This problem was given by my friend. I dont know how to approach it. Any suggestions are welcome I pick two numbers, randomly, and tell you one of them. You are supposed to guess whether this is the lower or higher one of the two numbers I picked. Can you come up with a method of guessing that does better than picking the response low or high randomly Thaanx in Advance. Naren. ==== > This problem was given by my friend. > I dont know how to approach it. > Any suggestions are welcome I pick two numbers, randomly, and tell you one of them. You are supposed > to guess whether this is the lower or higher one of the two numbers I > picked. Can you come up with a method of guessing that does better than > picking the response low or high randomly Thaanx in Advance. Naren. If by number you mean a positive integer and if you randomly tell me one of the two integers then I will always say the number that you told me is the lower number. Since if you tell me one of the numbers is X I then know that there are X-1 integers below X and infinitely many integers above X. ==== > This problem was given by my friend. > I dont know how to approach it. > Any suggestions are welcome I pick two numbers, randomly, and tell you one of them. You are supposed > to guess whether this is the lower or higher one of the two numbers I > picked. Can you come up with a method of guessing that does better than > picking the response low or high randomly Thaanx in Advance. > sure, if the number is positive, you say high, if the number is negative, you say low. then you assume the distribution of random numbers is symmetric over 0. Herc ==== |What I noted is the oddity of believing that a constant factor of a |polynomial, which f^2 is, could factor off as a function. | |It seems unlikely to me that a math expert would make that mistake. You've only guessed that the way f^2 divides off is constant. The mistake a mathematician wouldn't make is to just keep saying how odd it would be if their guess were wrong, and pretend that that's a logical argument. Accusing others of oddly believing otherwise is just posturing. It's not up to others to prove it's false; it's up to you to prove that it's true. Here's a related example. The two roots r1(m) and r2(m) of x^2+x+2m=0 multiply together to give 2m, which is divisible by 2. At m=0 one of them is divisible by 2 and the other one is coprime to 2. But there's no simple pattern in their common factors with 2: m=1: r1=(-1+sqrt(-7))/2, r2=(-1-sqrt(-7))/2. 2m=2=r1*r2. But that means that r1 and r2 are themselves the algebraic integer factors of 2 dividing r1 and r2. Neither r1 nor r2 is divisible by 2 in the algebraic integers. m=2: r1=(-1+sqrt(-15))/2, r2=(-1-sqrt(-15))/2. The algebraic integer factors that r1 and r2 have in common with 2 are sqrt(r1) and sqrt(r2). Neither r1 nor r2 is divisible by 2. m=-1: r1=1, r2=-2. The algebraic integer factors that r1 and r2 have in common with 2 are again 1 and 2. Keith Ramsay ==== > Starting with sequences with equal numbers of zeros and ones, I think > a new thing to consider are those sequences having 1/4, 1/8, 1/16, > etcetera ones, and then also 1/3, 1/5, 2/5, 1/6, 1/7, 2/7, 3/7, 1/8, > 3/8, 1/9, 2/9, 4/9, etcetera many ones, or zeros. Well, I got into the case where a quarter of the sequence elements are > ones or zeros. n! / ( (n/4)!(3n/4)! 2^n ) I problem here with getting an identity that converges to a value is > that (n/4)!(3n/4)! >> (n/2)!(n/2)!. I thought I might be able to use > Euler's formula for Gamma. > ... I want to determine a relation between (n/4)!(3n/4)! and (n/2)!(n/2)!, I think I went about it in an overly complicated way. (n/4)!(3n/4)! / (n/2)!(n/2)! = (n/4)!(3n/4)! / (n/2)!(n/2)! So then, when I use (n/4)!(3n/4)! in an expression where there was instead (n/2)!(n/2)! that appears to converge, then I should be able to multiply it be (n/2)!(n/2)! / (n/4)! (3n/4)! to get the value of the expression when it has (n/2)!^2 instead of (n/4)!(3n/4)!. The same should hold true for all the cases like (n/8)!(7n/8)! (which is greater than (n/4)!(3n/4)! and also (n/2)!^2) and for (n/3)!(2n/3)!. That is to say, if f(n) is a simple ratio of expressions and that so is g(n) where each is h(n) divided by the expression (n/2)!(n/2)! for f(n) and (n/4)!(3n/4)! for g(n) that f(n) = g(n) * (n/2)!(n/2)! / ((n/4)!(3n/4)!), and that that would be true for any expression for rational a/b < 1 that f(n) = g(a, b, n) * (n/2)!(n/2)! / ( ((a/b)n)! ((1-a/b)n)! ) where f(n) = n! / ((n/2)!(n/2)! 2^n) = g(n,1,2) and g(n,a,b) = n! / ( (a/b)n! (1-a/b)n! 2^n ). Thus I was surprised when that did not appear to be the case for the small finite values of n that I calculated. I guess it does, I may have had some other error in my program, I think I already tried it. f(4)= 1.500000000000000000000000000000000000000000000000000000000000 f(16)= 1.571044921875000000000000000000000000000000000000000000000000 f(64)= 1.589548059967470344452933339596256701042875647544860839843750 f(256)= 1.594211517956484839624950947759399703028478483734198325900710 f(1024)= 1.595379577150690797583016460816556042629587428727203086922211 f(4096)= 1.595671726561313225912210533040091324648758050191856710414061 f(16384)= 1.595744772287097827906598649182402930052438187131273212971257 I still need to understand the proof of why that value in the limit as n->oo is sqrt(8/Pi), as I haven't seen one yet, but it appears that it is very simple to generate the expressions for the expected number of sequences with a given fraction of ones or zeros. n! / ((n/2)!^2 2^n) ~= sqrt(8)/ sqrt(Pi*n) n! / ((a/b)n! (1-a/b)n! ! 2^n) ~= ( sqrt(8) (a/b)n! (1-a/b)n! ) / (n/2)!^2 sqrt(Pi * n) With this and Stirling's formula it is possible to determine n^n, e^n, (a/b)n!, 2^n, and n in terms of each other and Pi. Another thread today on sci.math is discussing probabilities that equal zero. Pick a number between zero and one, I'll probably bet it's one half, with probability at least infinitesimally greater than one half. After all, the average value of all of the reals between zero and one is one half. Of any characteristic of the amount of ones and zeros of any infinite string of coin tosses, it is more probable that it has nearly 1/2 ones and 1/2 zeros than any other fraction. Then again, on pick a number between zero and n for unknown n, I pick four. Could infinite monkeys each typing at random infinitely generate the works of Shakespeare? Easily, the work of Shakespeare is a finite string. How many cigarettes would they smoke? The probability of selecting a given sequence at random from the 2^n possible sequences is 1/ 2^n. That evaluates to zero for infinite n in terms of x, y, or z unrelated to n, but 2^n / 2^n evaluates to equal to 1 for any n. Ross ==== > handled by professor Chapman. Let m=[K:Q]. We can represent multiplication I'm not, and I never have been, a professor. In the U.S., wouldn't your equivalent position be professor? Or at least associate professor? I never understood the British system. How many levels are there anyways? Anyway, my point is that in the States, the normal form of address for someone in your position would be professor, so it's natural people are going to keep doing this. Are you going to keep correcting everybody who addresses you that way? I think this is the second time I've seen you do this recently. ==== NUMEROLOGY 5 if 2 is 100, 50 is 135, 1000 is 145 what is 5 billion? my guess is 733 or 469, at least 300 unless its only 216 this is where I'm trying to get a figure for the highest IQ, (2 year old post) 2 is 100 means if you are average (one in two) then your IQ is 100. If you are one in 50 (top 2% - mensa) then IQ is 135. with no knowledge about the extremities of the bell curve or IQ notice I am spot on here with 216, as most figures for highest IQ are just over 200 to 220, which ties in with the gist of the post that I can just guess things correctly. Although here is my best demo of guessing where I beat 81 to 1 odds live. http://tinyurl.com/j9vr Mmmmm-mmmm! Wholesome humble pie! I could try lying to save face but I have to admit it: Four out of four right. Herc ==== > Is there an algorithm to determine the minumum number of generators for a > finite group ? Clearly, for a cyclic group it is 1, for a symmetric > permutation group or a dihedral group it is 2 but what about any finite > group ? I asked John Conway the inverse question - Can every group be generated from a set of generators whose orders multiply to give the order of the group? and he assured me that they could, at least up to the first Janko group. This puts an upper limit on the number of generators needed. S4 needs all 4 {3x2x2x2=24). GAP can provide a set of generators via RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n)))), but this may not be minimal. Roger Beresford. Make things as simple as possible, but no simpler (A. Einstein.) ==== >> Is there an algorithm to determine the minumum number of generators for a >> finite group ? Clearly, for a cyclic group it is 1, for a symmetric >> permutation group or a dihedral group it is 2 but what about any finite >> group ? I asked John Conway the inverse question - Can every group be >generated from a set of generators whose orders multiply to give the >order of the group? and he assured me that they could, at least up to >the first Janko group. This puts an upper limit on the number of >generators needed. Yes, but this upper limit would be much too large in general. >S4 needs all 4 {3x2x2x2=24). I am not sure what you mean by that. S4 is a 2-generator group. >GAP can provide a set >of generators via RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(m,n)))), >but this may not be minimal. But the question was to find a minimal generating set. If you just want a generating set, take the whole of G. Derek Holt. ==== The Principal Of Induction which is: a statement P is true for P=1, and true for P=k+1 if k was true, then P is true Is it the axiom or theorem?? B4E ==== > The Principal Of Induction which is: a statement P is true for P=1, > and true for P=k+1 if k was true, > then P is true Is it the axiom or theorem?? > In my text it says it lies at a different level than the axioms, together with this rule. If F and F->G then G the rule of induction is like the rule of inference, they are added to the axioms. But I don't see any formal reason for its special case, at this point it looks likely as a step to complete formalism, but we are short on categorising many other proof mechanisms. Herc ==== > I can't seem to construct one. Is there such a mapping? (R is the real >> numbers) >Maybe this will help. I will show a bijection from the interval (0, 1) to >the interior of a unit cube. Let a point in (0 , 1) be .a_1a_2a_3.... This >corresponds to the unique point (.a_1a_4a_6..., .a_2a_5a_8..., >>a_2a_5a_8...) in the interior of the unique cube. Now given the point >(.a_1a_2a_3a_4...., .b_1b_2b_3..., .c_1c_2... ) in the interior of the unit >cube correspond to the unique point (.a_1b_1c_1a_2b_2c_2...) in the >interval ( 0, 1). Of course you must talk about numbers like .399999..... Is >it .39999.... or is it .4000...?? >Steve > A space-filling curve should do the trick here as well, no? -Davis ==== > N^n injects into N? How is this possible? Does this mean that Z^n also > injects into Z? > This seems ridiculous to me, but I will have to think about it more after I > read all of the > responses. [...] N^2 injects into N as follows: (a,b) |-> 2^a * 3^b . For larger n, one can use any n distinct primes. This bears a resemblance to Godel numbering. David Bernier ==== You are playing the following game. There are 3 players armed with a gun. Player #1 hits his target 1/3 of the time. Player #2s hits her target half of the times and player #3 never misses. 1st player #1 shoots, then player #2 shoots if she is still alive, then player #3 shoots if she is still alive and then back to #1,etc. You are player #1. What is your best strategy? We do assume that all shooters make the optimal decision for themselves and if you are shot you die, ie you are out of the game. Steve ==== Steven > You are playing the following game. There are 3 players armed with a gun. > Player #1 hits his target 1/3 of the time. Player #2s hits her target half > of the times and player #3 never misses. 1st player #1 shoots, then player > #2 shoots if she is still alive, then player #3 shoots if she is still alive > and then back to #1,etc. You are player #1. What is your best strategy? We > do assume that all shooters make the optimal decision for themselves and if > you are shot you die, ie you are out of the game. > Steve Such problems are taken quite seriously in some circles :) but there can't be any point in shooting at player #2. Or am I missing something? Most of us have probably seen some form of the Xtown Mathematics Examination, where Xtown is the name of some rough neighbourhood nearby. Question 1) If Billy sells 20 grams of cocaine per day, and makes $30 per gram, then how many days will he need to pay back the loan shark if ... Question 2) If an AK-47 holds 30 rounds per clip, and fires 4 rounds per second... You get the idea. Larry ==== VnkYa.28002$O42.10475@news02.roc.ny... > You are playing the following game. There are 3 players armed with a gun. > Player #1 hits his target 1/3 of the time. Player #2s hits her target half > of the times and player #3 never misses. 1st player #1 shoots, then player > #2 shoots if she is still alive, then player #3 shoots if she is still alive > and then back to #1,etc. You are player #1. What is your best strategy? We > do assume that all shooters make the optimal decision for themselves and if > you are shot you die, ie you are out of the game. This is a well-known problem. The best strategy is to pass one's turn, or to shoot at the sky. -- Julien Santini ==== I'll be back. ==== >>Yes, that is what I want to say. >>Take a very big n, for example, n = e^10^10^100: >>if you pick an integer at random between 1 and n, the probability that it is >>prime is almost 0, much less than the probability that you won the biggest >>monthly lottery prize in USA 10^80 times in a row buying only one ticket >>each month. I think that's why it's big news when they find huge primes of many >(thousands?) of digits. Because you can't just pick those at random >and have any reasonable hope of success. Not really. The largest prime found so far has around 4 million digits, which is far, far, smaller than the example above. In the millions of digits range, the odds of picking a prime at random are about the same as some of the easier lotteries (and an order of magnitude better if you sieve out small prime multiples). The rarity of primes only increases linearly with the number of digits, whereas the running time of the primality testing algorithm increases polynomially. If you're going to attribute the difficulty to any one thing it should be the computational cost of the algorithm. -- Erick ==== > If the percentage of primes as a fraction of integers was not 0, > it would be an interesting constant. And if my Aunt had balls... Still, there are a couple or three things you might or might not know. 1) There IS a constant limiting density (the type you like) of SQUARE-FREE numbers - no squared-prime factors (or cubed etc). It is 6/pi^2 . 2) There IS a constant limiting probability that TWO numbers will be CO-prime. if they are randomly chosen in almost any sensible way. It is the same! 3) Given a large prime chosen at random, the probabilities that it is of type 4n+1 or 4n-1 are half each. Ditto for 6n+1 and 6n-1. There are many other results of these types. All good clean fun. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- Watch out for my definitive series of books on Prime Numbers. Just coming into print now... volume I: The Even Primes ---------------------------------------------------------------------------- -- ==== This problem of baselines is why economists use logarithms, > and why they sometimes construct indexes that are based > on the logs of values -- instead of the natural prices. > AIUI the reason economist use log sis because the procesesses they are modeling have exponential drift. Eg they tend to grow propotionally to their current size. It is easier for economists to think in terms of this instantaneous growth than the absolute amount of growth. Forgive me if this is what you meant. However I think with most (serious) currencies the stochastic element outweighs the exponential so they tend to just graph values. > If you don't get more specific help, I suggest that you > browse the basic textbooks on economics in a library. > Index and logarithm might be key-words. > [ snip, rest] -- > Rich Ulrich, wpilib@pitt.edu > http://www.pitt.edu/~wpilib/index.html > Taxes are the price we pay for civilization. Justice Holmes. ==== > I think that the curves formed by two surfaces rolling (without sliding > or twisting) should be the geodesic curves on the two surfaces, but I am > not sure. Is this right? are there any references on this topic? This is not true. Consider, for example, the sphere rolling on the plane. The position and orientation of the sphere can be described by a point in the manifold M = R^2 x SO(3) (R^2 representing the location of the point of contact, SO(3) representing the rotation of the sphere). Let u(t) be the curve in M that represents the path traced by the point of contact. To capture the non-slip and non-twist conditions, you need to look at the lift of u into the tangent bundle TM = R^2 x R^2 x SO(3) x so(3). The no-twist condition reallly means that the sphere is not allowed to rotate around the vertical axis; this means that the so(3)-component of the velocity vector u'(t) looks something like [ 0 0 -a2] [ 0 0 a1] [ a2 -a1 0] . Now, let u(t) = (x,R) and u'(t) = (x,v,R,A) . Then the no-slip condition is v = r [ a1 ; a2 ] where r is the radius of the sphere. You can use Chow's theorem to show that any point in M can be reached by a no-slip and no-spin path, which is certainly not true of geodesic motion. Note that the no-slip and no-spin conditions are linear constraints on the tangent vector; that is, an admissable tangent vector must lie in a certain linear sub-space of the tangent space. This collection of sub-spaces forms a sub-bundle of the tangent bundle, often called a distribution. (Not to be confused with the distributions in analysis such as the so-called Dirac delta function.) This topic is studied extensively in both differential geometry and control theory/mechanics. Keywords: non-holonomic constraints; Frobenius theorem; Pfaffian systems; distributions; derived flag; Chow theorem. Kevin. ==== > [snip] See the section ``Delta > 0 is Square'' in Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0 - PDF File at http://hometown.aol.com/jpr2718/ Either methods in that section or elsewhere in the file should cover what you > ask, depending on what you mean by ``this form of equation,'' and what you mean > by `` very large.'' As another poster noted, your equation has no solutions. As written, the lhs > is odd and the rhs is even. By 'very large' I mean that cofficients and variables may have several hundred digits. But if factoring is required... I guess I'm out of luck, right? ==== >My question now is how we can >integrate over a sphere using spherical coordinates when they are >badly behaved at several points, they don't constitute an atlas so >under this coordinate regime the sphere isn't a manifold, what is it, >why do integrals still work. One approach is to use a partition of unity. We can find two coordinate charts and functions f and g that are nonzero only on each chart separately, but where f+g = 1 everywhere. Then the integral of a function u can be computed as the integral of f*u + g*u, where the integral of f*u can be evaluated using the one coordinate chart, and the integral of g*u can be evaluated using the other one. For other manifolds more coordinate charts may be needed of course. Keith Ramsay ==== |I presume that Petry wants to subordinate the free enquiry of mathematics |to the commericial/governmental/military sponsored |interests of the scientists. Luckily he has no power or influence. I'm a little bit surprised at how popular the point of view is, that says what science and mathematics need is to be more firmly under the thumb of those sponsors. Keith Ramsay ==== |But is there any reason to think that constructive mathematics |*helps* in applied mathematics? A constructivist once told me that it did. Of course, one can shrug this off as bias, but I don't have any particular reason to think that he was claiming this because he held constructivist opinions, rather than that he held constructivist opinions because he'd discovered such things as it's having advantages in applicability. He also investigated this kind of thing more thoroughly than most people. It's always a little tricky because the people who do the most investigation tend to be self-selected for having come to believe that what they're investigating is good stuff, but I haven't seen good reason to think he's mistaken about this. By comparison, I can only offer a relatively off-the-cuff impression. My hunch is that what is good about it for applications (whether it actually makes it better than classical mathematics or not) springs from the same source as all its other kinds of goodness, rather than because of an extra advantage toward application. I say this partly because the latter notion seems to be a popular target for criticism, as if what we're doing when we weigh constructive and classical mathematics against each other is trying to decide whether the advantage in applicability is enough to outweigh losses thought to exist in other areas. |I don't think it really makes |much difference. Constructively, you can try to prove that (for |example) a differential equation has a solution, and the proof |itself will provide an algorithm for computing the solution. |Classically, one would break it into two different tasks: |(1) Deciding whether the equation has a solution, and |(2) Finding an algorithm for computing the solution. If |the classical mathematician manages to complete task 2, |then hasn't he accomplished the same practical goal as |the constructive mathematician? Backtracking to determine whether the end result is computable doesn't strike me as conducive to complete illumination of what is really going on. Aesthetically this is perhaps the main thing about what little constructive mathematics I've been able to do that I find most appealing: it seems much more like a complete laying bare of the nature of a mathematical situation, not by heroic efforts at incorporating computability considerations afterward, but simply in the natural course of development of the area. |The algorithms might turn out to be different, because |the constructive mathematician will come up with an algorithm |that is provably correct *constructively*, while the classical |mathematician only needs to prove that it is correct *classically*. The algorithms might turn out to be different, of course, but the difference between finding a constructive proof and a classical proof of its correctness is perhaps different from what you think. For a fairly broad class of results of this kind, any classical proof provides one with a constructive proof in an automatic way: if for each m, there exists an n such that P(m,n) where m and n are integers and P is primitive recursive has a classical proof, then it has a constructive proof. The slogan used (by certain logicians) is that classical and constructive mathematics have the same provably recursive functions. So the difference hinges on more subtle differences between the two standpoints. I think an example I mentioned not so long ago might serve again here. In the book of Pour El and Richards, they consider different senses of computability for solutions of a wave equation. In constructive mathematics two of them correspond naturally to two ways of defining L^2(R^3). The usual definition in classical mathematics is that they are the square integrable functions on R^3, modulo the ones that are zero except on a set of measure zero. Alternatively it can be defined as the completion in the L^2 norm of a space of smooth test functions. During a constructive development, the two definitions are kept distinct because they aren't constructively equivalent. You are naturally led to see the differences between them. Doing a classical development seems to me to be a little misleading, because you prove along the way that the two definitions are the same (i.e., nonconstructively equivalent) and thus are liable not to respect the distinction. Then it reappears in a more confusing guise when you try to work out some computability theory related to them. I think the people who've cited it as an example of potentially uncomputable physics, for example, have essentially been conned by classical mathematics' own misleading appearance. So one possible difference in how a mathematician working from a classical standpoint would develop an algorithm, and how a mathematician working from a constructive standpoint would, is that the terms in which the classical mathematician is considering the problem are to the constructive mathematician a bit like chimeras, combinations of animals from distinct species thought of as if they were simply individual animals. A proof might in its course might cultivate an element of L^2 (a sheep) and then make off with it as a square integrable function modulo equality almost everywhere (a goat) without being aware of what kind of information is being thrown out in doing so. Keith Ramsay ==== [snip] > I don't think that any mathematicians are saying that ZFC > is the *only* way to think about real numbers. They are > just saying that it is an extremely powerful way to think > about them. But it isn't. It is the best way to avoid any further thoughts or deeper analysis into the nature of real numbers. It's _the_ way to free math of philosophical considerations. Originally, mathematicians thought that one's philosophical assumptions couldn't possibly influence the math. This view is still popular in math. But it isn't true (intuitionism is one example, but perhaps more will be found, when we start looking). When it comes to foundations, things simply can be much more subtle and interesting than is suggested by FOL/ZF. And that insight is overwhelmingly absent. > Are there actual examples of results that > could not be obtained using ZFC? Sure. But mainstream maths can't appreciate these as 'actual results'. But let me ask another question: Is every result provable from ZFC a sensible result of mathematics? (I'm not saying useful, or applicable: 'sensible'.) For example, what do we ever do with the set P(P(P(P(P(P( N )))))) ? Then why do large sets play such a large role in ZFC? It seems to me that maths could do with much _less_ than ZFC, and it would require quite some (worthwhile) effort to find out how. >BTW, i know that intuitionism often leads to irritation >on the side of classical mathematicians. >They think that intuitionists sort of want to replace >classical math with something else. >But did non-Euclidean geometry 'replace' Euclidean geometry? >No, it just made an end to its exclusive monopoly. >And that only made math richer, not poorer. Right? I don't have any problem > with including intuitionist mathematics. I only have a > problem with David's wanting to *exclude* classical > mathematics. It is questionable whether the current foundations of math are really representing 'classical mathematics'. To me, ZFC is assuming far too much. Another problem with the current foundations is, that in practice it is an alibi to stop thinking about foundations. And a third problem is: ZFC makes us all sound like mediaeval wizards from the past, talking in their own language, about weird fantasy objects, only known by the in-crowd, after years of study. And, be fair: can anyone become a mathematician, nowadays, without being turned into a ZFC-talker? Herman Jurjus ==== |And, be fair: can anyone become a mathematician, nowadays, |without being turned into a ZFC-talker? As often as people on usenet refer to ZFC, I think the typical mathematician knows very little about it, nor does it make much difference to them. Keith Ramsay ==== |And, be fair: can anyone become a mathematician, nowadays, >|without being turned into a ZFC-talker? As often as people on usenet refer to ZFC, I think the typical >mathematician knows very little about it, nor does it make >much difference to them. Yes, I think you are right. It occurred to me recently that I have almost no instinct for what can be proved without using AC. For example, can you prove that there is no group G with |Aut(G)| = 3 (a problem that arose recently on sci.math) without using AC ? You reduce easily to an infinite abelian group G in which all elements have order 2 or, equivalently, an infinite vector space over the field of order 2. Without AC, can you say anthing at all about Aut(G) ? Derek Holt. ==== This problem has nothing to do with school, work, or anything like that. It's just a problem that occurred to me one day, and has been driving me nuts. Perhaps somebody here can provide me the solution, or at least give me some pointers. Here's the problem. Consider a game with the following rules: You are given an initial score and a time limit. We'll denote the time remaining by t, and your score at that time by x[t]. Your score changes over time as a random walk with a fudge factor that attracts it to zero: x[t-dt] = x[t]*e^(-dt) + Z*sqrt((1-e^(-2*dt))/2) where e is the bast of natural logarithms and Z is a standard normal random value (mean zero, variance 1). (Note that, for dt<<1, x[t-dt]-x[t] is approximately -x[t]*dt + Z*sqrt(dt).) You may stop the game at any time. If you stop the game, your final score is the score you have when you stop it. If you do not stop the game, your final score is the score you have when time expires. QUESTION #1: What is the optimal strategy? (I trust that the meaning of strategy and optimal are clear. I can post definitions if necessary.) QUESTION #2: Given an initial score x0 and time t0, and assuming that you play the optimal strategy, what is the expected value of the game? ==== > currently, the politic does to the people what they can't do back. I know this, 100,000 people in Townville know this and will > report to you that the government tortures adam of the bible > The population of Townsville is closer to 130,000 - which brings up the question, what else characterizes the 30,000 or so people of Townsville that do not know this thing you assert (namely, that the politic does to the people what they can't do back)? the politic does for the people what they can't do for themselves, > as far as I am concerned, this means the same as Seperate Church and State! > So, the politic does to the people what they can't do back, and also the politic does for the people what they can't do themselves. And these are not necessarily related, I suppose, anyway I think I'm following you so far, although the significance of what you are trying to say is failing me. Let Adam decide when he wants to be filmed, stop the Truman Show. > Adam is here for the future of humanity, not your personal training. > My personal training? Wait - what do you know of me? I'm not one of the 130,000 some residents of Townsville. But I accept your stipulation that there is an Adam, presumably a resident of Townsville, and that he is here for the future of humanity. You are not explicit about what being here for the future of humanity means, but I'll assume Adam means to do humanity some good. > I give you all this chance to do for yourselves, if the people > can take religion then the government will be YOUR audience. > If taking religion means realising that religion plays an important role to a subset of humanity, I'm not sure I understand how this realization leads to the condition that then the government becomes my audience, or what the significance of this is. > If the people can't do for themselves the seperation of church > and state then the government is not going to let go of a free > nuclear blast shield that is Adams current immortality, and > their free infinite knowledge they spy on all his life. > Mr. Cooper, are you currently keeping up with your medication? Can you recommend any way for the average reader to be able to tell, from your posts, which ones were posted while you are stable and medicated, and which ones are done whilst in the dark grip of a serious mental illness? > If the people can't seperate church and state, then you are > forever to live a life that the politic will do to you anything > that you can't do back. God is on your side, be on his. phone any of 100,000 people in Townsville Australia to > verify, just like the following 4 posts, that the truman show > is running live there, and the truman is constantly tortured > by a satelite that penetrates buildings and a team of narcisistic > agenst that work around the clock to persecute Adam. ME aus.tv know The Truman Show is true > _____________________________ > I'm from Townsville and YOU ARE the Truman! > http://tinyurl.com/iky5 > I was in Townsville over the weekend, and I heard him. > Very spooky! > http://tinyurl.com/iky8 >phone someone in Townsville, half of you must know someone there, >every day I go out people say THERES THE TRUMAN > I'm in Townsville. We're sick of you. > http://tinyurl.com/iky9 > http://tinyurl.com/iky4 > You rule Truman! ==== received didn't help me enough to figure out a proof. I am not a > student, so please don't feel that you are helping me cheat on > homework or something. I am using this group precisely because I > don't have the advantage of a professor to ask questions of. The question: Let P(A) denote the power set for some set A. > A subset B of P(A) is called an antichain if no element > of B is a subset of any other element of B. With N denoting the set > of natural numbers, prove that P(N) contains an uncountable antichain. The hints I recieved said to look for an uncountable antichain in > either P(Q) or P(QxQ), where Q is the set of rational numbers. Since > Q and QxQ have the same cardinality as N, this should solve the > problem. As most of these types of exercises use a diagonal argument, > I expect that one is probably required. So, what is the answer? There is a Theorem to the effect that there is a subset of P(N) of the > power of the continuum such that each element of A is infinite and if X > and Y belong to A then X / Y is finite (and consequently neither is > contained in the other). For each real r, 0 < r < 1 let > U(r) = { 2^n*(floor(n*r) +1) : n = 1, 2, ...}. > A = {U(r) : 0 < r < 1, r real}. ==== >> Well, that was too easy because it's already upper triangular. >> But if that 0 in the second column wasn't there, what you'd do ... >So this example is too special, that's why it can be done in one step, >instead of Strang's three steps? You said: But if that 0 in the second >column wasn't there, ... >Which zero? U1? In the previous sentence, I said because it's upper triangular. I'm referring to the 0 in the second column that makes A1 upper triangular. >So can you explain more clearly when can I use the simplified GUESSING, >when I need to use Strang's method? OK, let's try another example. [ 3 -2 8 ] A = [ -4 6 6 ] [ 1 -4 11 ] This has an eigenvalue 5 with normalized eigenvector [ 2/3 ] [ 2/3 ] [ 1/3 ] which we complete to the unitary matrix [ 1/2 ] [ 5 ] [2/3 ---- 0 ] [ 3 ] [ ] [ 1/2 1/2 ] U1 = [ 4 5 5 ] [2/3 - ------ ---- ] [ 15 5 ] [ ] [ 1/2 1/2] [ 2 5 2 5 ] [1/3 - ------ - ------] [ 15 5 ] Then [ 1/2 1/2] [ 11 5 74 5 ] [5 - ------- - -------] [ 5 15 ] U1^(-1) A U1 = [ ] [ -14 ] [0 29/5 --- ] [ 15 ] [ ] [0 -18/5 46/5 ] Note the first column has 0's below the diagonal, but the second column doesn't, so it's not upper triangular yet. Take the lower right 2 x 2 submatrix [ -14 ] [29/5 --- ] A2 = [ 15 ] [ ] [-18/5 46/5] This has an eigenvalue 10 with normalized eigenvector [ 1/2 ] [ 2 85 ] [ ------- ] [ 85 ] [ ] [ 1/2] [ 9 85 ] [- -------] [ 85 ] Complete this to a 2 x 2 unitary matrix [ 1/2 1/2] [ 2 85 9 85 ] [ ------- -------] [ 85 85 ] U2 = [ ] [ 1/2 1/2] [ 9 85 2 85 ] [- ------- -------] [ 85 85 ] which is the lower right 2 x 2 submatrix of the 3 x 3 unitary matrix [1 0 0 ] [ ] [ 1/2 1/2] [ 2 85 9 85 ] [0 ------- -------] V2 = [ 85 85 ] [ ] [ 1/2 1/2] [ 9 85 2 85 ] [0 - ------- -------] [ 85 85 ] Note that U2^(-1) A2 U2 is upper triangular. Then you want [ 1/2 1/2 1/2 1/2 ] [ 2 5 85 3 5 85 ] [2/3 ------------ ------------ ] [ 255 85 ] [ ] [ 1/2 1/2 1/2 1/2] U = U1 V2 = [ 7 5 85 2 5 85 ] [2/3 - ------------ - ------------] [ 255 85 ] [ ] [ 1/2 1/2 1/2 1/2] [ 2 5 85 2 5 85 ] [1/3 ------------ - ------------] [ 51 85 ] and lo! [ 1/2 1/2 1/2 1/2] [ 8 5 85 89 5 85 ] [5 ------------ - -------------] [ 17 255 ] U^(-1) A U = [ ] [0 10 8/3 ] [ ] [0 0 5 ] Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2