mm-384 === Subject: : Re: Question about Incompleteness (Godel)[...]> 2. What results are there about minimum complexity requirements for> a statement to be potentially undecidable in a consistent formal> system. (For instance, trivial statements like x+y > x for all x>0> are certainly not undecidable in PA, and this is because (as I> understand) such a statement is expressable in a much weaker formal> system which is consistent and complete.)[...]Gregory Chaitin defined the number Omega, 0I want to find x(n)>x(n+2) + 2*x(n+1) + c(n)*x(n) = 0>were c(n) is a given sequence and x(0) and x(1) are given.>thanks.Try transform methods, analogous to solving a 2nd order ODEusing Laplace transforms.=== === === Subject: : Re: Do Prime Algebraic Numbers even exist?Dik T. Winter Dik.Winter@cwi.nl > However, when m > 0 it is not easy to always find a unit.>What I meant here is that it may take a lot of work. There are>deterministic methods to get a unit (or the fundamental unit),>but the computational complexity becomes pretty big. I have done>extensive calculations with discriminant < 10000. (The discriminant>is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one>I did not complete was D = 409 (m = 409) where in the fundamental>unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the>range D < 1000. (Yes, I could have gone to some bignum package, but>did not feel inclined to do so...)This seems to be a reply to my post. Apparently you did not look at thewebpage I suggested.It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409. Thelength of the period of the continued fraction expansion of sqrt(409) is only21. So 21 iterations of the PQa algorithm (see my webpage if you're notfamiliar with this well-known simple algorithm) suffice to find the minimalpositive solution of x^2 - 409y^2 = +-1. There are only 5 numbers that need tobe computed at each iteration, three of the numbers never get bigger than 24,and each computation is pretty simple.Microsoft Excel does integer arithmetic to an accuracy of about 15 digits. Soeven in Excel it takes a fraction of a second to find x = 111,921,796,968, y =5,534,176,685 as the minimal positive solution.There are 19 more what?The length of the period of the continued fraction expansion of sqrt(D) is lessthan 4*sqrt(D)*ln(D) (at least for squarefree D). For D <= 10,000, this isless than 3,700. Maybe a lot by hand, but easy on any computer.One also frequently sees the comment that solving x^2 - Dy^2 = N is verycomplicated. This is hardly more difficult than solving x^2 - Dy^2 = +-1. Again, see my webpage at http://hometown.aol.com/jpr2718/John Robertson=== === Subject: : Re: Do Prime Algebraic Numbers even exist? > Dik T. Winter Dik.Winter@cwi.nl > However, when m > 0 it is not easy to always find a unit. >What I meant here is that it may take a lot of work. There are >deterministic methods to get a unit (or the fundamental unit), >but the computational complexity becomes pretty big. I have done >extensive calculations with discriminant < 10000. (The discriminant >is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one >I did not complete was D = 409 (m = 409) where in the fundamental >unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the >range D < 1000. (Yes, I could have gone to some bignum package, but >did not feel inclined to do so...)See my last remark above... > It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409.Yes, I know it is easy to find, consider however the constraints I give... > It is very easy to find the fundamental unit in Q(sqrt(D)) for D = 409. The > length of the period of the continued fraction expansion of sqrt(409) is only > 21. So 21 iterations of the PQa algorithm (see my webpage if you're not > familiar with this well-known simple algorithm) suffice to find the minimal > positive solution of x^2 - 409y^2 = +-1.Actually I *did* use it. > Microsoft Excel does integer arithmetic to an accuracy of about 15 digits. > So even in Excel it takes a fraction of a second to find > x = 111,921,796,968, y = > 5,534,176,685 as the minimal positive solution.As I have no access to Microsoft Excel... As I said, I did not feelinclined to go to a bignum package, so the results are limited to (in mycase) what a double precision number could handle. And I remained a biton the safe side, limiting the numbers to 10^12. > The length of the period of the continued fraction expansion of sqrt(D) is > less than 4*sqrt(D)*ln(D) (at least for squarefree D). For D <= 10,000, > this is less than 3,700. Maybe a lot by hand, but easy on any computer.Computational complexity not only involves the number of operations, butalso the complexity of operations. Operations on large numbers are morecomplex than operations on small numbers. When you look at it from a puremathematical viewpoint it is a bit different. But when you are actuallyimplementing stuff the largeness of numbers gets to play a big role.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : QuestionsI am beginner in algebraic topology and I want to understand theclassifying spaces,so I read from Steenrod's book and Milnor's paperthen I compared them with Husemoller's book and I could understand byit, but I could not solve some exercises (Ex.11 and 13, page58,Husemoller'sbook Fiber Bundles), so would you be so kind as tosolve them ? These exercises are:11.If u: G ---> H is a continuous group morphism, define a map B(u):BG ->BH. Under what circumstances do you have a functor ?13. Let $=(X,P,B) be a numerable principal G-bundle. Prove that $ isuniversal if and only if X is contractible Thank you very much Best wishes Rabab,=== === Subject: : Re: divergent series>S_1^inf [1-(ln(n))/n]^n = inf>how to prove this?>I assume you mean> oo> --- ln(n) n ( 1 - ----- ) [1]> --- n> n=1>-Yes, that's exactly what I mean.> When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n,> multiplying by n, and exponentiating, we get > ln(n) n> ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n)> n> 1 1> - - [2]> n 2 >For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1.>Since each term of [1] is greater than 1/(2n), compare with the harmonic>series.>your inequalities?For the inequality, log(1-a) > -a-a^2 when a < 1/2, I looked at theMaclaurin series for log(1-a) = -a - a^2/2 + O(a^3), and figured it hadto be bigger than -a - a^2 for some values of a around 0. Using thederivative of log(1-a) + a + a^2, I found that it is greater than 0 whena < .6838, so definitely when a < 1/2.For the inequality ln(n)/n < 1/2, I needed that inequality, so I justused the derivative of ln(n)/n to find the extrema of ln(n)/n and foundthat it is always less than 1/e which is less than 1/2. Of course nmust be greater than 0.Rob Johnson take out the trash before replying=== === Subject: : Re: divergent seriesLet me keep the thread going and pose: does the series sum_n 1/2^sqrt(n)converge or diverge? Curiously, I don't recall ever seeing this on a> calculus exam.--Ron Bruckconverges, by the integral test>>Sounds good, but offhand I don't see how to integrate 1/2^sqrt(x). Is>>it really integrable by elementary functions?> | sqrt(x)> | 2 dx> |> |> = | 2^y d(y^2)> |> |> = | 2y 1/log(2) d(2^y)> |> |> = 2/log(2) y 2^y - | 2/log(2) 2^y dy> |> = 2/log(2) y 2^y - 2/log(2)^2 2^y> = 2/log(2)^2 2^y (log(2) y - 1)> sqrt(x)> = 2/log(2)^2 2 (log(2) sqrt(x) - 1)Oops; while the integration above is correct, it does not converge asx->oo. Change 2^sqrt(x) to 2^-sqrt(x): | -sqrt(x) | 2 dx | | = | 2^-y d(y^2) | | = | 2y -1/log(2) d(2^-y) | | = -2/log(2) y 2^-y + | 2/log(2) 2^-y dy | = -2/log(2) y 2^-y - 2/log(2)^2 2^-y = -2/log(2)^2 2^-y (log(2) y + 1) -sqrt(x) = -2/log(2)^2 2 (log(2) sqrt(x) + 1)This does converge as x->oo.>>On the other hand, it's clear that 2^sqrt(n) is larger than, say, n^2,>>for all sufficiently large n. (Take logs, so need to know sqrt(n)>>grows faster than log(n).) Hence the series converges by the>>comparison test with sum of 1/n^2.>>More generally, if C > 1 and e > 0 are constants, then the sum of>>1/C^(n^e) converges. Proof is the same.Rob Johnson take out the trash before replying=== === Subject: : Re: How to prove this probability Inequality>X,Y,Z are random variablesTry to proveE[EX-E(X|Z)]^2>=E[EX-E(X|Y,Z)]^2>When is your homework due?It is a homework, it is a problem my friend asked me he is preparing qualifying examWe've already solved thatI don't like this kind of words When is your homework due?please don't waste your time on typing this.sorry for the offencethough maybe you are a professor=== === Subject: : Re: Cubic FormulaSo how would you go about solving x^3 + 9x^2 + 9x + 8 = 0?Long ago from self-study of an old College Algebra book (by C.H > Lehmann, published in 1962) I found, I learned the following method.By Descartes' Rule of Signs, the given equation has no positive root and > either three or one negative root.If it has a rational root, then this is of the form p/q where p is a > divisor of the constant term and q is a divisor of the leading > coefficient. Thus, in this case, the only possible rational roots are > -1, 2, -4, and -8. It just occurred to me that finding any rational number solutionsto polynomial equations using the method described above. Using thequadaratic, cubic, or quartic formulas would only be necessary to findroots which are not rational numbers. Michael=== === Subject: : Re: . The hardest of all hard facts .Dear Peter:> Dear David,Surely the anisotropy decreased over the 100 years from your firstcitation>to your last. The first citation considers the anisotropy of the speed of light in> air and other gases, the last the anisotropy of the speed of light in> vacuum. These are two different things.Yes and no. If the gas molecules are imaged on the aether, or areotherwise dependent on the aether for EM-moderated repulsion, which is theconcensus, then they are the same thing. You still have a secular decay inthe aether drift.So you can choose to believe that you have poked a hole in the bottom ofthe aetherist's boat, or it is simply that one cannot assert certainty atthe resolution limits of a given apparatus. I vote for the latter, but youcan certainly claim the former if you wish.David A. Smith=== === Subject: : Pigeonhole Speed-UpConsider the following inference, based on the Pigeonhole Principle: (i) There are at least m objects (ii) There are exactly n A's (iii) The f of each object is an ATherefore, (iv) There are two distinct objects a, b such that f(a) = f(b).For any particular choices of m and n, we can express (i)-(iv) asfirst-order formulas (note that m and n are not variables. When weexpress (i) as a first-order formula, it just contains a sequence of mexistential quantifiers, followed by conjunction of inequalities between thevariables).If m>n, this argument is valid in FOL. (When n = 0, premises (ii) and (iii)are inconsistent, so the argument is still valid.)There is a deductive proof of (iv) from {(i), (ii), (iii)} in your favouriteproof system for FOL. In a sense, this is a mathematics-free proof. At nopoint does one refer to numbers, sets or functions.What interests me is the unfeasible length of this mathematics-free proof,when m, n are not tiny.Consider the formula [(i) & (ii) & (iii)] -> (iv)Call this PHP(m,n). If m>n, then PHP(m,n) is true. The informal settheoretical proof of the fact that if m>n, then PHP(m, n) is true is not toolong.Problem 1 (Technical): Given m and n such that m>n, how long is a FOL proofof PHP(m,n)?I don't know too much about complexity theory, where such matters arestudied. I do know that people have studied the complexity of proofs ofpropositional formulations of PHP, giving detailed bounds.As a piece of experimental mathematics, I set up my SPASS theorem proveron my PC to prove PHP(m,n), for some small m,n. PHP(2,1), PHP(3,1), PHP(4,1)were very quick (less than a second). For various small m, PHP(m, 2) wasvery quick also, only a few seconds for PHP(7,2). Curiously, PHP(4,2) tookless than PHP(3,2), but then it got longer again.Letting n=3, PHP(5,3) and PHP(6,3) were both several minutes.However PHP(4,3) took 1 hour and 16 minutes!I suspect that PHP(5,4) would take a rather large amount of computing time.(The real-world example below concerns PHP(10^9, 180).)Problem 2 (osophical):We all know that PHP(m,n) is true, if m>n. The problem is *how* we know thatit's always true.In particular, consider an application of this. E.g., (i) There are at least one ion people (ii) There are exactly 180 countries (iii) Each person is in exactly one country,So, (iv) There are at least two distinct people in the same country.Presumably everyone agrees that (iv) follows logically from (i),(ii),(iii).But in order to justify this belief, one must appeal to genuine mathematics,as one cannot actually perform the maths-free computation feasibly. If thespeed-up of such proofs goes as fast as I suspect it does, then it'spossible that a first-order proof of (iv) from (i),(ii),(iii) may have moreHowever, informal mathematical reasoning, about a *set* of people, and a*set* of countries, and their cardinal numbers, and a *function* from peopleto countries, provides a quick proof of (iv) from (i),(ii),(iii). In otherwords, there seems to be a problem for a very strict formalist aboutmathematics: they cannot accept the above reasoning, since it directlyappeals to numbers, sets and functions.--- Jeff=== === Subject: : are Doubly-Infinites the points of Lobachevskian Geometry Re: cracks in Euclidean Geometry and why Reals are fake(most snipped)Research: All I need is hard-core evidence that Doubly-Infinites are> all negative numbers. If I can get that evidence, then all of my above> thoughts would be confirmed.> Let us look at a few P-adic of a 10-adic such as .....333333. and....67 and.....101100111000. And the question of sign value does not become anissue for they are all of one sign. They are all of a positive numbersign value even though some of them can resemble a negative value suchas ....99998 which can be constrewed as -2. Not that it is -2 but hasthe resemblance of -2. But all the P-adics thus have this *positivevalue characteristic*.But now take a look at Doubly-Infinites, can we likewise unravel whatsign value these numbers possess? Are they all a value opposite ofthat of P-adics? Are they all *negative sign value* whereas p-adicsare *positive sign value*??Let us write some Doubly-Infinites such as .....333333.67676767.....Another Doubly-Infinite is .....99998.101100111000.........The main question then is whether that infinite string rightwardsupsets the sign value of the infinite string leftwards so very much sothat the sign value of the entire number entity in question changesfrom positive to negative.P-adics are infinite strings leftward and are all positive valuedspecimens. But what is the sign value of Doubly-Infinites? Does thatrightward infinite string force so much change upon the leftwardinfinite string that the number specimen in question is transformedinto a negative number?I am trying to get a handle on this question by contrasting actualphysical objects. And in contrast with Speed and Velocity andAcceleration in Physics. If we are allowed to speed ahead withvelocity and with acceleration this would be the infinite leftwardstring. But in our speeding and accelerating we are forced to payattention to little things, little distances would slow us down. Ingoing from 1 to 2 to 3 to 4 we can keep the speed. But if forced tolook at 1.3 between 1 and 2 and then 1.4 between 1 and 2 we are sloweddown.Another physical model is the shape of a convex lens to a concave lensin contrast to infinite leftward strings compared to doubly-infinites.With no infinite rightward string to slow down the leftward string itbecomes convex in shape. But when details of numbers between twoconsecutive rightward string numbers is demanded then a concavesurface is traversed.What I am trying to convey is that P-adics are infinite leftwardstrings and they are all positive numbers. But in Doubly Infinites,because you have a infinite rightward string attached to a infiniteleftward string, that the rightward string concaves the path. So thatwhere the P-adic would form a convex lens surface, the Doubly-Infiniteforms the opposite surface of concave and thus the opposite sign ofnegative vice positive.Perhaps there is a better physics analogy to what a rightward minutiadetail string does to the macro large string that goes leftward. Orperhaps mathematics itself can decide this case without any need torefer to physics. But I doubt it because all of mathematics is a tinysubset of physics.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies(www.iw.net/~a_plutonium) website of the science of AP under revisionwhat used to be my old science websitewww.newphys.se/elektromagnum/physics/LudwigPlutonium from years 1993=== === Subject: : Re: Zorn's Lemma Question>Let S be the real numbers (0,1)>>Since it is a set of real numbers it is partially ordered and everychain is>>obviously bounded by 1>>Yet (0,1) does not have a maximal element.>>What am I missing here?>The chain (0,1), for example, does *not* have a least upper bound *in S.*> More relevant to a question about Zorn's lemma is the fact that (0,1)> does not have an upper bound in S.Historical aside...1. Kuratowski published Zorn's lemma in 1922, Zorn in 1935. Why didn't K'sdiscovery catch on? Maybe because his paper is 34 pages long, and Zorn'sonly four pages :)2. I'm not certain, but I think that both of those authors spoke of _least_upper bounds, and someone later noticed that any upper bound will do.3. For an ordered set to have a maximal element, it is sufficient that any_well-ordered_ subset have an upper bound. Again I'm not sure who was first,but it might have been someone in the Bourbaki group.LH=== === Subject: : Re: Zorn's Lemma Question> 3. For an ordered set to have a maximal element, it is sufficient that any> _well-ordered_ subset have an upper bound.Can we prove that is sufficient without using the Axiom of Choice? Ina discussion of Zorn's Lemma, one might like to avoid using AC.=== === Subject: : Re: Zorn's Lemma Question _well-ordered_ subset have an upper bound.> Can we prove that is sufficient without using the Axiom of Choice?Unlikely. Zorn's Lemma is *equivalent* to the Axiom of Choice, so Iwouldn't expect to prove it in ZF, even in the usual (ostensiblyweaker) form with chain in place of well-ordered subset.=== === Subject: : Re: Zorn's Lemma Question> 3. For an ordered set to have a maximal element, it is sufficient that any> _well-ordered_ subset have an upper bound. Can we prove that is sufficient without using the Axiom of Choice?Unlikely. Zorn's Lemma is *equivalent* to the Axiom of Choice, so I> wouldn't expect to prove it in ZF, even in the usual (ostensibly> weaker) form with chain in place of well-ordered subset.> OK, what I meant was: Can we prove without AC that: (1) every totally ordered subset has an upper boundis equivalent to (2) every well ordered subset has an upper bound=== === Subject: : Re: Zorn's Lemma Question <0402261144220.21351-100000@gandalf.math.ukans.edu OK, what I meant was: Can we prove without AC that:> (1) every totally ordered subset has an upper bound> is equivalent to> (2) every well ordered subset has an upper boundMaybe this is a more natural counterexample. I believe the followingaxioms are simultaneously consistent with ZF:(1) any countable union of countable sets is countable;(2) no uncountable subset of R is well-orderable.Let W be the set of all well-orderings of N. For S in W, let o.t.(S)be the order type of S, some countable ordinal. Fix an injective mapf:W --> R. Define a linear ordering of W as follows: for S, T in W,let S < T if either o.t.(S) < o.t.(T), or else o.t.(S) = o.t.(T) andf(S) < f(T).Now every well-ordered (or even well-orderable) subset of W iscountable, and every countable subset of W has an upper bound in W, soevery well-ordered subset of W has an upper bound in W; but W itselfis totally ordered, and has no upper bound in W.=== === Subject: : Re: Zorn's Lemma Question <0402261144220.21351-100000@gandalf.math.ukans.edu OK, what I meant was: Can we prove without AC that:> (1) every totally ordered subset has an upper bound> is equivalent to> (2) every well ordered subset has an upper boundOh. I'm not sure, but I don't think you can prove that without AC. Infact, I don't think you can prove that a nonempty set of real numbers,with no greatest element, necessarily contains a strictly increasingsequence of real numbers.Now suppose S is a nonempty set of real numbers, with no greatestelement, and suppose that S contains no strictly increasing sequence.(E.g., take an infinite Dedekind-finite set of real numbers--if such athing exists--and take a finite number of points off the top.) Everywell-ordered subset of S, being finite, has an upper bound in S; but Sitself is a totally ordered set with no upper bound in S.=== === Subject: : Re: Zorn's Lemma Question>Let S be the real numbers (0,1)>Since it is a set of real numbers it is partially ordered and every>chain is>obviously bounded by 1>Yet (0,1) does not have a maximal element.What am I missing here?>The chain (0,1), for example, does *not* have a least upper bound *in S.*>> More relevant to a question about Zorn's lemma is the fact that (0,1)>> does not have an upper bound in S.>Historical aside...>1. Kuratowski published Zorn's lemma in 1922, Zorn in 1935. Why didn't K's>discovery catch on? Maybe because his paper is 34 pages long, and Zorn's>only four pages :)>2. I'm not certain, but I think that both of those authors spoke of _least_>upper bounds, and someone later noticed that any upper bound will do.Well that's curious, if true. Raises an obvious question... hmm.Ok, let's call that LZL, and the currently standard Zorn's lemma ZL.In fact LZL is as strong as ZL, because LZL allows you to well-orderany set by showing that there is a maximal well-ordered subset,and you can use that to prove ZL.Otoh there are situations where ZL applies but LZL cannot beapplied directly, for example if we were using ZL to show that(0,1) union (1,2] has a maximal element. Of course that's anabsurdly phony example, since there's no way to verify thehypothesis of ZL before you notice that there's a maximalelement. Hence the question: Are there any actual (ornatural) applications of ZL where not every chain has aleast upper bound?>3. For an ordered set to have a maximal element, it is sufficient that any>_well-ordered_ subset have an upper bound. Again I'm not sure who was first,>but it might have been someone in the Bourbaki group.>LH=== === Subject: : Re: Zorn's Lemma Question> Larry Hammick>Historical aside......>2. I'm not certain, but I think that both of those authors spoke of_least_>upper bounds, and someone later noticed that any upper bound will do.> Well that's curious, if true. Raises an obvious question... hmm....> Otoh there are situations where ZL applies but LZL cannot be> applied directly, for example if we were using ZL to show that> (0,1) union (1,2] has a maximal element. Of course that's an> absurdly phony example, since there's no way to verify the> hypothesis of ZL before you notice that there's a maximal> element. Hence the question: Are there any actual (or> natural) applications of ZL where not every chain has a> least upper bound?Not that I can think of. A typical application of ZL involves atotally ordered (by inclusion) set of subsets of something, andit turns out that the union of them is in the set, and is of coursea least upper bound. I think it would be easy, especially backthen, to overlook the fact that the hypotheses in LZL can beweakened.LH=== === Subject: : Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings)>The other was>that the technical people didn't communicate well. In particular, I>recall that they produced a scatterplot of temperature versus amount>of o-ring erosion FOR LAUNCHES WHERE THE O-RINGS WERE ERODED, which>showed no clear correlation. Not shown on the plot were the launches>with NO EROSION, which were all at warm temperatures. A more suitable>plot would have made clear that there was plenty of evidence that>erosion was greater at low temperatures than at high temperatures, and>that launching at a temperature substantially lower than any previous>launch was madness.Your summary is correct. Dropping the 0's (because they didn't addany more information) was a little more than a failure to communicate.The statistical reference for this is Dalal, Fowlkes and Hoadley, 1989.Risk analysis of the space shuttle: Pre-Challenger prediction of failure.JASA 84:945--957-- My real email address ismcintosh ##at## research ##dot## telcordia ##dot## com=== === Subject: : MotivateCan someone please explain to me the usage of the word motivate, as in arecent posting Motivation for e?The context implies that the word is being used as a synonym for explain,but I have never seen this usage in a dictionary or thesaurus.Has it recently come into the language? Have I mis-understood the usage?Jim Brown=== === Subject: : Re: MotivateJim Brown asked,> Can someone please explain to me the usage of the word motivate, as in a> recent posting Motivation for e?This appears to be American mathematician's argot. It is unfamiliar to atleast some of the British mathematicians who post here, and thus is probablyfairly recent. I have seen this usage of the word motivate in quite a few> The context implies that the word is being used as a synonym forexplain,> but I have never seen this usage in a dictionary or thesaurus.The particular meaning of motivate in this kind of context appears to beprovide background for understanding why this notation/problem/conjecturewould be interesting to mathematicians. In other words, to explain to acallow undergraduate, who hasn't developed enough mathematical maturity toknow why the issue is important to mathematicians, what got mathematiciansinterested in the issue in the first place.> Has it recently come into the language? Have I mis-understood the usage?It must be quite a recent usage, and it appears to be confined entirely tomathematics, and yet to be universal even among mathematicians. I haven'tused it myself yet, as best I can recall, but I rather like it and think itmay spread, at least to discussions of pedagogy in, e.g., science andhistory and so forth.I hope this motivates the usage of the term motivate among somemathematicians [wry grin]. And I would, of course, defer to the professionalmathematicians here--which I am not--for further discussion of this usageamong mathematicians, which I will read with interest.-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn in Freedom (TM) http://learninfreedom.org/kmbunday AT earthlink DOT net (preferred email address)=== === Subject: : Re: Motivate> Jim Brown asked,Can someone please explain to me the usage of the word motivate, as in a> recent posting Motivation for e?This appears to be American mathematician's argot. It is unfamiliar to at> least some of the British mathematicians who post here, and thus is probably> fairly recent. I have seen this usage of the word motivate in quite a fewThe context implies that the word is being used as a synonym for> explain,> but I have never seen this usage in a dictionary or thesaurus.The particular meaning of motivate in this kind of context appears to be> provide background for understanding why this notation/problem/conjecture> would be interesting to mathematicians. In other words, to explain to a> callow undergraduate, who hasn't developed enough mathematical maturity to> know why the issue is important to mathematicians, what got mathematicians> interested in the issue in the first place.Has it recently come into the language? Have I mis-understood the usage?It must be quite a recent usage, and it appears to be confined entirely to> mathematics, and yet to be universal even among mathematicians. I haven't> used it myself yet, as best I can recall, but I rather like it and think it> may spread, at least to discussions of pedagogy in, e.g., science and> history and so forth.I hope this motivates the usage of the term motivate among some> mathematicians [wry grin]. And I would, of course, defer to the professional> mathematicians here--which I am not--for further discussion of this usage> among mathematicians, which I will read with interest.As you say, it means to explain why anybody actually cares about whatis about to be presented. I first heard this usage in a slightlydifferent form. In the late sixties some mathematics textbooks werecriticized as presenting motivationless or unmotivated mathematics. Since I only studied mathematics in the United States, I can't tellyou that it is an Americanism, but it was certainly used in America atleast 37 years ago.Achava=== === Subject: : Re: MotivateJim Brown asked,>> Can someone please explain to me the usage of the word motivate, as in a>> recent posting Motivation for e?>This appears to be American mathematician's argot. It is unfamiliar to at>least some of the British mathematicians who post here, and thus is probably>fairly recent. I have seen this usage of the word motivate in quite a fewIt is not mathematician's argot, it is educationists' argot. There is this belief among those who do not understand concepts that the only way to learn somethingis to be sufficiently motivated. Learning something nowbecause it may be useful later is not acceptable.-- 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 Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Motivate|Karl M. Bunday Jim Brown asked,||>> Can someone please explain to me the usage of the word motivate, as in a|>> recent posting Motivation for e?||>This appears to be American mathematician's argot. It is unfamiliar to at|>least some of the British mathematicians who post here, and thus is probably|>fairly recent. I have seen this usage of the word motivate in quite a few||It is not mathematician's argot, it is educationists' |argot. There is this belief among those who do not |understand concepts that the only way to learn something|is to be sufficiently motivated. Learning something now|because it may be useful later is not acceptable.herman is being silly here; motivation is indeed mathematicians'sargot. i first ran across it around 1980 when i was trying to getinto a math course at suny stony brook for which i didn't have theprerequisites; the professor was worried at first that i might nothave the motivation to do well in the course and was amused when iresponded by trying to demonstrate my earnestness about learningmathematics. it was obviously already pretty old argot by then.-- [e-mail address jdolan@math.ucr.edu]=== === Subject: : Re: Public Notice: Copyright Violations> It seems to me that this would not be a copyright issue. I have never> heard of a product being put out for all to use and then, years later,> lawsuits are filed regarding the use of that product. It would be> hard for James to prove that he provided adequate information for the> general public to know that the material wasn't freely placed into the> public domain. He put no copyright notices on his messages, nor did> he notify anybody until years later.> Copyright notices are not needed according to the convention of Berne> (to which the US is also a signatory, albeit a quite late one).> Everything is born copyrighted.> unlimited copying for republication (to all newsservers around the> world). Whether this permission is only for newsservers remains to be> seen. When you copy something on paper in a book it is good usage to> ask the original poster for permission. The main reason is that the> it is debatable whether the form of publication changes or not.Seems to be covered the fair use clause, even if it were marked 'copyright':from: http://www.copyright.gov/title17/92chap1.html#107 107. Limitations on exclusive rights: Fair use38Notwithstanding the provisions of sections 106 and 106A, the fair use of acopyrighted work, including such use by reproduction in copies orphonorecords or by any other means specified by that section, for purposessuch as criticism, comment, news reporting, teaching (including multiplecopies for classroom use), scholarship, or research, is not an infringementof copyright. In determining whether the use made of a work in anyparticular case is a fair use the factors to be considered shall include -(1) the purpose and character of the use, including whether such use is of acommercial nature or is for nonprofit educational purposes;(2) the nature of the copyrighted work;(3) the amount and substantiality of the portion used in relation to thecopyrighted work as a whole; and(4) the effect of the use upon the potential market for or value of thecopyrighted work.The fact that a work is unpublished shall not itself bar a finding of fairuse if such finding is made upon consideration of all the above factors.KeithK> -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,+31205924131> home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/=== === Subject: : Re: Public Notice: Copyright Violations <4Ne%b.4390$Zp.3684@fed1read07> Discussion, linux)> Seems to be covered the fair use clause, even if it were marked 'copyright':> from: http://www.copyright.gov/title17/92chap1.html#107> Notwithstanding the provisions of sections 106 and 106A, the fair use of a> copyrighted work, including such use by reproduction in copies or> phonorecords or by any other means specified by that section, for purposes> such as criticism, comment, news reporting, teaching (including multiple> copies for classroom use), scholarship, or research, is not an infringement> of copyright. In determining whether the use made of a work in any> particular case is a fair use the factors to be considered shall include -> (1) the purpose and character of the use, including whether such use is of a> commercial nature or is for nonprofit educational purposes;> (2) the nature of the copyrighted work;> (3) the amount and substantiality of the portion used in relation to the> copyrighted work as a whole; andYou seem to be overlooking the third clause. Entire books (ornot clear whether entire Usenet posts can be reproduced for the samepurpose. (Note: I know that many classes involve a substantial number ofpractice.)> (4) the effect of the use upon the potential market for or value of the> copyrighted work.> The fact that a work is unpublished shall not itself bar a finding of fair> use if such finding is made upon consideration of all the above factors.> -- If anything is true in general about Usenet, it's that people can goon and on about just about anything. -- James Harris speaks thetruth.=== === Subject: : Re: Public Notice: Copyright Violations> postings here are intended (and expected) to be 'public', i.e. this is a> public forum. You cannot expect to retain copyrights on material which> you have given into the public domain.> Hrm. According to Dutch law I *cannot* put something into the public> domain.Interesting...> There is only one way for me to lose copyright: give it to> some other legal person or institution.So, according to Dutch law, it is not possible for a living person to place any of hisworks into the public domain? That's not only interesting, it is extremely odd.> domain 70 years after my death ;-).Well, with James' work we can hope for the best.--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=== === Subject: : Re: Public Notice: Copyright Violations... > There is only one way for me to lose copyright: give it to > some other legal person or institution. > So, according to Dutch law, it is not possible for a living person to > place any of his works into the public domain? That's not only > interesting, it is extremely odd.Why is it so very odd? The concept of public domain with all thatit entails is a pure US thing, I do not know which countries have thesame concept, but as far as I know, the UK has not, to name one.For instance, I may transfer copyright to somebody else, that does *not*mean that he can do anything with it he wants. There are some rights Iretain and can not give away (the so-called neighbouring rights).still not the right to mutilate it, I have still the right to prohibityou from doing so. This is even true 70 years after my death with therights of my heirs.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/=== === Subject: : Re: Public Notice: Copyright Violations> ...> There is only one way for me to lose copyright: give it to> some other legal person or institution.> So, according to Dutch law, it is not possible for a living person to> place any of his works into the public domain? That's not only> interesting, it is extremely odd.> Why is it so very odd?See below.> The concept of public domain with all that> it entails is a pure US thing, I do not know which countries have the> same concept, but as far as I know, the UK has not, to name one.In the US, at least, many creative works which are commissioned by or sponsored by thegovernment are automatically and specifically in the 'public domain' and notcopyrightable. This is because the are paid for by public funds. Who owns governmentworks (paid by public funds) in other countries?> For instance, I may transfer copyright to somebody else, that does *not*> mean that he can do anything with it he wants. There are some rights I> retain and can not give away (the so-called neighbouring rights).> still not the right to mutilate it, I have still the right to prohibit> you from doing so. This is even true 70 years after my death with the> rights of my heirs.I consider this odd because if creates a 'right' which I cannot disclaim, abandon orforfeit in the manner of my own choosing. As such, it is not really a right at all, butis now an obligation. I am obliged to possess the thing whether I choose to do so ornot, unless I dispose of it in a manner prescribed by law. Under this concept, I candestroy my own works, but I cannot disown them. Don't you think that is odd?--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=== === Subject: : Re: Public Notice: Copyright Violations> As such, it is not really a right at all, but is now an obligation. I> am obliged to possess the thing whether I choose to do so or not,It's not a thing that you own, it's a set of rights that you have,some of which you can transfer to someone else. You can choose not toexercise those rights - for example, by allowing other people to makecopies of your work - so what obligation does having those rightsimpose on you?-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules!=== === Subject: : Re: Public Notice: Copyright Violations> As such, it is not really a right at all, but is now an obligation. I> am obliged to possess the thing whether I choose to do so or not,> It's not a thing that you own, it's a set of rights that you have,> some of which you can transfer to someone else. You can choose not to> exercise those rights - for example, by allowing other people to make> copies of your work - so what obligation does having those rights> impose on you?Before answering your question, let me acknowledge that the term 'rights' is usedin many different ways, not all of which are equivalent. We hear about women'srights, children's rights, welfare rights, smoker's rights, minority rights,etc. I don't think the concept of rights, per se, is easily discussed with theprecision expected of mathematical terms.Having said that, let me recall that the US Constitution authorizes the grantingof limited duration copyrights for the purpose of promoting the arts. If thecurrent law limits my ability to benefit from my copyrights, I am not well-servedby the law. For example, if I can transfer my copyrights to another individual ororganization I may expect to get paid for the transfer. But if I am under someprior legal obligation to retain some level of ownership which I cannot transfer,the value of the sale to the buyer is diminished. That is, the buyer ispurchasing an inferior set of rights and only partial ownership, so his benefitin 'owning' the copyright is clearly less than mine. Hence, I cannot turn this'benefit' (permanent ownership of some rights) into any tangible reward throughthe sale.The obligation that this kind of 'right' imposes is one of permanent ownership: I*must* possess it, whether I choose to or not, and therefore I have lost anyright to sell or dispose of it as I see fit.It appears that there is a sizable support group for this state of affairs. I amnot a member of that group.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com=== === Subject: : Re: Public Notice: Copyright Violations>> The concept of public domain with all that>> it entails is a pure US thing, I do not know which countries have the>> same concept, but as far as I know, the UK has not, to name one.>In the US, at least, many creative works which are commissioned by or sponsored>by the>government are automatically and specifically in the 'public domain' and not>copyrightable. This is because the are paid for by public funds. Who owns>government>works (paid by public funds) in other countries?Right words but quite wrong statments. As I understand it the works of aUS civil servant can not be copyrighted. But that does not extend to worksdone by others even if the money comes from the US governement. In the UK,Canada etc the crown has copyright of the works by the civil servants. This is just the usual notion of the work for hire belonging to whoever hired it. Hiring may have a technical meaning in this context.An example is the fancy logo for US airports which is copyrighted and the copyright owned by the US Dept of Transport because they paid someone todevelop the logo and then bought the copyright to it. They had the work by an outside group so that it could be copyrighted and then purchased bythem in order to retain control, so the story goes.But you should only believe advice from a lawyer and not various amateurs.>> For instance, I may transfer copyright to somebody else, that does *not*>> mean that he can do anything with it he wants. There are some rights I>> retain and can not give away (the so-called neighbouring rights).>> still not the right to mutilate it, I have still the right to prohibit>> you from doing so. This is even true 70 years after my death with the>> rights of my heirs.In French law, if I understand the vague things I have seen, there is aright of authors which is distinct from copyright which allows the authorto have a more lasting control of the use of his work. The place this showsup in the popular press is when a sculpture is used in a form which offendsthe original artist. Under English common law type rules it is tough as theartist sold the copyright and that is the end of the story. But underFrenchstyle rules the artist can object even after the copyright has been sold.Again you get professional advice on these varius concepts and where theyapply.=== === Subject: : Re: Public Notice: Copyright Violations>In the US, at least, many creative works which are commissioned by or sponsored>by the>government are automatically and specifically in the 'public domain' and not>copyrightable. This is because the are paid for by public funds. Who owns>government>works (paid by public funds) in other countries?> Right words but quite wrong statments. As I understand it the works of a> US civil servant can not be copyrighted.I'm not sure what was wrong with my statement, which was: In the US, at least,many works which are commissioned or sponsored by the government are automaticallyand specifically in the 'public domain'. Such works include a large body ofscientific software developed at LANL and the (then) National Bureau of Standards(now NIST). LINPACK and STARPAK come to mind. Perhaps it appeared I was referring to*any* work, rather than *many* works, since I did add that the rationale for publicownership rested with the use of public funds. I did not mean to imply that anythingthe government pays for is in the public domain, only that there are many specificworks which fit this model.> Again you get professional advice on these varius concepts and where they> apply.Agreed -- and check back frequently, because copyright law is a moving target.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com=== === Subject: : Re: Public Notice: Copyright Violations>There have also since the beginning of news been a wide variety of ways>to access it.Yes, for example you could do more /usr/spool/news/general/*.Until the mid-90s, people posted with the expectation that they wereusing a transitory medium. It was known that certain newsgroups werearchived, but no-one expected that (say) prospective employers would-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules!=== === Subject: : Re: Public Notice: Copyright Violations>>There have also since the beginning of news been a wide variety of ways>>to access it.Yes, for example you could do more /usr/spool/news/general/*.Until the mid-90s, people posted with the expectation that they were> using a transitory medium. It was known that certain newsgroups were> archived, but no-one expected that (say) prospective employers wouldI know I was assuming back in 1992 that someone would eventually pullall the archives together and make a searchable whole.It was a logical extension, and just an extension of degree from whatwas then availble.=== === Subject: : Re: Public Notice: Copyright ViolationsBy posting communications on or through the Service, you automatically >grant Google a royalty-free, [...]>I'm thinking similar stipulations apply no matter which vehicle you choose >to post to Usenet.Why on earth would that be so? Who would make these stipulations, andhow would posters agree to them?It's conceivable that a Usenet provider (an ISP for example) mighthave a contract with its customers that allows them to do suchthings, but it's hardly likely to allow Google to do them.And of course not all of us post through commercial providers. We'vehad Usenet access for 20 years, and I'm not aware of any suchagreements with the sites that our news servers peer with.Furthermore, Google obtained the historical Usenet archives withoutany possibility of the original posters having agreed to it, sinceno-one considered the possibility at that time.-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules!=== === Subject: : who canhi,who can solve my problem? x^2*y+x*y'+y=0 ,f(x)=y if x->tan(t) now write x^2*y+x*y'+y=0 with t,f(t) thank you for your help hupo=== === Subject: : Re: who can>hi,who can solve my problem?> x^2*y+x*y'+y=0 ,f(x)=y if x->tan(t) now write x^2*y+x*y'+y=0 with t,f(t)> thank you for your help> hupoThis is an Euler differential equation. It is usually solved bychanging the independent variable to t by the equation t = ln(x) so dy/dx = dy/dt * dt/dx = (dy/dt)*(1/x) andd^2y/dx^2 = -(1/x^2)*(dy/dt) + (1/x)*(d^2y/dt^2)*(1/x)= (1/x^2)*((d^2y/dt^2)* - dy/dt) Plugging these in leads to a constant coefficient equation in t whichis easily solved. Then put x = e^t.--Lynn=== === Subject: : Re: who can>hi,who can solve my problem?> x^2*y+x*y'+y=0 ,f(x)=y if x->tan(t) now write x^2*y+x*y'+y=0 with t,f(t)Maple can. I get / 2 | d | | /d --- Y(t) | | 2 |-- Y(t)| tan(t) 2 | 2 | dt / dt | tan(t) |- ------------------ + -----------| /d | 2 2| |-- Y(t)| tan(t) 1 + tan(t) 1 + tan(t) / dt / -------------------------------------------- + ---------------- 2 2 1 + tan(t) 1 + tan(t) + Y(t) = 0which simplifies somewhat: / 2 /d |d | 1/4 |-- Y(t)| sin(4 t) + (-1/8 cos(4 t) + 1/8) |--- Y(t)| + Y(t) = 0 dt / | 2 | dt /Why would you want to do this?Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Saxon Math critique> I would like information regarding the Saxon Math program. How well does it meet national math standards, etc.?> Having had to teach (in a semi-independent environment) Saxon's Algebra I, I felt that my life was being drained away by a horde of invisible leeches. If you're trying to use the texts in a 'home school' environment, then the amount of repetition will only bore one (or however many children you have) to a semi-comatose state. Using the texts to try to teach a dozen or so bright kids was a form of cruel and unusual punishment. I finally took an entire week-end and edited the idiot thing's exercises to remove the problems based on lessons about a hundred pages back from that set of problems.Now, don't mistake me: As a mathematician, I was handed a shock the like of which I'd not felt since 'Nam when I first ran into that institutionalized idiocy of new new math, quite erroneously called 'College Prep Math' and its attendant stupidities of using calculators to avoid having the little darlings learn to do simple calculations on their own, and the relegation of math teachers to nannies wiping the intellectual drool from their charges' alleged minds. Saxon was one of the first to oppose this damnfoolishness, and his texts reflect this opposition. The repetition in his system is an effort to make sure that the student will learn the methods in the text, BUT if you are dealing with one or more really bright student(s), the repetition will not only bore them, it might well convince them not to enjoy math. That would be a definite tragedy.As to the so-called national standards, the NCTM (to take an extremely disgusting example) is all in favor of the CPM rubbish. I'd take a look at the requirements of your state (or province, or...) and then decide if the Saxon texts will cover the material. Here in California, the land of the Governator (Goddess help us!), Saxon's two algebra books do a pretty good job. However, YMMV, this is all IMPO, beware the phase or the Moon, the Ides of March, &c.Bottom line: If you're trying to educate one or more students on your own, then Saxon's books seem to be the best of the lot. If you're looking for a supplementary text for a student that is bright but is being bored silly by what passes for 'public' 'education', then they might well serve the turn. I hope you have access to a copy, so that you might judge for yourself.HTH,DrieuxPS Note that the spelling of my name in the email is part of the spam trap.=== === Subject: : Re: Saxon Math critiquethank you, mister Drieux! I should have known,what a dysservice I have done, all these yearsof going to schoolboard meetings,without investigating this Scholastic ****. maybe they can get Dame Jo and Company [*] to investin Agnlosaxon Math Futures!--* Industrial Smoke & Mirrors, Ltd. Parents Left Behind, Inc. Mel Gibson and His Dawg, Tops Corp. > unusual punishment. I finally took an entire week-end and edited the > idiot thing's exercises to remove the problems based on lessons about a > hundred pages back from that set of problems.Now, don't mistake me: As a mathematician, I was handed a shock the > like of which I'd not felt since 'Nam when I first ran into that --Give Earth a Trickier Dick Cheeny -- OBNOXICO, out of office?http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac=== === Subject: : Re: 3 Squares Covering 1 Circle:-)))=== === Subject: : Help proving a lemmaGiven this situation: You have a circle tangent to a line. The points ofthe midpoint of circle are given and the equation of the line is given.Find the radius.I found the radius by using the distance formula and substitution.My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt (A^2+ B^2) ]I have to prove it.I dont know where to start. Can someone show me how to prove such a lemma?=== === Subject: : Re: Help proving a lemma> Given this situation: You have a circle tangent to a line. The points of> the midpoint of circle are given and the equation of the line is given.> Find the radius.> I found the radius by using the distance formula and substitution.Good.> My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt(A^2> + B^2) ]To me this looks like *the* distance formula. I assume Ax+By+C=0 is theequation for the line, and x,y is a point. Then r is the distance from thepoint to the line, if I am not mistaken.I wonder what distance formula you used...-Michael.=== === Subject: : Re: Help proving a lemma> Given this situation: You have a circle tangent to a line. The points of> the midpoint of circle are given and the equation of the line is given.> Find the radius. I found the radius by using the distance formula and substitution.Good.My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt> (A^2> + B^2) ]To me this looks like *the* distance formula. I assume Ax+By+C=0 is the> equation for the line, and x,y is a point. Then r is the distance from the> point to the line, if I am not mistaken.I wonder what distance formula you used...-Michael.D = | Au + Bv + C | / [ sqrt(A^2+ B^2) ] is a fairly standard formula for the distance from point (u,v) to the line with equationAx + By + C = 0. There are a variety of ways of validating the formula.Probably the easiest is in terms of vectors. Then vector [A, B] is a vector perpendicular to line Ax + By + C = 0, and if [x, y] is any point on the line one must have that [A,B]*[x,y] = -C, where * indicates dot product, and then the distance from [u,v] to the line is the magnitude of the projection of [u-x, v-y] perpendicular to the line (e.i., parallel to [A,B].Filling in the details should give you a proof.=== === Subject: : Re: Help proving a lemmaIgor Y> Given this situation: You have a circle tangent to a line. The points of> the midpoint of circle are given and the equation of the line is given.> Find the radius.> I found the radius by using the distance formula and substitution.> My teacher has given us this llama though: r = | Ax + By + C | / [ sqrt(A^2> + B^2) ]> I have to prove it.> I dont know where to start. Can someone show me how to prove such alemma?If the line is Ax + By + C = 0 and the _center_of the circle is given, say (u,v), then the point oftangency (s,t) will satisfyAs + Bt + C = 0B(s-u) - A(t-v) = 0the distance from (u,v) to (s,t). That distance isthe radius.(The formula for the radius will not involve x and y.Maybe your lemma is only part of the problem?)LH=== === Subject: : Re: Representing the ordering of the perception of colour?How can the order in the perception of colour be represented?Please note that chromaticity spaces like CIE are chosen tosatisfy useful linearity rules and *not* to make distancerelate to perceptual distance. For the latter, google searchfor LUV-star, though you may need to be an advanced googler as theusual spelling (LUV*) uses Google's meta-symbol!> I'm asking this because I'm designing the colour layout in a CAD programme> which needs to be determined in a logically consistent manner.> You might start by requiring that the B&W intensities be identical or> not. Then worry about hue and saturation, etc.Choose a few hues (e.g. red, yellow, green, blue) for the categoriesand then adjust luminance for the variation within a category.To make this work right you'll need to add some redness into the bluecolors, etc.Beware: The usual luminance formula (.59 green + .30 red + .11 blue)is calibrated for daylight-white, not CRT-white.James=== === Subject: : Re: Representing the ordering of the perception of colour?> The reason red-purple-blue seems like a reasonable sequence has to do with> the response of the color receptors in the eye, not anything to do with the> light wavelengths themselves.Yep. And, particularly with green, there are two quite different visualinputs that can lead to the same perceived colour. One can be lookingat a green whose wavelength mix stimulates mostly your green receptor cones,or alternatively at one which stimulates just the correct mixture of redcones and blue cones, and get the same effect. Side by side you'd swearthey were the same hue, saturation, luminosity, everything.There's another peculiarity. Every colour input stimulates all threecolour receptor types at least a little bit. Even the purest red affectsnot only your red cones but a tiny bit of green and maybe even blue.Of course your brain sorted all this out long ago, and the reddest redyou can ever see is the one whose differential stimulus is the max.HOWEVER - I have read that though they don't yet exist, there could becone-specific neuro-blockers that, injected into the retina could(temporarily!) block the blue and green receptor cones. Then, everythingwould look reddish, but really red objects would look SUPER-red, (!),because your brain wouldn't be dealing with the customary small backgroundnoise of reds and blues. What a sensation that might be!..................Anyway, I thought it might be time to re-post one of my old ones,about colour co-ordinates...===============================I've always been intrigued by the two standard painters' ways of orderingcolors, and their mathematical relationship. (And actually, insofar ascolors are physical, I think the math is genuine, not pseudo.)Both ways have one dimension being hue; arranged naturally in a circle fromred through orange, yellow, green, blue, purple, & back to red. The other twodimensions are then Cartesian and Polar versions of the same color array! Thewhole arrangement is on a double cone (with two point poles and an equator),with the axis running from black to white (the 2 poles) through shades of gray.And each triangular half-slice, from axis out to exterior, are shades of a single hue as follows.The HLS mentioned above, is the polar, with the origin as the pure color,the radius as the saturation (actually its complement), and the angle asthe luminance. The angle is one quadrant only, and the radius is in [0,1].Then there is the cartesian style, HWB, Hue-%White-%Black, (more natural forcommercial paint mixers!) The pure hue is at the origin again, and the amountsof black and white mixed in are along the two co-ordinates.To get it just right, we must flatten the quarter-circle arc of totalde-saturation (100% black & white in various proportions) into a straightline, and turn the diagram on its side. Here it is all pictured for hue yellow;(and if you disagree with some of my color words just insert your own; andthis same slice could probably also take beige, sandy, khaki...) <--- de-saturation --- black | ___ | | | | % of black added | | | | brown | | | dark luminance gray tan | | | | | mid blonde | gray yellOw <---(O is the origin as usual!) V | buff / | / | / | fawn / / pale / / gray cream / % of white added | / | / | / |__ | / | / |/ whiteThis is the luminance/saturation ( = black/white) slice for hue = yellow.To make the polar/cartesian correspondence stand out, the left-hand lineof grays should really be a left-bulging arc, of course. But(i) this would be terrible in ascii, and(ii) all the other hues can now be neatly added in by joining them all alongthe gray-line, so as to get a nice little bi-conical booklet of colors, withpages from yellow through orange, red etc and back to yellow, (with as many huegradations as suits you).So color is thus naturally presented in 3-dimensional cylindrical co-ordinates!Either as HLS, (circular hue crossed polar shades), or as HBW, (circular hue crossed cartesian b/w mixes).I've always thought this polar/cartesian correspondence was particularly cool!--------------------------------------------------------- ---------------------- Taylor wft@math.canterbury.ac.nz------------------------------------- ------------------------------------------ We're pushing back the frontiers of knowledge! Yes; but from which side?--------------------------------------------------------- ----------------------=== === Subject: : Re: Representing the ordering of the perception of colour?> The reason red-purple-blue seems like a reasonable sequence has to dowith> the response of the color receptors in the eye, not anything to do withthe> light wavelengths themselves.> Yep. And, particularly with green, there are two quite different visual> inputs that can lead to the same perceived colour. One can be looking> at a green whose wavelength mix stimulates mostly your green receptorcones,> or alternatively at one which stimulates just the correct mixture of red> cones and blue cones, and get the same effect. Side by side you'd swear> they were the same hue, saturation, luminosity, everything.Isn't that being colour-blind? :-)It seems strange that the brain should equate two different sets of neuralstimulations. Is that a social artifact?> HOWEVER - I have read that though they don't yet exist, there could be> cone-specific neuro-blockers that, injected into the retina could> (temporarily!) block the blue and green receptor cones. Then, everything> would look reddish, but really red objects would look SUPER-red, (!),> because your brain wouldn't be dealing with the customary small background> noise of reds and blues. What a sensation that might be!Sounds like LSD to me :-)-Michael.=== === Subject: : Re: Representing the ordering of the perception of colour?>> Yep. And, particularly with green, there are two quite different visual>> inputs that can lead to the same perceived colour. One can be looking>> at a green whose wavelength mix stimulates mostly your green receptor cones,>> or alternatively at one which stimulates just the correct mixture of red>> cones and blue cones, and get the same effect. Side by side you'd swear>> they were the same hue, saturation, luminosity, everything.>It seems strange that the brain should equate two different sets of neural>stimulations. Is that a social artifact?I don't think it IS two different sets of neural stimulations. And it'scertainly not a social artifact; it's a mathematical question, and itruns something like this. (I may have the biological details a bit off.)You've got three kinds of cones, each sending a neural impulse whosestrength is an integral (over all possible wavelengths) of the incomingmix of light, multiplied by a kernel which is a sort of bell curve witha peak at some frequency or another. That is, shine a pure coherent redlight at a red cone, and it sends a signal to the brain; let the light'swavelength shift away from red (while maintaining intensity) and thecone's response diminishes, but does not immediately drop to zero.So suppose a ray of light enters the eye having intensity I(lambda)at frequency lambda. After the cones (at one point in the retina) respond, the brain will get 3 impulses with strengths int I(lambda) K_i (lambda) d lambda i = 1, 2, 3which actually are inner products < I, K_i > in L^2(R).Now the biological question is, can you have different signals Iwhich give the same set of three numbers to the brain; the correspondingmathematical question is, can you have different functions whose inner products against the three K_i are equal. And of course theanswer is yes: your inner products just project an infinite-dimensionalvector space down to a 3-dimensional one, so of course there's a kernel.Sight is a fascinating set of things to think about, IMHO, especially whencompared to the sense of hearing. If you send a mix of two sound waves tomy ear, I will in fact hear a single tone which is mid-way between them,but I will hear a beat frequency as the intensity varies. That is: sin( (x+h)t ) + sin( (x-h)t ) = 2cos( ht ) sin( xt )But realistically we only perceive this mix of signals in the way Idescribed it verbally when h is very small. That's when you realize thatsight and hearing are two very different senses; among other differences,one of them is sensitive to minute differences in frequency but only ina narrow range, while the other is not as sensitive in the small but responds to a great range of frequencies. You can't see and octave,for example. (You also have a retina full of cones but only one pairof eardrums; you can't hear a landscape.)dave=== === Subject: : Re: Representing the ordering of the perception of colour?tres etrange, yet apparently experimentally validatedby monsieur Land et al! if LSD really does that, then count me outof that party. seriously,he's just talking about vacance de coleur, butthat would *seem* to be dependentupon the trichromacie apparentment de vision(je ne me parle pas ****, especially French .-) > at a green whose wavelength mix stimulates mostly your green receptor> cones,> or alternatively at one which stimulates just the correct mixture of red> cones and blue cones, and get the same effect. Side by side you'd swear > It seems strange that the brain should equate two different sets of neural> stimulations. Is that a social artifact?HOWEVER - I have read that though they don't yet exist, there could be> cone-specific neuro-blockers that, injected into the retina could> (temporarily!) block the blue and green receptor cones. Then, everything> would look reddish, but really red objects would look SUPER-red, (!),> because your brain wouldn't be dealing with the customary small background> noise of reds and blues. What a sensation that might be!--Give Earth a Trickier Dick Cheeny -- OBNOXICO, out of office?http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac=== === Subject: : Re: Representing the ordering of the perception of colour?>It seems strange that the brain should equate two different sets of neural>stimulations. Is that a social artifact?According to Roger Shepard, it's evolved behavior.@INCOLLECTION{Shepard1992, author = {Roger N. Shepard}, editor = {Jerome H. Barkow and Leda Cosmides and John Tooby}, year = 1992, title = {The Perceptual Organization of Colors: An Adaptation to Regularities of the Terrestrial World?}, booktitle = {The Adapted Mind}, pages = {495--532}, publisher = {Oxford}, address = {New York}}Lee Rudolph === === Subject: : Re: Representing the ordering of the perception of colour?Uncle Al How can the order in the perception of colour be represented?> Ie. in this ordering, red and orange are close to one another whle redand> blue are far apart. Light red > medium red > dark red. The perceptionof> colour is continuous.> Color spaces historically have been defined in various ways, e.g., a> CIE chromaticity diagram and tristimulus coordinates, or Pantone> chips.> http:/www.pantone.com/> I'm asking this because I'm designing the colour layout in a CADprogramme> which needs to be determined in a logically consistent manner. So for a> number of groups of objects, I need the colours to be as different as> possible, but the objects within that group have to look visuallyrelated,> by shades of colour as an example.> You might start by requiring that the B&W intensities be identical or> not. Then worry about hue and saturation, etc. Edwin Land (of> Polaroid) was obsessed with Mondrians and how adjacent colors changed> perceived values. You could thus be mathematically on target and> perceptually out to lunch.I have an old question about color vision; maybe Unsubtle Al knows theanswer.Human vision has R,G,B components in some sense, and two or threecorrepsonding varieties of color blindness are possible. But, last I heard,no anatomical difference between one retinal cone and another had ever beenfound. Theory: instead of differing cells or even differing pigments in oureyes, maybe we've got a _tunable_ pigment. Different kinds of colorblindness might still have an explanation, involving enzymes, like anymetabolic disease.One other bit, doubtless familiar to experts in the room. Take a look at thevitamin A molecule, which participates in BW vision. It is immediate thatthat molecule will absorb radiation, because the long string of conjugateddouble bonds is a _conductor_. (Conjugated double bonds can pass electronsin either direction.)LH=== === Subject: : Re: Representing the ordering of the perception of colour?> Uncle Al> How can the order in the perception of colour be represented?> Ie. in this ordering, red and orange are close to one another whle red> and> blue are far apart. Light red > medium red > dark red. The perception> of> colour is continuous. Color spaces historically have been defined in various ways, e.g., a> CIE chromaticity diagram and tristimulus coordinates, or Pantone> chips. http:/www.pantone.com/ > I'm asking this because I'm designing the colour layout in a CAD> programme> which needs to be determined in a logically consistent manner. So for a> number of groups of objects, I need the colours to be as different as> possible, but the objects within that group have to look visually> related,> by shades of colour as an example. You might start by requiring that the B&W intensities be identical or> not. Then worry about hue and saturation, etc. Edwin Land (of> Polaroid) was obsessed with Mondrians and how adjacent colors changed> perceived values. You could thus be mathematically on target and> perceptually out to lunch.> I have an old question about color vision; maybe Unsubtle Al knows the> answer.> Human vision has R,G,B components in some sense, and two or three> correpsonding varieties of color blindness are possible. But, last I heard,> no anatomical difference between one retinal cone and another had ever been> found.You last heard wrong. There are three different types of cones, eachwith a different spectral sensitivty.> Theory: instead of differing cells or even differing pigments in our> eyes, maybe we've got a _tunable_ pigment. Different kinds of color> blindness might still have an explanation,The three different types of cones feed into encoders (this reducesthe number of nerve fibers needed in the optic nerve). The mostcommon kinds of color blindness are due to disfunction of encoders(usually the red-green encoders).Paul Cardinale=== === Subject: : The coin problemSuppose you need to draw a prise between 3 people with equal chance toall. Can it be done using one unfair coin (that is the prob. of headsand tails is to your specification, not 1/2)? === === Subject: : re:The coin problemOK, I apologize for not being explicite enough. The number of tossesmust be bounded, procedures that mightgo on indefinitely are not allowed. === === Subject: : Re: The coin problem* Alexey007@hotmail-dot-com.no-spam.invalid> OK, I apologize for not being explicite enough. The number of tosses> must be bounded, procedures that might> go on indefinitely are not allowed.Then no procedure is ever possible unless the coin is fair and thenumber of participants is a power of two. (Actually, there might be other solutions for a given unfair coin withknown probability of head. But I don't know if it is so, and my gutfeeling is no..)-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92=== === Subject: : Re: The coin problem> OK, I apologize for not being explicite enough. The number of tosses> must be bounded, procedures that might> go on indefinitely are not allowed.Even if the probability of going on forever is 0?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: The coin problem> Suppose you need to draw a prise between 3 people with equal chance to> all. Can it be done using one unfair coin (that is the prob. of heads> and tails is to your specification, not 1/2)?> ----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **> ---------------------------------------------------------- > There are 2 issues here:(1)removing bias(2)using a coin, which is two-sided, to draw among three items(1)Removing biasBias can be removed by repeated tosses. If you toss a coin twice andyou change the call to alike or different instead of heads ortails, some of the bias of a single coin toss will have been removed(the same applies for 2 different coins). More bias can be removed bydoing more tosses.(2)Drawing among 3 itemsFor this you have to toss the coin twice. With the composite coin theactual number of tosses will be more. First arbitrarily define yourtosses as follows:aa no result: toss againab 1ba 2bb 3This is really binary arithmetic.Note that I have used aa so that it can mean either heads heads oralike alike, depending on whether you are doing single or compositetosses.=== === Subject: : Re: The coin problemSo I'm allowed to specify p=prob(heads) to be, say, a root> of p^4 + (1-p)^4 = 1/3,> or p=0.2421306902?Then I will hazard a guess of yes. :)Don CoppersmithOk, you cracked this one.Next one: show that it can be done with one coin for any number n ofpeople and find a good upper bound on the minimal number of requiredtosses. === === Subject: : Re: The coin problem>Suppose you need to draw a prise between 3 people with equal chance to>all. Can it be done using one unfair coin (that is the prob. of heads>and tails is to your specification, not 1/2)?and later>Next one: show that it can be done with one coin for any number n of>people and find a good upper bound on the minimal number of required>tosses.There are several possible schemes. I don't think it is too hard to determine either the minimal number of tosses or (more importantly) the expected number of tosses to complete the contest in each time, but I don't have time right now to type it out (in the margins :-). I'll do it later if no one else has done so.1) Flip the coin once; call its outcome the target. On each round, flip the coin for each remaining player. Eliminate all players who do not have the target, unless none do. Repeat until one player, the winner remains. Minimum number of tosses: n+1.2) If possible, assign each player a unique specific pattern of ordered n tosses such that each has the same number of heads. For each round, make n ordered tosses; the first person whose pattern appears wins. Minimum number of tosses: n. The total number of tosses has the distribution of n times a geometric.3) A very slow one: In each round, flip the coin once for each player. If one player has an outcome which is different from all the others, that person wins. Otherwise, repeat. Minimum number of tosses: n. The total number of tosses is n times a geometric with parameter n p (1-p) [p^(n-2) + (1-p)^(n-2)] , whence the expected number of tosses is {p (1-p) [p^(n-2) + (1-p)^(n-2)]}^(-1).4) In each round, flip the coin once for each remaining player. Unless all outcomes are the same or there are equally as many heads as tails, eliminate all players in the majority. (So with seven players, if there are four heads and three tails, eliminate all the players with heads.) The last player remaining wins. One has to use a different rule when only two players remain. Or alter the rules such that that cannot happen: e.g., when m > 4 players remain and the group splits 2, m-2, eliminate the 2 rather than the m-2. Or, as a variation, eliminate the even group (rather than the majority) when an odd number of players remain. Minimum number of tosses: n.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: The coin problem>Next one: show that it can be done with one coin for any number n of>people and find a good upper bound on the minimal number of required>tosses.This seems to be true at least if the k'th row in the Pascal trianglehas all values divisible with (n-1) except the first and the last(being ones). Then the number of required throws is the first such rowk. A requirement is also that the polynomial p^n + (1-p)^n = 1/3 havezeros in (0, 1).For primes it's true that if the second element of any row is prime,then all numbers on that row are divisible by that prime. So thetheorem is true at least if n prime.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : re:The coin problemOf course there should be only finitely many tosses. And you canspecify the prob. of heads/tails to be any constant quantity youwant. === === Subject: : Re: The coin problem>Of course there should be only finitely many tosses. And you can>specify the prob. of heads/tails to be any constant quantity you>want.> In the schemes presented, the process will end in finite time (with probability 1), as long each outcome (heads or tails) has a positive probability. It is just that the possible number of trials is unbounded. (Cf. flipping a coin until the first head appears.)-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: The coin problem>Suppose you need to draw a prise between 3 people with equal chance to>all. Can it be done using one unfair coin (that is the prob. of heads>and tails is to your specification, not 1/2)?Ever hear of the game called odd-man-out?Flip the coin for each person. If all three get heads or all three gettails, flip again. Otherwise, one person has gotten something differentfrom the other two. The person with the unique result wins.Rob Johnson take out the trash before replying=== === Subject: : Re: The coin problemIn sci.math, Rob JohnsonSuppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?> Ever hear of the game called odd-man-out?> Flip the coin for each person. If all three get heads or all three get> tails, flip again. Otherwise, one person has gotten something different> from the other two. The person with the unique result wins.Assume a biased coin. Then:HHH = h^3 -- tossHHT = h^2tTTH = ht^2TTT = t^3 -- tossThe probability that person A will win is(ht^2 + h^2t) / (3*(ht^2 + h^2t)) = 1/3since the coin is biased but can't change its biasbetween flips; the probabilities for B and C areidentical. Therefore odd man wins is extremelyfair, assuming the *coin tosses* are done fairly (i.e.,no palming, coin-switching, distraction ploys, etc.) witha single coin.Of course if h = 0.99, then t = 0.01 and it might be awhilebefore one gets a useful result, as the probability ofall heads is about 0.97. This means that it might takea few flips... :-)> Rob Johnson In sci.math, Rob Johnson>all. Can it be done using one unfair coin (that is the prob. of heads>and tails is to your specification, not 1/2)?>> Ever hear of the game called odd-man-out?>> Flip the coin for each person. If all three get heads or all three get>> tails, flip again. Otherwise, one person has gotten something different>> from the other two. The person with the unique result wins.>Assume a biased coin. Then:>HHH = h^3 -- toss>HHT = h^2t>TTH = ht^2>TTT = t^3 -- toss>The probability that person A will win is>(ht^2 + h^2t) / (3*(ht^2 + h^2t)) = 1/3>since the coin is biased but can't change its bias>between flips; the probabilities for B and C are>identical. Therefore odd man wins is extremely>fair, assuming the *coin tosses* are done fairly (i.e.,>no palming, coin-switching, distraction ploys, etc.) with>a single coin.>Of course if h = 0.99, then t = 0.01 and it might be awhile>before one gets a useful result, as the probability of>all heads is about 0.97. This means that it might take>a few flips... :-)In the odd-man-out procedure that both Stephen Herschkorn and I havesuggested, there is a possibility that the number of iterations canget large since there is a p^3+(1-p)^3 probability of needing to tryagain for each round (where p is the probability of heads). This leadsto an average duration of 1/(3p(1-p)). The minimum average duration is4/3 for a fair coin (p = 1/2). The more unfair the coin is, the longerthe average duration is.Rob Johnson take out the trash before replying=== === Subject: : Re: The coin problem> Suppose you need to draw a prise between 3 people with equal chance to> all. Can it be done using one unfair coin (that is the prob. of heads> and tails is to your specification, not 1/2)?> ----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **> ---------------------------------------------------------- > it can be done, but not if there is a limit on the number of tosses of the coin that will be allowed.For example:HH person A winsHT person B winsTH person C winsTT start over.=== === Subject: : Re: The coin problem>>Suppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?>>---------------------------------------------------------->> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>>---------------------------------------------------------- >>>it can be done, but not if there is a limit on the number of tosses of >the coin that will be allowed.>For example:>HH person A wins>HT person B wins>TH person C wins>TT start over.No, then each does not have an equal chance, since the coin is not fair.Here's a way:Flip the coin once for each player. If one outcome is different from the other two, that person's toss wins. Otherwise, repeat.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: The coin problemSuppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?>>----------------------------------------------------- ----->> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>>---------------------------------------------------------- >>> it can be done, but not if there is a limit on the number of tosses of >the coin that will be allowed.For example:>HH person A wins>HT person B wins>TH person C wins>TT start over. No, then each does not have an equal chance, since the coin is not fair.Here's a way:> Flip the coin once for each player. If one outcome is different from > the other two, that person's toss wins. Otherwise, repeat.Sorry, Didn't see the un.=== === Subject: : Re: The coin problem>Suppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?>>----------------------------------------------------- ----->> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>>---------------------------------------------------------- >> it can be done, but not if there is a limit on the number of tosses of>the coin that will be allowed.For example:>HH person A wins>HT person B wins>TH person C wins>TT start over. No, then each does not have an equal chance, since the coin is not fair.I guess I don't see that. To me all four events above are equally likely,and each pair of throws is independent of the previous pair. You arenot suggesting that a penny's unbcedness should be taken intoaccount, are you?=== === Subject: : Re: The coin problem>>Suppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?------------------------------------------^^^^^^^^------- ------------------ >>---------------------------------------------------------->> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>>---------------------------------------------------------- >> >it can be done, but not if there is a limit on the number of tosses of>the coin that will be allowed.>For example:>HH person A wins>HT person B wins>TH person C wins>TT start over.> No, then each does not have an equal chance, since the coin is not fair.I guess I don't see that. To me all four events above are equally likely,> and each pair of throws is independent of the previous pair. You are> not suggesting that a penny's unbcedness should be taken into> account, are you?Why not? The original question so specifies, doesn't it?=== === Subject: : Re: The coin problem>>Suppose you need to draw a prise between 3 people with equal chance to>>all. Can it be done using one unfair coin (that is the prob. of heads>>and tails is to your specification, not 1/2)?>>---------------------------------------------------------->> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>>---------------------------------------------------------- >> >it can be done, but not if there is a limit on the number of tosses of>the coin that will be allowed.>For example:>HH person A wins>HT person B wins>TH person C wins>TT start over.> No, then each does not have an equal chance, since the coin is not fair.I guess I don't see that. To me all four events above are equally likely,> and each pair of throws is independent of the previous pair. You are> not suggesting that a penny's unbcedness should be taken into> account, are you?Certainly, at least the way I read it (can it be done with an unfaircoin, if we've already found a method to do it with a fair one?),especially since you can give the probabilities to yourspecification, presumably making them as badly biased as you want. Ithink what's being asked for is a method that is fair, whateverprobabilities we throw at it.One thing I'm wondering - the few methods suggested are recursive. Thedifficulty seems to be that 3 is not a power of 2, which would allowus to do a sort of binary search on the players. This is fixed byadding a fourth player called try again to change the 3 into 4,something we know how to deal with easily. Can we find a methodwithout recursion, though? Are there methods which have a necessarilybounded number of flips?=== === Subject: : Re: The coin problem* Gus Gassmann>HH person A wins>HT person B wins>TH person C wins>TT start over.> No, then each does not have an equal chance, since the coin is not fair.I guess I don't see that. To me all four events above are equally likely,> and each pair of throws is independent of the previous pair. You are> not suggesting that a penny's unbcedness should be taken into> account, are you?Why shouldn't it? If the probability of head is 0.99 and tail is0.01 I certainly would want to be person A.-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92=== === Subject: : Re: The coin problem> * Gus Gassmann>HH person A wins>HT person B wins>TH person C wins>TT start over.> No, then each does not have an equal chance, since the coin is not fair. I guess I don't see that. To me all four events above are equally likely,> and each pair of throws is independent of the previous pair. You are> not suggesting that a penny's unbcedness should be taken into> account, are you?> Why shouldn't it? If the probability of head is 0.99 and tail is> 0.01 I certainly would want to be person A.My reference was to an experiment in which two high school kidsestablished experimentally (by tossing ten US 1 cent coins 10,000times each) that the probability of Heads is not equal to 0.5.But surely the difference is too small to matter in this context.What if in the OP's question there are only two candidates?Would you seriously suggest that instead of one person choosingHeads, the other Tails they continue tossing the coin twice untilthe two outcomes are different?=== === Subject: : Re: The coin problem* Gus Gassmann> My reference was to an experiment in which two high school kids> established experimentally (by tossing ten US 1 cent coins 10,000> times each) that the probability of Heads is not equal to 0.5.> But surely the difference is too small to matter in this context.Well, the problem in question specifically pointed out an unfaircoin. What if in the OP's question there are only two candidates?> Would you seriously suggest that instead of one person choosing> Heads, the other Tails they continue tossing the coin twice until> the two outcomes are different?Yes. A wins with HT and B with TH.-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92=== === Subject: : Re: The coin problem>> * Gus Gassmann>>HH person A wins>>HT person B wins>>TH person C wins>>TT start over.> No, then each does not have an equal chance, since the coin is not fair.I guess I don't see that. To me all four events above are equally likely,>> and each pair of throws is independent of the previous pair. You are>> not suggesting that a penny's unbcedness should be taken into>> account, are you?>> Why shouldn't it? If the probability of head is 0.99 and tail is>> 0.01 I certainly would want to be person A.My reference was to an experiment in which two high school kids> established experimentally (by tossing ten US 1 cent coins 10,000> times each) that the probability of Heads is not equal to 0.5.> But surely the difference is too small to matter in this context.What if in the OP's question there are only two candidates?> Would you seriously suggest that instead of one person choosing> Heads, the other Tails they continue tossing the coin twice until> the two outcomes are different?Since the OP _explicitly_ said that the coin was unfair, it isentirely appropriate to assume that the coin is unfair and to dealwith the resulting issues.Yes. Given an unfair coin and a requirement for an unbiaseddecision, I would suggest that they continue tossing the coin twiceuntil the outcomes are different. I would also suggest checkingfor a double-headed coin. It's tough to harvest much entropy whenonly one outcome is possible.I would not suggest the use of fancy techniques such as treating the TT and HH pairs as an even more badly biased generatorand using TTHH and HHTT (or TTTTHHHH and HHHHTTTT) as equiprobableevent pairs. John Briggs=== === Subject: : Fundamental Group of the circleI am looking over a proof relating to this.Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)I = [0,1] RThey then state:a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 inp^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0.They then state:c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x{0}, then there is a unique map ~F: Y x I -> R lifting F and restricting tothe given ~F on Y x {0}.Finally they state that a) is the special case of c) where Y is a point.I'm having trouble seeing this relation.Also, can I assume that lifting F and restricting to the given ~F on Y x{0} means that the image of ~F: Y x I equals the image of ~F : Y x {0}?=== === Subject: : Re: Fundamental Group of the circle> I am looking over a proof relating to this.> Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)> I = [0,1] R> They then state:> a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 in> p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0.> They then state:> c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x> {0}, then there is a unique map ~F: Y x I -> R lifting F and restrictingto> the given ~F on Y x {0}.> Finally they state that a) is the special case of c) where Y is a point.> I'm having trouble seeing this relation.I think I see this. I will omit the Y domain, since a) only requires asingle element. Then the path f:I->S^1,f(0)=x0 in S^1 from a) is aninstance of the function F from c). ~F:Yx{0}->R can be a constant functionsince |Y|=1, so it is simply ~f:{0}->R, ~f(0) = ~x0 from a). Using p froma) we see that (p o ~f) : {0}->S^1 = f|{0}->S^1, so ~f is a lift of f|{0} asstarting at ~x0 since we designed ~f(0) = ~x0 above.> Also, can I assume that lifting F and restricting to the given ~F on Y x> {0} means that the image of ~F: Y x I equals the image of ~F : Y x {0}?I think this means simply that ~F:YxI = ~F:Yx{0} on Yx{0}, but ~F:YxI's fullimage may be larger than ~F:Yx{0}'s.=== === Subject: : Re: series summation>Let Bin(k,m) be the binomial coeffient k choose m.>The problem is to find the sum:> Sum(k=m to infty) Bin(k,m)x^k>where 0 C(k,m) x y --- --- m=0 k=m oo k --- --- k m = > C(k,m) x y --- --- k=0 m=0 oo --- k k = > x (1+y) --- k=0 1 = -------- 1-x(1+y) 1 1 = --- ---------- 1-x 1-xy/(1-x) oo 1 --- x m m = --- > ( --- ) y 1-x --- 1-x m=0Equating the coefficients of y^m, we get that oo --- k x^m > C(k,m) x = ----------- --- (1-x)^{m+1} k=mRob Johnson take out the trash before replying=== === Subject: : Re: series summation>Let Bin(k,m) be the binomial coeffient k choose m.The problem is to find the sum: Sum(k=m to infty) Bin(k,m)x^kwhere 0 --- --- k m> > C(k,m) x y> --- ---> m=0 k=m oo k> --- --- k m> = > C(k,m) x y> --- ---> k=0 m=0 oo> --- k k> = > x (1+y)> ---> k=0> We can assume that 0<=xy/(1-x)<1. Then 0 1> = --------> 1-x(1+y) 1 1> = --- ----------> 1-x 1-xy/(1-x) oo> 1 --- x m m> = --- > ( --- ) y> 1-x --- 1-x> m=0Equating the coefficients of y^m, we get that oo> --- k x^m> C(k,m) x = -----------> --- (1-x)^{m+1}> k=mTo justify this refer to Rudin's Principles of Mathematical Analysis, 3rd ed, Theorem 8.5: It says: Suppose the series oo oo --- k --- k > a_n y and > b_n y --- --- k=0 k=0converge in the segment S=(-R,R). Let E be the set of all y in S at which oo oo --- k --- k > a_n y = > b_n y. --- --- k=0 k=0If E has a limit point in S, then a_n=b_n for n=0,1,2,....By Rudin, Theorem 3.39, the series oo 1 --- x^m k ---- > -------- y (1-x) --- (1-x)^m k=0 has a radius of convergence (-(1/x - 1), 1/x - 1).For the series oo oo --- --- k m > C(k,m) x y --- --- m=0 k=mfirst note that C(k,m) = k^m + O(1); that is, C(k,m)-k^m is bounded. By the ratio test oo --- m k > k x --- k=mconverges if |x|<1, so if |x|<1, oo oo oo --- k --- k --- m k > C(k,m) x = > (C(k,m)-k^m)x + > k x --- --- --- k=m k=m k=mconverges, as the last two series each converge. We're only concernedwith 0N implies | oo | | --- k| | > C(k,m) x | <= 1 | --- | | k=m | So | oo |1/m | --- k| limsup | > C(k,m) x | <= 1 . m | --- | | k=m | So again by Rudin, Theorem 3.39, oo oo --- --- k m > C(k,m) x y --- --- m=0 k=mhas a radius of convergence of at least (-1,1). Above weassumed that x<=x(1+y)<=1 which holds iff 0<=y<=1/x - 1. So with b=min{1, 1/x - 1} we have that for y in (0,b) oo oo oo 1 --- x^m k --- --- k m ----- > -------- y = > C(k,m) x y . (1-x) --- (1-x)^m --- --- k=0 m=0 k=mWith S=(-b,b), we have that E contains (0,b) so that E has a limit point in S; thus, Theorem 8.5 says that oo --- k x^m > C(k,m) x = -----------. --- (1-x)^{m+1} k=m=== === Subject: : Re: series summation>Let Bin(k,m) be the binomial coeffient k choose m.>The problem is to find the sum:> Sum(k=m to infty) Bin(k,m)x^k>where 0 So given that the degree I am pursuing is called Masters in *APPLIED*> Statistics, I thought the the courses would be heavily applied and taught in> the context of solving real world problems. No dice. Both courses I took in> the first semester (part-time evening program) had heavy theory. The> Bayesian class was not even as bad as the other one (Mathematical> Statistics). There was essentially no attempt on the part of the professor> to relate the theory to real world programs or to even give real world> examples to illustrate the concepts. It was formula, theory, formula,> theory, theory, formula, etc. I asked him about that and he said there's no> way around the theory. I too recently took a Math Statistics and introductory Bayesian classfrom a department that calls itself Applied Math and Statistics. Butbefore I took Math Statistics--an introductory class, I tried to takeanother more advanced class because that was all that was on offer atthe time. I found that I would have benefitted more from the advancedclass where applied problems were discussed, had I had some exposureto the concepts covered in the Math Statistics class. So I can see howthere is no getting around the theory. Perhaps your professor couldnot think of any examples, perhaps he didn't know of any (Math Statprofessors are often theoreticians), perhaps he was just trying tocover the syllabus and examples had not been designed in.Some of these introductory classes try to do too much. The class Itook had students from economics, psychology, astronomy, physics,math, bioinformatics, computer science and engineering as well as MSand PhD Applied Statistics students. I personally don't think even oneof these groups was well served and this was reflected in the gradedistribution which was multi-modal. The course itself tried to fit ayears worth of material in a ten-week quarter. The professor barelyhad time to catch his breath, let alone give relevant examples. Minewas a bad class but I suspect your class too had to accomodatedifferent kinds of students, not just those who were getting aMaster's degree in Applied Statistics.One of the challenges of being a graduate student is to cope with andfind a way to learn and get something out of a process and system thatis not very efficient.-neel=== === Subject: : Farey series= equivalent statement to Riemann Hypothesis Huh?Mime-version: 1.0Content-transfer-encoding: 7bitIn various recent popular mathematics books they write of the equivalentstatement to RH that relates to the difference in the order of the fractionsin the farey series and the ordered value of the fractions.Could someone translate this?Just what property is being tested?Wouldn't the order of the farey series be arbitrary depending on whatdirection(s) you want to start from. Eg up down left or right or even a treespanning algorithm. Or a mix?Anyone? === === Subject: : Re: Farey series= equivalent statement to Riemann Hypothesis Huh?> In various recent popular mathematics books they write of the equivalent> statement to RH that relates to the difference in the order of the fractions> in the farey series and the ordered value of the fractions.Could someone translate this?> Just what property is being tested?> Wouldn't the order of the farey series be arbitrary depending on what> direction(s) you want to start from. Eg up down left or right or even a tree> spanning algorithm. Or a mix?The Farey series of order n is the fractions a/b between zero and one, inclusive, in lowest terms, with b not exceeding n, in their natural order. E.g., the Farey series of order 5 is 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1, in that order. For large values of n this series (it should really be called a sequence, but we're stuck with the name) is very evenly distributed in the interval [ 0, 1 ]. The phrase in quotes can be made mathematically precise. Part of that precision involves introducing an error term (to measure how far away the series is from being perfectly evenly distributed). There is a conjecture on that error term which is equivalent to the Riemann Hypothesis. You can probably find the details in Kuipers and Niederreiter, Uniform Distribution of Sequences.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)=== === Subject: : Re: Farey series= equivalent statement to Riemann Hypothesis Huh?Holy Mackerel(go take a long swim in a very short desert oceanto the Age of Aquarius ... to paraphrase the emandationof Helen keller (first, non-morse .-)) anyone, else? > Wouldn't the order of the farey series be arbitrary depending on what> direction(s) you want to start from. Eg up down left or right or even a tree> spanning algorithm. Or a mix?--Give Earth a Trickier Dick Cheeny -- OBNOXICO, out of office?http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac=== === Subject: : Re: Complete, but not Baire> Can you remind the definition of a Baire space and a uniform space?Seehttp://en.wikipedia.org/wiki/Baire_spaceandhttp:// en.wikipedia.org/wiki/Uniform_spaceBest regards,Jose Carlos Santos=== === Subject: : Re: Continuum hypotheses and Lebesgue measure> Actually I'm familiar with the Cantor-Benedixon theorem in the exact> form I stated it. I'm quite sure that's the way Cantor saw the> result. The generalization to Polish spaces was probably done later> (perhaps by Benedixon).1) As I have already written in my previous post, it's Bendixon, notBenedixon.2) If what you meant was Cantor's theorem, you should have said so,instead of writing Cantor-Benedixon theorem.3) I don't know exactly what's the statement of Cantor's theorem,but it cannot be simply that any uncountable set of real numbershas a prefect subset. That's trivial: simply take a subset with asingle element.Best regards,Jose Carlos Santos=== === Subject: : Re: Continuum hypotheses and Lebesgue measure>> Actually I'm familiar with the Cantor-Benedixon theorem in the exact>> form I stated it. I'm quite sure that's the way Cantor saw the>> result. The generalization to Polish spaces was probably done later>> (perhaps by Benedixon).> 1) As I have already written in my previous post, it's Bendixon, not> Benedixon.> 2) If what you meant was Cantor's theorem, you should have said so,> instead of writing Cantor-Benedixon theorem.> 3) I don't know exactly what's the statement of Cantor's theorem,> but it cannot be simply that any uncountable set of real numbers> has a prefect subset. That's trivial: simply take a subset with a> single element.A singleton set has no accumulation points and therefore can't beperfect.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica> SIX ALL-PURPOSE TRICK QUESTIONS THAT WILL KEEP YOUR IGNORANCE FROMnoi tutti siamo in guerracon chi cerca di sfruttarcinon possiamo arrendercisenza lottare, la posta ingioco .8f troppo alta' j=== === Subject: : Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematicati guardo. ti assorbo.mi ipnotizzi.occhi senza parlpebresempre fissi su di te.' j> ROTFL!!!!> keroppi=== === Subject: : Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in Matematica geej> ti guardo. ti assorbo.Brutta oloturia lasciami stare!=== === Subject: : Re: JSH: Apology to Ramsay, why I post|> So I apologize to Keith Ramsay for questioning his honesty here.|> |> Thank you. I can't remember any time I've lied about mathematics, and I|> suspect it's because I just haven't ever.|> |> Excuse me if I say a little about this before letting it go, though.|> |> I've read that a compete canonical apology contains five parts. One, a|> statement of what was done wrong. Two, agreement that it was wrong. Three,|> a statement of the principle violated by it, i.e. why it was wrong. Four,|> an explanation of why it happened. Five, reason to believe it won't|> happen in the future. (Probably the main reason to say why one did it is|> to support one's explanation for why the other person doesn't need to|> worry about one's doing it again.)|> ||Are you sane?Yes. Do you see anything suggesting insanity in the above? I'mnot saying that every apology should be formal, by the way, ifthat's what's bothering you. Usually most of those componentsare implicit. But it's not quite an apology if they aren't at leastimplicitly there.It may seem like a paradox that a person would explain why theydid something in order to help convince someone they won't do itagain, but it makes perfect sense. Often the only way to be surenot to do something is to understand yourself well enough to knowwhy you had a tendancy to do it in the first place.As far as I can see, however, you don't have this kind of self-knowledgewhen it comes to your accusing others of lies. Certainly I'll consideryour latest accusation against me settled, but I don't have muchconfidence that you wouldn't do it again if similar circumstancesarose. I think it usually seems quite clear to you that we're lying toyou when you make such an accusation. You're just not looking atthat squarely. You kid yourself.Keith Ramsay=== === Subject: : Re: How big can a manifold be?> I read some of David Guald's paper and settled my conjectures about> 1-manifolds.> Thank you Zundark for providing the link.I don't know if you're still interested in the original question, butI was, and I've now managed to answer it:> Is there any limit to the cardinality of a connected n-manifold, that> is a connected Hausdorff space every point of which has a> neighborhood homeomorphic to R^n?Any such manifold has the cardinality of the continuum (assuming it's not empty, and n is nonzero).More generally:Let k be a cardinal greater than 1. Let X be a connected sequential space such that every point of X has a neighbourhood of cardinality atmost k^aleph_0, and each sequence in X has at most k^aleph_0 limits.Then |X| is at most k^aleph_0.(A sequential space is a topological space in which sequentially closed sets are closed. A sequentially closed set is a set D thatcontains all limits of sequences in D. Manifolds are first countable,and therefore sequential.)Sketch of proof:Put t = k^aleph_0. For each x in X, let U_x be an open neighbourhood of x of cardinality at most t.Assume X is noempty, and take some y in X. Define A_g for eachordinal g <= w_1 (where w_1 denotes the first uncountable ordinal) by transfinite recursion, as follows. Let A_0 = U_y. For each countable ordinal g, let B_g be the set of all limits of sequencesin A_g, and define A_(g+1) to be the union of all U_x with x in B_g.For each limit ordinal g <= w_1, define A_g to be the union of all theA_h with h < g.Then show that A_w_1 is open, closed (because sequentially closed) andnonempty (as it contains y). So X = A_w_1, because X is connected.Then show that |A_w_1| <= t.=== === Subject: : Balls and probability againHi all!Just wondering...(not homework)Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 ballsnumbered 2 and so on. So we have 4n balls.Now supose we extract balls from the bag 1 by 1.How many balls do we have to extract so that the probability of havingextracted 2 balls with the same number is 50%?N.=== === Subject: : Re: Balls and probability again>Just wondering...(not homework)>Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls>numbered 2 and so on. So we have 4n balls.>Now supose we extract balls from the bag 1 by 1.>How many balls do we have to extract so that the probability of having>extracted 2 balls with the same number is 50%?Suppose you extract k balls. The number of ways to do this is(4n choose k). Now if you wanted to make sure you had k different numbers, you could first choose the k numbers out of n (if k <= n)and then for each number chosen you choose one of the 4 balls withthat number. So the number of ways to choose balls with k differentnumbers is 4^k (n choose k), and the probability that you getk different numbers is f(k,n) = 4^k (n choose k)/(4n choose k). This can also be written as product_{j=1}^{k-1} (1-3j/(4n-j)). Now as n -> infinity, if k = t sqrt(n) we have (according to Maple)ln(f(k,n)) = -3/8 t^2 + O(n^(-1/2)), which will be ln(1/2) + O(n^(-1/2)) for t = 2/3 sqrt(6 ln(2)). I checked for n = 1 to 200and in each case, if k_n = round(2/3 sqrt(6 ln(2) n)), f(k_n-1,n) > 1/2 and f(k_n+1,n) < 1/2. So it looks like the answer is always either the closest integer to 2/3 sqrt(6 ln(2) n), or differs from it by at most 1.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Balls and probability again> Hi all!Just wondering...(not homework)Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls> numbered 2 and so on. So we have 4n balls.Now supose we extract balls from the bag 1 by 1.How many balls do we have to extract so that the probability of having> extracted 2 balls with the same number is 50%?> N.Let x = number of balls extracted. Then there will be 3*x ballsremaining which match the balls drawn. There will be 4*(n-x) ballswhich do not match.When 3*x >= 4*(n-x), the odds will be >= 50%.regards B=== === Subject: : Re: Balls and probability again>Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls>numbered 2 and so on. So we have 4n balls.>Now supose we extract balls from the bag 1 by 1.>How many balls do we have to extract so that the probability of having>extracted 2 balls with the same number is 50%?After extracting the first ball, the second ball has a 4(n-1)/(4n-1)chance of having a different number than the one chosen. If thathappens, the next ball has a 4(n-2)/(4n-2) chance of having a differentnumber than the others chosen. At stage k, if all the balls chosen havea different number, then ball k+1 has a 4(n-k)/(4n-k) chance of having adifferent number than the others chosen.Thus, on draw k+1, we have a 4(n-1)/(4n-1)...4(n-k)/(4n-k) chance ofhaving different colors on all balls = 4^k C(n-1,k)/C(4n-1,k).As soon as 4^k C(n-1,k)/C(4n-1,k) < 1/2, k+1 is the number of balls toextract.Using the fact that log(4(n-k)/(4n-k)) = log(1-k/n) - log(1-k/(4n)),we can show that when k is on the order of sqrt(n), we have -log(4^k C(n-1,k)/C(4n-1,k)) = 3k^2/(8n) + 3k/(8n) + 5k^3/(32n^2) - ...Therefore, as n gets bigger, k should be about sqrt(8n/3 log(2)) to geta probability of 1/2 that there is a pair of balls with the same number.Rob Johnson take out the trash before replying=== === Subject: : Re: Balls and probability again> Hi all!Just wondering...(not homework)Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls> numbered 2 and so on. So we have 4n balls.Now supose we extract balls from the bag 1 by 1.How many balls do we have to extract so that the probability of having> extracted 2 balls with the same number is 50%?> N.Hi Nestor, I see no problems with your English, and I am fairly picky. Nowto the problem at hand. I am thinking as I type, so please forgive alittle bit of wandering. If you choose n+1 balls, then you arecertain to have 2 balls with the same color. Let's look at theinverse problem. How do you choose balls all of whose colors aredifferent. For one ball this is a certainty. The second ball must beof a different color, so the probability is(4n-4)/(4n-1). For the third ball, the probability is (4n-4)(4n-8)/(4n-1)(4n-2) = 4^2(n-1)(n-2)/(4n-1)(4n-2). For the j'thball, the probability of having balls of all different colors is(4^(j-1))(n-1)(n-2)...(n-j+1)/(4n-1)(4n-2)...(4n-j+1).I don't suppose you usually get exactly 1/2 out of this fraction formost n, but if you choose the first j so that this expression is lessthan 1/2, you have the smallest j so that your probability of getting2 (or more) balls with same color is at least 50%.Hope this helps,Achava=== === Subject: : Re: Balls and probability againgest.nuria.telefonica-data.net:> Hi all!Just wondering...(not homework)Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls> numbered 2 and so on. So we have 4n balls.Now supose we extract balls from the bag 1 by 1.How many balls do we have to extract so that the probability of having> extracted 2 balls with the same number is 50%?> N.> Here's a hint, if you extract 5 balls, you will have at least two of one number. Look up the pigeon hole prinicple.I'll let you figure out the probabilities on your own, but here is how you should start...Pick one ball. It doesn't matter what number it is.Pick a second ball, there are two possiblities: It matches the first ball and you are done It doe snot match the first ball and you must draw againIf it does not match, pick a third ball: It matches the first ball, you are done It matches the second ball, you are done It does not match either and you must draw againIf it does not match either, pick a fourth ball: It matches the first ball, you are done It matches the second ball, you are done It matches the third ball, you are done It is the other ball, you are done because the next ball pick will have to match one of themFind the probabilities for each scenario. -- === === Subject: : Re: Balls and probability again> gest.nuria.telefonica-data.net:> Hi all! Just wondering...(not homework) Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls> numbered 2 and so on. So we have 4n balls. Now supose we extract balls from the bag 1 by 1. How many balls do we have to extract so that the probability of having> extracted 2 balls with the same number is 50%?> N. Here's a hint, if you extract 5 balls, you will have at least two of one> number. Look up the pigeon hole prinicple.This is incorrect as pointed out. However, given n sets of K balls (n*Kballs), if you extract n+1 balls, you must have at least two of the samekind, for, at least, integers K > 1, n > 0.=== === Subject: : Re: Balls and probability againTim Brauch wibbled:> gest.nuria.telefonica-data.net:Hi all!Just wondering...(not homework)Supose we have a bag full of numbered balls. 4 balls numbered 1, 4 balls> numbered 2 and so on. So we have 4n balls.Now supose we extract balls from the bag 1 by 1.How many balls do we have to extract so that the probability of having> extracted 2 balls with the same number is 50%?> N.> Here's a hint, if you extract 5 balls, you will have at least two of one > number. Look up the pigeon hole prinicple.I'll let you figure out the probabilities on your own, but here is how you > should start...Pick one ball. It doesn't matter what number it is.> Pick a second ball, there are two possiblities:> It matches the first ball and you are done> It doe snot match the first ball and you must draw again> If it does not match, pick a third ball:> It matches the first ball, you are done> It matches the second ball, you are done> It does not match either and you must draw again> If it does not match either, pick a fourth ball:> It matches the first ball, you are done> It matches the second ball, you are done> It matches the third ball, you are done> It is the other ball, you are done because the next ball> pick will have to match one of them> Find the probabilities for each scenario. And if ball 1 is a 1, ball 2 is a 2, ball 3 is a 3, ball 4 is a 4, ball 5 is a 5, ball 6 is a 6... ?-- 44 please=== === Subject: : Re: Balls and probability againI may be wrong, but I think:p(ball2 matches) = 3/(4N-1)p(ball3 matches) = (4N-4)/(4N-1) * 6/(4N-2)p(ball4 matches) = (4N-4)/(4N-1) * (4N-8)/(4N-6) * 9/(4N-3)etcEasy to say, I suspect it's some hideously mutated form of Pascal's Triangle.And you're looking for the sum of those terms >= .5 (50%)I have to change my son's nappy now...=== === Subject: : Re: Balls and probability again> gest.nuria.telefonica-data.net:> Here's a hint, if you extract 5 balls, you will have at least two of one> number. Look up the pigeon hole prinicple.mmmm? I dont think so. Look:n=100; 4*n=400 ballsball 1: number 4ball 2: number 23ball 3: number 7ball 4: number 62ball 5: number 20ball 6: number 15ball 7: number 2...ball i: number 7 ----> match!=== === Subject: : Re: Balls and probability againgest.nuria.telefonica-data.net:>> gest.nuria.telefonica-data.net:>> Here's a hint, if you extract 5 balls, you will have at least two of one>> number. Look up the pigeon hole prinicple.mmmm? I dont think so. Look:n=100; 4*n=400 ballsball 1: number 4> ball 2: number 23> ball 3: number 7> ball 4: number 62> ball 5: number 20> ball 6: number 15> ball 7: number 2> .> .> .> ball i: number 7 ----> match!> Sorry, I meant to say, Let's look at an example where you have 4 different kind of balls. Then what I said makes sense.To generalize, if you pick n+1 balls, you have to have a match. The procedure I gave works for more than 4 balls, if you generalize it. -- === === Subject: : Re: Balls and probability againSure, but I was looking for some kind of 'formula' for determining thenumber of balls to extract for a given probability (or the probability giventhe number of balls extracted), not a procedure.Unfortunately my knowledge in this field is quite limitated, but Imdeveloping a small program to aproximate it empiricaly unless anybody herecould find the solution.> gest.nuria.telefonica-data.net:>> gest.nuria.telefonica-data.net:>> Here's a hint, if you extract 5 balls, you will have at least two ofone>> number. Look up the pigeon hole prinicple. mmmm? I dont think so. Look: n=100; 4*n=400 balls ball 1: number 4> ball 2: number 23> ball 3: number 7> ball 4: number 62> ball 5: number 20> ball 6: number 15> ball 7: number 2> .> .> .> ball i: number 7 ----> match! Sorry, I meant to say, Let's look at an example where you have 4different> kind of balls. Then what I said makes sense.> To generalize, if you pick n+1 balls, you have to have a match. The> procedure I gave works for more than 4 balls, if you generalize it.> -- === === Subject: : Re: Balls and probability again> Sure, but I was looking for some kind of 'formula' for determining the> number of balls to extract for a given probability (or the probabilitygiven> the number of balls extracted), not a procedure.> Unfortunately my knowledge in this field is quite limitated, but Im> developing a small program to aproximate it empiricaly unless anybody here> could find the solution.Well, given n sets of K balls (n*K balls), if you extract n+1 balls, youmust have at least two of the same kind, for, at least, integers K > 1, n >0. So n+1 tells you the number of balls that need to be picked for there tobe a 100% chance of a double. So given n*K balls, can't you try tocalculate P(NUM_EXTRACTED = x), x in {0..n}, at least for small n? Maybeyou'll see a pattern.=== === Subject: : analysis problem...hello.......p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)show that if n = odd , p(x) has a root at least.-----------------------------my book is contained complex solution for this problem.but i think.........p(00) = 00p(-00) = -00thus by mean value theorem , p(x) has a root at least.it has logicality ??how do you think about it?=== === Subject: : Re: analysis problem...> hello.......p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)show that if n = odd , p(x) has a root at least.-----------------------------my book is contained complex solution for this problem.but i think.........p(00) = 00> p(-00) = -00thus by mean value theorem , p(x) has a root at least.it has logicality ??how do you think about it?Mmmm.... well, I think you're assuming that you can do some sort ofmath that says that oo^n > oo. You're attacking it the right way,though, by using the idea of very large numbers. Try limits if youwant to make it formal.You could show that a * x^n eventually dominates any given polynomialof degree n-1 or less, which would give you the tool you needed foryour result. How? Think about p(x) / a * x^n.=== === Subject: : Re: analysis problem...Content-transfer-encoding: 8bit> hello.......p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)show that if n = odd , p(x) has a root at least.-----------------------------my book is contained complex solution for this problem.but i think.........p(00) = 00> p(-00) = -00thus by mean value theorem , p(x) has a root at least.it has logicality ??how do you think about it?Your statementws p(00) = 00, p(-00) = -00 (assuming a_n > 0) conceal alot, and I suspect that this is the source of the book's complexsolution.Note, for example, that p(x)/x^n = a_n + a_{n-1}/x + ... + a_0/x^n --> a_n as x --> infinity or x --> -infinity. Since a_n is positive this meansthat for |x| sufficiently large, p(x)/x^n is positive. Since n is odd,this means p(x) > 0 for x sufficiently large, while p(x) < 0 for xsufficiently negative.--Ron Bruck=== === Subject: : Re: analysis problem...In sci.math, hot-girl: > hello.......> p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)> show that if n = odd , p(x) has a root at least.> -----------------------------> my book is contained complex solution for this problem.> but i think.........> p(00) = 00> p(-00) = -00> thus by mean value theorem , p(x) has a root at least.> it has logicality ??> how do you think about it?If one divides by a_n:x^n + (a_{n-1}/a_n)*x^{n-1} + (a_{n-2}/a_n)*x^{n-2}) + ... + a_0/a_n = 0and one can then take x_min to be n * minus the maximum of abs(a_i/a_n),0 <= i < n, and x_max would be n * the maximum of abs(a_i/a_n).Note that x_min = -x_max.In short, x^n should swamp all other terms. Once one has shown that,then p(x_min) < 0, p(x_max) > 0, and therefore, for some x betweenthe two, p(x) = 0.IMO, one problem with using p(oo) and p(-oo) is that(oo - oo)/2 (the mean) is undefined.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: analysis problem...hot-girl escribi.97 en el mensaje> hello.......> p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)> show that if n = odd , p(x) has a root at least.> -----------------------------> my book is contained complex solution for this problem.> but i think.........> p(00) = 00> p(-00) = -00> thus by mean value theorem , p(x) has a root at least.> it has logicality ??> how do you think about it?I think that the Mean Value Theorem (anyone of them) has nothing to do here.But Bolzano's Theorem said:If f:[a, b] ---> R is continue en [a, b] and f(a)*f(b) < 0, there is avalue c in (a, b), such as F(c) = 0This is very intuitive, but not obvious and require a proof. With thatresult is inmediate that p(x) has at least one real root.It isn't very important here, but observe that can be a_n < 0, and the signsof p(x) in +/- inf are the opposites that you said.-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: analysis problem...>hot-girl escribi.97 en el mensaje>> p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)>> show that if n = odd , p(x) has a root at least.If f:[a, b] ---> R is continue en [a, b] and f(a)*f(b) < 0, there is a>value c in (a, b), such as F(c) = 0>This is very intuitive, but not obvious and require a proof. With that>result is inmediate that p(x) has at least one real root.Or for those who like to shoot flys with cannons, the complex zeros ofa polynomial with real coefficients come in conjugate pairs. FTA thensays that if n odd then at least one zero is real.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: analysis problem...> I think that the Mean Value Theorem (anyone of them) has nothing to do here.But Bolzano's Theorem said:> Well hot-girl's native language is not English, so when she sawsomething called in her text intermediate value theorem shetranslated it as mean value theorem.=== === Subject: : Re: analysis problem...> hot-girl escribi.97 en el mensaje>> hello.......>> p(x) = a_0 + (a_1)x + ..........+ (a_n)(x^n) (a_n != 0 , a_i in R)>> show that if n = odd , p(x) has a root at least.>> ----------------------------->> my book is contained complex solution for this problem.>> but i think.........>> p(00) = 00>> p(-00) = -00>> thus by mean value theorem , p(x) has a root at least.>> it has logicality ??>> how do you think about it?I think that the Mean Value Theorem (anyone of them) has nothing to do here.But Bolzano's Theorem said:Also known as Intermediate Value Theorem; that's probably what the OPmeant.Best regards,Jose Carlos Santos=== === Subject: : finite trig sumIf r is an odd number >=3 then(*) sum[k=1 to r-1] 1/sin^2 (2pi k/r) = (r^2 -1)/3.I can get that from(**) pi^2 / sin^2 (pi x) = sum[k in Z] 1/(k-x)^2 (any x not an integer)but a more elementary proof would be nice. Also, has anyone seena deduction of (**) from (*)? It seems feasible in view oflim[r to infinity] r^2 sin^2 (pi/r) = pi^2.TIA,Larry=== === Subject: : Re: finite trig sumEn el mensaje:Cfl%b.79693$Hy3.43755@edtnps89,Larry Hammick escribi.97:> If r is an odd number >=3 then> (*) sum[k=1 to r-1] 1/sin^2 (2pi k/r) = (r^2 -1)/3.> I can get that from> (**) pi^2 / sin^2 (pi x) = sum[k in Z] 1/(k-x)^2 (any x not an> integer) but a more elementary proof would be nice. Also, has anyone> seen> a deduction of (**) from (*)? It seems feasible in view of> lim[r to infinity] r^2 sin^2 (pi/r) = pi^2.> TIA,> LarryS = Sum(1/sin^2(2pi k/r), k, 1, r-1) = 2Sum(1/sin^2(2pi k/r), k, 1, (r-1)/2)(r is odd)Let x = 2pi/r and t = sin^2(x). You has that1 + 0*i = (cos(rx) + i*sin(rx)) = (cos(x) + i*sin(x))^rTaking the imaginary part on both sides,0 = C(r, 1)cos^(r-1)(x)*sin(x) - C(r, 3)cos^(r-3)sin^3(x) + ... +(-1)^((r-1)/2)sin^r(x)dividing by sin(x),0 = C(r, 1)cos^(r-1)(x) - C(r, 3)cos^(r-3)sin^2(x) + ... +(-1)^((r-1)/2)sin^(r-1)(x) (#1)All exponents of cos(x) and sin(x) are even. Then the RHS can be write as apolynomial in t. Let s = (r-1)/2. Then you get0 = A(0) + A(1)t + A(2)t^2 + .... + A(s)t^sThe roots of this polynomial are the values of sin^2(2pik/r), with k = 1,..., s = (r-1)/2. Note that sin^2(2pk/r) = sin^2(2pi(r-k)/r).Let t = 1/u and multiply by u^s0 = A(0)u^s + A(1)u^(s-1) + A(2)u^(s-2) + ...+ A(s-1)u + A(s)The roots of this polynomial are the values of 1/sin^2(2pik/r), with k = 1,..., s. By Cardano-Vieta relations, the sum of all roots isS = 2*(- A(1)/A(0))A(0) = C(r, 1) = rA(1) is the exponent of sin^2(x) in #1A(1) = - C(r,1)(r-1)/2 - C(r, 3) = - r(r + 1)(r - 1)/6Then S = (r + 1)(r - 1)/3 = (r^2 - 1)/3 (q.e.d.)-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: derivative in the sense of distributions>> Is it true that a function in L^2 is always differentiable in the sense of>> distributions? In other words, for any f in L^2, does it exists a function>> g in L^1_loc such that:>>> - = for all v in C_0^infty>As David pointed out, every L^2 function (like every distribution) has a >distibutional derivative, but the latter need not be in L^1 locally. For >example, let f be the characteristic function of [0,1]. Then - = >v(0) - v(1). So the distributional derivative of f is the singular measure >delta_0 - delta_1, and there's no way this is going to be given by a g in >L^1_loc.Or more generally, if the distribution derivative is in L^1_loc thenf must be continuous (or f = g ae where g is continuous), in factlocally absolutely continuous.=== === Subject: : ANN: MathRevolt announces Infinity 1.2.6For immediate releaseContact: Julia RoweCompany: MathRevoltTitle: Public Relations ManagerE-mail: julia@mathrevolt.comMathRevolt announces Infinity 1.2.6breaking math software package that allows solving non-trivial mathanalysis tasks. Infinity is advanced math modeling application thatenables anyone who can express his task in language of equations analyzetechnology utilized in this software, it is capable of extremely precisecalculations. This feature is crucial not only for scientificcalculations but also for applied math modeling in business andengineering.Infinity is intended for broad range of computer users who possess basicmath knowledge and would like to solve complex math analysis tasks on aPC. The application suits perfect for serious scientific calculationswhere it beats any competing product both as to precision and speed ofoperation. On the other hand Infinity can be used in financial planning,exchange rates and stock market prediction, expected revenue evaluationetc. Ease of use and strong visualization module allow for on-the-flytask conditions altering and math model updating.As we have studied the market, we found there are two types ofsolutions: those that are overloaded with functionality that is neveractually used and those that have nothing to offer, - said PavelMikhailov, CEO of MathRevolt. When creating Infinity, our goal was tocome up with a software solution that would offer both speed andprecision. Besides that, the application should have been easy to useand priced reasonably! According to Mr. Mikhailov MathRevolt did its very best to presentall-inclusive math application that can be used for fast high precisioncalculations at an affordable price!About MathRevolt Founded in 1998, MathRevolt specializes in developing scientificsoftware applications and applied math solutions for corporate market.The company unites IT-specialists, scientists and software engineersdedicated to bringing latest technology advancements to the mass market.Besides developing applications, MathRevolt provides consulting servicesand product training.For more information, contact MathRevolt at: E-mail: julia@mathrevolt.comCorporate web site: http://www.mathrevolt.comProduct web site: http://www.mathrevolt.com/overview.html=== === Subject: : Re: ANN: MathRevolt announces Infinity 1.2.6> For immediate releaseContact: Julia Rowe> Company: MathRevolt> Title: Public Relations Manager> E-mail: julia@mathrevolt.comMathRevolt announces Infinity 1.2.6> breaking math software package that allows solving non-trivial math> analysis tasks. Infinity is advanced math modeling application that> enables anyone who can express his task in language of equations analyze> technology utilized in this software, it is capable of extremely precise> calculations. This feature is crucial not only for scientific calculations> but also for applied math modeling in business and engineering.Infinity is intended for broad range of computer users who possess basic> math knowledge and would like to solve complex math analysis tasks on a> PC. The application suits perfect for serious scientific calculations> where it beats any competing product both as to precision and speed of> operation. On the other hand Infinity can be used in financial planning,> exchange rates and stock market prediction, expected revenue evaluation> etc. Ease of use and strong visualization module allow for on-the-fly task> conditions altering and math model updating.As we have studied the market, we found there are two types of solutions:> those that are overloaded with functionality that is never actually used> and those that have nothing to offer, - said Pavel Mikhailov, CEO of> MathRevolt. When creating Infinity, our goal was to come up with a> software solution that would offer both speed and precision. Besides that,> the application should have been easy to use and priced reasonably!According to Mr. Mikhailov MathRevolt did its very best to present> all-inclusive math application that can be used for fast high precision> calculations at an affordable price! Hmm... An MS Windows-only program - you have just lost many browniepoints for that.=== === Subject: : Fundamental group of the circle - question on uniqueness of lifts> I am looking over a proof relating to this.> Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)> I = [0,1] R> They then state:> a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 in> p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0.b) For each homotopy f_t : I -> S^1 of paths starting at x0 and each ~x0 inp^(-1)(x0) there is a unique lifted homotopy ~f_t : I -> R of paths startingat ~x0.> c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x> {0}, then there is a unique map ~F: Y x I -> R lifting F and restrictingto> the given ~F on Y x {0}.They state that b) is obtained by applying c) with Y = I in this way: Thehomotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) as usual.A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)gives a unique lift ~F:IxI->R. The restrictions ~F|{0}xI and ~F|{1}xI arepaths lifting the constant path at x0, ...Now this is where I get lost. They then say:... hence they must also be constant by the uniqueness part of a).Why? Why does uniqueness imply they must be constant? (p o ~F)(0,t) = x0for all t in [0,1] as long as for all t in [0,1] : ~F(0,t) in p^(-1)(x0),which has an infinite number of elements. I can see someone saying it mustbe constant in order to be continuous, but I don't see what uniqeness per sehas to do with it.This type of reasoning is also used in showing that the function G : Z ->pi(S^1) is surjective. G(n) = [w_n], where [w_n] is the homotopy class ofw_n(s). w_n(s) = (p o ~w_n)(s), where ~w_n:I->R, ~w_n(s) = ns : the lift ofw_n. So to refresh w_n(s) = (cos 2 pi (ns), sin 2 pi (ns)). To prove thatG is surjective they first assume G(m) = G(n), which means w_m = w_n. Letf_t be a homotopy from w_m = f_0 to w_n = f_1. By b) this homotopy lifts toa homotopy ~f_t of paths starting at 0. ** The uniqueness part of a)implies that ~f_0 = ~w_m and ~f_1 = ~w_n. **Again, I understand that by unique they mean a single distinct lift, but Idon't understand how one is supposed to know/prove that _these functions inparticular_ are the unique lifts implied by a).=== === Subject: : Re: Fundamental group of the circle - question on uniqueness of lifts - 2nd tryI am looking over a proof relating to this.----------Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)I = [0,1] RPropositionsa) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 inp^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0.b) For each homotopy f_t : I -> S^1 of paths starting at x0 and each ~x0 inp^(-1)(x0) there is a unique lifted homotopy ~f_t : I -> R of paths startingat ~x0.Theoremc) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Y x{0}, then there is a unique map ~F: Y x I -> R lifting F and restricting tothe given ~F on Y x {0}.----------Both a) and b) should be derivable from c).They state that b) is obtained by applying c) with Y = I in this way: Thehomotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) as usual.A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)gives a unique lift ~F:IxI->R.Now this is where I get lost. They then say:The restrictions ~F|{0}xI and ~F|{1}xI are paths lifting the constant pathat x0, hence they must also be constant by the uniqueness part of a).First, why does ~F(1,t) lift the constant path at x0? All b) specifies isthat F(0,t) = x0, not that F(1,t) = x0 as well. Assume F(1,t) = x2, forarguments sake.Also, why does uniqueness imply ~F(0,t) and ~F(1,t) must be constant (giventhe both lift the constant path at x0)?(I note that from c) alone, one must prove ~F is a homotopy).b) implies ~F(0,t) = ~x0, so it is constant. But this doesn't seem torequire uniqueness.We know that F(1,t) = p(~F(1,t)) = x2, so theoretically, as long as it wascontinuous, all we need is that ~F(1,t) in p^(-1)(x2) for all t in [0,1].So again, why is x2 = x0, and why is the correct 'unique' lift in this casenecessarily a constant function?Another 'uniqueness' argument is used in showing that the function G : Z ->pi(S^1) is surjective. G(n) = [w_n], where [w_n] is the homotopy class ofw_n(s). w_n(s) = (p o ~w_n)(s), where ~w_n:I->R, ~w_n(s) = ns (the lift ofw_n). So to refresh w_n(s) = (cos 2 pi (ns), sin 2 pi (ns)). To prove thatG is surjective they first assume G(m) = G(n), which means w_m = w_n. Letf_t be a homotopy from w_m = f_0 to w_n = f_1. By b) this homotopy lifts toa homotopy ~f_t of paths starting at 0. They then state: ** The uniquenesspart of a) implies that ~f_0 = ~w_m and ~f_1 = ~w_n.**I understand that by unique they mean a single distinct lift, but I don'tunderstand how one is supposed to know/prove that _these functions inparticular_ are the unique lifts implied by a). It seems like ~f_0 shouldwork as long as it was homotopic to ~w_m.---=== === Subject: : Re: Fundamental group of the circle - question on uniqueness of liftsmessage> I am looking over a proof relating to this. Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)> I = [0,1] R They then state: a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0in> p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0. b) For each homotopy f_t : I -> S^1 of paths starting at x0 and each ~x0in> p^(-1)(x0) there is a unique lifted homotopy ~f_t : I -> R of pathsstarting> at ~x0. c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Yx> {0}, then there is a unique map ~F: Y x I -> R lifting F and restricting> to> the given ~F on Y x {0}. They state that b) is obtained by applying c) with Y = I in this way: The> homotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) asusual.> A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)> gives a unique lift ~F:IxI->R. The restrictions ~F|{0}xI and ~F|{1}xI are> paths lifting the constant path at x0, ...> Now this is where I get lost. They then say:> ... hence they must also be constant by the uniqueness part of a).> Why? Why does uniqueness imply they must be constant? (p o ~F)(0,t) = x0> for all t in [0,1] as long as for all t in [0,1] : ~F(0,t) in p^(-1)(x0),> which has an infinite number of elements. I can see someone saying itmust> be constant in order to be continuous, but I don't see what uniqeness perse> has to do with it.Actually, let me replace this with:-------They state that b) is obtained by applying c) with Y = I in this way: Thehomotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) as usual.A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)gives a unique lift ~F:IxI->R*The restrictions ~F|{0}xI and ~F|{1}xI are paths lifting the constant pathat x0, hence they must also be constant by the uniqueness part of a).*-----First, why does ~F(1,t) lift the constant path at x0? All b) specifies isthat F(0,t) = x0, not that F(1,t) = x0 as well. Assume F(1,t) = x2, forarguments sake.Also, why does uniqueness imply ~F(0,t) and ~F(1,t) must be constant? ~F isa homotopy, so by definition ~F(0,t) = ~x0 (is constant at the point ~x0 inparticular by b)) and ~F(1,t) = k, k in p^(-1)(x2). Uniqueness is notrequired, only that ~F is a homotopy.=== === Subject: : Re: Fundamental group of the circle - question on uniqueness of liftsOriginator: grubb@lola>They state that b) is obtained by applying c) with Y = I in this way: The>homotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) as usual.>A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)>gives a unique lift ~F:IxI->R>*The restrictions ~F|{0}xI and ~F|{1}xI are paths lifting the constant path>at x0, hence they must also be constant by the uniqueness part of a).*----->First, why does ~F(1,t) lift the constant path at x0? All b) specifies is>that F(0,t) = x0, not that F(1,t) = x0 as well. Assume F(1,t) = x2, for>arguments sake.Thus, t->po~F(0,t) is the constant path at x0., I.e. ~F(0,t) liftsthe constant path at x0. Do you know a path that *also* lifts theconstant path at x0? What does uniqueness then tell you?As for ~F(1,t), are you sure that F isn't a homotopy of *loops* at x0rather than of paths starting at x0?--Dan Grubb=== === Subject: : Re: Fundamental group of the circle - question on uniqueness of lifts>They state that b) is obtained by applying c) with Y = I in this way:The>homotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) asusual.>A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)>gives a unique lift ~F:IxI->R>*The restrictions ~F|{0}xI and ~F|{1}xI are paths lifting the constantpath>at x0, hence they must also be constant by the uniqueness part of a).*> ----->First, why does ~F(1,t) lift the constant path at x0? All b) specifiesis>that F(0,t) = x0, not that F(1,t) = x0 as well. Assume F(1,t) = x2, for>arguments sake.> Thus, t->po~F(0,t) is the constant path at x0., I.e. ~F(0,t) lifts> the constant path at x0. Do you know a path that *also* lifts the> constant path at x0? What does uniqueness then tell you?- I agree F(0,t) = p(~F(0,t)) = x0. This is because I see that b) directlydefines F and ~F this way.- By uniqueness, this other path must be equal to ~F(0,t) then. I justhaven't gotten to the proof of that yet.> As for ~F(1,t), are you sure that F isn't a homotopy of *loops* at x0> rather than of paths starting at x0?I'm pretty sure it is, but I can't seem to see how it's implied. b) definesF simply as a homotopy of paths starting at x0. So in order to prove b)from c), one must show c) implies b) works for more than just F's that arehomotopies of loops. Perhaps the other properties of b) force F to be aloop homotopy? Check out page 29, Section 1.1, theorem 1.7, of thefollowing text for the full treatment. Maybe (probably) I have left someimportant information out that will clear this all up.http://www.math.cornell.edu/~hatcher/AT/ATch1.pdfThis is just chapter 1 and contains the relevant portion.(An initial complain about this book: he _barely_ defines what a lift is(ie: he defines a specific lift function and then expects the reader toextrapolate the general concept of lift from this) and then goes on to useit in various forms as in a lifted homotopy. The book's definitions areusually mixed into the prose of the book. Definitions usually in Englishwith a bare minimum of notation, which as in the so-called 'definition' oflift, can be frustrating for the reader new to the subject (for me atleast)).--------a) b) and c):a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0 inp^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0.b) For each homotopy f_t : I -> S^1 of paths starting at x0 and each ~x0 inp^(-1)(x0) there is a unique lifted homotopy ~f_t : I -> R of paths startingat ~x0.c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R liftingF|Y x {0}, then there is a unique map ~F: Y x I -> R lifting F andrestricting to the given ~F on Y x {0}.=== === Subject: : Re: Fundamental group of the circle - question on uniqueness of liftsmessage> I am looking over a proof relating to this. Let p : R -> S^1, p(s) = (cos 2 pi s, sin 2 pi s)> I = [0,1] R They then state: a) For each path f: I -> S^1 starting at a point x0 in S^1 and each ~x0in> p^(-1)(x0) there is a unique lift ~f : I -> R starting at ~x0. b) For each homotopy f_t : I -> S^1 of paths starting at x0 and each ~x0in> p^(-1)(x0) there is a unique lifted homotopy ~f_t : I -> R of pathsstarting> at ~x0. c) Given a map F : Y x I -> S^1 and a map ~F : Y x {0} -> R lifting F|Yx> {0}, then there is a unique map ~F: Y x I -> R lifting F and restricting> to> the given ~F on Y x {0}. They state that b) is obtained by applying c) with Y = I in this way: The> homotopy f_t in b) gives a map F:IxI->S^1 by setting F(s,t)=f_t(s) asusual.> A unique lift ~F:Ix{0}->R is obtained by an application of a). Then c)> gives a unique lift ~F:IxI->R.> The restrictions ~F|{0}xI and ~F|{1}xI are> paths lifting the constant path at x0, ...I also don't get why ~F|{1}xI lifts the constant path at x0, since from b) Fis stated to only start at x0, not also end at x0. Hence we could only say~F|{0}xI lifts the path x0.=== === Subject: : Re: Trying to unify axioms.> A axiom is a stated> unprovable assumption, It is indefensible for being an axiom - and> can be falsified by a single reproducible counterdemonstration. Falsifiability, indefensible - this has nothing to do with it. It's> simply a fact that in logic, every axiom of a theory T is provable in T.Only one line can be drawn parallel to a given line through an> exterior point. Everybody knows that from high school geometry. It> is obviously true as Euclid's Fifth Postulate (here restated as> Playfair's Axiom). There is only one glitch: It is empirically> wrong. There are no lines parallel to a given line on the Earth'> surface. Pookie pookie. The are an infinite number of lines that can be drawn> parallel to a given line through an exterior point on a hyperbolic> surface. Euclid is incomplete for his fifth postulate.Listen to the man Uncle Al, listen to the man. It IS a matter ofterminology and of the theory in question. Why don't you agree on anaxiom must not necesarily be provable? It really doesn't matterwether there are axioms that can be proven after a fashion, or thereare theories in which axioms are provable by definition. The point is,that everything else in the theory must be derivable from the set ofits axioms, and he is not contesting that. Now for the parallel lines,come on, you can't expect the postulate to hold in a non euclideangeometry. Of all people here you should know better than that.By the way, how was that molten brie with chipotle the other day? Youguys really do weird things to food in California. Personally, I likethe more basic stuff. My favourite is is the the chilli cheese dogwith the chilli cheese fries and a bottle of rootbeer at Pink's. Nofancy Wolfgang Puck concoction can beat that.regards.=== === Subject: : Re: Trying to unify axioms.> You could, I suppose, distinguish between the *proper* consequences> of a theory and the *improper* consequences---a proper consequence> is a statement that is not an axiom, but is logically deducible from> the axioms. However, that's really not a very useful distinction. Also, those who consider it objectionable that an axiom A isprovable in a theory in which it is an axiom, would most likelynot find it less objectionable that A&A is provable. In mathematics, we don't speak of axioms as provable, but we dospeak of theorems having immediate proofs from the axioms. This finedistinction has no role to play in logic. One would like to have alogical theory of the complexity of proofs, but present day prooftheory cannot define or analyze informal distinctions between proofs,such as a proof being a mere variant of another, or containing anessentially new idea, or being simpler than another proof, or moreinformative.=== === Subject: : Re: Trying to unify axioms. <0q4%b.127219$FO1.2443626@weber.videotron.net> Discussion, linux)>> You could, I suppose, distinguish between the *proper* consequences>> of a theory and the *improper* consequences---a proper consequence>> is a statement that is not an axiom, but is logically deducible from>> the axioms. However, that's really not a very useful distinction.> Also, those who consider it objectionable that an axiom A is> provable in a theory in which it is an axiom, would most likely> not find it less objectionable that A&A is provable.So then we could say that the improper consequences are the least setcontaining the axioms and closed under the rules of propositionallogic.And then you'll complain that P(17) isn't a proper consequence of (A x)P(x). And I doubt my proposed solution of that problem isworkable, but we could do it this way. For each axiom A, define thetheory of the axiom th(A) (so the theory consists of those thingsprovable by a single axiom). Then the improper portion is the unionof all the theories th(A), maybe with some additional closure underpropositional rules.It's an ugly hack with no prior justification for the definition.It's an unmotivated attempt to capture all of the examples of trivialconsequences of the axioms without getting other stuff and it'scertainly of no particular interest to proof theorists. But I wonderwhat those who want a distinction between proper and improperconsequences would think.Probably, it contains way too much. It's pretty clear that A & A isso obvious that it should be included, but should arbitrarily complexproofs involving a single axiom (with an unlimited number ofapplications of the quantifier rules) really be called improper? Idoubt it. I bet it wouldn't take much work to give an example inwhich the conclusion is not at all obvious from the axiom. One might fixing the number of applications of the axiom in a proof tofix his definition of Th(A). Since A & A is provable underpropositional rules, this wouldn't affect it. It *would* mean thatP(17) & P(18) is not improper, depending on how and when we takeclosure under deductive rules.Besides, I guess, the folks that don't think axioms are deducible fromthemselves are going to balk when we start tossing in conditionals ofthe form P -> Q where ~P is any improper consequence.> In mathematics, we don't speak of axioms as provable, but we do> speak of theorems having immediate proofs from the axioms. This fine> distinction has no role to play in logic. One would like to have a> logical theory of the complexity of proofs, but present day proof> theory cannot define or analyze informal distinctions between proofs,> such as a proof being a mere variant of another, or containing an> essentially new idea, or being simpler than another proof, or more> informative.Oh, sure, modern proof theory can't do it, but I bet I can come upwith a solution this morning. Watch.-- C is for Cookie. That's good enough for me. Cookie Monster=== === Subject: : Re: missing pages from Spanier's Algebraic Topology>I ordered a copy of Spanier's Algebraic Topology, and it arrived>today, minus page 13-14. (...)>Springer. Instead, I was wondering if someone with access to a>scanner would just scan these two pages (page 13 and 14 in the 1991>printing--basically the first two pages of chapter 1) and email me the>scans at npenton@hotmail.com. (...) I have a scanner and the book, but it's a much older edition. If this> does not make it useless for you, I can send you such a scan as> a GIF image, but tell me first with what begins page 15 of your book,> so that I can better guess what content you are missing. In my book> the half-empty first page of chapter 1 is a short introduction, that> continues on the next page before section 1; do you have this> introduction ?Page 15 begins with and it is called the identity morphism of Y. Given morphisms f:X->Y and g:Y->X such that gf=1_X, g is called a leftinverse of f and f is called a right inverse of g. A two sidedinverse...Page 12 is just a title page, reading Chapter one: Homotopy and theFundamental GroupSo no, I do not think I have the introduction to chapter one. If youcould send me whatever lies between these in your edition, I would begrateful. My understanding is that he only made minor revisions andcorrections between editions.Nathan=== === Subject: : Lebesgue measure extensionDoes anyone know if there exists countably additive extension ofLebesgue measure onto the sigmafield of all subsets of R? Of course itwould have to be not isometry invariant.regardsolej=== === Subject: : re:Lebesgue measure extensionYou can't disprove the CH either. The question is whether it's possible to disprove the possibility ofthis extension. === === Subject: : Re: Lebesgue measure extension> Does anyone know if there exists countably additive extension of> Lebesgue measure onto the sigmafield of all subsets of R? Of course it> would have to be not isometry invariant.That would mean c is a real-valued measurable cardinal.Sounds good to me. I like measurable cardinals.But (of course) you cannot prove it exists in ZFC.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : re:Lebesgue measure extensionNot if the continuum hypothesis holds, that's for sure. === === Subject: : Re: determinantX-Filename: sci.math/A103RTFG fi[Hyphen]zmachava@hotmail.com (Achava Nakhash, the Loving Snake)nG> This is a skew-symmetric matrix. It has the property that all eigenvalues> are purely imaginary.> As an added hint, you might take note that iA is Hermitian, and> Hermitian matrices necessarily have non-negative eigenvalues. What is ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Let A be nxn Hermitian maatrix. If c is an eigenvalue of A, => there is x != 0 such that Ax=cx. c = = = = = = conjugate(c) => c = conjugate(c). => c is real. Hou do you show that c is non-negative?> the relationship between the eigenvalues of A and those of iA?> Achava--=== === Subject: : Re: determinant fi[CapitalEth]zmachava@hotmail.com (Achava Nakhash, the Loving Snake)n264G> This is a skew-symmetric matrix. It has the property that all eigenvalues> are purely imaginary.> As an added hint, you might take note that iA is Hermitian, and> Hermitian matrices necessarily have non-negative eigenvalues. What is> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Let A be nxn Hermitian maatrix.> If c is an eigenvalue of A, => there is x != 0 such that Ax=cx.> c = = = = = = conjugate(c) c = conjugate(c).> => c is real.> Hou do you show that c is non-negative?You don't, because it's false.Example: A = [-1].I think Achava is confused with 'positive definite'.Wilbert=== === Subject: : What is freedom ?Hi Starblade Darksquall, You suggested, nature IS progress. The future is not set, but the means by which we must get to the future is essentially set. However, it also possesses degrees of freedom. .Just because we don't perfectly know what the future will be that can't make nature somehow incomplete.Nature is complete by definition.So although the future is imperfectly known, the future is sure to be perfectly set. ( i.e. Time is perfectly spatial, not directional )Q. What is progress ? A. Control is the goal. And nature never worries about that. ( or anything else )Q. What is freedom ?A. It's facing the same trivial choices day after day ... And then feeling a false sense of freedom because of that.=== === Subject: : principal bundles in algebraic geometryI'm looking for a nice and introductory exposition of the concept ofprincipal bundles P-->X in algebraic geometry, that explains me inmodern language e.g. why mumford's definition (in GIT) is the same assaying that it should be locally trivial in the etale topology (isit?), that every principal GL bundle is already locally trivial in theZariski topology (I know, there is always Serre), and theconstructions of associated geometric and/or algebraic (meaninglocally free sheaf) vector bundle. I found some helpful stuff inMilne's Etale Cohomology, but I would appreciate a more extensivetreatment especially concerning the constructions of frame bundleassociated to a vector bundle and vector bundle associated to aprincipal bundle.I'm thankful for any hints,markus=== === Subject: : Help on curvature and tangentsHiI'd really appreciate any help in these problems1. Prove that a(s) is a straight line iff all its tangent lines areparallel.2. Let a(s) be a curve with curvature k!=0, prove that a(s) lies in aplane iff there is a point x0 in R^3 such that every osculating planepasses through x0.Carlos=== === Subject: : Re: Help on curvature and tangents> Hi> I'd really appreciate any help in these problems> 1. Prove that a(s) is a straight line iff all its tangent lines are> parallel.Suppose a(s) is a continuous curve that is not a straight line. Are the tangents parallel?Suppose the tangents are not parallel. Can a(s) be a straight line?2. Let a(s) be a curve with curvature k!=0, prove that a(s) lies in a> plane iff there is a point x0 in R^3 such that every osculating plane> passes through x0.What does the osculating plane tell you? What does the curvature not equal to 0 tell you?-- email: wtwentyman at copper dot net=== === Subject: : Re: Help on curvature and tangents> Hi> I'd really appreciate any help in these problems> 1. Prove that a(s) is a straight line iff all its tangent lines are> parallel.> 2. Let a(s) be a curve with curvature k!=0, prove that a(s) lies in a> plane iff there is a point x0 in R^3 such that every osculating plane> passes through x0.Only in advance apparently.You *were* welcome.Maybe you could start learning the theory before you try yourhomework.=== === Subject: : Re: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with fixed thicknessActually i want to know the intercept length when a random lineintersect with a convex shell with fixed thickness. how many methodsthe random line can be choosed? Can you get me the details? thank youin advance!> En el mensaje:477a8527.0402180800.756b523e@posting.google.com,> ChenHS escribi.97:> If a random line intersects with a ellipsoid shell or parellelepiped> shell with a fixed thickness, how much is the statistical intercept> length?You must define in a precisse way how the random line is choosed. The answer> depend on it and it can be very different.=== === Subject: : Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QDtCq05123; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1Q6hsi01874 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1Q6hrP23219;>>O.k. then , as long there is no conflict with :>>Re(e^[iPi]) = Re(-1+i[0]) = -1 >That has never been in doubt. Neither has the observation that>Im(e^[i pi]) = Im(-1+i[0]) = 0.>The statement that caused all the problem was the statement that>e^[i pi] = 0.>>Actualy stated e^[ipi]=i0=0>Which is exactly the same statement in the long run. And I have not>yet seen you retract the statement, as you should have by now.>>[REF:>> A) e^[ipi]=-1 the real part solution and >> B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.]>Arguments which you have now acknowledged as invalid (they are either>both valid or both invalid, and the false conclusion in B from the >true premise that exp[i pi] = -1+i[0] demonstrates that the argument>used in B is invalid).>David McAnallyThe following was agreed:Re(e^[iPi]) = Re(-1+i[0]) = -1 ANDIm(e^[i pi]) = Im(-1+i[0]) = 0.However,let me point out that this theory ofcomplex notationbeing ,admitedly ,helpful for solving complicated engineering [not a panacea however ]problems[taking care simply angles of unity]in engineering it should not interfere with Eucledian Geometry.The Classical osophers were very strict ,in specifying that the only tools that should be allowed to be used,in solvinf Geometric Problems ,should be theungraduated straight edge and the compass [maximum].I, shall add this too.When the resultant of the two components of the complexnumber is calculated ,then -i , and +i are ignored .[I, would rather say that only +i is ignored ]Panagiotis Stefanides=== === Subject: : Re: e is transcendental> However,let me point out that this theory of> complex notation> being ,admitedly ,helpful for solving > complicated engineering [not a panacea however ]problems> [taking care simply angles of unity]> in engineering it should not interfere with > Eucledian Geometry.There is only one theory of complex notation.Where a, b, c, and d are real numbers, the following two lines areequivalent:a + bi = c + dia = c AND b = dThings like a = d or b = c (or even a + bi = d) don't enter intoit.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W=== === Subject: : re:determinantHint: if n is odd expand the determinant by the definition and showthat it's zero. === === Subject: : Re: determinant by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QEWU910487; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1QE2Li07574 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1QE2LC08005;>Hello!>Please help me to solve the following determinant:>| a1+a1' b1+b1' c1+c1' d1+d1' e1+e1' | >| a2+a2' b2+b2' c2+c2' d2+d2' e2+e2' |>| a3+a3' b3+b3' c3+c3' d3+d3' e3+e3' |=?>| a4+a4' b4+b4' c4+c4' d4+d4' e4+e4' |>| a5+a5' b5+b5' c5+c5' d5+d5' e5+e5' |>1. is there some shortcut to get the full solution to this determinant, i mean rather than solve it brute force...._____________________________________________________ _____Wanting to get it without brute force is good, but I hope you realize thisproblem is easily within the range of computer algebra systems. I did it withFermat, it took 0.133 seconds (on a two year old Mac), and the answer has 3840 terms. I didn't try other systems.Robert H. Lewishttp://www.bway.net/~lewis/=== === Subject: : Re: determinantHint: det (a1+a1' , b1 + b1' ; a2 , b2) = det (a1 , b1 ; a2 , b2) + det(a1' , b1' ; a2 , b2)Robert>Hello!Please help me to solve the following determinant:| a1+a1' b1+b1' c1+c1' d1+d1' e1+e1' |>| a2+a2' b2+b2' c2+c2' d2+d2' e2+e2' |>| a3+a3' b3+b3' c3+c3' d3+d3' e3+e3' |=?>| a4+a4' b4+b4' c4+c4' d4+d4' e4+e4' |>| a5+a5' b5+b5' c5+c5' d5+d5' e5+e5' |1. is there some shortcut to get the full solution to this determinant, imean rather than solve it brute force....> __________________________________________________________> Wanting to get it without brute force is good, but I hope you realize this> problem is easily within the range of computer algebra systems. I did itwith> Fermat, it took 0.133 seconds (on a two year old Mac), and the answer has3840 terms. I didn't try other systems.> Robert H. Lewis> http://www.bway.net/~lewis/=== === Subject: : Re: Full Beal Conjecture (revised)> According to great interest at this question,> I would like to discuss following classification: Once generally Beal Conjecture dissmiss x^a + y^b = z^c> for x;y;z integers of gcd=1 and for a;b;c natural numbers >2 1.) For a;b;c as odd numbers we'll have some fall of Beal's> conjecture, which might be proved once using similar method> as aplicable for FLT conjecture for p=3 and bigger primes.> ( this method uses also factorisation of numbers shaped as 3M+2> into factors as square numbers, what is applicated as 6n+1> shapes and turns into squares of all primes once beginning > with 5 . I used to write down first lines of it at ca 17.02.04 ) 2.) For one, two or even three of exponents a;b;c as even numbers> but bigger than 4 and so on containing some odd number bigger or equal to 3 it will be possible to turn such fall into (1)-st one: a=a1*a2; b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 = (z^c2)^c1 where a1;b1;c1 odd numbers.> 3.) For a;b;c as even numbers but discribed exclusively as 2^k or with the part a2;b2;c2 as 2^k where k=2 or bigger natural numbers so such fall is to collect into x^4 + y^4 = z^4 conjecture 4.) For only few combinations once one of a;b;c is equal to 4 but others odd or two of a;b;c are equal to 4 but the one resisting odd there is not possible to use the classification and method for odd exponents. 5.) Finally for a;b;c as even numbers but shaped in that way,> that it is possible to rewrite conjectures:> x^2 + y^4 = z^4 or x^4 + y^4 = z^2> You can see, that conjectures x^4 + y^4 = z^4> and x^2 + y^4 = z^4 or x^4 + y^4 = z^2 were already determinated> firstly by P.Fermat. Once it is some hope for method for odd exponents so the resisting small gap can be closed too and so> on it will be possible to express that any sum of powers bigger> than 2 can not be substituted with the power bigger than 2.> (Does Fermat expressed, that exponents of such powers are equal ?)> Enjoy super games of powers RoI appreciate your intent to help but I can't understand your (math)grammer. Sorry and thanks again. Kerry=== === Subject: : Judging convergence of Monte Carlo methodDear list,Trying to establish a continuous chirality measure for planar structures Iam intensively dependend on the Monte Carlo method for detaileddiscretisations. When calculating the standard deviation of the singleresults, I encountered a much more rapid decrease than expected, whencomparing the mean values each say 10 million calculations:The means are still fluctuating in the second and third digit, but therelative standard deviation is already around 1E-6.Does anyone else have experience using this method in general and might havean idea about a proper convergence paramater.Thank you in advance!Alexander-- Direct answers can be sent to A (period) S (period) Schwanecke (at)soton (period) ac (period) uk.Direkte Antworten an: A (punkt) S (punkt) Schwanecke (bei)soton (punkt) ac (punkt) uk.=== === Subject: : Re: Judging convergence of Monte Carlo methodWith Monte Carlo, it sounds like you have a solution to the problem at hand,3 significant digits is reasonable, good.The SD may have many significant digits, but if it is about 1/10 or morecompared to the mean, then the mean will continue change, and the number ofSD significant digits is misleading.Some other problems have a large SD and the results do not settle down withlarger n, the mean continues to shifts around.I think the weak law of large numbers is involved here, worth looking itup.Depends on the problem (number of dimensions, underlying relationships, timeframe, process stability, color of the Noise, etc)//////////////////////////////////////////////////> Dear list,> Trying to establish a continuous chirality measure for planar structures I> am intensively dependend on the Monte Carlo method for detailed> discretisations. When calculating the standard deviation of the single> results, I encountered a much more rapid decrease than expected, when> comparing the mean values each say 10 million calculations:> The means are still fluctuating in the second and third digit, but the> relative standard deviation is already around 1E-6.> Does anyone else have experience using this method in general and mighthave> an idea about a proper convergence paramater.> Thank you in advance!> Alexander> -- > Direct answers can be sent to A (period) S (period) Schwanecke (at)> soton (period) ac (period) uk.> Direkte Antworten an: A (punkt) S (punkt) Schwanecke (bei)> soton (punkt) ac (punkt) uk.=== === Subject: : Re: How far can one go with self-study nowadays?>> My personal opinion: I think that, for many learning university>> mathematics for the first time, the biggest hurdle is mastering the>> rigorous and formal style of mathematical language, notation, and>> thought.> I'd agree with this... when someone has been self-teaching mathematics >to themselves for 10 years, they don't really need to be proving things to >other people, and so sufficiently proving theorems to themselves usually >suffices.If they are self-teaching mathematics, and not computation,they need to be proving the theorems. The teaching of realmathematics needs to start in elementary school; college isFAR too late. In the old days, the 100% proof oriented high school geometry (Euclid) course at least did this. Once they get to an education level that challenges their >knowledge level, they have to start getting formal all of a sudden.> For instance, I was differentiating functions while I was in grade 9 and >integrating functions in grade 10, and felt pretty confident about solving >calculus problems because I never came across any I couldn't solve (as a >kid) and ended up with something stupid like 98% in high school calculus. >Once I hit university and slept through most classes, the midterm was a >shocker: Epsilons? Prove the least-upper-bound property of reals? (why >prove that? seemed obvious enough) Prove the sequential characterization >of limits? Prove Bolazano-Weierstrauss? (or, after that midterm, >Bozo-WiseAss)What you had up to that point was not mathematics, just plug-and-chug manipulation. NO mathematical concepts wereexplicitly taught, etc.Variables belongs with beginning reading, and good formallogic, including the predicate calculus, HAS been taughtin elementary school. Limits belong not later that theintroduction of decimals, and the Peano Postulates and development from them earlier.When I was on my post-doc, some of us tried to come up with a definition of a mathematician. The final resultwas Someone who might not KNOW 2+2, but who can prove it is 4.-- 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 Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Generating random numbers with specific properties>> Generate a series of pseudo-random numbers. Then use a filter to throw out>> those that do not satisfy your condition.>Is this considered the standard general approach to creating data with>arbitrary parameters -- the one to fall back on when there isn't a>more direct generator algorithm?>Are there other such general approaches, or is this one by far the>best (again, in cases where there isn't a more direct approach)?One should use random numbers if possible. But the acceptance-rejection process is not what is stated above.For example, to generate random variables with probabilityC/n^2, (it is not necessary to know C), one can take thereciprocal of a uniform (0,1) random variable, take itsinteger part K, and use K for the answer with probability(K+1)/(2K). This is NOT particularly good, but is easilyunderstood.It is easy in many cases to do better, as numerical analysisis an art, but Devroye's book is one of the best.-- 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 Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Generating random numbers with specific properties> Is there any book, website, or whatever that has a good, detailed> description of the algorithms needed to generate sequences of random> numbers with a given distribution, mean, variance, higher-order> moments (kurtosis, skew, etc.), autocorrelation, cross-correlation> with other such series, etc? In other words, you figure out the> properties and then you can either look up or figure out the algorithm> to generate it.It is not difficult to simulate from a distribution if you have anexplicit form for its PDF of CDF. The problem is that a finite set ofmoments, such as the mean, variance, skew, and kurtosis, does notuniquely specify the probability distribution, so there are manypossibilities.A flexible way to derive a probability distribution satisfying almostany set of moment conditions is to use a finite mixture of Gaussians.Finding the mixture coefficients (the weights, means, and variances)given the moments amounts to solving a set of nonlinear equations. Atwo-component normal mixture has 6 parameters, so you could fitmoments 0 to 5. If you want to simulate skewed data with highkurtosis, a mixture of Student t distributions could be tried.The skewed t distribution can fit data with skew and excess kurtosis.A reference is Financial Data and the Skewed Generalized tDistribution,, by P Theodossiou, Management Science, Vol. 44, No:12-1, December 1998, pp. 1650-1661.A similar idea is studied in the paper An asymmetric generalizationof Gaussian and Laplace laws, by T. Kozubowski and A. Ayebo, Journaldownloadable at http://unr.edu/homepage/tkozubow/0skeexp1.pdf .There is a Pearson family of distributions that can be used to fitmoments. One book that discusses its practical use is MathematicalStatistics with Mathematica, by Rose and Smith. A recent paperstudying the Pearson distribution, along with the first paragraph ofthe abstract, isJournal of Time Series AnalysisNon-Gaussian Filter and Smoother Based on the Pearson DistributionSystemby Yuichi Nagahara The Pearson distribution system can represent wide class ofdistributions with various skewness and kurtosis. We develop apractical approach of using all types of its distribution systemincluding the type-IV distribution which was difficult to implement. for more info.Another paper along the same lines is by Michael Rockinger and EricJondeau and can be downloaded fromhttp://campus.hec.fr/profs/rockinger/Publications.htm .The citation of the published paper is Volume 106, Issue 1, 01-January-2002 Journal Of Econometrics Vol. 106 (1) pp. 119-142Entropy densities with an application to autoregressive conditionalskewness and kurtosisa Michael Rockingerb Eric JondeauI have not addressed the time-series properties you mentioned, but Ithink the papers do.=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?Nathaniel Deeth> Age 11> Math GeniousIf you are really a math genius, you would know that 11+5 = 16.It's axiomatic that anyone who describes himself as a genius almost certainly isn't.=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?Considering this past posts:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe= off&selm=b96453c2.0311040110.73ab30fe%40posting.google.comhttp ://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&selm= b96453c2.0308160413.7f44ff58%40posting.google.comAnd particulary this one:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe= off&selm=b96453c2.0311121133.79790fc6%40posting.google.comIs safe to assume that this poster is just pulling our leg?Arraitz----------------------------------------------- That's free enterprise, friends: freedom to gamble, freedom to lose.And the great thing -- the truly democratic thing about it -- is thatyou don't even have to be a player to lose.-Barbara EhrenreichI have to confess, this is homework, so I am not looking for a full> answer, just some hints where to look.> I'm taking a math history class and the assignment is to write a paper> on whether Cantor was just misinformed and ultimately innocent despite> the harm he did mathematics... or whether he intentionally strove to> murder mathematics.> So far I've read into his biography and I am leaning toward the 2nd> case, since when he was young he was sexually abused by a math> teacher. That leads me to believe he had a big grudge against> mathematics in general.> If I adopt that route, one of my main arguments in this paper will be> that it would be impossible for anyone to be SO wrong about infinite> numbers purely be coincidence. I mean, he'd have to be really tripping> on some hard stuff if we're to believe he REALLY believed the nonsense> he published about infinite numbers.I'm sure most of you have written a paper to this effect some time or> other in one of your mathematics history classes :-)Nathaniel Deeth> Age 11> Math Genious> Welcome back, Nathaniel. You've been an 11-year-old math genius for> some years now, so I presume that you were born on Feb 29. I am too> lazy (see previous posts) to figure out when you were born using this> information.The truth is that Cantor intentionally strove to murder mathematics. > He had been laughed at by mathematicians because, even as a child, he> had a lousy singing voice in spite of the fact that his last name was> Cantor. This rankled all the more as he was not Jewish. His family> had converted some years before.He was also utterly incensed by the fools who insisted that .999999...> was acually equal to 1.0, a pseudo-fact that any fool can see is> false. He determined to go them one better. He started out trying to> show that it was not possible to represent every number as a sum of at> most n cubes (for any n), but he found the effort wearing, so he moved> on to other pursuits. He published several proofs of Goldbach's> conjecture and Fermat's last theorem using only the mathematics of> Fermat's and Goldbach's day, but these efforts were met with scorn and> derision. Finally he decided to get even. He invented a confusing> and mystical doctrine of the infinite that has devolved into chaos and> anarchy, just as had planned. He spent the remainder of his days> addicted to zornication, and he finally died insane as a result.Achava=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar? <1489d39f.0402260039.25b22eec@posting.google.com> Discussion, linux)> Is safe to assume that this poster is just pulling our leg?Duh.What's remarkable is the number of respondents who point out(repeatedly) that Nathan the Great should be considerably older than11 by now. Personally, I always find a joke much funnier when it'sbeen patiently explained to me. Repeatedly.-- you, but just remember, I'm the guy who proved Fermat's Last Theoremin just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?Nathaniel Deeth> Age 11> Math GeniousIf you are really a math genius, you would know that 11+5 = 16.Whatever capability Deeth may declare himself to have in mathematics, he wisely does not claim for his spelling.=== === Subject: : Re: The coin problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QGli023681; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1QGTfi21952 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1QGTfg17948;So I'm allowed to specify p=prob(heads) to be, say, a rootof p^4 + (1-p)^4 = 1/3,or p=0.2421306902?Then I will hazard a guess of yes. :)Don Coppersmith>Suppose you need to draw a prise between 3 people with equal chance to>all. Can it be done using one unfair coin (that is the prob. of heads>and tails is to your specification, not 1/2)?----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>---------------------------------------------------------- > http://www.usenet.com=== === Subject: : Re: The coin problem>So I'm allowed to specify p=prob(heads) to be, say, a root>of p^4 + (1-p)^4 = 1/3,>or p=0.2421306902?Let's see:p^3 * (1-p) = a ~= 0.010758p^2 * (1-p)^2 = b ~= 0.033674p * (1-p)^3 = c ~= 0.1053984a + 6b + 4c = 2/3 and 2a + 3b + 2c = 1/3. So make it four tosses andlet the events be assigned as:Player 1: {E_1111, E_2222}Player 2: {E_1112, E_1121, E_1122, E_1212, E_1221, E_1222, E_2122}Player 3: {E_1211, E_2111, E_2112, E_2121, E_2211, E_2212, E_2221}Then each player will have 1/3 odds. Neat.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Dumb logarithm questionGiven an exponential relationship:X^Y=ZY can be found from X and Z.logX(Z)=YHow do you find X given Y and Z?=== === Subject: : Re: Dumb logarithm questioncaden@sdsu.edu asks this:>How do you find X given Y and Z?>Given an exponential relationship:>X^Y=ZChange-of-base formula. G C=== === Subject: : Re: Dumb logarithm question> Given an exponential relationship:> X^Y=ZY can be found from X and Z.> logX(Z)=YHow do you find X given Y and Z?> X^Y=Z so Z^(1/Y)=X=== === Subject: : Re: Dumb logarithm questionX=Z^(1/Y)egX^2=ZX=Z^(1/2)=sqrt(Z)> Given an exponential relationship:> X^Y=Z> Y can be found from X and Z.> logX(Z)=Y> How do you find X given Y and Z?=== === Subject: : Re: Dumb logarithm question charset=Windows-1252X-ID: VxwejEZJYecz2paRhkHNHh2WKMZmdFt2z0R3uucg8M3B6g3OjVpd4NX=Z^(1/Y )> Does that also work for complex numbers?I mean if X, Y, Z are from |C.Karl=== === Subject: : Re: Dumb logarithm question> Does that also work for complex numbers?I mean if X, Y, Z are from |C.In that case, you have to remember that you aren't dealing withsingle-valued functions.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W=== === Subject: : Re: convolution of measures>>Suppose M and N are distributions (measures), and it's given that>>M * N = d_1, >>where * indicates convolution and d_1 indicates the delta measure>>concentrated at 1. How do you show that either M = d_c or N = d_c for>>some constant c? This should probably follow from some really simple>>manipulations with convolutions, but I lamentably don't see how to do>>it.>Here is a solution which works for finite measures; perhaps it can be >modified by approximation to all measures?>The problem is equivalent to showing that if X and Y are independent >random variables and X + Y is almost surely 1, then either X or >Y is almost surely constant. Actually, they both are, since>Var(X + Y) = Var(X) + Var(Y) = 0.If X+Y is almost surely 1, then if X <= c with positive probability, Y >= 1-c with probability 1. The rest ofthe proof can be completed from this.For arbitrary measures, as the total measure of the convolution must be the product of the total measures,one can reduce it to probabilities.For Schwarz distributions, the theorem is false; positivityof both is required.-- 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 Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Mathematics library for generating labeled trees? I am looking for a C/C++/Java library that would generate the labeledtrees for a given number of nodes. I could work on my own library,but I was wondering if there is already libraries that do these?=== === Subject: : about neglecting small terms in calculushi all,the question may appear vulguar or stupid but let me assure that itsmy genuine confusion or doubt.calculus is nothing but the rate of change of a function with thevariable when the variable approaches to zero.and while calculating itwe often eliminate the smaller term..eg. let y=x**3;the y + dy = x**3 + 3 x **2 dx + 3x dx**2 + dx **3so we neglect dx**2 and dx **3 termhence dy/dx = 3x**2.i understand it mechanically but i am not convinced why to neglect theabove two terms however small they may be..suppose the above function represents the rate of sperm depositioninside a woman's clitoris, any of the above neglected quantities iscapable of producing a fertile egg inside.why to neglect the term???regds,yogesh=== === Subject: : Re: about neglecting small terms in calculus> hi all,> the question may appear vulguar or stupid but let me assure that its> my genuine confusion or doubt.> calculus is nothing but the rate of change of a function with the> variable when the variable approaches to zero.and while calculating it> we often eliminate the smaller term..> eg. let y=x**3;> the > y + dy = x**3 + 3 x **2 dx + 3x dx**2 + dx **3> so we neglect dx**2 and dx **3 term> hence dy/dx = 3x**2.> i understand it mechanically but i am not convinced why to neglect the> above two terms however small they may be..> suppose the above function represents the rate of sperm deposition> inside a woman's clitoris, any of the above neglected quantities is> capable of producing a fertile egg inside.> why to neglect the term???> First, I'd suggest that you review mammalian reproductive biology.Second, review the limit definition of the derivative.Rick=== === Subject: : Re: about neglecting small terms in calculushi all,> the question may appear vulguar or stupid but let me assure that its> my genuine confusion or doubt.> calculus is nothing but the rate of change of a function with the> variable when the variable approaches to zero.and while calculating it> we often eliminate the smaller term..> eg. let y=x**3;> the> y + dy = x**3 + 3 x **2 dx + 3x dx**2 + dx **3> so we neglect dx**2 and dx **3 term> hence dy/dx = 3x**2.This is not the way to prove that d(x^3)/dx = 3*x^2. _Either_ useepsilon-delta techniques, or learn how to use infinitesimals. Nothinggets arbitrarily neglected in either approach.-- G.C.=== === Subject: : Re: Interesting problem>perhaps that isn't what is meant by multiplicative closure - that>would seem to be ideal closure, but I believe it can be extended to>show the answer. >If ab is in B, and 1/b is in B, then so is a#contradiction assuming a in A and ab in B, that is...>so 1/b is in A. Now wlog 1 is in B, and hence so is 2, thus 1/2 is in>A, hence 1/2+1/2=1 is in A# No, you could only say thus 1/2 is in A if there was some a in A such > that 2a is in B. Of course that can't happen. Why can't 1/2, 1, 2 all> be in B? In fact if 1 is in B, then all positive rationals are in B. > Moreover, all rational powers of rationals are in B.> But there are lots of other numbers, e.g. perhaps some transcendentals, > that could be in A. > There will have to be _lots_ of other numbers in A. If t is in A, notonly must we have that 1/t is in B, but for all b in B, we must havethat t/b is in A.We've only been considering multiplication; but we also must have thatfor all b in B with b < t, t-b in A.I'm still not convinced that a non-empty A can exist...Cheers - Chas=== === Subject: : Re: Interesting problem>I'm still not convinced that a non-empty A can exist...Neither am I. But I'm also not convinced that it can't exist.Note that if you only had to worry about addition, not multiplication,there would be examples: e.g. take an additive but nonlinear function f on the reals (f(x+y) = f(x)+f(y) for all x,y in R, but not f(x) = cx.Such exotic beasts are consequences of the Axiom of Choice (e.g.using a Hamel basis for the reals over the rationals). Thentake A = {x > 0: f(x) >= 0} and B = {x > 0: f(x) < 0}. However,these won't satisfy the multiplicative requirements.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: The coin problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QJG0Z06413; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1QJF2i06317 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1QJF1a28786;If there are n people, select k satisfying 2^k geq (k+1)(n-2)n (so k grows like 2*log_2(n)).Define Bin(k,i)=k!/i!(k-i)!,and define a(i),b(i) by Bin(k,i)=a(i)+(n-1)*b(i)with 0 leq a(i) leq n-2.Player 1 gets a(i) instances of (i heads, k-i tails) for each i.Each other player gets b(i) instances for each i.For p between 0 and 1 define f(p) = sum_i a(i)p^i (1-p)^{k-i}.If p is the probability of heads, f(p)=1/n,since that will be Player 1's share.Clearly f(0)=1,and f(1/2) = sum_i a(i)/2^k leq (k+1)(n-2)/2^k leq 1/n,so there is some p between 0 and 1/2 for which f(p)=1/n.For this value of p, player 1 gets his fair share, and we can see that the others do as well.Don Coppersmith>So I'm allowed to specify p=prob(heads) to be, say, a root>> of p^4 + (1-p)^4 = 1/3,>> or p=0.2421306902?>> Then I will hazard a guess of yes. :)>>> Don Coppersmith>Ok, you cracked this one.>Next one: show that it can be done with one coin for any number n of>people and find a good upper bound on the minimal number of required>tosses.----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>---------------------------------------------------------- > http://www.usenet.com=== === Subject: : surreal numbersI'm having a few problems with the surreal numbers. My main problem isthat in the process of defining the surreal numbers you are using thenatural numbers as the day that a number was made, or the ordinalnumbers if you complete the process. So you are assuming the existenceof the ordinals to begin with. How do you get around this?=== === Subject: : Re: surreal numbersOriginator: dmoews@ccrwest.org (David Moews)|I'm having a few problems with the surreal numbers. My main problem is|that in the process of defining the surreal numbers you are using the|natural numbers as the day that a number was made, or the ordinal|numbers if you complete the process. So you are assuming the existence|of the ordinals to begin with. How do you get around this?It is not necessary to consider birthdays at all. Once you have proved that 0, 1, 2, ... are all surreal numbers, it follows immediately thatomega is a surreal number. In a similar way, you can go on to construct all the ordinals.-- David Moews dmoews@xraysgi.ims.uconn.edu=== === Subject: : What does Inverse of a Matrix mean?I have been using matrix formulas for calculating regression coefficientsetc.It is simple to understand what we are doing when we add, multiply,transpose a matrix (in terms of the underlying data). However, what exactlyare we doing when we take the inverse of a matrix? Can someone explain insimple terms - if possible with a practical example.thanks.=== === Subject: : Re: What does Inverse of a Matrix mean?> I have been using matrix formulas for calculating regression coefficients> etc.> It is simple to understand what we are doing when we add, multiply,> transpose a matrix (in terms of the underlying data). However, what exactly> are we doing when we take the inverse of a matrix? Can someone explain in> simple terms - if possible with a practical example.thanks.An identity (necessarily square) matrix of order n, has 1's on its principal diagonal (upper left to lower right) and zeros elsewhere.If I is an identity matrix and A is any matrix of the same size, then I*A = A*I = I, where * indicates the usual matrix multiplication.Not all square matrices have inverses, but the inverse of a (necessarily square) matrix, A, of any size, is a square matrix, B of the same size such that A*B = B*A = I.The square 2 by 2 matrix, A = [[a b][c d]], will have an inverse if and only the determinant e = ad - bc is not zero, and that inverse will be then be B = [[ d/e -b/e][ -c/e a/e]], as you may easily verify by doing the multiplications A*B and B*A.The form of the inverses, when they exist, for larger square matrices is more complicated, but they will exist if and only if the determinant of the matrix is not zero.If you know, or can learn, about the Gauss-Jordan row reduction method of solving a system of linear equations, one can find the inverse of a matrix A by the row reduction of the augmented matrix [A|I] to the form [I|B], where I is the appropriate identity matrix and B will be the desired inverse. If the process cannot be completed ( and you have not made any errors in executing it), then the inverse does not exist.=== === Subject: : Re: What does Inverse of a Matrix mean?> I have been using matrix formulas for calculating regression coefficients> etc.> It is simple to understand what we are doing when we add, multiply,> transpose a matrix (in terms of the underlying data). However, what exactly> are we doing when we take the inverse of a matrix? Can someone explain in> simple terms - if possible with a practical example. thanks.An identity (necessarily square) matrix of order n, has 1's on its> principal diagonal (upper left to lower right) and zeros elsewhere.If I is an identity matrix and A is any matrix of the same size, then> I*A = A*I = I, where * indicates the usual matrix multiplication. ^ meaning A...-- G.C.=== === Subject: : Re: What does Inverse of a Matrix mean?I have been using matrix formulas for calculating regression coefficients> etc.> It is simple to understand what we are doing when we add, multiply,> transpose a matrix (in terms of the underlying data). However, what exactly> are we doing when we take the inverse of a matrix? Can someone explain in> simple terms - if possible with a practical example.If I is the n-by-n matrix with 1's down the main diagonal and 0'severywhere else, and if A is an n-by-n matrix, then an n-by-n matrix Bsuch that A*B = B*A = Iis the inverse of A.Example A = [ 2 1 ] [ 5 3 ]inverse of A = [ 3 -1 ]. [ -5 2 ]-- G.C.=== === Subject: : Re: Metric Engineering, Dark Energy, Quintessence, Phantom Energy> re: http://qedcorp.com/destiny/> www.stardrive.org/starship.shtmlwww.geocities.com/zcphysicsms/ chap12.htmof his metric engineering section> not understanding the w = -1 property of zero point energy and the > (energy density)(1 + 3w) active source in Einstein's GR from the > /zpfguv term of the effective exotic vacuum stress-dark energy/matter > tensor field. Hence he gets it wrong saying positive zero point energy > density gravitates. In fact it anti-gravitates from the domination of > the negative exotic vacuum quantum pressure. However, this error is > easily fixed.Note w = 0 for ordinary matter at slow speeds.w = + 1/3 for EM radiation like the CDMw = - 1 for zero point vacuum fluctuationsw = -1/3 to > -1 for quintessent fieldsw < - 1 for phantom energy fieldsMy vacuum coherence theory makes the strong Popper falsifiable > prediction w = -1 on the nose, exactly. But it doesn't matter, since Popper was a osopher. The people with poitive intellect has only referred to dark energy, not dark matter. Since dark matter was invented by an idiot called Newton, not Einstein.=== === Subject: : Re: Metric Engineering, Dark Energy, Quintessence, Phantom EnergyJack -- the Jack that we knowfrom his interminable, ne'er-replying abusenet leavings -- isin the midstream of the Copenhagen Scholastic,without a life-preserver ... unless you're talkingabout the Ring over the Tutti-Fruiti Flavor! seriously,I have to wonder about Popperian falsifiabilitation,weakly, strongly, ultimately, or vacuuous. this would also beof interest to George Soros, and other Friends of Trickier Dick-- see my sig! on the other hand,Jack might rescue Universe from Eisntein's Constant,X = it's running away from you as fast as it possibly can,thus *requiring* the ubiquitous talents of the Leagueof Extraordinary Metric Engineers ... just a thought. > the negative exotic vacuum quantum pressure. However, this error is > easily fixed. > My vacuum coherence theory makes the strong Popper falsifiable > prediction w = -1 on the nose, exactly.dark matter, i.e. dark matter detectors forever silent as a matter of > principle except for false positive errors.--Give Earth a Trickier Dick Cheeny -- OBNOXICO, out of office?http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanac=== === Subject: : Re: Metric Engineering, Dark Energy, Quintessence, Phantom EnergyIn message , Jack >TYPO>w = + 1/3 for EM radiation like the CDM>should be>w = + 1/3 for EM radiation like the CBR (Cosmic Background Radiation)You don't like Cadbury's Dairy Milk?-- Richard Herring=== === Subject: : Re: Bounds on Moments> Suppose x is a non-negative random variable. Given that the first>> moment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What are>> the bounds upon m3=E(x^3)?Why should we expect such a bound? For example, let f(x) = c/(1+x^4), where >c is chosen so that the integral of f over (0,oo) is 1. Then for the >probability density function f we have E(x^1) and E(x^2) < oo, but E(x^3) = >oo.No upper bound, but there is a lower bound. For example, for any > p >= 0 we have Q(t) = t (t-p)^2 >= 0 for all t >= 0, so> E[Q(x)] = m3 - 2 p m2 + p^2 m1 >= 0, i.e. m3 >= 2 p m2 - p^2 m1.> In particular, take p = m2/m1 to get m3 >= m2^2/m1.> Conversely, consider a distribution with mass m1^2/m2 at > m2/m1 and 1-m1^2/m2 at 0 (noting that 0 < m1^2/m2 <= 1):> we have E[x] = m1, E[x^2] = m2 and E[x^3] = m2^2/m1, so that> bound is best possible.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2As suggested, after examining your previous post on this matter fromfully understand.To find the minimum m3 given {m2, m1}, I need solve for the smallestm3 such that the determinant of some moment matrix M (a Hankelmatrix?) is non-negative?What form will the relevant matrix have? Is it[m0 m1][m1 m2]or[m0 m1 m2][m1 m2 m3][m2 m3 m4]How can either be useful, since one doesn't include m3 and the otherhas m4? I must have somthing wrong.=== === Subject: : Re: Bounds on Moments> Suppose x is a non-negative random variable. Given that the first> moment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What are> the bounds upon m3=E(x^3)?>As suggested, after examining your previous post on this matter from>fully understand.>To find the minimum m3 given {m2, m1}, I need solve for the smallest>m3 such that the determinant of some moment matrix M (a Hankel>matrix?) is non-negative?The previous post was about measures on the real line, whereas here(because you're talking about nonnegative random variables) the measures are restricted to be on the nonnegative reals. There is no lower boundon m_3 in terms of m1 and m2 without some restriction.The Stieltjes moment condition says that the sequence {m_0, m_1, m_2, ...}are the moments of a positive measure with support on [0,infinity) iffthe Hankel matrices [ m_0 m_1 ... m_k ] [ m_1 m_2 ... m_k ][ m_1 m_2 ... m_{k+1} ] [ m_2 m_3 ... m_{k+1} ][ ... ... ] [ ... ... ][ m_k m_{k+1} m_{2k} ] and [ m_k m_{k+1} m_{2k-1}]are all positive semidefinite. So in the case of m_3, you want [ m_1 m_2 ]det [ m_2 m_3 ] >= 0.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation): Umberto Eco is never about self indulgent auto adoration, much as he: enjoys his little games. If you were his characters he would give you: some mean punishment, as he did to Belbo and Diotallevi for messin': around where they shouldn't been a messin'. You particularly: Galathaea, remind me of that kraut neurotic bitch at the umbanda: ceremony.You should see me then in the industrial-punk outfits I used to wear in mymore angst-filled youth.: Anyway a totally different question. Are you very attractive? I mean,: is there a REAL reason why these creeps start drooling all around the: place every time you say anything?Since I am more interested in intelligence and creativity, you would noteven get a first date if that gives you any idea as to league comparisons.Nobody has drooled over me here. Some people have asked questions anddiscussions were started. Others have been rude, and flames were started.Just another day on the usenet.: I'm still waiting for your explanation of how the proof of halting the: problem relates to heyting particularly non boolean structures. I hope: it's something more than the similitude between halting and: heytingI pointed to the inconsistency used in the proof to show that one could notalways solve a halting question as a true / false in any finite time. You,like others, wanted to talk about the metamodel instead of the model. Ifyou are really interested in the topic of computation and Heytingstructures, you should really read the reference on the Curry-Howardisomorphism I gave in my more focus on the constructive discussion thread(its available online), or if you want me to explain things you can stop theinsults.-- ===-=-=-=-=-=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation)[snippa lotta selfadoring prose gooey astermakergrown man cry]> That was really beautiful. Thank you for that.Hey girls, I sure would like you to move this conversation to the> womens loo or some such more appropiate place. I mean, get real!Yeah ... like I'm going to be completely crushed being insulted by> somebody supposedly named Guenther von Knakspott.> You're not? Darn! I was counting on the brutal finality of onedecisive blow. Now it's me who is crushed. Back to square 1 for me,doh!=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) <40325A2C.21936CE9@hate.spam.net> <103513r3gafocff@corp.supernews.com> <40328829.1E5C341D@hate.spam.net> <1035esdt30h4d38@corp.supernews.com> <1039dr5kpjloif1@corp.supernews.com> <+No$awZ0HMNAFw0n@baesystems.com> <2a0cceff.0402210820.773d478d@posting.google.com> In message , Edward >> In message <2a0cceff.0402210820.773d478d@posting.google.com>, Edward>...>>There: I've used ontology in a sentence, and it wasn't so bad, was>>it? ;-)>>(Richard Herring vigorously spits and loudly calls for a pint).>> Oh, I'd call for a pint whatever you said ;-)>> I think what I mostly don't like about the word is that it evokes>> Anselm's argument for the existence of God. One false step and you'll>> believe your model really exists.>Oh well ... I am ignorant of Anselm's argument, so I can't fall into>_that_ trap, can I? ;-)I'm sure a Google search on ontological argument will tell you all you need. Then you can play spot the fallacy to your heart's content.>...>> I think you're really talking about epistemology.>Perhaps. But then I suppose I'd have to own to the opinion that any>meaning of ontology beyond one given to it by our episto-thingie was>meaningless.>(There ... I've said, if you strategically snip out some intervening>words, that ontology ... was meaningless. _That_ should make you>happy. ;-)Depends on your hermeneutic.-- Richard Herring=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) Shouldn't that be G.9fnther von Arschloch?> You particularly Galathaea, remind me of that> kraut neurotic bitch at the umbanda ceremony. > How darest thou, thou Teutonic blackguard, insult the fair damsel> in such blatantly _ad puellam_ terms? My Gallic blood boils and> gallantry makes me grow wings so that I may fly to her rescue. > Anyway a totally different question. Are you very attractive?> Er... on second thoughts... we might have here a pot-bellied old> geezer with snot in his moustache, half his breakfast still on> his shirt front (don't egg yolks make for a lovely colour?)...That sounds like G.9fnther.> you take first go, mate. We Frogs have been at peace with you> for more than half a century now, let's not waste that over a> piece of stale tail.> I mean, is there a REAL reason why these creeps start> drooling all around the place every time you say anything? > It takes all kinds, all kinds. De gustibus non disputandum.> If the editor of Maledicta was reading this, I'd feel justified> in posting here a few URLs of porn sites that would rid> your stomach of its contents forthwith and without delay.> Without his blessing, I shall refrain.Good man, mon cher M. Guy. With an emetic like G.9fnther around, whoneeds porn to puke? > In my view, satire and ridicule is the most effective way> of dealing with his, her, or its, kind. And satyre too.Vive l'esprit fran.8dais! > Would you talk metaphysics with a broken 78rpm record?> Or even... strawberry-pie recipes, or baby-bum wiping> techniques? I don't.Mais pourquoi parlez-vous avec un trou du cul comme G.9fnther?-- Reinhold (Rey) AmanM A L E D I C T AP.O. Box 14123Santa Rosa, CA 95402, USAhttp://www.sonic.net/maledicta/=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation)> Shouldn't that be G.9fnther von Arschloch?Wow, to be insulted by THE authority in insults himself ! I feelhonoured, really. It's not a great deal of course, I just got twoassholes and one emetic but it's a start after all. How does one getpromoted to something more fancy, more sophisticated?> You particularly Galathaea, remind me of that> kraut neurotic bitch at the umbanda ceremony.> How darest thou, thou Teutonic blackguard, insult the fair damsel> in such blatantly _ad puellam_ terms? My Gallic blood boils and> gallantry makes me grow wings so that I may fly to her rescue.> Anyway a totally different question. Are you very attractive?> Er... on second thoughts... we might have here a pot-bellied old> geezer with snot in his moustache, half his breakfast still on> his shirt front (don't egg yolks make for a lovely colour?)...That sounds like G.9fnther.No buddy no. I don't mean to start a flame war against you, you arethe expert in this things, but take a look at yourself at:http://www.sonic.net/maledicta/aman.htmland picture yourself in a ragged yellowed t-shirt and with a snottymoustache, it fits the description perfectly.> In my view, satire and ridicule is the most effective way> of dealing with his, her, or its, kind. And satyre too.Vive l'esprit fran.8dais!> Would you talk metaphysics with a broken 78rpm record?> Or even... strawberry-pie recipes, or baby-bum wiping> techniques? I don't.Mais pourquoi parlez-vous avec un trou du cul comme G.9fnther?There you go ladies and gentlemen, thats your run of the mill krautnazi schizophrenia if you mark my meaning. One second he iscelebrating the libertarian french spirit, the next he is censoringhis french chum and questioning his actions.What was that thing about the first amendment on your homepage?Since I don't even understand what bug bit you, I respectfully bid myfarewell for now.regards.=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) > Mais pourquoi parlez-vous avec un trou du cul comme G.9fnther?Des fois, je me fais tellement chier que je lis m.90metoutes les conneries qui passent ici. Des fois, jeme fais tellement, tellement chier... que j'y r.8eponds!But hey, I also occasionally find TRUE LOVE. Galathaea!*smooooooooch* sweet'art!Have one of your favourite drinks in remembrance ofour late friend D.C.L.=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation): But hey, I also occasionally find TRUE LOVE. Galathaea!: *smooooooooch* sweet'art!Mais Jacques! Je t'aime aussi!I knew you were a softy...-- ===-=-=-=-=-=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation)Forget it Jacques, I'm not dueling you over the yolk dripping fatsoeeeeeK!> In my view, satire and ridicule is the most effective way> of dealing with his, her, or its, kind. And satyre too.Would you talk metaphysics with a broken 78rpm record? > Or even... strawberry-pie recipes, or baby-bum wiping > techniques? I don't.It sounds like fun to me, but don't forget that subtlety has beenabsent from the kraut mind for decades. I believe it was forbidden bythe nazis and no one ever cared to restore it to legalityafterwards...=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues> -=-=-=-=-= the spectre continues =Now, with this long history of logical analyses of language, I always find> it strange that there are comments like:Attention being devoted here entirely to the classical two-valued theories> of truth-functions, quantifiers, and identity, with syntax, semantics, and> pragmatics built upon their basis, there will be no concern with alternative> forms of logic, so-called three-valued (or, more generally, n-valued)> logics, modal logics, intuitionist logics, and the like. The view is that> whatever is valuable in these alternatives can be achieved more readily> within the classical framework by suitable extensions. But, it is known that that supposition is not only wrong, it is dead wrong. Since classical logic is *not* based on *truth-functions*. It's based on *causality*. And since the readiliness that osphers achieve that end-framework it based not on logic, but on another retarded invention of theirs called over-priced tuition, it is also readily discarded by most up-to-date people in intelligence as a mere knee-jerk historical extension of their's from suburbia Cairo & Boston into Greenwick Village and their San Franciso head, highway rest-stop known as Southern California. in the introduction to Semiotics and linguistic structure by R. M. Martin,> where such a representation is not faithful to actual usage or expression.> Now, I just recently posed some questions on the newsgroups concerning> nonclassical logic, and certain linguists and physicists were actually quite> confrontational about the idea of educating about these nonclassical logics.> Expressibility was always proffered as a reason, even though no one can even> claim that boolean structures possess a uniqueness in this regards (lambda> calulus is rich enough). Since Physics doesn't even have anything to do with *Classical Logic*, their moron opinions about modern logic are even more irrelevent than their dip Black Holes are to the subject.=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues: > Oh, did you miss the thread on modality in language as well?:: What, Andrew Patterson's nonsense? That was (a) independent: of your posturing and (b) obviously received largely with: indifference.You see, that is the great evil laying at the heart of all your anger withmy post. There are people out there besides you, Brian, with many variedinterests. As long as we stay on topic to a particular group, everyoneshould have the right to post their interests and questions. The usenetdoes not follow office politics. Newcomers have all the same rights asthose who have been posting for years.: > Expense?:: Yes, expense. You are, for example, directly responsible: for cluttering sci.lang with off-topic mathematics and: complaints about the lack of physical content in your posts: from sci.physics.One collection of people upset that I am posting mathematics, another upsetthat there is not enough. Maybe it is these two groups that should bearguing between each other and not be including me at all, but I think themore prudent action would be for those who don't find content they areinterested in to just skip my threads.: > I took the time to write several long posts concerning a topic: > relevant to all newsgroups posted to.:: In your opinion. The relevance is not apparent to many of: those better qualified than you to judge.Are we to start wearing qualification badges now? I'll post all of mine andeveryone else can post theirs and we can walk around comparing lengths...: [...]:: > As for sci.lang, obviously there are problems in understanding reasoningon: > topological trees (and linguistic phylogeny) that have been evidenced: > several times recently in that forum.:: Jacques erred in calling his three trees topologically: equivalent (though in fact two of them are -- as unlabelled: trees), but I doubt that he was trying very hard: you've: offered nothing to suggest that you should be taken any more: seriously on the subject of linguistic phylogeny than, say,: Cavalli-Sforza, who is a linguistic ignoramus.I know... I know...The one making the errors is extremely intelligent, and the one who pointsthem out an ignoramus. Isn't that always the case?: > Plus the whole cognitive origins of: > language, semiotics and natural language models, etc. features of my: > exposition that I've been willing to discuss in depth make yourstatement: > patently false.:: Bollocks. I've not seen you offer anything substantive on: any of those topics, which with the possible exception of: the first are in any case not topics of great interest to: most of the regulars here, judging by the eight or so years: that I've been reading the group.I haven't touched on Barry Smith's work in mereotopology yet (though brought it up at one point), and haven't explored Gardenfors as deeply as Iwould like, but I have given a list of references concerning cognitiveorigins in linguistics that explore nonclassical logics. The very firstanticlassicalist explored nonclassical negation in natural language invarious schools of linguistics. This received a nice response with severalof the questions still being explored. And then in this anticlassicalistthat this response falls under, I explore comments in a book on semioticswhich make explicit a certain classical bias that I am trying to explore.various discussions in a manner that would have some clarity yet explorethings deeper. Besides the more mathematical forays I must take intoKripke-Joyal semantics, the ell-Benabou language, convexity, andoperationalist approaches to quantum mechanics, I also hope to explore theformation of language in children (with the evidences of early logics),questions of deixis and identity, indexicals and model theoretic semantics,and other areas of natural linguistics that involve the introduction ofnonclassical logics in their interpretation. I also don't mind exploringthe logic of topologies and proper reasoning about phylogenies, since thatseems to be a particular trouble encountered by a couple of the louderparticipants on sci.lang.I would agree, though, that _you_ haven't contributed anything to theexploration of these topics.[...]: You still don't get it, do you? What's on topic is: determined by the group itself, especially when it has a: core of regulars. If you'd paid the slightest attention,: you'd have recognized that sci.lang runs to historical: linguistics (about which you've demonstrated considerable: ignorance) and sociolinguistics, with occasional forays into: syntax. (Oh, and food.)You still don't get it, do you? What's on topic is established when thegroup gets created and is done to help people find discussions in topicsthey are interested in. That you cast off the posts of those like AndrewPatterson who do not share your same interests yet are quite on topic isevidence of your need to control what is not yours to control. That youcome into the discussion I have started and to spew forth your bile andoffer no constructive talk is evidence that the bias infecting you has quitea hold.I never forced you to read the threads I started. I don't go aroundstarting fights with anyone, though I will play out any manifestations ofthe male alpha problem as they appear. You start attacks, mister Scott, andthat puts you in error here.[...]-- ===-=-=-=-=-=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues[...]|> Even on the level of predicate|> calculus, is anybody seriously using something in between intuitionist|> logic and classical logic?||I want to say yes to this...Can you name names? Remember that I was referring to actual doing ofmathematics *in* such a logic, not just doing metamathematics of it.[...]|> It seems to me that we need|> both to know more constructive mathematics, and to have better reasons for|> what we think it's pros and cons are. The lack of experience leaves our|> general idea of the relationship between constructive and classical|> mathematics resting on too little.||And a lack of historical perspective. There are entire lines of inquiry|that are ignored.Excuse me if I word-wrap your text a bit. What's an example of an entireline of enquiry being ignored? I don't doubt that they exist, but I don'tknow what kind of thing you have in mind.|The general sense of|mathematics as some fixed collection of truths does not reflect the|various debates.It's not clear to me that thinking of it as a fixed collection of truthsis a problem per se. I can see that ignoring all the difficulties ofinterpretation or identification of truths can cause problems.|I just ran across a reference on John Baez' home page|to a osopher of mathematics calling for a new foundational approach|that reflects what mathematicians do. There is no need for that.I don't see the connection between this remark and the previous one.[...]|Where you ask about someone using something between intuitionist and|classical logic, I need to redirect to semantics. Tarski initiated|algebraic semantics with cylindrical algebras. When you pursue this line|of inquiry, you get to Halmos' polyadic algebras and finally to the|representation theorem I recently found in|Balbes and Dwinger,||If L is a closure algebra, then L^o is a Heyting|algebra under the partial ordering of L and is also|a D01-subalgebra of L||More remarkable is the fact that every Heyting|algebra can be represented this wayI doubt this is constructively valid, is it?| The closure algebras are Boolean algebras,With more structure.|A Boolean algebra L with an additive closure|operator ^c in which 0^c=0 is called a closure|algebra. An element (a e L) is called open if|(a-bar e L^c). The set of open elements is|denoted by L^o||So, there is a close relationship between Heyting algebras and Boolean|algebras that seems to be ignored.Are you sure you're correctly gauging what's being ignored? If theclosure operator is the familiar operation of topological closure, thenthe L^o are just the examples I mentioned already, the open sets oftopological spaces ordered by inclusion. This is familiar and also coversmany of the cases most often used by logicians.|There|is additional structure involved that relates the two. There are semantics|that do not seem to interest many in|questions as they apply in foundations.Well, you certainly can find Heyting algebras that aren't Boolean, andthat satisfy additional logical laws beyond intuitionist logic. Thatgives you a way of defining an intermediate logic. But then what do youplan to do with it once you have it?|Yet, if you are going to embed a Heyting algebra in such a way that|its specification induces a D01-subalgebra, one might ask if the|representation deserves to first be interpreted as a system rather than|in the pieces that can be syntactically represented in isolation. In|other words, is one man's D01-subalgebra another man's bilattice?Well, this kind of mixed structure is a lot like one of the logics inwhich there are both constructive and classical operators or quantifiers.It's all fine and all, but I think you need some idea of what you intendto do with it.|This is what I mean when I ask why anyone should consider a new attempt|at foundations. During my argument with George Greene this year I|learned a lot by looking for sources to understand his objections. In|large part, it seems as if bibliographies are partitioned. It is not at|all clear to me that there is any debate left. I really believe there|is enough information for a coherent integration of these questions.What questions? Really, you need to pose more definite questions.|You talk about pretending that the law of excluded middle is not true|while implicitly depending upon it.Usually it's pretty overt dependence on it, in the sense that no attemptis made to make the theory constructive.|I think I need to see a formal specification where the natural language|used is not depending on AND and OR in the sense of classical logic. The|best paper I have found thus far is from Jacek Malinowski analyzing|Strawson presupposition. Getting to classical logic from something that|is not classical logic simply does not seem to make sense. But, this is|so far removed from constructive mathematics, that someone who learns|Bishop or Markov with an interest in *mathematics* certainly has no|interest in the regress to presupposition via negation.I don't see how you think such a formal specification will help you.Formal specification of what, by the way? Why doesn't a formalspecification done constructively count, or would it?|And, then there is someone like me who stumbled on de Morgan fields and|de Morgan isomorphisms without any clue and without being able to explain|the skepticism in any coherent fashion.Skepticism about what?|The questions are so disjointed.Then you have to address them separately.It seemed for a moment like your thesis was going to be that there were,contrary to my impression, people doing mathematics in intermediatelogics (i.e., between intuitionist and classical), but now I can't tell.Give sticking to one thesis a try.|I mentioned the historical perspective just because of my own experience.|I have been all over the place. I do not have any particular|disagreement with Frege's ontology of number.I don't think Frege's ontology of number is a phrase that would beimmediately understood by many people.|But, it is not at all clear that that is|the concept of number applicable under algebraic semantics.I don't know what concept you mean.|I intuitively understood that term in the same|sense that others refer to descriptive set theory.You've lost me. I have some idea of what descriptive set theory is,an inkling of algebraic semantics, and have read a few introductorydescriptions of Frege's work. I don't see that it makes any sense tosay that you understand algebraic semantics to mean descriptiveset theory.|And, I have finally found a modern osopher who has|discussed the distinction between Frege's resolution of sematnical|applicability and the problem of descriptive applicability.||In my own researches, I can find particularly similar statements|concerning two fixed values in Dedekind's writings, Abraham Robinson's|note on threshold logic, and the papers I recently found from the Steklov|Institute of Mathematics in Russia. All of this work seems to relate to|the algebraic semantics.||And, just for kicks... I would love to be able to even talk to you about|the Jordan-Holder theorem. But,... :-) I started out thinking|classically and ended up defending intuitionism with which I did not even|agree from the standpoint of a Heyting algebra.What does it mean to say you agree with intuitionism from the standpointof a Heyting algebra?|I am not even close to thinking about this in terms of a chain of groups.|But, I have done a lot of diving even if it seems otherwise.||So, I started this part by saying that no one took my posts seriously.|I cannot say that I blame anyone for that. Nevertheless, the issues are|so spread out over so many topics, it is hard to be coherent until you|figure out what someone else knows or does not know.||:-)I don't really buy this complaint.It's one thing to be hoping that your audience (some of them) have heardof algebraic semantics, Frege, descriptive set theory, representationtheorems for distributive lattices, and Strawsonian presupposition. It'squite a different thing to pretend you're in doubt as to whether thethread you think you see running through all that stuff can be properlyconveyed to your audience by this kind of loose wandering around betweenthem.A rule of thumb is that if you can't express your issue as a thesisof some kind, it probably won't come off as a coherent thesis to youraudience.Keith Ramsay=== === Subject: : Re: 'erf' function in C> For large arguments one might want to switch to asymptotics.> I think, that is what Genz makes in his approx for cdf normal.I found the following algorithm (using Google)http://tigger.smu.edu.sg/software/mnp-stuff/stat.ubc.ca /pnorms2.cdeveloped by W.D. Cody (there is a full reference in the link). It worksfor the entire domain, and produces Phi(-37.5) = 4.60535e-308 withdouble precision. I don't have access to the paper by Cody, butaccording to the comments in the text, the algorithm is accurate to 18significant decimal places.-- Jesper Lund=== === Subject: : Re: Clean coordinates on a unit sphere> Hi - I'm a computer science student working in the field of vision and > graphics. I'm looking for a clean way to represent a _direction_ in 3d > space (this can also be imagined as a position on a unit sphere).I don't like the traditional methods of using a 3d vector or spherical > coordinates for the following reasons:- 3d Vector e.g. (x, y, z): This is of course very common - can be > rotated with matrices, quaternions... but it also describes length, > which I don't need.- Spherical coordinates e.g. (theta, phi): This is more appropriate as > it does not describe length. But it still appears ugly to me since > multiple coordinate pairs can describe the same point e.g. (x, pi/2).Is there a better way? I'm sure I'm not the first who asked this > question. It seems similar to the problem of creating a 2d map of earth, > so it's probably been thought about before... And since there doesn't > seem to be a _popular_ solution, it's probably hard - but that's what > makes it FUN !!!My ideal solution would map a unit sphere with a uniform density (if > that makes any sense to any one). Could a person use some type of > tesselation of a sphere with an arbitrary density???Any ideas are good ideas! Please keep the language simple - I'm not as > smart as you.> Nathandefined obviously, mostly because I'm not even sure what exactly I want. One way I was thinking about it was, that by increasing the parameter(s) with a constant step, and drawing the according points on the sphere, the resulting points should be uniformly distributed.A strange, possibly dumb question: Are at two or more parameters required? Is it possible to map the surface with a single parameter (or maybe 1.5 ?) Maybe real numbers are not appropriate for what I'm striving towards?Here's a _crazy_ idea that I had along the way (trying to map the sphere surface with single parameter):Imagine the south pole being represented by 0, the north pole being represented by 1. Now wind a path from one pole to the other around the sphere in a spiral shape. With an infinite tight winding, every point on the sphere is covered. The point corresponding to the parameter value of 0.5 is on the equator and so on. _Obviously_ we can now map all points on the sphere with a single parameter. Where's my mistake?Hope you have some fun with my insanity,Nathan=== === Subject: : Re: Clean coordinates on a unit sphere> Hi - I'm a computer science student working in the field of vision and> graphics. I'm looking for a clean way to represent a _direction_ in 3d> space (this can also be imagined as a position on a unit sphere). I don't like the traditional methods of using a 3d vector or spherical> coordinates for the following reasons: - 3d Vector e.g. (x, y, z): This is of course very common - can be> rotated with matrices, quaternions... but it also describes length,> which I don't need. - Spherical coordinates e.g. (theta, phi): This is more appropriate as> it does not describe length. But it still appears ugly to me since> multiple coordinate pairs can describe the same point e.g. (x, pi/2). Is there a better way? I'm sure I'm not the first who asked this> question. It seems similar to the problem of creating a 2d map of earth,> so it's probably been thought about before... And since there doesn't> seem to be a _popular_ solution, it's probably hard - but that's what> makes it FUN !!! My ideal solution would map a unit sphere with a uniform density (if> that makes any sense to any one). Could a person use some type of> tesselation of a sphere with an arbitrary density??? Any ideas are good ideas! Please keep the language simple - I'm not as> smart as you.> Nathan> defined obviously, mostly because I'm not even sure what exactly I want.> One way I was thinking about it was, that by increasing the parameter(s)> with a constant step, and drawing the according points on the sphere,> the resulting points should be uniformly distributed.> A strange, possibly dumb question: Are at two or more parameters> required? Is it possible to map the surface with a single parameter (or> maybe 1.5 ?) Maybe real numbers are not appropriate for what I'm> striving towards?> Here's a _crazy_ idea that I had along the way (trying to map the sphere> surface with single parameter):> Imagine the south pole being represented by 0, the north pole being> represented by 1. Now wind a path from one pole to the other around the> sphere in a spiral shape. With an infinite tight winding, every point on> the sphere is covered. The point corresponding to the parameter value of> 0.5 is on the equator and so on. _Obviously_ we can now map all points> on the sphere with a single parameter. Where's my mistake?> Hope you have some fun with my insanity,Your mistake is with your infinite tight winding idea. I don't think thatthe idea can be made rigorous enough to work. Either it goes north by ZEROfrom one turn to the next... which is not very useful. Or, it goes north bysome positive amount... and we miss some points.However, if you _really_ want to do it with just a single parameter... justdo it with two parameters somehow... and then interleave their decimaldigits. So that a point such as (2.345678..., 0.987654....) is mapped to thenumber 20.394857667584...-- Clive Toothhttp://www.clivetooth.dk=== === Subject: : Re: measure of stuck-togetherness>>So the earth's surface is about one-quarter land. That land could be>>distributed across the surface of the globe in different ways - I can>>imagine at one extreme, all the land stuck together in one roundish>>continent, and at the other extreme, all the land spread out evenly, in>>equally-spaced tiny islands. Is there a measure, preferably just one>>number, that captures these differences? If so, how is it calculated?>> How about: The mean great circle distance between two points uniformly and independently distributed over the land area.For the more general case, somehow add in:The total land area.The radius of the earth.-- ------------------------Mark Jeffrey Tilfordtilford@ugcs.caltech.edu=== === Subject: : re:measure of stuck-togethernessThat doesn't work. === === Subject: : Re: Factoring RSA And the Reversible Multiplier(comments appreciated)Btw, for all problems where m = p*q , where p and q are primes. The multipliercircuit and a comparator are perfectly reversible. This special case doestake advantage of the fact that p and q are primes. The reason being is pis not a prime, there will be more factors and more uncertainty ininformations. In a way, the number of prime factors is directly related tothe information uncertainty at the inputs.-suresh> The Lord of the Rain( Suresh __NoJunkMail kumar)>I have been looking into Edward Fredkin's Fredkin Gate built usingard>balls.>Here's something that came to my mind.>The RSA number to be factored m = p * q>p and q are primes.>[ stuff deleted ]The procedure would work best, if p != q. For this case notice thatone of>the number in (p,q) have to be greater than other. So, the MSB of one,is at>a higher index than the other. By forcing the MSBs of A and B to takevalues>1 or 0, we will force just one ordering of (p,q) to appear at theinput.[will explain later]. How does the idea sound? Not good. Fredkin gates have extra outputs beyond the values of theBoolean> functions they're implementing. To factor N-bit integers by reversing> a multiplier composed of Fredkin gates, you'd have to know the values of> (I believe) O(N^2) extra outputs. Since you only have 2N outputs (i.e.,> the product), you would have to do some kind of search on the extraoutputs> that is probably no easier (probably harder) than the search you have to> do to factor large integers.> The above comments are incorrect.> As was explained in the original Conservative Logic paper, it is> always possible to build a circuit of Fredkin gates or iard Ball> Model gates that has only the informationally required number of extra> outputs (or inputs). In the case of a circuit designed to factor the> product of 2 primes, the only informational uncertainty is which> output gets the larger factor. The design of the circuit can take> into account such cases as pq=1 pq, pq=-p -q, pq=-1 -pq etc. Of> course, if the circuit cannot take advantage of the fact that both> factors are primes, then the informational uncertainty very is much> greater. My comments are not merely theoretical as I long ago> programmed a large Mathematica system that implemented factoring by> means of a reversible multiplier. Because it actually simulated the> reversible digital logic, it was very slow. My conclusion was that,> for the algorithm I implemented, a hardware version was not quite> competitive with the best existing factoring algorithms. However, I> still think that the general approach could be successful.> The circuit had the property that if you put 35 into the outputs and 5> into one of the inputs, the other input became 7. (A Fredkin gate> really has no inputs or outputs, it is perfectly symmetrical and works> in either direction; the same is true of any well formed circuit of> Fredkin gates.)> Kumar's idea is not quite right either. If p>q, a number of the most> signifigant bits of both can still be the same. However, as Kumar> suggests, one can take advantage of what knowledge one has about the> possibilities related to MSB's and also LSB's.> The LSB's of the product also enforce constraints on the LSB's of the> factors. The problem is that such ideas are only useful for a few of> the leading or trailing bits. It's all the bits in the middle that> pose the big problem!> Ed F=== === Subject: : Re: Factoring RSA And the Reversible Multiplier(comments appreciated)Hello Dr. Fredkin, long time no talk. It is possible to plug in a reversible comparator and expect it to be. Thatwould minimize the uncertainty in information. Since RSA 'n' number to befactored, can be writtern as either p*q or q*p. If p (p>q) is forced to,always appear on one side of the inputs, there is little to no uncertainty.-suresh> The Lord of the Rain( Suresh __NoJunkMail kumar)>I have been looking into Edward Fredkin's Fredkin Gate built usingard>balls.>Here's something that came to my mind.>The RSA number to be factored m = p * q>p and q are primes.>[ stuff deleted ]>The procedure would work best, if p != q. For this case notice thatone of>the number in (p,q) have to be greater than other. So, the MSB of one,is at>a higher index than the other. By forcing the MSBs of A and B to takevalues>1 or 0, we will force just one ordering of (p,q) to appear at theinput.>[will explain later]. How does the idea sound? Not good. Fredkin gates have extra outputs beyond the values of theBoolean> functions they're implementing. To factor N-bit integers by reversing> a multiplier composed of Fredkin gates, you'd have to know the values of> (I believe) O(N^2) extra outputs. Since you only have 2N outputs (i.e.,> the product), you would have to do some kind of search on the extraoutputs> that is probably no easier (probably harder) than the search you have to> do to factor large integers.> The above comments are incorrect.> As was explained in the original Conservative Logic paper, it is> always possible to build a circuit of Fredkin gates or iard Ball> Model gates that has only the informationally required number of extra> outputs (or inputs). In the case of a circuit designed to factor the> product of 2 primes, the only informational uncertainty is which> output gets the larger factor. The design of the circuit can take> into account such cases as pq=1 pq, pq=-p -q, pq=-1 -pq etc. Of> course, if the circuit cannot take advantage of the fact that both> factors are primes, then the informational uncertainty very is much> greater. My comments are not merely theoretical as I long ago> programmed a large Mathematica system that implemented factoring by> means of a reversible multiplier. Because it actually simulated the> reversible digital logic, it was very slow. My conclusion was that,> for the algorithm I implemented, a hardware version was not quite> competitive with the best existing factoring algorithms. However, I> still think that the general approach could be successful.> The circuit had the property that if you put 35 into the outputs and 5> into one of the inputs, the other input became 7. (A Fredkin gate> really has no inputs or outputs, it is perfectly symmetrical and works> in either direction; the same is true of any well formed circuit of> Fredkin gates.)> Kumar's idea is not quite right either. If p>q, a number of the most> signifigant bits of both can still be the same. However, as Kumar> suggests, one can take advantage of what knowledge one has about the> possibilities related to MSB's and also LSB's.> The LSB's of the product also enforce constraints on the LSB's of the> factors. The problem is that such ideas are only useful for a few of> the leading or trailing bits. It's all the bits in the middle that> pose the big problem!> Ed F=== === Subject: : Re: Full Beal Conjecture (revised) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1QLiBG21476; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1QLg8i21336 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1QLg8J16187; I'll cut the grammer and hope You'll anderstand: >> Beal conjecture dissmiss x^a + y^b = z^c>> for x;y;z integers of gcd=1 and for a;b;c natural numbers >2 1.Fall): For a;b;c as odd numbers can be proved once using such method, wchich is also correct for FLT conjecture for p=3 and bigger primes.>> ( this method uses also factorisation of numbers shaped as 3M+2>> into factors as c^2 (squares), what is applicated to 6n+1>> shapes and turns into squares of all primes once beginning >> with 5 . I used to write down first lines of it at ca 17.02.04 ; see topic Square factors of the odd values of shape 3M+2)>> 2.Fall) For one, two or three of exponents a;b;c as even numbers but bigger than 4 and so on containing some odd number bigger or equal to 3 it will be possible to turn such fall into (1.Fall) one: a=a1*a2; b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 = (z^c2)^c1 where a1;b1;c1 odd numbers.>> 3.Fall) For a;b;c as even numbers but discribed exclusively as 2^k or with the part a2;b2;c2 as 2^k where k=2 or bigger natural numbers it will turn into x^4 + y^4 = z^4 conjecture, solved firstlyby P. Fermat>> 4.Fall) For only few combinations once one of a;b;c is equal to 4 but others odd or two of a;b;c are equal to 4 but the one resisting odd there is not possible to use the method for odd exponents. It turns into conjectures: x^4 + y^b = z^c where b;c equal 3 or bigger prime numbers x^4 + y^4 = z^c where c equal 3 or bigger prime numbers with possible inversions of x;y;z values 5.Fall) Finally for a;b;c as even numbers but shaped in that way, that it is possible to rewrite it into conjectures:>> x^2 + y^4 = z^4 or x^4 + y^4 = z^2 what used to be profed also by P. FermatOnce it is some hope for method for odd exponents so the resisting small gap can be closed too and so on it will be possible to express that any sum of powers bigger than 2 can not be substituted with the power bigger than 2. (Does Fermat express, that exponents of such powers are equal ?)>>> Enjoy super games of powers>>> Ro>I appreciate your intent to help but I can't understand your (math)>grammer. Sorry and thanks again. Kerry So do I for Your interest and apologise for my first edition grammar poverties. I hope, that You can understand my position litlle bit better... Ro=== === Subject: : Re: Documents about multiplicative order in general and mersenne numbers> Maybe it would work better if you let us know what it is that you > actually want to know about multiplicative order. Then perhaps > someone could tell you whether or not what you want to know is known, > and, if it is, where to find it.Hello Gerry,thank you for your reply.The reason why I asked this is, that I think that the multiplicativeorder plays an important role in number theory and that it should bepossible to find out much more than what is usually written inintroductory books about number theory.I recognized that there is a strong relationship between mersennenumbers and the multiplicative order in a way that you can useknowledge about the mersenne numbers for statements about themultiplicative order in general.For example I would be very interested in documents which tellanything about factors of mersenne numbers.I think that most people concentrate on mersenne numbers whoseexponent is prime, but for me also the mersenne numbers with compositeexponents are interesting.Here is what I know about them2^p | 2^(p*q)-12^q | 2^(p*q)-1where p and q are prime.but 2^(p*q)-1 has some factors besides the ones in 2^p and 2^q.What is known about this factors?Of course I am not just interested in exponents which are composed ofexactly two exponents, but in all composed exponents in general,because i know now that the divisors of mersenne numbers with primeexponents are Cunningham numbers, right?But I hardly know anything about the factors of mersenne numbers whichare composite but have no gcd with one of the mersenne numbers whichare divisors of the first mersenne number with the composite exponent(described above).Kind regardsJ.9frgen=== === Subject: : Re: Fastest-converging algorithm to logarithm?Ron,I have used the following or similar code for natural log in the past - itis faster than the built-in log function on some compilers.Hope it may be of some use,Best wishes, Edmund Dunlop======================================================== ======================= program mylog_example implicit none integer,parameter :: dp = selected_real_kind(15,307) real(kind=dp) :: x!10 write(*,(1x,Enter x >,$)) read(*,*,end=20,err=20)x write(*,*) write(*,*) x = ,x write(*,*) write(*,*) ln(x) (exact) = ,log(x) write(*,*) ln(x) (approx.) = ,mylog(x) write(*,*) go to 10!20 stop! contains! function mylog(x) result(log_x) implicit none integer,parameter :: dp = selected_real_kind(15,307) real(kind=dp),intent(in) :: x real(kind=dp) :: y,fy,fy2,log_x integer :: iy(2),iy2,iexp real(kind=dp),parameter :: f3 = 1.d0/3.d0 real(kind=dp),parameter :: f5 = 0.2d0 real(kind=dp),parameter :: log2 = 0.6931471805599453d0 equivalence (y,iy(1)),(iy2,iy(2))!! Algorithm based on the series! ln((1+x)/(1-x)) = 2.*(x + (x**3)/3. + (x**5)/5. +(x**7)/7. + ... )! This series converges when abs(x) < 1.!! Assumptions include: x is in IEEE T_floating format, >0.d0,! normalised, not NAN, etc.! Tested on a PC (Intel chip), using double precision arithmetic! y = x iexp = ibits(iy2,20,11)-1022 ! Must choose appropriately for precisionetc. call mvbits(1022,0,11,iy2,20) ! So y = x/2.d0**iexp; 0.5 <= y <= 1.0 fy = (1.d0-y)/(1.d0+y) ! 1.d0/3.d0 <= fy <= 0.d0, series belowgoes fy2 = fy*fy ! in powers of 2 => goodconvergence fy = fy+fy*fy2*(f3+fy2*f5) ! fy = fy + (fy**3)/3.d0 + (fy**5)/5.d0 +.... log_x = real(iexp,dp)*log2-(fy+fy) return end function mylog! end program mylog_example================================================= ==============================> What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-)> Is there a quadratically converging method? I seem to recall there IS> one.> And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function).> --Ron Bruck=== === Subject: : Re: Fastest-converging algorithm to logarithm?says...> What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-)Is there a quadratically converging method? I seem to recall there IS> one.And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function).--Ron Bruck> First scale the log by writing x = e^n y, n an integer ande^(-1)< y <= 1. y can be approximated very accurately by alow degree polynomial or even a rational function.The answer is then n + ln(y).=== === Subject: : Re: Fastest-converging algorithm to logarithm?(x-1)-1/2(x-1)^2+1/3(x-1)^3-...... =ln(x) where 0 What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-)> Is there a quadratically converging method? I seem to recall there IS> one.> And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function).> --Ron Bruck=== === Subject: : Re: Fastest-converging algorithm to logarithm?(x-1)-1/2(x-1)^2+1/3(x-1)^3-...... =ln(x) where 0 For x close to 2, this is probably the _slowest_ known method to calculateln x.> If one uses a series, one should use that for log((1+x)/(1-x)); but this> is not enough to give the requested quadratic convergence.> Arnold Neumaier=== === Subject: : Re: Fastest-converging algorithm to logarithm? >What's the fastest-converging algorithm to compute ln(x)? And to avoid >answers like, Compute log_3(x) and multiply it by an appropriate >constant, the only acceptable operations in the algorithm are >addition, subtraction, multiplication, division, and, say sqrt. I >suppose we can allow rational constants and x, too :-) >Is there a quadratically converging method? I seem to recall there IS >one. >And how do you know, in general, whether there's a quadratically >convergent method or not (not just for the ln problem, but for a >general function). >--Ron BruckI assume you want an algorithm which produces arbitrary precision results.I don't know whether it is possible to device an algorithm for thiswithout fixing the maximum attainable precision beforehand but here some ideas (these for the binary representation of x given, for a decimal oneit goes analogously) write x=2^n*z 1<=z<2 then it remains to compute ln 2 and ln zwith the maximum precision fixed ahead you could consider ln 2 as a givenconstant. for ln z use ln z = 2 * phi(z)* sum_{k=0 to infty} phi(z)^{2k}/(2*k+1) with phi(z)=(z-1)/(z+1) (<=2/3 for z in [1,2] )hence that series converges rather fast, you get about one bit for each summandin the worst case. you could go a step further in looking for i in{0,1,2,3,4,5,6,7} such that z in [2^(i/8),2^((i+1)/8)] (with these constants also computed and stored beforehand) then using log2 z = i/8 + log2 (z*2^(-i/8)) with the new argument now in [1,2^(1/8)] and using the same series, nowgaining now about 9 bits with every summand. there is also a continued fraction representation of ln z in form of a jacobifraction also involving the phi(z) above which gives even faster convergence. but you might also go quite another way: using the integralrepresentation and rombergs method. this one is easily applied to int_{1 to x} 1/t dt and gives along the diagonal of the scheme a superlinearconvergence and since in this situation roundoff plays no essential role(use halving the grid) you might fix the precision of the total computationknowing the precision wanted.hthpeter=== === Subject: : Re: Fastest-converging algorithm to logarithm?> What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-)> Is there a quadratically converging method? I seem to recall there IS> one.> And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function).> --Ron BruckA method for calculating Ln(x) to order O(x^-4)is to iterate the following. Lnx can be 1 forthe first iteration. Lnx = pi*x*(1+4/(x^2))*(1-1/Lnx))/(2*AGM(x,4))AGM represents the arithmetic-geometric mean.If your x value is >1 but not large, calculateLn(y) for a reasonable n, say n=20 y = x*2^nand then Ln(x) = Ln(y) - n*Ln(2)to get Ln(2), calc Ln(2^n) Ln(2) = (1/n)*Ln(2^n)For 0 < x < 1, calculate the log of it's reciprocalfirst.All of this can be done with add, sub, mul, div,and sqrt.Search Google for rpb028 for more detail.- Brian=== === Subject: : Re: Fastest-converging algorithm to logarithm?Originator: hack@watson.ibm.com (hack)> What's the fastest-converging algorithm to compute ln(x)? ...What follows does not answer the question above, but a related one:what's a fast and easy way to compute a logarithm? What follows isa fixed-loop computation, so convergence is not involved.Here is a method I figured out (in 1980) for computing decimal logarithm inan environment where (software) decimal arithmetic with arbitrary precisionis supported (e.g. some interpretive languages). To compute the n-placelogarithm of X, raise X to the power 10**n using 2n-digit precision andexponential representation. The result will then be Y*10**Z = X**(10**n),with Y between 1 and 10. Taking Logs: Z+Log(Y) = 10**n Log(X)Raising X to such a large power (via repeated squaring) is quite fast.(When X is close to 1 one has to be careful, of course, and rescale.)Michel.=== === Subject: : Re: Fastest-converging algorithm to logarithm?A method for calculating Ln(x) to order O(x^-4)> is to iterate the following. Lnx can be 1 for> the first iteration. Lnx = pi*x*(1+4/(x^2))*(1-1/Lnx))/(2*AGM(x,4))AGM represents the arithmetic-geometric mean.The arithmetic mean of x and y is (x+y)/2 and the geometric mean is(x*y)^(0.5). Could you please explain what you mean by thearithmetic-geometric mean?=== === Subject: : Re: Fastest-converging algorithm to logarithm?> A method for calculating Ln(x) to order O(x^-4)> is to iterate the following. Lnx can be 1 for> the first iteration. Lnx = pi*x*(1+4/(x^2))*(1-1/Lnx))/(2*AGM(x,4)) AGM represents the arithmetic-geometric mean.> (x*y)^(0.5). Could you please explain what you mean by the> arithmetic-geometric mean?I can explain the recipe better than its implications. To findthe value of AGM(a,b), start with a0 = a, b0 = b, and do thefollowing iterations until ai = bi, which shouldn't take too manyiterations. See the MathWorld entry below for more info.http://mathworld.wolfram.com/ Arithmetic-GeometricMean.htmlExample: Find Ln(1000000)x = 1000000AGM(1000000,4)a(i+1) = (1/2)(ai + bi)b(i+1) = (ai*bi)^(1/2)(ai,bi)1000000.000000 4.000000500002.000000 2000.000000251001.000000 31622.839847141311.919924 89091.887535115201.903729 112204.035922113702.969825 113693.089255113698.029540 113698.029433113698.029486 113698.029486113698.029486 113698.029486Ln(i=0) = 1Ln(i+1) = pi*x*(1+4/(x^2))*(1-1/Lni))/(2*AGM(x,4))(Lni) 1.00000000000013.81551055801613.81551055806713.815510558067Exce l's intrinsic function call givesLn(10^6) = 13.815510557964- Brian=== === Subject: : Re: Fastest-converging algorithm to logarithm?> Is there a quadratically converging method? I seem to recall there IS> one.Shouldn't Newton's Method on a^x - b = 0 give you x = log_a(b) withquadratic convergence? > And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a general> function).I believe Newton's Method is guaranteed to give you quadratic convergencenear any root of a function whose first derivative (or Jacobian forfunctions from R^n->R^n) is invertible at the root and differentiable ina neighborhood of the root...Google seems to think I'm right; there's a statement (with slightly weakerconditions) and a short proof here:---Roy Stogner=== === Subject: : Re: Fastest-converging algorithm to logarithm?>> Is there a quadratically converging method? I seem to recall there IS>> one.Shouldn't Newton's Method on a^x - b = 0 give you x = log_a(b) with> quadratic convergence?Never mind; the implications of the only acceptable operations youlisted originally just sunk in. For Newton's Method to work you need tobe able to approximate your function value and derivative within tighterand tighter tolerances as you get close to the root, and that means thecost of evaluating a^x will increase in later steps. I'll try to improvemy reading and following directions skills in the future. ;-)---Roy Stogner=== === Subject: : Re: Fastest-converging algorithm to logarithm?> Shouldn't Newton's Method on a^x - b = 0 give you x = log_a(b) with> quadratic convergence?> Never mind; the implications of the only acceptable operations you> listed originally just sunk in. For Newton's Method to work you need to> be able to approximate your function value and derivative within tighter> and tighter tolerances as you get close to the root, and that means the> cost of evaluating a^x will increase in later steps. I'll try to improve> my reading and following directions skills in the future. ;-)Nevertheless, your idea may work. I mean, e^x may be calculated to arbitraryprecision using the Taylor series.-Michael.=== === Subject: : Re: Fastest-converging algorithm to logarithm?What's the fastest-converging algorithm to compute ln(x)?Here's one from LANL you could use as a reference against whatever otherimplementations you can find. http://netlib.org/fn/alog.f--Dr.B.Voh------------------------- -----------------------------Applied Algorithms http://sdynamix.com=== === Subject: : Re: Fastest-converging algorithm to logarithm?http://groups.google.com/groups?q=mathematica+crank+ log+group:sci.*&hl=en&lr=&ie=UTF-8&group=sci.*&selm=3ee26405% 240%243938%24b45e6eb0%40senator-bedfellow.mit.edu&rnum=1Here's the actual messageHi everyone, recently, i found a way to compute Exp/Log very quickly, even much muchquicker than taylor series approximations. But i am not sure, if they aretotally new, or they have already been discovered. Pointers would help.The following files were written in mathematica(* Fast way to compute Log/Exp root *)(* Preliminaries *)one = SetPrecision[1, 400]; (* Creates one with 400 digits of precision *)(* Pow of number c to integer power n *)Clear[powAtomicTable];(* The methods for computing Exp/Log are based on this method for computing pow *)pow[c_, 0] := 1; pow[c_,1]:= c;pow[ c_, i_]:= pow[c,i ] = If[EvenQ[i], pow[c, i/2]^2 , c * pow[c, (i-1)/2]^2];(* Log taylor series creator *)logSeries[k_, 0, x_] := x * logSeries[k,1,x];logSeries[k_ ,n_, x_] := If[ n < k, 1 - x *n/(n+1) * logSeries[k, n+1, x],1];(* Exp taylor series creator *)expSeries[k_, 0, x_] := expSeries[k, 1, x];expSeries[k_, n_, x_] := If[ n < k , 1 + (x/n)*expSeries[k, n+1, x], 1];(* Suppose, we through the trouble of computing an atomic Exp exactly *)n = 32; (* n controls the number of terms *)expAtomic = expSeries[ n, 0, one / 2^(2*n) ];(* Suppose we build a table for quick intermediate processing *) powAtomicTable[ x_, 0] :=1; powAtomicTable[x_,1] = x;powAtomicTable[ x_, i_] :=powAtomicTable [x, i] = powAtomicTable[ x, i-1]^2; (* Create Tables for fast Computation *)pow[2,2*n];powAtomicTable[ expAtomic, 2*n];(* Computes Log quickly for numbers between 1 and 2 *)logFastA[ x_, 0] := 0;logFastA[ x_, i_]:= If[ x > powAtomicTable[expAtomic, i], pow[2,i-1] + logFastA[ x/ powAtomicTable[expAtomic,i], i-1], logFastA[ x, i-1]];logFast[ x_, n_] := Module[{k,z},k = logFastA[x,n]; z = x/pow[expAtomic, k] - 1; k/pow[2,n] + logSeries[ n, 0, z]](* Computes Exp quickly for numbers between 1 and 2 *)expFast[y_, n_] := Module[ {k,p}, p = pow[2,n]; k= Round[ p * y]; pow[expAtomic,k] * expSeries[n, 0, y -k/p]] ;logFast[12*one/10, 2*n]Out[94]= 0.1823215567939546262117180251545146331973893379144869839427264 516567089274806459178493452037169711655300052070648129885763295 192323639741784518560049386476449817107025145695567161878812659 111083175957652385595296989949867047298122092208505910785048337 805319105785465962939054918744598668342618212386743368391937951 743250138964782408990094383508558205553202751165542776156602364 77321974209263562976958In[92]:=Log[12*one/10]Out[92]= 0.1823215567939546262117180251545146331973893379144869839427264 516567089274806459178493452037169711655300052070648129885763295 192323639741784518560049386476449817107025145695567161878812659 111083175957652385595296989949867047298122092208505910785048337 805319105785465962939054918744598668342618212386743368391937951 743250138964782408990094383508558205553202751165542776156602364 77321974209263562976958In[102]:=expFast[9/10 *one, 2*n]Out[102]= 2.4596031111569496638001265636024706954217723064400830207485457 366574665529433658608704971852759308015023147342824690012118633 771207543752629611846982441019238731189807525535462060855114839 157755657466338779445701713104413428070392400118009500678776532 765788933267408406757286069763530707024830787976497135853645366 779479888576435077529202762644094367456361742973388290754582071 3144111178082100098288In[103]:=Exp[9/10*one]Out[103]= 2.4596031111569496638001265636024706954217723064400830207485457 366574665529433658608704971852759308015023147342824690012118633 771207543752629611846982441019238731189807525535462060855114839 157755657466338779445701713104413428070392400118009500678776532 765788933267408406757286069763530707024830787976497135853645366 779479888576435077529202762644094367456361742973388290754582071 31441111780821000982879Exp[ one/2^(2*n)] 1.0000000000000000000542101086242752217018420079820249449562765 347033800146352241731262843668671718530414924774809003762287597 777522529744319253011556136658978935624994712265976530020372297 928670940012070838413785631346030564920099516333692913570710459 160861486270280352080328828434270435647785739926264420089269209 706187587624299802883578503554038850120765084372377899605846958 234532561529809130679285660739901849218669> What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-)> Is there a quadratically converging method? I seem to recall there IS> one.> And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function).> --Ron Bruck=== === Subject: : Re: Fastest-converging algorithm to logarithm?here's a nice and fast way to compute logRemember that given a number n = exp( sum_i a_i 2^k y )So, suppose i can calculate x = exp( y )for y close to 1x^2 = exp( 2 y)x^2^2 = exp( 4 y)x^2^2^2 = exp( 8 y)Suppose i want to calculate log of number alog(a) = sum_i a_i 2^i yFirst i can calculate exp(y) from known power series. Then, if a > 1i can do the followingset w[1] = exp(y), n = 1, b[1] = 1;for(;;)if ( (a / w[n]) > 1) { n++; w[n] = w[n-1]*w[n-1]; b[n] = b[n-1]*2;}else break;This way, i can compute the highest i in the sum sum_i a_i 2^i yOnce i have computed that, i can, construct the log(a) algorithm in thefollowing recursive fashionlog(a) = log(a,n);log(a, n) = if( n = 0, 0, if( a / w[n] > 1, b[n]*y + log(a/w[n],n-1),log(a/w[n],n-1) ))The alogrithm will compute log(a) to n decimal places n iterations, foralmost any a.The Lord of Chaos (Suresh Devanathan)http://groups.google.com/groups?q=mathematica+crank +log+group:sci.*&hl=en&lr=&ie=UTF-8&group=sci.*&selm=3ee26405% 240%243938%24b45e6eb0%40senator-bedfellow.mit.edu&rnum=1> Here's the actual message> Hi everyone,> recently, i found a way to compute Exp/Log very quickly, even much much> quicker than taylor series approximations. But i am not sure, if they are> totally new, or they have already been discovered. Pointers would help.> The following files were written in mathematica> (* Fast way to compute Log/Exp root *)> (* Preliminaries *)> one = SetPrecision[1, 400]; (* Creates one with 400 digits of precision*)> (* Pow of number c to integer power n *)> Clear[powAtomicTable];> (* The methods for computing Exp/Log are based on this method forcomputing> pow *)> pow[c_, 0] := 1; pow[c_,1]:= c;> pow[ c_, i_]:=> pow[c,i ] = If[EvenQ[i], pow[c, i/2]^2 , c * pow[c, (i-1)/2]^2];> (* Log taylor series creator *)> logSeries[k_, 0, x_] := x * logSeries[k,1,x];> logSeries[k_ ,n_, x_] :=> If[ n < k, 1 - x *n/(n+1) * logSeries[k, n+1, x],1];> (* Exp taylor series creator *)> expSeries[k_, 0, x_] := expSeries[k, 1, x];> expSeries[k_, n_, x_] :=> If[ n < k , 1 + (x/n)*expSeries[k, n+1, x], 1];> (* Suppose, we through the trouble of computing an atomic Exp exactly *)> n = 32; (* n controls the number of terms *)> expAtomic = expSeries[ n, 0, one / 2^(2*n) ];> (* Suppose we build a table for quick intermediate processing *)> powAtomicTable[ x_, 0] :=1; powAtomicTable[x_,1] = x;> powAtomicTable[ x_, i_] :=powAtomicTable [x, i] = powAtomicTable[ x,i-1]^2> ;> (* Create Tables for fast Computation *)> pow[2,2*n];> powAtomicTable[ expAtomic, 2*n];> (* Computes Log quickly for numbers between 1 and 2 *)> logFastA[ x_, 0] := 0;> logFastA[ x_, i_]:=> If[ x > powAtomicTable[expAtomic, i],> pow[2,i-1] + logFastA[ x/ powAtomicTable[expAtomic,i], i-1],> logFastA[ x, i-1]];> logFast[ x_, n_] := Module[{k,z},k = logFastA[x,n];> z = x/pow[expAtomic, k] - 1; k/pow[2,n] + logSeries[ n, 0, z]]> (* Computes Exp quickly for numbers between 1 and 2 *)> expFast[y_, n_] :=> Module[ {k,p}, p = pow[2,n]; k= Round[ p * y];> pow[expAtomic,k] * expSeries[n, 0, y -k/p]] ;> logFast[12*one/10, 2*n]> Out[94]= 0.1823215567939546262117180251545146331973893379144869839427264 5165670892748> 064591784934520371697116553000520706481298857632951923236397417 845185600493864> 764498171070251456955671618788126591110831759576523855952969899 498670472981220> 922085059107850483378053191057854659629390549187445986683426182 123867433683919> 379517432501389647824089900943835085582055532027511655427761566 023647732197420> 92> 63562976958> In[92]:=> Log[12*one/10]> Out[92]= 0.1823215567939546262117180251545146331973893379144869839427264 5165670892748> 064591784934520371697116553000520706481298857632951923236397417 845185600493864> 764498171070251456955671618788126591110831759576523855952969899 498670472981220> 922085059107850483378053191057854659629390549187445986683426182 123867433683919> 379517432501389647824089900943835085582055532027511655427761566 023647732197420> 92> 63562976958> In[102]:=> expFast[9/10 *one, 2*n]> Out[102]= 2.4596031111569496638001265636024706954217723064400830207485457 3665746655294> 336586087049718527593080150231473428246900121186337712075437526 296118469824410> 192387311898075255354620608551148391577556574663387794457017131 044134280703924> 001180095006787765327657889332674084067572860697635307070248307 879764971358536> 453667794798885764350775292027626440943674563617429733882907545 820713144111178> 08> 2100098288> In[103]:=> Exp[9/10*one]> Out[103]= 2.4596031111569496638001265636024706954217723064400830207485457 3665746655294> 336586087049718527593080150231473428246900121186337712075437526 296118469824410> 192387311898075255354620608551148391577556574663387794457017131 044134280703924> 001180095006787765327657889332674084067572860697635307070248307 879764971358536> 453667794798885764350775292027626440943674563617429733882907545 820713144111178> 08> 21000982879> Exp[ one/2^(2*n)] 1.0000000000000000000542101086242752217018420079820249449562765 3470338001463> 522417312628436686717185304149247748090037622875977775225297443 192530115561366> 589789356249947122659765300203722979286709400120708384137856313 460305649200995> 163336929135707104591608614862702803520803288284342704356477857 399262644200892> 692097061875876242998028835785035540388501207650843723778996058 469582345325615> 29> 809130679285660739901849218669> What's the fastest-converging algorithm to compute ln(x)? And to avoid> answers like, Compute log_3(x) and multiply it by an appropriate> constant, the only acceptable operations in the algorithm are> addition, subtraction, multiplication, division, and, say sqrt. I> suppose we can allow rational constants and x, too :-) Is there a quadratically converging method? I seem to recall there IS> one. And how do you know, in general, whether there's a quadratically> convergent method or not (not just for the ln problem, but for a> general function). --Ron Bruck=== === Subject: : Re: Difficulty of calculus vs. discrete math1st calculus is not hard, what you are calling discrete can be new ideas formany starting out.Could be getting used to the notation, and concepts.Advance Calculus can be quite difficult, proof to proof, and then discreteMath such as usedin transforms (like image processing) can be difficult but has newdiscoveries, such as Turbo Coding in the last few years.> I have taken college courses both in calculus and in discrete> mathematics. What surprised me was the difficulties that other> students were having with discrete mathematics. For some reason, they> found calculus much easier.> The discrete math was baby stuff: formal logic, divisibility,> combinatorics, and the like.> Why would one find calculus easy and discrete math difficult?=== === Subject: : Re: Difficulty of calculus vs. discrete math> I have taken college courses both in calculus and in discrete> mathematics. What surprised me was the difficulties that other> students were having with discrete mathematics. For some reason, they> found calculus much easier.> The discrete math was baby stuff: formal logic, divisibility,> combinatorics, and the like.> Why would one find calculus easy and discrete math difficult?I find that discrete objects often have less structure than continuousobjects. Hence one will often be stuck with problems with no analyticalsolution more often in discrete math which makes calculations and proofsmore difficult and time consuming. Also, one will find a lot less 'tricks'to solve huge classes of problems in discrete math, while such mechanicaltricks abound in continuum math. Hence students often find discrete math asubject where almost every problem requires one to essentially 'start fromscratch'. I remember reading just such a post from a student learning graphtheory.Note, this is from my own personal experience.=== === Subject: : Re: Difficulty of calculus vs. discrete math> I have taken college courses both in calculus and in discrete> mathematics. What surprised me was the difficulties that other> students were having with discrete mathematics. For some reason, they> found calculus much easier.> The discrete math was baby stuff: formal logic, divisibility,> combinatorics, and the like.> Why would one find calculus easy and discrete math difficult?Well, calculus is - in some weird sense - just the continuum limit ofdiscrete maths. It's like - again in some very warped analogy - thedifference between classical mechanics and quantum mechanics. The latter hasdiscrete energy levels, the difference between them being proportional toPlanck's constant.Just to give an example of what I mean. You could pretty quickly figure outthe definite integral of 1/x from x=4 to x=1000, say. But say you needed thesum of 1/n from n=4 to n=1000. It's certainly possible, but would probablytake more time to figure out.The same happens in probability. The Binomial distribution can - in somecases - be approximated with the Normal distribution, and the latter isoften much easier to work with.So my conjecture is that calculus is easier to work with, because of thelact of discreteness. Of course, the continuum limit possesses its ownquirks, including (but not limited to) the fact that integration anddifferentiation is not always interchangeable.Another reason might be that calculus is more intuitive. It seems prettyeasy understanding integration as a means of calculating area, anddifferentiation as a means of calculating slope. But combinatorics is oftenquite surprising, even though in principle it is just a matter of counting.-Michael.=== === Subject: : Re: Difficulty of calculus vs. discrete math|> Why would one find calculus easy and discrete math difficult?i think that there's some potential for talking at cross-purposes in adiscussion about this question, because there are both some goodreasons and some bad ones, the bad ones relating to the fraudulentcorrupt nature of the educational systems in many places around theworld. i don't feel like getting into a discussion of the badreasons at the moment; instead i just want to make a minor commentabout some of the good ones, which are mostly very different fromthe bad ones.|Well, calculus is - in some weird sense - just the continuum limit of|discrete maths. It's like - again in some very warped analogy - the|difference between classical mechanics and quantum mechanics. The latter has|discrete energy levels, the difference between them being proportional to|Planck's constant.||Just to give an example of what I mean. You could pretty quickly figure out|the definite integral of 1/x from x=4 to x=1000, say. But say you needed the|sum of 1/n from n=4 to n=1000. It's certainly possible, but would probably|take more time to figure out.||The same happens in probability. The Binomial distribution can - in some|cases - be approximated with the Normal distribution, and the latter is|often much easier to work with.||So my conjecture is that calculus is easier to work with, because of the|lact of discreteness. Of course, the continuum limit possesses its own|quirks, including (but not limited to) the fact that integration and|differentiation is not always interchangeable.||Another reason might be that calculus is more intuitive. It seems pretty|easy understanding integration as a means of calculating area, and|differentiation as a means of calculating slope. But combinatorics is often|quite surprising, even though in principle it is just a matter of counting.it's true that sometimes formulas look simpler in the continuum limitbecause annoyingly complicated high-frequency terms may vanishthere. however, as for this business about calculus allegedly beingmore intuitive, that's highly arguable (both for and against, butespecially against), and it's probably better to describe thesituation rather differently: in calculus, you pat yourself on theback whenever you manage to solve _any_ problem (for examplecalculating the area under a curve), because from an intuitive pointof view it's actually surprising that you can accomplish _anything_ inthe continuum limit. solving the discrete analog of the same problemis often so ridiculously simple (in the area-under-the-curve case it'ssimply a matter of counting discrete boxes) from a conceptualviewpoint that it doesn't give you much of a sense of accomplishment.thus generally in discrete mathematics all the easy problems havealready been done practically even before you get started and what'sleft is the hard problems, whereas in continuum mathematics it'spossible to impress people to a certain extent while working only onproblems that are secretly actually surprisingly easy.another phenomenon is this: discrete means zero-dimensional andcontinuous means higher-dimensional, and naively it seems likehigher-dimensional problems should be more difficult thanlower-dimensional ones. but sometimes dimension enters into aproblem in a negative way, so that the higher-dimensional versionof a problem is actually easier than the lower-dimensional one. forexample according to what i've heard, the study of dynamical systemshas some tendency to get more difficult as the dimension of the orbitspace gets higher, but the dimension of the orbit space gets lower asthe dimension of the time semigroup gets higher.i said i wouldn't comment on the bad reasons, but i changed my mind:another phenomenon is that in discrete mathematics formulas are oftenrather complicated though easy to logically derive, while in thecontinuum limit the analogous formulas are often simpler though moredifficult to logically derive. thus at a certain stage of learning,formula-derivers may enjoy discrete mathematics more whileformula-memorizers enjoy continuum mathematics more.-- [e-mail address jdolan@math.ucr.edu]