constructionhere doesn't allow for a decomposition *in general* in the ring ofalgebraic integers.So why do I emphasize in general? That's because you may indeed'nd for some particular x that you can get an algebraic integerdecomposition, but you can't 'nd 'nd a de'nition like for a_1(x),and a_2(x), where they're *guaranteed* to be algebraic integers foralgebraic integer x because they're the roots ofa^2 - (x - 1)a + 7(x^2 + x).Checking at x=0 reveals that the actual constant terms of thefactorization are 7 and 2, where Decker picked a_1(0) = 0 at x=0.Now then, consider what happens if you divide both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2where the b's are roots of some unknown quadratic, though the 'rstand last coef'cients ARE known:b^2 + ? b + (x^2 + x).If you're a little confused on that point, remember that I hada_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x),where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,so that you could havew_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x)and now dividing both sides by 7 givesb_1(x) b_2(x) = x^2 + x.If you assume that the b's like the a's before them are algebraicinteger functions (that common sense again) then there should be ade'ning quadratic *in the ring of algebraic integers* like there ISfor the a's, as for the a's that de'ning quadratic isa^2 - (x - 1)a + 7(x^2 + x).Now then Decker probably picked his example not for me to outline someof the 'ner points of Advanced Polynomial Factorization Theory (justmade that up, being a discoverer is cool).As I mentioned in my other thread on his example, he probably pickedit because at x=1, you have *both* a's with sqrt(7) as a factor.It turns out that the ring over which you can get that factorization,starting from initial steps, is a 'eld, where 7 is a unit.To understand that fact, you need to understand my work a LOT morethoroughly, like how I found my own factorization. My guess is thatDecker just 'ddled with things until he found something that hethought suited his purposes!But mathematics is about consistency and logic. There is a reason foreverything in mathematics. And when you make changes that seem simpleenough to you, underneath a lot of mathematical machinery may be atwork.Hopefully by talking to you about de'ning algebraic integer functionsI've helped you to get some sense of mathematical ideas clearly at thelimits of human ability to understand.Want more? Then read various threads on sci.math that I've started,and be sure to check out my blog archives:James Harris =For other counterexamples, investigate the ultrametric case.In the p-adic numbers, the open unit ball U = {y : d(y,0)<1}is a closed set, even though there are many points z with d(z,0)=1. => >Is there a suf'cient condition so that the closure of every open ball>of a metric space is the closed ball of same center and radius? The>closure of an open ball is always a subset of the closed ball of same>center and radius, but it can be a proper subset.> > Necessary and suf'cient is that for every point p in the space, the > only local minimum of the function x -> d(x,p) is at x=p.>Yes, that's right. Fix p in the space S and de'ne f:S->R by f(x) =d(x,p). Suppose the only local minimum of f is at p. Let B(r) be theopen ball centered at p and with radius r and C(r) the correspondingclosed ball. If x is in in C(r), then, since x is not a local minimumof f, every neigborhood of x contains an element y such that f(y) =d(y,p) < f(x) = d(x,p) = r, which shows y is B(r). This implies thatevery neighborhood of x intersects B(r), so that x is in the closureof B(r). We conclude F(r) is a subset of cl B(r) and, since theconverse is always true, it follows that F(r) = cl B(r).For the converse, suppose some q<>p in S is a local minimum of f andlet r = d(q,p), so that q is in F(r). Since q is a local minimum, qhas a neighborhood V on which f(x) = d(x,p) > f(q) = d(q,p) =r forx<>p. Since q is not in B(r), we see no point of V is in B(r) and,therefore, x is not in cl B(r). So, in this case, cl B(r) is a propersubset of F(r).Since p is an arbitrary element of S, your proposition follows. =Is there a suf'cient condition so that the closure of every open ball>of a metric space is the closed ball of same center and radius? The>closure of an open ball is always a subset of the closed ball of same>center and radius, but it can be a proper subset. Necessary and suf'cient is that for every point p in the space, the> only local minimum of the function x -> d(x,p) is at x=p. Yes, that's right. Fix p in the space S and de'ne f:S->R by f(x) => d(x,p). Suppose the only local minimum of f is at p. Let B(r) be the> open ball centered at p and with radius r and C(r) the corresponding> closed ball. If x is in in C(r), then, since x is not a local minimum> of f, every neigborhood of x contains an element y such that f(y) => d(y,p) < f(x) = d(x,p) = r, which shows y is B(r). This implies that> every neighborhood of x intersects B(r), so that x is in the closure> of B(r). We conclude F(r) is a subset of cl B(r) and, since the> converse is always true, it follows that F(r) = cl B(r).Do you intend F(r) = C(r) ?> For the converse, suppose some q<>p in S is a local minimum of f and> let r = d(q,p), so that q is in F(r). Since q is a local minimum, q> has a neighborhood V on which f(x) = d(x,p) > f(q) = d(q,p) =r for> x<>p. Since q is not in B(r), we see no point of V is in B(r) and,> therefore, x is not in cl B(r). So, in this case, cl B(r) is a proper> subset of F(r).> Since p is an arbitrary element of S, your proposition follows. points, then there are open balls whose closure is not the> corresponding closed ball, right?Almost. Just one will suf'ce provided space is multi-point.Let p be isolated point. Pick x /= p, let r = d(x,p).p not in B(x,r); p not in cl B(x,r); p in closed ball B[x,r]---- => Is there a suf'cient condition so that the closure of every open ball>of a metric space is the closed ball of same center and radius? The>closure of an open ball is always a subset of the closed ball of same>center and radius, but it can be a proper subset. Necessary and suf'cient is that for every point p in the space, the> only local minimum of the function x -> d(x,p) is at x=p. Yes, that's right. Fix p in the space S and de'ne f:S->R by f(x) => d(x,p). Suppose the only local minimum of f is at p. Let B(r) be the> open ball centered at p and with radius r and C(r) the corresponding> closed ball. If x is in in C(r), then, since x is not a local minimum> of f, every neigborhood of x contains an element y such that f(y) => d(y,p) < f(x) = d(x,p) = r, which shows y is B(r). This implies that> every neighborhood of x intersects B(r), so that x is in the closure> of B(r). We conclude F(r) is a subset of cl B(r) and, since the> converse is always true, it follows that F(r) = cl B(r).> > Do you intend F(r) = C(r) ?Oh! Yes, sure! Sorry, I ended up using 2 different notations for thesame thing. I should have used either F(r) or C(r), not both. It was amistake.> > For the converse, suppose some q<>p in S is a local minimum of f and> let r = d(q,p), so that q is in F(r). Since q is a local minimum, q> has a neighborhood V on which f(x) = d(x,p) > f(q) = d(q,p) =r for> x<>p. Since q is not in B(r), we see no point of V is in B(r) and,> therefore, x is not in cl B(r). So, in this case, cl B(r) is a proper> subset of F(r).> Since p is an arbitrary element of S, your proposition follows. points, then there are open balls whose closure is not the> corresponding closed ball, right?> > Almost. Just one will suf'ce provided space is multi-point.> Let p be isolated point. Pick x /= p, let r = d(x,p).> p not in B(x,r); p not in cl B(x,r); p in closed ball = >> isolated points, then there are open balls whose closure is not >> the corresponding closed ball, right?> >> Almost. Just one will suf'ce provided space is multi-point. >> Let p be isolated point. Pick x /= p, let r = d(x,p). >> p not in B(x,r); p not in cl B(x,r); p in closed ball B[x,r]Small oversight, your proof of the proposed theorem excellent.---- =William Elliot scribbled the following:> >> Do you intend F(r) = C(r) ?> >Oh! Yes, sure! Sorry, I ended up using 2 different notations for the> >same thing. I should have used either F(r) or C(r), not both.> Do I detect a European accent to your math? As I recall,> Europeans use F for closed sets and O for open sets.In Helsinki, Finland, at least, F is generally used for closed sets andU for open opens. F probably comes from ferm.8e (French for closed)but I don't know where U comes from. I've thought of it as a smalluniverse, but it could also come from Umgang(sp?), German forneighbourhood or something.-- /-- Joona Palaste (palaste@cc.helsinki.') ------------- Finland ---------- http://www.helsinki.'/~palaste --------------------- rules! --------/I will never display my bum in public again. - Homer = >> Do you intend F(r) = C(r) ?> >Oh! Yes, sure! Sorry, I ended up using 2 different notations for the> >same thing. I should have used either F(r) or C(r), not both.> Do I detect a European accent to your math? As I recall,> Europeans use F for closed sets and O for open sets. In Helsinki, Finland, at least, F is generally used for closed sets and> U for open opens. F probably comes from ferm.8e (French for closed)> but I don't know where U comes from. I've thought of it as a small> universe, but it could also come from Umgang(sp?), German for> neighbourhood or something.>Don't know. To avoid O from being confused with 0 ?I avoid using l as it confuses with 1 upon fast readingand in the days of typerwriters, l was used for 1. => Is there a suf'cient condition so that the closure of every open ball> of a metric space is the closed ball of same center and radius? The> closure of an open ball is always a subset of the closed ball of same> center and radius, but it can be a proper subset.> Here's a conjecture: For every point x in a metric space, there isradius r(x) such that if 0 < a < r(x), the closure of B(x;a) isthe closed ball of radius a about x. => Here's a conjecture: For every point x in a metric space, there is> radius r(x) such that if 0 < a < r(x), the closure of B(x;a) is> the closed ball of radius a about x.Take the metric space X = {0} U {1/n; n positive integer} with the usualmetric and let r be a positive real, N a positive integer such that 1/N Is there a suf'cient condition so that the closure of every open ball> of a metric space is the closed ball of same center and radius? The> closure of an open ball is always a subset of the closed ball of same> center and radius, but it can be a proper subset.> > It seems to me this is more a metric property than a topologicalproperty. For example, consider the two metrics d1(x,y) = |x-y| and d2(x,y) = min(|x-y|,1) on the reals. Both these metrics arecompatible with the usual topology on R. In both cases, an open ballof radius 1 is an open interval of length 2. With d1, the closedball is the closure of this interval, whereas with d2, the closed ballis the entire space.Stephen J. Herschkorn => Is there a suf'cient condition so that the closure of every open ball> of a metric space is the closed ball of same center and radius? The> closure of an open ball is always a subset of the closed ball of same> center and radius, but it can be a proper subset.> > It seems to me this is more a metric property than a topological> property. For example, consider the two metrics d1(x,y) = |x-y| and> d2(x,y) = min(|x-y|,1) on the reals. Both these metrics are> compatible with the usual topology on R. In both cases, an open ball> of radius 1 is an open interval of length 2. With d1, the closed> ball is the closure of this interval, whereas with d2, the closed ball> is the entire space.> Yes, sure. And we see that if radius <1 then the closure and theclosed ball are always the same.Amanda =I say a,b are factors of c iff c is equivalent to some word generatedby a and b. For example, a and b would be factors of c ifc=ababababaabaaab. For a given non-singular matrix, is there any wayto determine the set of factor pairs? Is there an algorithm todetermine if two matrices are factors of diagonal matrix?Let a,b,c be 2x2 non-singular matrices. Is there a way to determinefor what values of n, or for how many values of n,c^n*a*c^-n,c^n*b*c^-n are factors of a diagonal matrix? Anyinformation or references relating to these questions or moregeneralized ones would be greatly appreciated.Gershon Bialer =>Today, at the Barnes and Noble Bookstore, I saw the book The Colossal>Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by>Martin Gardner. The book says that>e^[pi*sqrt(163)] = 262,537,412,640,768,744>which is an integer exactly.>Speci'cally, the book says that Srinivasa Ramanujan(1887-1920)>computed the number manually to 262,537,412,640,768,743.999999, but>could not go further; then someone in France computed two million>digits after the decimal point which are all 9; then someone>ingeniously used the Euler's Constant to prove that the result is>exactly the integer as given above.>A quick search in the Internet revealed that the above claim is not>true and it 'rst showed up as a April's Fool's joke by Martin>Gardner, and he subsequently admitted that it was a joke.In fact, after readint the Scienti'c American April Foollooked up Ramanujan's paper cited, and from that I couldprove that it was not an integer.>Then how come it still got into the book?-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University =>> I checked out a book called All the Mathematics You Missed [But Need to>> Know for Graduate School] from the library and was surprised by its>> contents. The book is divided into 16 sections that I am supposed to>> know before I get into graduate school. This is my last year and I>> can check off very little.>> Here are the sixteen topics that I need to know along with whether or>> not I will have completed them by the end of the year:>> 1. Linear Algebra - YesOf MODERATE importance. The computation, no.>> 2. Real Analysis - YesYes.>> 3. Differentiating Vector Valued Functions (jacobians, inverse>> function theorem) - No (nothing like this taught at my school)Of mild importance; a week in a good course.>> 4. Point Set Topolgy - No (not offered here)If it is general topology. Usually undergraduate coursesare not.>> 5. Classical Stokes Theorems - Yes>> 6. Differential Forms and Stokes Theorem - No (nothing like that here)>> 7. Curvature for Curves and Surfaces (differential geometry) - No>> (not offered)>> 8. Geometry - No (only course offered is one for future high school>> teachers and was advised not to take it)These are not basic courses at all.>> 9. Complex Analysis - No (schedule con§icts last year and this year)Not basic, but useful.>> 10. Countability and the Axiom of Choice - No (not offered but I have>> looked into it a bit)Too specialized. A strong course in set theory is needed.>> 11. Algebra - YesABSTRACT algebra.>> 12. Lebesgue Integration - No (not undergrad here)>> 13. Fourier Analysis - No (I thought this was for engineers)>> 14. Differential Equations - YesNone of these are essential, and are mostly cookbook.>> 15. Combinatorics and Probability - No (combinatorics not offered;>> probability only after calc-based statistics is taken)If the probability course has a statistics requirement, it islikely that both courses are of poor quality. EVERY statisticscourse should be based on probability, which does not requireany statistics whatever.>> 16. Algorithms - No (the closest thing to what is described here is a>> mid-level computer science course).You need to get away from the idea that mathematics consistsof calculating; get the concepts, which requires suf'cientabstraction to get away from using real world connectionstoo much.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University =>>Message-id: [...]>>If you are aiming for a ph.d, then I would suggest getting into the best M.S>>program you can, and then apply to a great ph.d program when you get your>>M.S.. This is essentially what I am doing, for I am in the same situation.>>Good luck!>This is basically what I plan on doing. Is it bad to get a masters at a school>with a PhD program and then transfer to another school afterwards, even if you>get good recommendations? I'm applying to a few pretty low-ranked PhD programs>and one masters program. I just don't know of any really GREAT masters>programs but I'm sure they are out there. Do you know of any?Watch out! Many schools with MS programs have essentiallycomputational programs, and what you need for a PhD programis the ability to do abstract mathematics and proofs.Unfortunately, most undergraduate schools, as well as highschools, have almost completely concentrated on computationalcourses. Whether you can calculate derivatives and integralsis of little importance; do you know what they are, and canyou prove theorems? I suggest you 'nd a STRONG abstractalgebra course to start out, and proceed by looking for themost general abstract material you can 'nd. I have had a student tell me that the biggest problem he hadwith general topology was that he had had metric topology;this is the usual method. It is unfortunate that we use thechronological instead of logical method, and increase theamount of abstraction. Fully abstract is more understandable,as there is less you can use.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University hrubin@odds.stat.purdue.edu (Herman>Watch out! Many schools with MS programs have essentially>computational programs, and what you need for a PhD program>is the ability to do abstract mathematics and proofs.I presume by computational you mean non-rigorous cookbookmathematics, not actual scienti'c computation and assorted numericalmethods. =Bushnell,> Do you want tp say that any university outside US (Madrid, Beijing inyour> examples) are mediocre places?I know enough of the Chinese-speaking world to be quite sure that anundergraduate degree in mathematics from Beijing University is world-classby anyone's standards. A math major from Beijing who ends up at a Big Tenstate university for a master's before going to a more prestigious Ph.D.program was probably limited by economics (ability to pay the tuition withavailable assistanceships) or English ability as measured by foreign studentadmission tests much more than by math ability. Presumably, the TOP studentsfrom Beijing easily get into the top graduate programs in the United Statesin the 'rst place, but the also-rans from Beijing would be very attractivecandidates to most graduate schools in the United States, based on all theinformation I have been able to obtain about current patterns in foreignstudent §ows. =Nothing is necessary,or suf'cient! =I have been analysing this problem but can't prove the last bits. Iwould be grateful for some help.We have a equilateral triangle pointing upwards. The triangle iscomposed ofa number of dots. By moving some of the dots this triangle can be madetopoint downwards.Example: O O O (a) O O (b) OTriangle (a) has 2 rows and is composed of 3 dots. By moving the upperdotwe get triangle (b). What is the connection between the number of rows and the minimumnumber ofdots that must be moved to turn the triangle?Start by de'ning the number of dots A(n) the nth-triangle has.The triangle can also be illustrated as a right triangle: O O O a = 1 + 2 1 2 O O O a = 1 + 2 + 3 O O O 1 2 3We easily realize that the nth triangle has 1 + 2 + 3 + ... + n dots,This sum can be written: n(n + 1) 1 + 2 + 3 + ... + n = -------- (1) 2Equ (1) can easily be proved by induction on n. Therfore: n(n + 1)A(n) = -------- , n > 1 2The function A(n) constitutes for the natural numbers the thirddiagonal inPascals triangle.O is dots that has to be moved to turn the triangle. O X X X X O X O X O X X X X O X O X O X X O X X O X X X X X X O X X O X X O O O O X X X O X X X O X X X X X X X X O X X X O X X X O O ONotice how the 'xed dots are symetrical.By observing practical experiments i have created the table below.B(n) isthe number of dots to move.n A(n) B(n)-----------1 1 02 3 13 6 24 10 35 15 56 21 77 28 98 36 12By looking at this table I assume that B(n)=§oor(A(n)/3). But how caniprove this?It seems that B(n) also is some kind of sum:n B(n)--------------------------------1 02 0 + 13 0 + 1 + 14 0 + 1 + 1 + 15 0 + 1 + 1 + 1 + 26 0 + 1 + 1 + 1 + 2 + 27 0 + 1 + 1 + 1 + 2 + 2 + 28 0 + 1 + 1 + 1 + 2 + 2 + 2 + 3n 0 + 1 + ... + §oor((n+1)/3)Intrestingly the numbers seems to be repeated 3 times before theyincreasewith 1. = [...] What is the connection between the number of rows and the> minimum number of dots that must be moved to turn the triangle?> > [...]> > O is dots that has to be moved to turn the triangle.> > O > X X X X > O> > X O X O > X X X X > O X O X > > O > X X O X X O> X X X X X X > O X X O X X > O> > O > O O > X X X O X X X O> X X X X X X X X > O X X X O X X X > O O> O> > > Notice how the 'xed dots are symetrical.> > By observing practical experiments i have created the table below.> B(n) is the number of dots to move. n A(n) B(n)> -----------> 1 1 0> 2 3 1> 3 6 2> 4 10 3> 5 15 5> 6 21 7> 7 28 9> 8 36 12I get B(6) = 8, and your pattern implies B(n) = (n-3)(n-2)/2 + 2for n > 2.------------------------------------------------------------ ---------------John R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen =>A book I'm reading asserts If H and K are two subgroups of a group G,>and if every element g of G can be uniquely expressed as g=xy, where x>is in H and Y is in K, then H and K are both normal in G.>I haven't been able to show this. I think I might be missing something>really simple.I do not think you are missing anything. I believe it isIwasawa's Theorem that a semi-simple non-compact Lie group Ghas a compact subgroup K and a solvable subgroup S such thatevery element g of G has a unique expression g=ks. Clearly,neither K nor S can be normal.A special case of this is the theorem that an n-dimensionalmatrix of determinant 1 can be represented as an orthogonalmatrix times a matrix which is 0 below the diagonal. Ifthe diagonal elements of the triangular matrix are requiredto be positive, the representation is unique, and theconstruction is straightforward.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University => I'm looking for an easy but rigorously correct introduction to the> kalman 'lter for a 'nance application. Either a document (available> on internet) or a textbook would be good. Any suggestions?For speci'c topics, a monograph (a book on *one* topic) is good placeto start. TryBrian D. O. Anderson and John B. Moore, _Optimal Filtering_ 1979 ISBN0-13-638122-7. This has no economic examples but does explain Kalman'ltering clearly and concisely in the 'rst chapter or three. Thebulk of the book is variations and tricks which I have found to bealso of value.Masanao Aoki is a good author and has written, e.g., _Optimization ofStochastic System_ which promises application to economic theory andeconometrics, but the economics is light and the sysengr =Let me translate a nice question from de.sci.mathematik,for which we still don't have an answer.Given a quadrilateral Q with corners A,B,C,D in thatorder, i.e. with sides AB, BC, CD, DA.How to 'nd the largest rectangle R, such that R iscompletely within Q?My feeling is, that one has to distinguish betweena lot of cases, regarding the shape of Q. So weneed clever algorithms for shape-classi'cation,followed by some rules for the computation in eachclass.There are strong time constraints, as I learned.The classi'cation idea itself is mathematical stuff,I believe. The rest will be better sought for in somecomp.sci.xxxx?Rainer Rosenthalr.rosenthal@web.de =Only a partial answer besides the point, but may furnish insight ; forspecial case when diagnols cut at right angles, it is the join ofcenter points of sides AB, BC, CD, DA. :) .. When diagnols cutobliquely,the parallogram so formed has maximum area,half of the areaof Q. > Let me translate a nice question from de.sci.mathematik,> for which we still don't have an answer.> > Given a quadrilateral Q with corners A,B,C,D in that> order, i.e. with sides AB, BC, CD, DA.> How to 'nd the largest rectangle R, such that R is> completely within Q?> > My feeling is, that one has to distinguish between> a lot of cases, regarding the shape of Q. So we> need clever algorithms for shape-classi'cation,> followed by some rules for the computation in each> class.> > There are strong time constraints, as I learned.> The classi'cation idea itself is mathematical stuff,> I believe. The rest will be better sought for in some> comp.sci.xxxx?> > Rainer Rosenthal> r.rosenthal@web.de => Somewhere in the IRS forms [and this is no joke, it's true]You may the thinking of the following, included as a 'ller in one ofthe MAA journals some years back. [Anyone have the exact reference?]Someone dies, and his will makes some bequests, then ends by sayingafter taxes are paid, any remaining money should be donated tocharity. Well, if charitible donations are tax-deductible, then theamount of the donation effects the taxes. What to do? The I.R.S.supposedly has a form where you do this computation, but it amountsessentially to trial and error. In fact the problem can be solved byhigh-school algebra. (Solution of a linear equation.) Or, not eventhat: the problem can be solved using the method of false position:that method is described in the Rind Papyrus, which dates from maybe2650 B.C.And some people say the I.R.S. is behind the times...-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ => I recall having been taught not use operations such as +, -, /, x that will> effect both sides(at the same time) of an identity to be proven. And so I> told my students to follow this rule. One student respectfully and> thoughtfully said why and went on to pose that the relationship between the> two expressions we are trying to show are identical must be either >, >=, =,> <= or < . So with the exception of multiplying or dividing by a negative> value, the unknown relationship will stay the same(if we operate on both> expressions at once).> > For example (trivial but it demonstrates the question) suppose you want to> prove that 1 + sin(-x) = 1 - sin(x) .> Normally we would apply the cofunction identity sin(-x) = - sin(x) the left> hand expression and be done. But why not 'rst subtract -1 from both> expressions and then apply the cofunction identity?> > If the rule I remember being taught is valid there should exist some pair of> trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical> but would appear to be identical if one operates on both expressions> simulataneously. For example suppose one were trying to prove Expr1(x) => Expr2(x). In the process he/she divided both sides by cos(x)and found that> the modi'ed expresssions were in fact equal. I can't come up with such a> pair of expressions that would validate the rule I remember. Can anyone help?> I sould say there is nothing wrong with the method. BUT when it iswritten and handed in, so that when it is read the writer is not thereto explain it, the explanations should be included on the page.Don't just write a squence of formulas. Include also text that tellshow these formulas are related.Example: Prove that 1 + sin(-x) = 1 - sin(x). This would be true if we knew sin(-x) = -sin(x) by adding 1 to bothsides. We know sin(-x) = -sin(x) because ... Therefore 1 + sin(-x) = 1 - sin(x).Bad: Prove that 1 + sin(-x) = 1 - sin(x). Subtract 1 from both sides, sin(-x) = -sin(x), done. =I thought of a simple problem and i suspect that it has already been worked on, but I haven't been able to 'nd any discussion of it in intro Graph Theory books. Does anyone know where to look for a treatment of this problem?The speci'c example I thought of is:Supposing that every person in the United States can name what state they live in and all of its adjacent states. What is the smallest group of people required to name all 50 states, and what states are they from?I can show by simple arguments (involving only the degree of the vertices) that there must be at least 9 states. And i have found several solutions with 13 states through random trial and error.Clearly it can be generalized to graphs unrelated to the United States. But is there a solution with 12 states?Steve = I thought of a simple problem and i suspect that it has already been> worked on, but I haven't been able to 'nd any discussion of it in intro> Graph Theory books. Does anyone know where to look for a treatment of> this problem? The speci'c example I thought of is: Supposing that every person in the United States can name what state> they live in and all of its adjacent states. What is the smallest group> of people required to name all 50 states, and what states are they from? I can show by simple arguments (involving only the degree of the> vertices) that there must be at least 9 states. And i have found several> solutions with 13 states through random trial and error.>If you will send me (or post) a list of the states together with theadjacent states, I will see whatI can do with the question. Preferred form: a list of lists of the form[i, j1, j2,...,jk]where i is a number between 1 and 50 representing the i-th state and j1, j2,...,jk are the numbers of the adjacent states. (I have worked on similarproblems.)--Edwin ClarkPS If you want to check out the literature yourself do a Google orMathSciNet search on dominating set or domination number(s) =...> Supposing that every person in the United States can name what state> they live in and all of its adjacent states. What is the smallest group> of people required to name all 50 states, and what states are they from? I can show by simple arguments (involving only the degree of the> vertices) that there must be at least 9 states. And i have found several> solutions with 13 states through random trial and error.> [...] is there a solution with 12 states?> > If you will send me (or post) a list of the states together with the> adjacent states, I will see what I can do with the question. Preferred > form: a list of lists of the form [i, j1, j2,...,jk]> where i is a number between 1 and 50 representing the i-th state and j1, j2,> ...,jk are the numbers of the adjacent states. ...Yes, a posted list would be good, to clarify for example whether youcount corner adjacencies (like UT and NM, or AZ and CO), and which underwater boundaries (as between HI and AK, MN and MI, or RI and NY) you treat as adjacencies. Without specifying a list you probably won't get useful answers.-jiw => > I thought of a simple problem and i suspect that it has already been > worked on, but I haven't been able to 'nd any discussion of it in intro > Graph Theory books. Does anyone know where to look for a treatment of > this problem?> > The speci'c example I thought of is:> > Supposing that every person in the United States can name what state > they live in and all of its adjacent states. What is the smallest group > of people required to name all 50 states, and what states are they from?> > I can show by simple arguments (involving only the degree of the > vertices) that there must be at least 9 states. And i have found several > solutions with 13 states through random trial and error.> > Clearly it can be generalized to graphs unrelated to the United States. > But is there a solution with 12 states?> > > Steve> This is called Graph Covering in general, and what you're looking for is a solution for the Minimum dominating set. Unfortunately, it's an NP problem.http://www.nada.kth.se/~viggo/wwwcompendium/node11. html =The consistent way to say, There is no absolute truth (How about this one?) ;o)THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.George Buyanovsky => > The consistent way to say, There is no absolute truth (How about this one?) ;o)> > THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.> > George BuyanovskyNo it isn't George.De'ne statement.De'ne absolutelyDe'ne true.De'ne your premises.Then take your meds.I know that you're supposed to be on meds, because according to you itcannot be true that George doesn't take meds.:)Richard =how do you de'ne absolute?> The consistent way to say, There is no absolute truth (How about thisone?) ;o) THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT. George Buyanovsky => The consistent way to say, There is no absolute truth (How about this> one?) ;o)> > THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.> > George Buyanovsky1. THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.2. Statement 1. is true:) => >This is not fair, hinting at the existence of juicy §amewars>elsewhere without saying where. a.a. is what? alt.a??? ?> > Not much of a §ame war. For some reason, maky spent a year or so in> alt.atheism during which time he called everyone an idiot, and at the> same time praising me (despite the fact that I had never replied to or> read any of his messages there). At one point, he posted claiming I> supported his position, which I did not, and I followed-up saying> so.come again? where? when? how?> Then he had the little silly interlude here where he claimed that> a page that described job opportunities in government whose job title> included the word mathematician was an accurate measure of the kind> of jobs that someone with a mathematics Ph.D. could get without> further training.still in denial?> You might even remember that.> > Since then, maky has been sending insults my way.> > -- > => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> => > Arturo Magidin> magidin@math.berkeley.edu>> Here is a shocking admission: I'm curiously growing weary of this>> fascinating exchange. Major snippage below.[...]>> > You used this claim to support the claim> that Arturo is incompetent. This is just utter bullshit.>> you have some reading comprehension problems. my initial follow up in>> no way claims that arture is incompetent.Explain this post is indeed either incompetent as a>> | > mathematician, or more likely a liar, or both.>> | >> | well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs>> | in academic caste system.>> | >> | so whomever you are, may well be right on this one...>> `----How is jstevh@yahoo.com right, given that Arturo is an adjunct?well, some who happen to be adjuncts happen to be incompetent - duh.> > You demonstrate your usual blunt reasoning, maky.> > For the record, your post was an insult to me, and you know it. That's> how I read it, that's how everyone reasonable read it.shoudn't you be insulted by your employers instead?please explain.> Just the latest in your campaign to attack me because I would not> support you in a.a.come again?> [.snip.]> > -- > => Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A man's capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of 'gures few readers can criticize. A great> many people are staggered to this extent, that they imagine there> must be the inde'nite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> => > Arturo Magidin> magidin@math.berkeley.edu> Here is a shocking admission: I'm curiously growing weary of this> fascinating exchange. Major snippage below.> > > > [...]> >> You used this claim to support the claim>> that Arturo is incompetent. This is just utter bullshit.>> you have some reading comprehension problems. my initial follow up in> no way claims that arture is incompetent.> > Explain this post is indeed either incompetent as a> | > mathematician, or more likely a liar, or both.> | > | well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs> | in academic caste system.> | > | so whomever you are, may well be right on this one...> `----> > How is jstevh@yahoo.com right, given that Arturo is an adjunct?>>well, some who happen to be adjuncts happen to be incompetent - duh.You demonstrate your usual blunt reasoning, maky.For the record, your post was an insult to me, and you know it. That's>> how I read it, that's how everyone reasonable read it.>No, but that's just because I 'nd your opinions to be beneathinsult, and your current claims a show of cowardice. You can't evenstick by your sly attempts at an insult.>shoudn't you be insulted by your employers instead?No. Your statement is plain. It is an attack on me.>please explain.may well be right can only refer to the fact that I was being called(through copying an old post of James Harris) incompetent as amathematician, or more likely a liar, or both. You are agreeing withthe possibility, which means you are stating your opinion that it mayvery well be true that I am an incompetent and/or a liar.>> Just the latest in your campaign to attack me because I would not>> support you in a.a.come again?For someone as ignorant as you are, you don't fake ignorance too =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu> >> Here is a shocking admission: I'm curiously growing weary of this>> fascinating exchange. Major snippage below.[...]>> > You used this claim to support the claim> that Arturo is incompetent. This is just utter bullshit.>> you have some reading comprehension problems. my initial follow up in>> no way claims that arture is incompetent.Explain this post is indeed either incompetent as a>> | > mathematician, or more likely a liar, or both.>> | >> | well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs>> | in academic caste system.>> | >> | so whomever you are, may well be right on this one...>> `----How is jstevh@yahoo.com right, given that Arturo is an adjunct? well, some who happen to be adjuncts happen to be incompetent - duh.> > How does that increase the likelihood that jstevh@yahoo.com is right?what makes you assume that's i claim i maintain? didn't you read my previous reply?> Some who happen to be named Arturo happen to be incompetent, too, I'm> sure. This fact doesn't lead one to conclude that Arturo Magidin is> likely to be incompetent or that jstevh@yahoo.com is likely right.but the name arturo is not in any way connected with academic ranks. ADJUNCT is.get it?> And in any case, what does the fact that some adjuncts are incompetent> matter, unless you were inferring that Arturo is more likely to be> incompetent? >> that you are not alleging that adjuncts are likely to be liars, so you must>> be insinuating that adjuncts are incompetent. geez, i would hate to assume that you are having problems with> existential quanti'ers. is that it?> > Well, I think I know a thing or two about logic, but please enlighten> me about what error I made regarding existentials.well, 'rst tell me how much logic you know. can you handle venn diagrammes?>> At least, the straightforward interpretation of your stupid words is>> that somehow the information that *Arturo* is an adjunct increases the>> probability that Arturo is a liar or incompetent. well, given that some adjuncts are indeed incompetent, i would say> that a randomly selected adjunct may well be incompetent. as to how> the probability that a randomly selected adjunct mathematician being> incompetent compares to the probability that a randomly selected> non-adjunct mathematician being incompetent, i'll leave that as an> open question for the moment being.> > If you leave it as an open question, that rather makes the point of> your comment (that jstevh@yahoo.com may be right) rather hard to> grasp.> > Did you mean only this?what else could i have meant?do you have further objections?> (1) Some adjuncts are incompetent.> (2) Arturo is an adjunct.> -------> Therefore, it is possible that Arturo is incompetent.> > If so, perhaps your grasp of modalities is a touch weaker than my> grasp of existential.> >> I eagerly await your explanation that shows my lack of reading>> comprehension skills. its becoming clear that you did not understand the content of my> post.> > Know what? I agree. I still don't. I have no idea *what* your post> meant,why the objections then? a bit emotional? what?> unless my bad argument above captures it.the argument where you are trying to patch poor reading comprehension?> you also made a tonne of assumptions about it. but worry not, i'll> gladly dissect it for you, should you need further assistance> undertanding it...> > Assist on.> > now, where are your answers to the questions i asked?> > Which ones? I won't respond to your fantasies regarding those slave> laborers known as adjuncts, as it's not particularly my interest.then, what are you doing in this discussion?> Your insinuation that Arturo is possibly/probably incompetent because> of his job title while simultaneously not calling his competence into> question, is a curiously more diverting topic.in that case, you are invited to exploit it further. <87y8swbnll.fsf@phiwumbda.org> <87fzf3ku82.fsf@phiwumbda.org> <87isjxvg7q.fsf@phiwumbda.org> <8765fwy57l.fsf@phiwumbda.org> =>> > Here is a shocking admission: I'm curiously growing weary of this> fascinating exchange. Major snippage below.> > > > [...]> > maintain that Magidin is indeed either incompetent as a> | > mathematician, or more likely a liar, or both.> | > | well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs> | in academic caste system.> | > | so whomever you are, may well be right on this one...> `----> > How is jstevh@yahoo.com right, given that Arturo is an adjunct?>> well, some who happen to be adjuncts happen to be incompetent - duh.How does that increase the likelihood that jstevh@yahoo.com is right? what makes you assume that's i claim i maintain? didn't you read my> previous reply?Well what the fuck did [you] may well be right on this one... mean,if it didn't mean that your comments increased the likelihood that theoriginal [im]poster was right?>> that you are not alleging that adjuncts are likely to be liars, so you must> be insinuating that adjuncts are incompetent.>> geez, i would hate to assume that you are having problems with>> existential quanti'ers. is that it?Well, I think I know a thing or two about logic, but please enlighten>> me about what error I made regarding existentials. well, 'rst tell me how much logic you know. can you handle venn> diagrammes?I have more than suf'cient background in logic, I am sure. Honest.[...]Did you mean only this? what else could i have meant? do you have further objections?> (1) Some adjuncts are incompetent.>> (2) Arturo is an adjunct.>> ------->> Therefore, it is possible that Arturo is incompetent.I just want to be clear: The argument above is the argument you knowclaim to have advocated? *This* is an argument that you think is agood argument? This from the man that wants to teach me aboutexistentials?You're adorable.>> Know what? I agree. I still don't. I have no idea *what* your post>> meant, why the objections then? a bit emotional? what?> unless my bad argument above captures it. the argument where you are trying to patch poor reading> comprehension?No, the argument with premises numbered (1) and (2) and withunlabeled conclusion. The one you seem to think is just peachy.> you also made a tonne of assumptions about it. but worry not, i'll>> gladly dissect it for you, should you need further assistance>> undertanding it...Assist on.now, where are your answers to the questions i asked?Which ones? I won't respond to your fantasies regarding those slave>> laborers known as adjuncts, as it's not particularly my interest. then, what are you doing in this discussion?Mostly, mocking a complete moron. Shame you hadn't noticed.-- ...you are around so that I have something else to do when I'm not'guring something important out. I was especially intrigued on thisiteration by cursing, which I think I'll continue at some later dateas it's so amusing. --- support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVdx31773; =you can also join this site:http://alephnulldimension.net/scgi-bin/ikonboard.cgimain topic headings:philosophy/ spiritualitymathematicsartsciencemiscellaneousthe site de'nitely needs more users. hey, maybe you can help form its future...also, physicsforums.com has a nice feature of being able to post tex with [tex][/tex] commands where $$ would normally be.cheersphoenix => > I've come across this equation which describes the dependency of> certain factors in the operation of bipolar transistors:-> > Ic = Is[exp(v/t)-1] > Ic is the transistor's collector current (typically from a few to> tens of mAmps)> Is is its saturation current (typically *very* small like 5 fAmps,> for instance.> > I'm confused about the term exp and what it refers to - does it mean> exponent or e^xyes> or something else? I need someone to show me a> worked example, please.> > Let Is = 5^-15> Let (v/t)=0.5> > Now, from the above, how would I 'nd Ic?> exp(0.5) you can do on your calculator, exp(0.5)=1.649, solc = ls*(exp(v/t)-1) = (5^(-15))*(0.649) = 2.13*10^(-11) approx.Now v/t will have to be a bit larger to get values like the ones yousuggest. Here, if ls=5^(-15) and v/t = 5, then we getlc = (5^(-15))*(exp(5)-1) = 4.86*10^(-9),v/t = 20 to get lc = 0.0159,v/t = 30 to get lc = 350. =It would seem that the Action integral for the world sheets of classicalstring theory are presently justi'ed as a higher dimensional version of theone dimensional case. But I wonder if this formulation would be a naturaldescription if we were 'rst given the geometry of a world-sheet. What wouldbe the most natural mathematical description of what is happening with aworld-sheet?I have a justi'cation for a type of world-sheet geometry, but I am not sureof the mathematics to descript it. At one instant of time there is a curvethrough space with a function evaluated at each point along the length ofthe curve. This string sweeps out a surface as time passes. If I don'tknow the path it takes or the function at each point on the surface, then itwould seem that the best I can do is to describe the situation with asurface integral of a scalar function of some unspeci'ed path. The integralis then a functional of the path. This is pretty much what the Actionintegral is. Then I would naturally apply requirements of invariance withcoordinate changes to come up with a Euler-Lagrange vector. Would I thenhave a reason to assume a vanishing variation to come up with Euler-Lagrangeequations (set to zero) and also Noether's theorem of conserved values? Isthis last step assuming conserved quantities to begin with? Or is avanishing variation in and of itself a type of symmetry or invariance thatone would naturally expect to employ due to some obvious intrinsic property?I wonder. Comment requested.See: http://www.sirus.com/users/mjake/StringTh.html =I have a question and would appreciate any help.Let (f_n)_ be a sequence of real valued functions de'ned on [a, b]that converges uniformly to f. For every x in [a,b], let F_n(x) =Integral (from a to x) f_n(t)dt. Then, (F_n) converges to F(x) =Integral (from a to x) f(t)dt. Is this convergence uniform on [a,b]?If not, is there any condition that assures uniformity?If each f_n is de'ned on [a, inf) and its in'nite integral exists,then does the sequence of such in'nite integrals converge to thein'nite integral of f?Artur =>I have a question and would appreciate any help.>Let (f_n)_ be a sequence of real valued functions de'ned on [a, b]>that converges uniformly to f. For every x in [a,b], let F_n(x) =>Integral (from a to x) f_n(t)dt. You have to say a little more than you have or this integraldoesn't even exist. Possibly you meant to assume that f_nwas continuous - that would do it.>Then, (F_n) converges to F(x) =>Integral (from a to x) f(t)dt. Is this convergence uniform on [a,b]?Yes.>If not, is there any condition that assures uniformity?If each f_n is de'ned on [a, inf) and its in'nite integral exists,>then does the sequence of such in'nite integrals converge to the>in'nite integral of f?No. Someone else has already given a counterexample.There's a condition that does give convergence here: If yousuppose that g >= 0, the integral of g is 'nite, and |f_n| <= gfor all n then the integral of f_n converges to the integralof f. This is the Dominated Convergence Theorem frommeasure theory.>Artur************************David C. Ullrich => I have a question and would appreciate any help.> Let (f_n)_ be a sequence of real valued functions de'ned on [a, b]> that converges uniformly to f. For every x in [a,b], let F_n(x) => Integral (from a to x) f_n(t)dt. Then, (F_n) converges to F(x) => Integral (from a to x) f(t)dt. Is this convergence uniform on [a,b]?> If not, is there any condition that assures uniformity?Yes, because | F_n(x) - F(x) | can be majorized by theproduct of (b-a) and the maximum difference between f_n(t) and f(t).> If each f_n is de'ned on [a, inf) and its in'nite integral exists,> then does the sequence of such in'nite integrals converge to the> in'nite integral of f?No. Consider a sequence of functions f_n which converge to f(t) = 0such that for each n, int (0,in'nity) f_n = 1, but theygrow shallower and wider ( e.g., lambda exp(-lambda t) , whenlambda = 1/n). Artur>This had better not have been homework :-).Best wishes, MikeX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft kramsay@aol.com (KRamsay) said:>As far as beauty goes, it seems to me that ordinarily proofs which>avoid proof by contradiction (and proof via the contrapositive) when>they don't need them *usually* are a little bit more beautiful than>the version written in an unnecessarily negative form, so if we're>going to talk about minor cosmetic blemishes, sure, it's ever so>slightly better (IMO) to write the proof in a forward direction:>the distance to zero is continuous, and since that's bounded the set>is bounded, etc.OTOH, if a reductio ad absurdum is short and sweet while the directproof involves a myriad of special cases, most Mathematicians wouldconsider the reduction ad absurdum to be more beautiful and moreelegant.-- Shmuel (Seymour J.) Metz, SysProg and tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft = at 02:54 PM, conesetter@btopenworld.com (conesetter) said:> To all: please excuse my not being good at following what others>have done.The problem was not your failure to follow; the problem was youracting as though you had followed it and claiming an imaginary errorin the material that you did not follow. Admitting ignorance andasking for clari'cation would have yielded a different thread.>But one may think that>branches are worth studying for there own sake, and then in context>just think of Riemann surfaces as a byproduct. Certainly, although I don't see the motivation for it. At 'rstreading your analysis in those terms seems correct.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <3fee0f44$21$fuzhry+tra$mr2ice@news.patriot.net> <3fef7511$8$fuzhry+tra$mr2ice@news.patriot.net> <3ff2665e$2$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft = at 12:16 PM, raynand@netzero.net (Jefferson Rourke) said:>Yes, I did answer honestly. No. You lied and you're still lying.>I lack a belief in gods.K3wl. However, your position goes beyond lacking a believe in a god;you actively believe that there is no god and insult people whodisagree with you. To describe your position is simply lacking abelief in a god is dishonest.>I on the other hand am honest and complex.One out of two isn't bad. Now if you could only cause the 'rst to betrue.>No you didn't read any false claims. Really. Then why do the dictionaries disagree with you?>You chose to misunderstand what was being said.No, you tried to write things and then pretend that you hadn't. Youtried to claim that words meant things that they patently didn't.>You just don't get THE pointI get the point that you are a bigot and a true believer. You feltdiscussing the issue of religion in general rather than the particularpoint he raised. Well, what's sauce for the goose is sauce for thegander. If you can address a peripheral part of his post rather thanwhat he wished to discuss, then I can address the parts of your postthat show you to be a dishonest fool with delusions of adequacy.>Other ideas were borrowed from other cultures such as the zero >which was brought to the west by muslims from India.India was another religious culture. So you admit that religouscultures have contributed something positive?>Then you are sadly lacking in your knowledge of historical facts.PKB. It is *you* that is exhibiting a lack of historical knowledge, alack of reading comprehension, or both. The phrase I was commenting onwas the human race entering the Dark Ages. Well, tonto, only onepart of the Human Race entered the Dark Ages; Asia did not and theMuslims you so despise did not. Or perhaps you don't consider themHuman?>You might want to do some research on how the Catholic church >burned whole libraries and thousands of priceless books.Don't presume to teach your grandmother to suck eggs. >Where did I say they are not human?the human race entering the Dark Ages would seem to exclude everysociety that did not enter the Dark Ages.>How did you come up with this anyway?By having an open mind and looking at the available evidence. Youshould try it sometime.>I never said the Dark ages were universal, Shmol.Nu, faygleleh, how is the human race not universal? Or are you incommunication with sentient life on another planet?>I was speaking in the context of the Catholic church and >christianity in Europe.Were that true you would have written Europe, not the human race.>So who's not paying attention, Shmol?You. You don't seem to even be able to use cut-and-paste correctly.>Nice try, but you do not de'ne who I am, Correct. You do. By your actions, including your words. Havingtransmitted them for all to see, you can't require people to pretendthat they don't exist.>I'm a true believer in mankind,The religous are part of mankind. Your choice of the terms mentalillness and madness show that you have a faith that they are wrong,deny it how you will.>EVERYBODY is uneducated about something.Yes, but you have such broad coverage of things to be ignorant about.>if you think you can talk down to me because I didn't know a phrase>you're laughable.No, I think that everybody here can talk down to you because you are afool. Your willful ignorance is just a symptom.-- Shmuel (Seymour J.) Metz, SysProg and tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft rsm109@york.ac.uk (rsm109@york.ac.uk) said:>It's not just iron that modern science believes was formed in stars>and later condensed to form the solar system. It's all elements>except H and possibly He. H, He, Li and Be, none of which were known to the ancients. For thatmatter, CHON were not known to the ancients, nor were any of the otherelements necessary for life with the exception of Fe and S.-- Shmuel (Seymour J.) Metz, <3fef7511$8$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft = at 04:39 PM, raynand@netzero.net (Jefferson Rourke) said:>Well then, I'll put it this way for you and Shmuel I lack a belief>in gods, therefore, when someone comes around with a religion or>other mystical god-concept and tells me it is reality without any>proof except faith - I don't believe them and I deny that it is truth>in reality.in (Seymour J.) Metz, SysProg and JOATnot reply to =Wheeler is correct, Einstein's GR is battle-tested bothphilosophically and experimentally.So is orthodox quantum theory.Simplistic frontal assaults on their well-established foundations willnever succeed.Jack,Not so. The term simplistic is too fuzzy. The previous statement,that Wheeler is correct, Einstein's GR is battle-tested bothphilosophically and experimentally. So is orthodox quantum theory -is a distorsion of Wheeler's views.JS: I do not understand your point here? Please give more details. I suggest that that we may never understand this strange thing,the quantum, until we understand how information may underliereality. Information may not be just what we *learn* about the world.It may be what *makes* the world. (J.A. Wheeler, ïGeons, Black Holesand Quantum Foam')JS: Recent work by Lee Smolin on the weak hologram principle adds a lot of detail to Wheeler's remark. Area as quantum information §ow. See also Smolin's pop book Three Roads to Quantum Dynamics.One issue in regard to quantum information §ow is signal locality vs. signal nonlocality.http://www.quantum'elds.com/469Maclay.pdfI say that quantum gravity needs signal nonlocality in violation of orthodox quantum theory with signal locality. This allows information inside the black hole to be available to the outside observer and may change Lee Smolin's distinction between the strong and weak hologram principles.LOCAL MACRO-QUANTUM theory, with signal nonlocality is to GLOBAL micro-quantum theory with signal locality as LOCAL General Relativity with curvature is to GLOBAL Special Relativity without curvature.AJ: Quantum theory needs to unify dy-namics and bi-namicshttp://www.cassiopaea.org/quantum_future/jadpub.htm# blaja95aarkJS: What is bi-namics?##################################################Dr Arkadiusz Jadczykhttp://www.cassiopaea.org/quantum_future/homepage.htmJS : Your website is very interesting Ark, but would require many hours to seriously comprehend.Can you give a kind of public synopsis of your key ideas, your motivation, your philosophy?Show us your hand in this Cosmic Poker Game. :-) You are a well-educated PhD theoretical physicist. =Paul Zielinski has attempted a defense of the Yilmaz-Puthoff type challenge toEinstein's gravity theory on the basis that there should be a LOCAL stress-energydensity tensor of the pure gravity 'eld in vacuum far from normal mass-energysources i.e. Tuv(Matter) = 0.In fact such a tensor exists trivially ~ Guv(Einstein) which is exactly zeroin ordinary non-exotic vacuum without any w = -1 zero point dark energyof negative pressure or dark matter of positive pressure. This conclusionis consistent with MTW(1973) before exotic vacua were conceived anddiscovered experimentally as we now know in the factthat TOTAL energy, momentum and angular momentum of thepure gravity 'eld are not simple integrals of LOCAL OBSERVABLES.Smolin below addresses this issue at least obliquely:arXiv:hep-th/0003056 v1 8 Mar 2000The strong and weak holographic principlesLee Smolin_Center for Gravitational Physics and GeometryDepartment of Physics, The Pennsylvania State UniversityUniversity Park, PA, USA 16802 andThe Blackett Laboratory,Imperial College of Science, Technology and MedicineSouth Kensington, London SW7 2BZ, UKMarch 1, 2000ABSTRACTWe review the different proposals which have so far been made for theholographic principle and the related entropy bounds and classify them intothe strong, null and weak forms. These are analyzed, with the aim of discoveringwhich may hold at the level of the full quantum theory of gravity.We 'nd that only the weak forms, which constrain the information availableto observers on boundaries, are implied by arguments using the generalizedsecond law. The strong forms, which go further and posit a bound on theentropy in spacelike regions bounded by surfaces, are found to su_er fromserious problems, which give rise to counterexamples already at the semiclassicallevel. The null form, proposed by Fischler, Susskind, Bousso andothers, in which the bound is on the entropy of certain null surfaces, appearsadequate at the level of a bound on the entropy of matter in a singlebackground spacetime, but attempts to include the gravitational degrees offreedom encounter serious dif'culties. Only the weak form seems capable ofholding in the full quantum theory.The conclusion is that the holographic principle is not a relationshipbetween two independent sets of concepts: bulk theories and measures ofgeometry vrs boundary theories and measures of information. Instead, it isthe assertion that in a fundamental theory the 'rst set of concepts must becompletely reduced to the second._ smolin@phys.psu.edu.....Quasi-local quantities in classical general relativityEven in classical general relativity, it is well understood that diffeomorphisminvariance and the equivalence principle forbid the possibility of local de'nitionsof the basic dynamical quantities such as energy, momentum andangular momentum. These kinds of quantities can only be de'ned in termsof integrals over two dimensional surfaces in the spacetime. When thosesurfaces are taken to the boundary, in non-cosmological spacetimes, thesebecome the well known asymptotic de'nitions of energy, momentum and angularmomentum. However, even in cosmological spacetimes where there areno boundaries one may de'ne what are called quasi-local observables[45, 46],in which the energy, momentum and angular momentum of an arbitraryregion are de'ned in terms of certain integrals over its boundary. SincePenrose's original suggestion[45] many different proposals have been madefor such quasi-local observables[46].If there are to be non-trivial notions of energy, momentum and angularmomentum in a quantum theory of cosmology then, these must be de'nedso that their classical limits are these quasi-local quantities. The simplestpossibility is that the hamiltonian in quantum gravity should itself be quasilocal,that is de'ned on two dimensional surfaces, which in the classical limitwill become spacelike surface embedded in spacetime. This implies someform of the holographic principle, for if the Hamiltonian is associated withsurfaces there must be many hamiltonians, each associated with a differentchoice of surfaces, and the same must be true of the algebra of observablesand the hilbert spaces on which they are represented.Relational approaches to quantum cosmologyAnother kind of argument for the importance of surface observables in aquantum theory of cosmology was given by Crane[3], even before the holographichypothesis of `t Hooft and Susskind was proposed. Crane noted thediffculties of de'ning a coherent measurement theory for a quantum stateof the whole universe and proposed instead that the division of the universeinto two parts-system and observer-that is basic to Bohr and Heisenberg'smeasurement theory might be relativised, so that there would be notone quantum state of the universe, but a system of observable algebras andhilbert spaces, one associated with every possible splitting of the universeinto two parts[3].To realize this idea, Crane proposed a categorical framework to describepositingfunctorial relationships between the category of cobordisms of manifoldsto topological quantum 'eld theory, as those theories can be formulated insuch categorical terms. As topological quantum 'eld theories are the onlyspaces,one may try to use them to construct examples of holographic theories[13].Furthermore, as Crane pointed out, it may be possible to extend these structuresto quantum theories of gravity because it is a fact that at both theclassical and quantum mechanical level, and for any dimension[47], generalrelativity and supergravity can be understood as deformed or constrainedtopological quantum 'eld theories[49, 23, 13, 19, 51, 50, 14, 15].Crane's proposal has been an inspiration for the development of whathave been called relational[48, 42, 43, 44] or pluralistic [19, 5] approaches toquantum cosmology. Using the fact that general relativity and supergravityare constrained topological 'eld theories, it has been possible to realize thisidea in the context of full formulations of quantum gravity and M theory[33, 34, 54].An even stronger version of Crane's argument was proposed recently byMarkopoulou[42, 43, 44], who noted that even in classical general relativitythe logic of propositions which can be given truth values by observersin a closed universe is non-boolean, because each observer can only assertthe truth of falsity of propositions about their past.Jack interjects: Of course quantum logic is a partially ordered non-boolean lattice of yes-no answersto simple quantum measurements in Von Neumann's sense. Smolin's remark only works if there issignal locality. So this is an alleged curious connection between Einstein's gravity and orthodox micro-quantum theory.LS: Rather than beinga boolean algebra, the algebra of propositions relevant for a classical cosmologicaltheory is a multivalued Heyting algebra[42]. When quantized, theresulting algebra of projection-like operators cannot be represented on a singleevery possible event at which observations are made[43]. As each observerreceives information from a distinct past, the algebra of observables theythey observe, must vary10. Given the conclusions reached in the precedingsections of this paper, this is framework is then appropriate for a formulationof the weak holographic principle[10].13 ConclusionsThe conclusion of the arguments we have given here is that the holographicbound and holographic principle can only survive in a quantum theory ofcosmology in their weak forms, proposed in [10]. While logically weaker,this form is more radical than the strong forms, in its implications for how10 Related structures have been studied also by Isham and collaborators [55], who notethe consistenthistories proposal[56] precisely. a measurement theory of quantum cosmology must be constructed. First,the weak forms require that causal structure exist even at the Planck scale.This most likely cannot be realized in a conventional formulation of quantumspacecontaining the physically allowed wavefunctions of the universe. Instead,such a description may have to be formulated along the lines proposed ina representation for an algebra of observables accessible to a single localobserver at an event or a local region of a spacetime history. These will berelated to each other by maps which re§ect the quantum causal structure.In such a spacetime, evolution becomes closely intertwined with the§ow of quantum information which also de'nes the causal structure at the Planckscale. Interactions have to do with the processing of the information atevents; as noted in [43, 44] a quantum spacetime then becomes very like aquantum computer that can dynamically evolve its circuitry.Jack: I note key role of signal locality in Smolin's analysis.LS: It is then dif'cult to escape the conclusion that the holographic principle,in its weak form, is telling us that nature is fundamentally discrete. JS: Is this hitting a §ea with a sledgehammer? ;-)LS: The 'niteness of the information available per unit area of a surface is to be takensimply as an indication that fundamentally, geometry must turn out to reduceto counting. Of course this conclusion has been reached independentlythrough other arguments coming from quantum gravity[1, 23, 24, 25, 5] andstring theory[52, 2]. But, as can be seen most clearly from the argument ofJacobson[53], the entropy bounds and holographic principle tell us that thedescription of nature in terms of classical spacetime geometry is not onlyanalogous to the laws of thermodynamics, it must be exactly the thermodynamicsof the fundamental discrete theory of spacetime.JS: I do not like the implicit assumption of thermal equilibrium at the sub-quantalspin network level of extra variables that ensures signal locality as shown byAntony Valentini.LS: What we learn from the analysis of this paper is that in such a theorythere is no room for the notion of a bulk theory, and hence no fundamentalrole for a bulk-boundary correspondence. There is instead a network ofscreen histories, which describe what possible observers might be able toobserve from particular events in their spacetime. By averaging over historiesa bulk description may emerge at the semiclassical level, but only as anapproximation in which the past of a particular observer can be described to'rst order in a perturbation expansion in terms of a particular _xed classicalhistory. Thus the proper role of a bulk-to-boundary map may be to serve asa correspondence principle to constrain the classical limit of a backgroundindependent quantum theory of gravity.To put it most simply: the holographic principle is not about a relationshipbetween two sets of concepts, bulk and screen and geometry andinformation §ow. It is the statement that the former reduce entirely to thelatter in exactly the same sense that thermodynamic quantities reduce toatomic physics. The familiar picture of bulk spacetimes with 'elds and geometrymust emerge in the semiclassical limit, but these concepts can playno role in the fundamental theory.Can this picture be used to construct a realistic quantum theory of gravitywhich addresses also the other problems in the subject? As mentionedin [10] an example of such a theory is provided by a class of background independentmembrane theories proposed in [33]. These extend the formalismof loop quantum gravity in a way as to provide a possible background independentform of string theory[34, 54]. So the answer is a very provisional,yes. Much work remains to be done, but the moral is that the holographicprinciple, in at least its weak form, is likely to feature signi'cantly in boththe mathematical language and the measurement theory of the future background => > Sorry, the ridiculous assertion is 0.999.... does NOT equal 1.> > It certainly does!> > Just try to subtract 0.999... from 1:> > 1 - 0.999... = 0> > Reason:> > There is no real number between 0.999... and 1, and, therefore, they> must be one and the same number!> > PHNeat proof, but you are playing foot loose with the de'nition of =. 0.999... is an in'nite series, a shorthand notation for .9, .99,.999, ... 1 is an integer. The relationship is that 1 is a (the)value for which, for every D>0 there is an N such that for all M>N thevalue of the Mth number in the series is between 1-D and 1+D.The problem occurs when people start saying that .999... equals 1.Charlie Volkstorf => Neat proof, but you are playing foot loose with the de'nition of =.> 0.999... is an in'nite series, a shorthand notation for .9, .99,> .999, ... 1 is an integer. The relationship is that 1 is a (the)> value for which, for every D>0 there is an N such that for all M>N the> value of the Mth number in the series is between 1-D and 1+D. The problem occurs when people start saying that .999... equals 1.But this is exactly the de'nition of an in'nite decimal. What doyou thing .999... means, if not the limit of the partial sums? => Neat proof, but you are playing foot loose with the de'nition of =.> 0.999... is an in'nite series, a shorthand notation for .9, .99,> .999, ... 1 is an integer. The relationship is that 1 is a (the)> value for which, for every D>0 there is an N such that for all M>N the> value of the Mth number in the series is between 1-D and 1+D.> > The problem occurs when people start saying that .999... equals 1.There are ways to associate a value with 0.9999... such that the valuedoes not equal 1. There are two immediate consequences, though:1) 0.3333... does not equal 1/3 .2) 0.9999... and 0.3333... (and other such values) are not real numbers.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W => There are ways to associate a value with 0.9999... such that the value> does not equal 1. There are two immediate consequences, though: 1) 0.3333... does not equal 1/3 . 2) 0.9999... and 0.3333... (and other such values) are not real numbers.To expand on this.If you assume 0.999... and 1.000... represent real numbers,then 0.999... = 1.000... by de'nition.Real numbers can be de'nedas the limit of a Cauchy sequence.The limit of the sequence represented by .999... is 1.You can claim 0.999... != 1.000... but you shouldn'tcall them real numbers. Real number has a standardde'nition.Russell- 2 many 2 count =In sci.logic, 22:31:37 -0500<1g6yh02.h3uybwz75oe4N%panoptes@iquest.net>:> >> Neat proof, but you are playing foot loose with the de'nition of =.>> 0.999... is an in'nite series, a shorthand notation for .9, .99,>> .999, ... 1 is an integer. The relationship is that 1 is a (the)>> value for which, for every D>0 there is an N such that for all M>N the>> value of the Mth number in the series is between 1-D and 1+D.The problem occurs when people start saying that .999... equals 1.> > There are ways to associate a value with 0.9999... such that the value> does not equal 1. There are two immediate consequences, though:> > 1) 0.3333... does not equal 1/3 .> > 2) 0.9999... and 0.3333... (and other such values) are not real numbers.Well, if x = 0.9999.... = 1-d, then (x/10 + 0.9)= 0.9999.... = 1-d/10. So either we have some veryweird set of numbers all with the same decimal expansion0.9999... (and another weird set of numbers all of whichare non-zero in'nitesimals), or d = 0 and they're allequal to 1. The second hypothesis is probably morestraightforward.Also, 0.999... can be taken as a sequence, which turnsout to be a Cauchy sequence, with limit 1.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. => Sorry, the ridiculous assertion is 0.999.... does NOT equal 1. It certainly does! Just try to subtract 0.999... from 1: 1 - 0.999... = 0 Reason: There is no real number between 0.999... and 1, and, therefore, they> must be one and the same number! PH Neat proof, but you are playing foot loose with the de'nition of =.> 0.999... is an in'nite series, a shorthand notation for .9, .99,> .999, ... 1 is an integer. The relationship is that 1 is a (the)2 - 1 is a function, 1 is a numberthey're still equal.What is the number between 0.99... and 1?Herc> value for which, for every D>0 there is an N such that for all M>N the> value of the Mth number in the series is between 1-D and 1+D. The problem occurs when people start saying that .999... equals 1. Charlie Volkstorf =There exists a bijection from the Cantor set (middle thirds removed) to[0,1]. (Ternary rep - 0's and 2's -> Binary rep 0's and 1's)Is this mapping continuous? It may be but I'm having trouble seeing whythis is so.Help?LW => > There exists a bijection from the Cantor set (middle thirds removed) to> [0,1]. (Ternary rep - 0's and 2's -> Binary rep 0's and 1's) Is this mapping continuous? It may be but I'm having trouble seeing why> this is so.Yes, it is continuous. Call it f. It is easy to see that, given x and yin the Cantor set, |f(x) - approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i02JJiX09965; papers published, or lecture notes written.keep us posted.> Here is my web site and also list of papers submitted for> publications.congratulations on your newly acquired adjunct BastiCincinnati, OH.>------------------------------------------------------->
http://hometown.aol.com/mbasti21/myhomepage/index.html Papers submitted for publication:1-Basti.M. Solving polynomials with differential equations. Submitted>for Polynomial. Submitted for publication to the>proceedings of publication to the>Mathematics of computation of the publication to the>electronic Research announcements of the Computation.6-Basti.M. New Methods Of Solving Riccati Journal on Computation.7-Basti.M. New Methods Of Solving Riccati Differential Equations II.>Submitted for publication Journal on applied Mathematics.9-Basti.M. New solutions of Bessel Differential Equations. Submitted>for publication to => I recommend you read > > S. Smale's review's of E. C. Zeeman's catastrophe theory book,> Bull. Amer. Math. Soc., Vol. 84, 1360-1368, 1978. > > S. Andreski, Social Science as Sorcery, St. Martin's Press, 1972.> > C. Truesdell, An Idiot's Fugitive Essays on Science, Springer, > 1984, p. 122ffAlso, try V. Arnold's book on Catastrophe Theory. He's a smart guy who has some interesting things to say.-- -- Lou Pecora My views are my own. =>a week, I guess a few more won't make much difference.If there's no con'dentiality clause, I'll post the problem (if any),>here.--------------------------------------------------- ------------------------>John R Ramsden (jr@adslate.com)>--------------------------------------------- ------------------------------You will post us or tell us you aren't / can't, right?--Lynn =The problem is this:You have 2 experiments both modeled by a simple regression model,Y = alpha + beta(x - xbar) + R with R~G(0,sigma) and independant.(note G represents the Gaussian distribution)The 'rst experiment has n1 samples, the second has n2 samples.We know Sxx1,Sxx2 (likewise for Syy,Sxy) least square estimates forbeta1, beta2, alpha1 and alpha2 and sigma1 and sigma2 (i will callleast square estimates beta1^ and beta2^ etc.)So the test for (beta1 - beta2) = 0 goes as follows (please correct mewhere i'm wrong)First, beta1^ ~ G(beta1, sqrt(sigma1/Sxx))likewise for beta2^.so beta1^ - beta2^ ~ G(beta1 - beta2, sqrt(sigma1/Sxx1 + sigma2/Sxx2))therefore, ((beta1^ - beta2^) - (beta1 - beta2))/sqrt(sigma1/Sxx1 +sigma2/Sxx2) ~ G(0,1)BUT ... we dont know sigma1 and sigma2, only their estimates.Normally, with the sigma that you dont know and turn thegaussian to a students t distro, but you have two sigmas under a rootsign which means you have to go back to the de'nition for sigma^which involves a double sum but I will try to write it down but itsbasically the square root of the sum of the residiules squared:sigma^ = sqrt(sum(i = 1 to 2)*sum(j = 1 to n(i))*r(i,j)^2/(n1 + n2 -2))now, 'rst is the denominator right? n - q = n1 + n2 - 2?? (q isnumber of independant parameters).now expanding the double sum is messy so i wont type the forumla outbut I cant simplify it to anything that uses beta^'s and Sxx1/2's,Sxy1/2's etc.I think, on this particular exam, the prof wanted us to just sub inthe estimates for the real values (and not go back to the de'nitionof sigma^) but I thought this was fundementally wrong.Any thoughts? Any need for clari'cation?Chris =>> http://www2.b3ta.com/hawking/>> --->> Tourette's bugger, post some content please, I just disconnected before I> clicked your post. ok, I just got the §ash show, not bad, but who was the young cronie besides> Einstein and hawking?>My guess is Bohr, crossposted to sci.math for the answer.Herc =I CHALLENGE |-|ERC, THIS IS A FOR ME for the 24 period we agree on beforehand, winneris person with most POSTS TO USENET DURING this beforehand agreed upontime period;3) In the event of TIE # OF POSTS we resume IT GOOD.Winner is person WITH MOST POSTS TO USENET IN THE 24 PERIOD WE AGREEUPON BEFOREHAND.RESULTS , Also open to negotiation WITHIN REASON:-Winner:Winner has LOSER pay for high bandwidth (50 gigs plus xfers per month)website FOR ONE YEAR , winner gets to put whatever on website andLOSER MUST SAY ONE OF1) IF YOU |-|ERC YOU WIN, I admit you HAVE PARANORMAL POWERS, DESERVETHE RANDY MILLIONS, AND ETC, I put it in writing for one of the webpages2) IF **I** WIN, YOU =CLAP CLAP to you says Lordjudging by the names of the 2 replies.CLAP CLAPClave informative broadcaster.But OK, starting soon, I don't want you slandering me thoughas part of your posts, what subject matter?And afterwards we collate the replies to measure my claim.Herc =I challenge him too.....TO SHUT THE FUCK UP BECAUSE HE'S A FUCKING LOONY!!!! =Spam King [bow down u ho both.PLONK => Kevin's logic is to the point, and I agree 100%. Just as an addendum,> however, I would clarify that when he talks about tangential force> he's talking about the relationship of the sides of the bolt (or> whatever) to the vector of rotation. Thinking about using force which> is most nearly parallel to the vector of rotation, you can see that> something like a wing nut (essentially a 2-sided head) is able to exert> the most force in the proper vector. A 3-sided head (which doesn't> exist, to my knowledge, as a standard screw, bolt or nut) would be next> best. After that, your square head would be next best able to exert> force to the vector of rotation.> > As Kevin points out, however, this has little to do with what actually> WORKS best.> > ïSporky'> > Consider heads of 2, 3, 4, 5, etc. sides. As the number of sides> increases, the cross-sectional area of the head also increases. It is> this cross-section that determines the shear strength of the head.> However, as the number of sides increases, the force acting on the> head become more tangential: with a pure circle it is strictly> tangential (but the surface area is the maximum). Note that the> goodness of these two properties vary in opposite directions: as one> gets better the other gets worse. Therefore, there must be some> optimum, and that could very likely be the square head.> > In the real world, however, idealization based on an engineering> principle is seldom the overall best. For example, a winch shaft is> round instead of splined or square (or any other §at-sided shape),> and the shaft is coupled using a strict friction-'t. This is because> the friction 't can give very nearly the same torque as a spline, but> the failure mode is much less disasterous (it simply slips instead of> shearing the splines from the shaft).> > At some point, a product will be measured by what it does not do, not> necessarily by how well it does what is advertised.> > KevinI trust Archie has a large bucket handy in his workshop- he'll need itfor all the broken (square shafted ) bits!!before breaking the bit!Jim G = I'm just a sophomore in college, so I don't know much. In our mathbook, it said the integral of sin(x^2) can't be integrated easily. What does it mean, ïeasily'? Is there a simple antiderivative ofsin(x^2), or is it some kind of in'nite thing? Has it been proveneither way? I'm just curious... John Savage =John> I'm just a sophomore in college, so I don't know much. In our math> book, it said the integral of sin(x^2) can't be integrated easily.> What does it mean, ïeasily'? Is there a simple antiderivative of> sin(x^2), or is it some kind of in'nite thing? Has it been proven> either way? I'm just curious...It's one of the two Fresnel integrals. To quote MathWorld, In this form,they have a particularly simple expansion in terms of spherical Besselfunctions of the 'rst kind. I don't like the sound of that:) The de'niteintegrals (from zero or -in'nity to +in'nity) are not hard to get, e.g. bya Fourier transform.http://mathworld.wolfram.com/ support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NgB05938; =I've come across this equation which describes the dependency of>certain factors in the operation of bipolar transistors:-Ic = Is[exp(v/t)-1]>Ic is the transistor's collector current (typically from a few to>tens of mAmps)>Is is its saturation current (typically *very* small like 5 fAmps,>for instance.I'm confused about the term exp and what it refers to - does it mean>exponent or e^x or something else? I need someone to show me a>worked example, please.Let Is = 5^-15>Let (v/t)=0.5Now, from the above, how would I 'nd Ic?>paul.>-- I expect history will be kind to me, since I intend to write it.> - Winston by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032Ngd05934; =If you don't mind, Im going to use your own proof, with some variations, in order to prove the same, but only with naturals.Let f: N -> N be a mapping from the naturals to the naturals. Note: we do not assume anything at all about f except that it is a mapping. We make no assumptions about whether f is injective (1-1) or not, whether it is surjective (onto) or not, or whether the range of f is 'nite or in'nite. The mapping f is arbitrary because we assume absolutely nothing about it except that it is a mapping from N to N.This mapping determines a list of naturals in the sense that we can write down all the naturals in the range of f in order: f(1) f(2) f(3) . . .Before going into the proof, we will carry out a neutral transformation of the list. Neutral transformation means that after it, the total amount of elements in the list has not changed, neither its value, neither its order. As the number of digits of the natural numbers increases as its value grows, we will add enough zeroes on the left of each natural in the list, in order to equal the amount of 'gures of the naturals with a bigger number of signi'cant digits in the list.This is a neutral transformation, and it will always be possible. Firstly because we do assume nothing about the list (it is arbitrary), and second because adding zeroes on the left is a variation of the bijection used by Cantor to count the naturals (i.e. a 1-1 correspondence between f(k) and f(B), being f(B) a natural with a bigger number of signi'cant digits). N' will be the transformed set of naturals. Therefore, after the neutral transformation we will have f: N -> NProposition: Let f: N -> N be given. Then f is not a surjection.Proof. We are to show that there exists n in N such that n is not in the range of f. That is, n != f(k) for any k in N.We do this by de'ning, for each k, the k-th digit in the natural representation of n. Given k > 0, we 'rst look at d_k, the k-th digit following the 'rst digit in the representation of f(k) from our list. We next de'ne the k-th digit of n, n_k, as follows: If d_k is a 1, set n_k = 2. If d_k is not a 1, set n_k = 1.Then the number n = (n_1)(n_2)(n_3)... is the required number. It is not in the list because for each k, n differs from f(k) in the k-th digit.Nicolas de la Foz>Let f: N -> R be a mapping from the naturals to the reals. Note: we do>not assume anything at all about f except that it is a mapping. We make>no assumptions about whether f is injective (1-1) or not, whether it is>surjective (onto) or not, or whether the range of f is 'nite or>in'nite. The mapping f is arbitrary because we assume absolutely>nothing about it except that it is a mapping from N to R.This mapping determines a list of reals in the sense that we can write>down all the reals in the range of f in order: f(1)> f(2)> f(3)> .> .> .Proposition: Let f: N -> R be given. Then f is not a surjection.Proof. We are to show that there exists x in R such that x is not in the>range of f. That is, x != f(k) for any k in N.We do this by de'ning, for each k, the k-th digit in the decimal>representation of x. Given k > 0, we 'rst look at d_k, the k-th digit>following the decimal point in the representation of f(k) from our list.>We next de'ne the k-th digit of x, x_k, as follows: If d_k is a 1, set x_k = 2.> If d_k is not a 1, set x_k = 1.Then the number x = .(x_1)(x_2)(x_3)... is the required number. It is>not in the list because for each k, x differs from f(k) in the k-th>digit.-- >Dave Seaman>Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.><http://www.commoncouragepress.com/index.cfm?action= support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NhR05963; =My initial question has had to run the gauntlet and did not come out unscathed (i.e., as a tough question... it turned out to be really easy). But, perhaps, there's stilla little life left in it, afterall?My Guess is as Good as Yours Theorem 1.2:Let (G,*) be a ('nite) non-abelian group with nontrivial center and assume it possible to 'nd a ring (R, *, +)such that G is a subgroup of R with respect to *. In addition, assume that for all x in G: -x(R) in G (where-x(R) is de'ned as the additive inverse of x). Then Gis of even order.My Guess is as Good as Yours Theorem 1.1 (initially froma reply to Arturo Magidin): -x(R) is independentof R, thus we may write -x for it.Happy support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NhX05946; =The following question did not need the additional complicationof R ring. Let G be a group. What is known aboutgroups with the property that there are elements x,y,z in G such that xy=yx, xz=zx, but yz != zy ? I admit,Im asking this question somewhat rhetorically- because I have found such a group. Btw., this property has very little to do with the reasons for me investigating thegroup found as of yet.Let R be a ring and [,]:R*R -> R, [x,y] = xy - yx the commutor function.Are those rings for which for all x,y,z in R:> [x,y] = 0 and [y,z] = 0 -> [x,z] = 0 somehow simpler than other rings? If so,>do any other of any such ring's properties follow >automatically from support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032Nh805977; => This is not my question, however. It is:>> (Idea- since {1,-1} is a nontrivial group, ...), does>> it follow that every group with nontrivial center >> has even order?C. DementNo. Trivially every abelian group has non-trivial centre , just pick one>that has odd order. anticipating the next one before I could ask it:>If you were going to consider non-abelian groups, try>a group of order p^n for p odd. The book should prove that at some point>this has non-trivial centre; this is the same as possessing at least two>conjugacy classes with approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NjF06063; = [.snip.]> This is not my question, however. It is:>> (Idea- since {1,-1} is a nontrivial group, ...),What does -1 mean in the context of an arbitrary group? Of course, you are correct here. I suspect one could saythe following: Given an arbitrary group (G, *), let (R, *, +)be any ring such that G is a subgroup with respect to * and in a way that every additive inverse of G is in G,then -1(R) is the additive inverse of 1 with respect to + and is in G. If no such ring exists then G does not have a -1.My Guess is as Good as Yours Theorem 1.1:the element -1 designated as above is independentof the encompassing ring (R, *, +). >And what makes you think that in your group, even if -1 makes sense,>-1 is not equal to 1?Using the technique above, it would follow that 1 is no longer a unit, am I right?> does>> it follow that every group with nontrivial center >> has even order?No. Every group whose order is the power of a prime has nontrivial>center (you will deduce it as a consequence of the class formula),>in particular, any group of order p^n, p an odd prime, n>0, (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NiR05997; =respect: 1) to feel or show honor or esteem for; hold in high regard...2)to show consideration for; avoid intruding upon or interfering with...>> > > But then, why does he pay attention to my posts at all, as that is a> sign of respect?> > What ever gave you that idea?>> Respect comes from Latin. It might help you to look up the word's etymology.Why? I'm interested in the term as it is used today, and why you>> think responses to your post are a sign of respect -- in the common,>> everyday meaning of the word respect.Etymology doesn't help answer that question.>Then look up the *current* de'nition.Why don't you do that and report back?>James support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i032NkI06081; =>> Here is a shocking admission: I'm curiously growing weary of this>> fascinating exchange. Major snippage below.[...]>> > You used this claim to support the claim> that Arturo is incompetent. This is just utter bullshit.>> you have some reading comprehension problems. my initial follow up in>> no way claims that arture is incompetent.Explain this post is indeed either incompetent as a>> | > mathematician, or more likely a liar, or both.>> | >> | well, he is an ADJUNCT assistant professor. ADJUNCTS are the pariahs>> | in academic caste system.>> | >> | so whomever you are, may well be right on this one...>> `----How is jstevh@yahoo.com right, given that Arturo is an adjunct?>>well, some who happen to be adjuncts happen to be incompetent - duh.> > You demonstrate your usual blunt reasoning, maky.> > For the record, your post was an insult to me, and you know it. That's> how I read it, that's how everyone reasonable read it.>>No, but that's just because I 'nd your opinions to be beneath>insult, and your current claims a show of cowardice. You can't even>stick by your sly attempts at an insult.>shoudn't you be insulted by your employers instead?No. Your statement is plain. It is an attack on me.well, it was not inteded as such. if apologies you seek,i apologise for insulting you.now onto the obvious topic of my post. don't you agree withmy assessment of the adjunct rank in academia?>>please explain.may well be right can only refer to the fact that I was> being called (through copying an old post of James> Harris) incompetent as a mathematician, or more likely a> liar, or both. You are agreeing with the possibility, which> means you are stating your opinion that it may> very well be true that I am an incompetent and/or a liar.what i said, does it have mean that?> Just the latest in your campaign to attack me because I would not> support you in a.a.>>come again? For someone as ignorant as you are, you don't fake ignorance> too well.-- > == ==>It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> =Arturo Magidin>magidin@math.berkeley.eduIs there a closed form for this sum:i=nSUM x^(gcd(i,n))i=1(Note: This arises in counting the number of essentially distinctcolorings of a directed cycle of length n).Siamak =This is a reponse to a comment something like of what use isabstract math.through a telescope.Van