In a fifty-year-old math paper about integrals of hermite polynomials the author uses this strange symbol (-m)_{k}, i.e. lower index notation, for what appears to be something like binomial coefficients (m, k not necessarily integers). Is that correct? Or what else does it mean? It's probably NOT binomial coefficients if my own calculations are correct. But it's definition for integers appears to be some quotient of factorials. Heiko Gimperlein ==== I would guess that (x)_k denotes x(x-1)(x-2)...(x-k+1). Does that make sense in this paper? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== Might be a rising or falling factorial power. For integer exponent, m to the rising power k is the product of the k factors m, m + 1, m + 2, ..., m + k - 1, and m to the falling power k is the product of the k factors m, m - 1, ..., m - k + 1. For k zero, the empty product is 1 as usual; for k negative, define so as to keep certain simple laws of exponents. Factorial of the natural number m is either 1 to m rising or m to m falling. I learnt this from Graham, Knuth, Patashnik: Concrete Mathematics. -- Jussi ==== > 1. what is the magnitude of the set of all groups? There is no set large enough to contain all groups. > 2. what is the magnitude of the set of all sets? There is no set large enough to contain all sets. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== If I did not make a mistake calculating the first few terms by hand, here is a recursively-defined sequence which is not in the EIS yet. a[0] = 1; and for m >= 1, a[m] = sum{j=0 to a[m-1](mod m)} a[j] Ascii-art: a[m-1](mod m) --- a[m] = / a[k] --- k=0 And, 0 <= a[m-1](mod m) <= m-1. The sequence begins (maybe): 1, 1, 2, 4, 1, 2, 4, 9, 2, 4, 9, 30, 15,... What can be said about this sequence? Does it have a closed-form (ie nonrecursive) representation? Also, other sequences can be based on the same idea: a partial sum somehow involving earlier terms, BOTH in the general term and in the limit of the indexes used in the sum. Leroy Quet ==== > No, this is your mistake. If both A and A-B are infinite, then there exist > permutations sigma of A for which sigma(B) is a proper subset of B. The > inverse of such a permutation would not satisfy sigma^-1(B) <= B, so > the set you define is not a subgroup. > > For example, A = integers, B = positive integers, sigma(x) = x+1. > Ahh....thanks. I forgot about infinite sets. Good to hear (from the other poster) that my proof works for finite ones at least. ==== msk ==== Bob, your explanation shows why the common presentation of multiplication of complex numbers as a free convention : (a,b)*(c,d) = (ac-bd, ad+bc) is insufficient because also (a,b)*(c,d) = (ac-2bd, ad+bc) will conform to the laws of algebra. As you showed its necessary to introduce the concept of metrics. L.Rodriguez ==== Can anyone suggest how to show that: If V is an inner product space and T is an orthogonal projection on V then T is self adjoint. The converse is also true, but I didn't have any problem proving it. dan ==== > Can anyone suggest how to show that: > > If V is an inner product space and T is an orthogonal projection on V > then T is self adjoint. > > The converse is also true, but I didn't have any problem proving it. = = + = since y-Py is orthogonal to the range of P. Similarly, = (or exchange x and y and take conjugates), leading to the desired = . --Ron Bruck ==== What's fascinating about math proofs is that they can be staring you in the face, hiding in plain sight, as it were. For instance, consider all the arguments about the factorization 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) and the question of factors in common with 5 among the a's. With the focus on sqrt(5) no one seemed to notice that the claims of counterexample against my work would work against *any* non-unit algebraic integer f which is a factor of any of the a's. To see that go ahead and solve for the roots like before to get a^3 + 12a^2 - 65 = 0. Now use a = fb, to get f^3 b^3 + 12f^2 b^2 - 65 = 0, and divide out f^2 to get f b^3 + 12b^2 - 65/f^2 = 0. For any f, where algebraic manipulations can give you a polynomial with integer coefficients, you will have a non-monic irreducible over Q, which proves that none of its roots are algebraic integers. That fact comes from noting that to get an integer with f, you'll need the others factors of 5 that some of you apparently presume must exist. That is, they form a factor ring, so unless f is a simple radical, like sqrt(5), you need the other factors, so algebraic manipulations won't do it. So then, you have that the polynomial f b^3 + 12b^2 - 65/f^2 = 0 which you might presume has one algebraic integer root, has been shown to have none because you can't form a polynomial with integer coefficients, let alone one that is monic with integer coefficients using less than an infinite number of algebraic manipulations. That problem results with irreducibles over Q, and is the problem I've been talking about for a while. Posters in arguing with me, were helped by my focus on f=sqrt(5), so people were being directed to focus on that case but then failed to realize that the objections would apply across the board for *any* non-unit algebraic integer f. As an example, where you can use algebraic manipulations consider x = (-1+sqrt(-3))/2, so 2x + 1 = sqrt(-3), and squaring both sides gives 4x^2 + 4x + 1 = -3, so 4x^2 + 4x + 4 = 0, and dividing off 4, you have x^2 + x + 1 = 0, which is the polynomial which shows that (-1+sqrt(-3))/2 is an algebraic integer. For some of you the question of algebraic manipulations may seem esoteric, so I've presented the question of the polynomial y^3 + sy^2 + ty - 13 that would result if you divided factors in common with 5 from the roots. For those of you still not convinced by the proof above, perhaps trying to find 's' and 't' will help you. So yes in fact the odd result is that the three roots r_1, r_2, and r_3, of a^3 + 12a^2 - 65 are in fact *each* coprime to 5 ***in the ring of algebraic integers***. A rough analogy is the case of 2 and 6 in the ring of evens, as 6 is prime in the ring of evens as it is coprime to 2. However, the ring of evens is still complete, while the problem I've described in the ring of integers leads to the contradiction that though r_1 r_2 r_3 = 65 r_1, r_2 and r_3 are each coprime to 5. The mathematics is here strange enough to confuse, but it needs to be considered carefully to see if any large errors have snuck into core mathematics along with it. After all, mathematics is extremely sensitive to flaws. Even a very small error can have huge repercussions. James Harris ==== ESB Consultancy is pleased to announce the latest release of ESBCalc Pro - a full featured, scientific calculator for Win32 Platforms. http://www.esbconsult.com/esbcalc/esbcalcpro.html ESBCalc Pro is an Enhanced Scientific Calculator for Win32 Platforms with Infix Notation, Full Exponential Notation support, Brackets, Scientific Functions (Trigonometric, Hyperbolic, Logarithmic - including Base 10, Base 2 & Natural - plus more), Memory, Paper Trail, Result Help. Now also includes Floating Decimal Point, optional start in last position and many other enhancements to the User Interface and Calculator Engine. Grab a trial version today :) -- ESB Consultancy, http://www.esbconsult.com Home of ESBPCS, ESBStats, ESBPDF Analysis & ESBCalc Kalgoorlie-Boulder, Western Australia (TeamND, TeamOE, Addict Support, eLists.org Management) They are making fun of you because they know you are schizophrenic. This isn't very nice of them, but there are people who will get attention for themselves out of the problems of others. Get help. You need to see a psychiatrist. -- Michelle Malkin (Mickey) http://questioner.www2.50megs.com atheist/agnostic list ordainer EAC Bible thumper thumper BAAWA Knight who says SPONG! ==== > lot of talk no action > haven't seen any replies with the answer next to the phone number yet! > its not that difficult, this whole town of 100,000 people are all in on it. No one responded to your first post because you are a COMPLETE FUCKING > LOON, and it made no sense. > the others have clarified it, > its going up your optic nerve but the signals not registering. > 1000s of 1000s of witnesses to I'm the truman, all have known > for ___over_1_year___, i'm getting impatient it hasn't gone public yet > so what can I do? make sense yet? you got anything to lose > by replying 'yes i checked your witnesses they didn't know any truman', > you got anything lose 'yes these 5 all said its real, you R the truman'. > > Herc Herc, What you say if the following townsville numbers didn't have a clue what on earth you were talking about: 4750 0700 4031 1107 4753 4444 4779 0233 4760 1380 4781 4182 4781 4301 ==== > What you say if the following townsville numbers didn't have a clue what on earth you were talking about: Let's take a look at these numbers (just out of curiosity): > 4750 0700 Great Barrier Reef Marine Park Authority - ermits Officer - Research > 4031 1107 NuShape Foods Outlet > 4753 4444 Australian Institute of Marine Science > 4779 0233 Townsville Vet Clinic > 4760 1380 Alamo Rent-A-Car - Townsville Airport > 4781 4182 James Cook University - School of Psychology > 4781 4301 James Cook University - Distance Education Interesting what you can come up with using Yahoo! for just a couple of minutes, eh? :) Have a nice day. -- David Loewen -- - http://www.degoo.com/index.php?refid=mitaka ==== > What you say if the following townsville numbers didn't have a clue what on earth you were talking about: Let's take a look at these numbers (just out of curiosity): > 4750 0700 > Great Barrier Reef Marine Park Authority - ermits Officer - Research 4031 1107 > NuShape Foods Outlet 4753 4444 > Australian Institute of Marine Science 4779 0233 > Townsville Vet Clinic 4760 1380 > Alamo Rent-A-Car - Townsville Airport 4781 4182 > James Cook University - School of Psychology 4781 4301 > James Cook University - Distance Education Interesting what you can come up with using Yahoo! for just a couple of > minutes, eh? :) Have a nice day. interesting what reading a post will reveal, 0700, 1107, 4444? so don't phone businesses with non business enquiries. I guarantee several of these will say Yes The Truman lives in Townsville and Yes we can hear him. I can post up more random numbers from the phone book or scan a few pages and upload that if its not enough witnesses. Try is anything weird happening in Townsville the last year? if you want to guarantee they're not being sarchastic. 47482160 > 47788360 > 47792822 > 47290642 > 47254486 > 47855847 > 47491445 > 47230018 > 47785779 > 47861843 > 47772731 > 47720161 > 47714484 > 47211929 > 47753611 > 47210420 > 47235886 > 47790345 > 47831124 The comments you will get will be exactly like the posts to aus.tv, the only people admitting they were from townsville ALL shared the same sarchasm approach, direct confirmation. Herc ==== The comments you will get will be exactly like the posts to aus.tv, >the only people admitting they were from townsville ALL shared the >same sarchasm approach, direct confirmation. Herc > You still can't accept that these people are putting you on. They > know that you are totally around the bend. By the way, tell the Powerpuff Girls that I said hi > on the evidence so far is unjustified. People might not speak up here because of group persecution, doesn't mean your cult has any validity. Yeah, I'm a newbie at sci.math, and to tell you the truth I don't understand what the fuss is about. Could you please take your time for a minute or so and tell me what is it that you're trying to say. Ok, this is what I know so far, you claim that there are camera's on you, and people from Townsville Australia admit this. And I saw those 4 links of those alleged Townsvillers. Ok *********** Seems you've jumped to the conlusion there is no such thing as a truman and worked your psychological assesment back from that. There's only one alias showing on the google search phrase i am the truman. Its a 5 billion to one, but had to be someone. I have an online proof at www.adamskingdom.com I have half dozen proofs but you've yet to pull your fingers out of your ears. Take several hours to understand the website proof if you're smart and open minded, several minutes to contact the 100,000 witnesses. They all listen to the truman company interrogate me and discuss the myriad of evidence, so coming from an authorative source they are open enough to believe it, noone in Townsville disputes it. Likely you will only open your eyes when a recognised newsreader presents the same facts I do. Herc ==== > Interesting what you can come up with using Yahoo! for just a couple of > minutes, eh? :) > > Have a nice day. > > -- > David Loewen Had to get them from somewhere! Doh! Thought I put a shrink on there.. ==== > > interesting what reading a post will reveal, 0700, 1107, 4444? > so don't phone businesses with non business enquiries. > > I guarantee several of these will say Yes The Truman lives in Townsville > and Yes we can hear him. I can post up more random numbers from > the phone book or scan a few pages and upload that if its not enough > witnesses. Try is anything weird happening in Townsville the last year? > if you want to guarantee they're not being sarchastic. > Why do you think residents would, but businesses wouldn't? Surely the people running the businesses are residents as well? ==== line always seems to be busy. Dennis -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > I am wondering if there are recommended guidelines concerning timed > math facts -- number of facts per minute (or set amount of time) for > each grade level? I'm wondering how anyone would know what guideline would be enough. Would someone set this up as a criterion-based set of standards (based on what theory of how mathematical ability develops?) or would one simply empirically observe age norms and then report those? How could we be sure that presentation by paper worksheets in vertical format, paper worksheets in horizontal format, flash cards with oral answers, computer-based drills with multiple-choice answers, computer-based drills with typed-in answers, are strictly comparable? How do we know that what is the age norm in one country is comparable to what is the age norm in another country? Part of the premise that underlies the question here, I fear, is the usual assumption of lock-step age-grading in school http://learninfreedom.org/age_grading_bad.html and part of it is the assumption that facility with math facts is IN ALL INSTANCES necessary for further advancement in mathematical reasoning, for which the famous story of Ernst Eduard Kummer http://dirphys.harvard.edu/Humour/Science/sciencejokes/sciencejokes.10of9.tx t provides a counterexample. Just wondering . . . . P. S. The EPGY program at Stanford University http://epgy.stanford.edu/ would have data in their files about math facts performance of a self-selected group of gifted children who use a computer-based distance-learning math course. I don't know how much of the research specifically related to their math races program has been published. My son used those materials to learn to solve conditional probability word problems mentally at fairly high speed while he was of fifth-grade age. I have no idea what is a reasonable expectation for the general population for solving that kind of problem, which is probably insoluble to most college graduates in the United States. -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > Problem 4. Labeled balls (as in #2 and #3). At step n, remove both > balls labeled n and 10n. Switch the labels these balls, then return > the ball newly labeled 10n back in the basket. Discard the ball newly > labeled n. I think JH will insist there are no balls left in the basket > at noon. But how does this differ from Problem 3? > > In problem #4, a ball initially labelled n will eventually be removed, > relabelled 10n and put back. However, every integer, no matter how > many times you increase it tenfold, remains finite. There are an > infinite number of times each ball is removed and an infinite number > of times it is put back (albeit with a new label)...a classic infinity > minus infinity problem. > > The solution, as before, is noting that any non-empty collection of > natural numbers has a least element. If a ball exists in set S = lim > Sn, it must have a label, and that label must have a positive finite > number. But Label n was removed at step n, and this is true with > every n. Therefore, the set of integers on the remaining labels has > no smallest value. So there are no labels, and thus no balls. The > set is empty. I agree that there are no labels, but I don't agree that there are no balls. It seems clear to me that any ball with an initial label not divisible by 10 will never be removed. So at noon, there will still be an infinite number of (unlabeled) balls remaining. It's not contradictory that the balls are labeled before noon but not after noon, just as in the original problem it's not contradictory that the bucket is non-empty before noon but not after noon. > Problem 5. Labeled balls. At the nth epoch, you remove one of the > remaining balls at random (uniformly amongst the remaining balls). In > this case, my probability calculations say that, with probability 1, the > basket is empty at noon. (My calculations may be erroneous - if you get > a different answer, ask and I will share. Basically, the probability > that any specific ball remains is 0. I don't *think* I have made an > unjustifiable reordering of limits anywhere.) Yet how does this differ > from the #1 (unlabeled balls)? > > I have not really examined this case, but I would agree. For example, > the probability that Ball #1 would be removed in Step 1 is 1/10, in > Step 2 is 1/19, in Step n is 1/(9n+1), the sum of these independent > trials appears to be 1. > > It differs from the unlabelled balls example because you have > specified a rule to determine which ball to remove. Given the rule > for each step n, we can determine a solution (even if it's only a > probability distribution of answers). I have no idea how to approach this. It might be easier to start with this more straightforward variant: You have a bucket, initially with no coins in it. For n>0, at time 1-2^(-n), you put a coin in the bucket (say coin n), flip all the coins in the bucket, and remove all the coins that show heads. What's the probability distribution for the number of coins in the bucket at time 1? It feels like the bucket should be empty with probability 1, but I haven't been able to put together a convincing argument. - Nate ==== > I agree that there are no labels, but I don't agree that there are no > balls. It seems clear to me that any ball with an initial label not > divisible by 10 will never be removed. So at noon, there will still > be an infinite number of (unlabeled) balls remaining. It's not > contradictory that the balls are labeled before noon but not after > noon, just as in the original problem it's not contradictory that the > bucket is non-empty before noon but not after noon. Since by the rules of the game, no ball is reentered without a label. I think that no labels = no balls (please, no jokes on this!) is therefore valid. Unless you can show me a natural number n such that 10n is not a natural number, I think the labeless ball hypothesis cannot happen. > You have a bucket, initially with no coins in it. For n>0, at time > 1-2^(-n), you put a coin in the bucket (say coin n), flip all the > coins in the bucket, and remove all the coins that show heads. What's > the probability distribution for the number of coins in the bucket at > time 1? > > It feels like the bucket should be empty with probability 1, but I > haven't been able to put together a convincing argument. Hmmm...interesting problem. For any coin n, the probability that the coin flips tails is 1/2, and thus the probability that it will never go to heads is lim n->oo (1/2)^n = 0. As this is true for each coin, I would think that the ending container is empty happen with probability 1. ==== > The balls-in-the-bucket problem defines what happens at every finite > step, but does not specify how to pass to the limit. Therein lies the > ambiguity. Let's try this version. Instead of a single bucket, we have an infinity each room has one ball and one bucket, and is responsible for placing the ball in the bucket and then removing it at the prescribed times. At noon the occupants are to phone their results to the front desk, reporting whether each ball is in its bucket or not. How many balls are in the buckets at noon? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > The balls-in-the-bucket problem defines what happens at every finite >> step, but does not specify how to pass to the limit. Therein lies the >> ambiguity. Let's try this version. Instead of a single bucket, we have an infinity >each room has one ball and one bucket, and is responsible for placing the >ball in the bucket and then removing it at the prescribed times. At noon >the occupants are to phone their results to the front desk, reporting >whether each ball is in its bucket or not. How many balls are in the buckets at noon? Here again you are assuming the pointwise convergence, and I agree it makes the most physical sense, and I have never disagreed that with that assumption there will be no balls at noon. I have said this numerous times. However, it is not necessary that this type of convergence be assumed, and by assuming another type of convergence, you might get a different limit. I am not saying that yours is not the answer I would give if asked this question. However, given that the question was asked about some other possible limits, I tried as best I could to give topologies that would lead to those limits. Rob Johnson take out the trash before replying ==== > The balls-in-the-bucket problem defines what happens at every finite > step, but does not specify how to pass to the limit. Therein lies the > ambiguity. >>Let's try this version. Instead of a single bucket, we have an infinity >>each room has one ball and one bucket, and is responsible for placing the >>ball in the bucket and then removing it at the prescribed times. At noon >>the occupants are to phone their results to the front desk, reporting >>whether each ball is in its bucket or not. >>How many balls are in the buckets at noon? > Here again you are assuming the pointwise convergence, and I agree it > makes the most physical sense, and I have never disagreed that with that > assumption there will be no balls at noon. I have said this numerous > times. However, it is not necessary that this type of convergence be > assumed, and by assuming another type of convergence, you might get a > different limit. No, I am not assuming convergence of any sort. If you were to mention idea what you were talking about. All he sees is one ball and one bucket. > I am not saying that yours is not the answer I would give if asked this > question. However, given that the question was asked about some other > possible limits, I tried as best I could to give topologies that would > lead to those limits. And I tried as best I could to explain why every answer that involves a topology on the space of functions a_n is irrelevant to the problem. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > The balls-in-the-bucket problem defines what happens at every finite >> step, but does not specify how to pass to the limit. Therein lies the >> ambiguity. Let's try this version. Instead of a single bucket, we have an infinity >each room has one ball and one bucket, and is responsible for placing the >ball in the bucket and then removing it at the prescribed times. At noon >the occupants are to phone their results to the front desk, reporting >whether each ball is in its bucket or not. How many balls are in the buckets at noon? > Here again you are assuming the pointwise convergence, and I agree it >> makes the most physical sense, and I have never disagreed that with that >> assumption there will be no balls at noon. I have said this numerous >> times. However, it is not necessary that this type of convergence be >> assumed, and by assuming another type of convergence, you might get a >> different limit. No, I am not assuming convergence of any sort. If you were to mention >idea what you were talking about. All he sees is one ball and one bucket. Yes, and he sees his bucket always empty after a certain time, so he sees his bucket (his point) converging to an empty bucket (under the discrete topology, where convergence is rather monotonous). Considering all the buckets, we get pointwise convergence of the bucket indicator functions. >> I am not saying that yours is not the answer I would give if asked this >> question. However, given that the question was asked about some other >> possible limits, I tried as best I could to give topologies that would >> lead to those limits. And I tried as best I could to explain why every answer that involves a >topology on the space of functions a_n is irrelevant to the problem. Your assumption is that if a bucket is always empty after a certain time before noon, then that bucket is empty at noon. This is the assumption of pointwise convergence. If you don't assume some sort of continuity, you cannot deduce the state at noon from the states before noon and all we are given is how to know the states of the buckets before noon. Rob Johnson take out the trash before replying ==== >>Let's try this version. Instead of a single bucket, we have an infinity >>each room has one ball and one bucket, and is responsible for placing the >>ball in the bucket and then removing it at the prescribed times. At noon >>the occupants are to phone their results to the front desk, reporting >>whether each ball is in its bucket or not. >>How many balls are in the buckets at noon? > Here again you are assuming the pointwise convergence, and I agree it > makes the most physical sense, and I have never disagreed that with that > assumption there will be no balls at noon. I have said this numerous > times. However, it is not necessary that this type of convergence be > assumed, and by assuming another type of convergence, you might get a > different limit. >>No, I am not assuming convergence of any sort. If you were to mention >>idea what you were talking about. All he sees is one ball and one bucket. > Yes, and he sees his bucket always empty after a certain time, so he sees > his bucket (his point) converging to an empty bucket (under the discrete > topology, where convergence is rather monotonous). Considering all the > buckets, we get pointwise convergence of the bucket indicator functions. Are you saying we have no balls left at noon *because* pointwise convergence holds, or are you saying that pointwise convergence holds *because* it describes the fact (independently verifiable via ZF) that no balls are left at noon? If the former, then you have not explained *why* pointwise convergence holds. > Your assumption is that if a bucket is always empty after a certain time > before noon, then that bucket is empty at noon. This is the assumption > of pointwise convergence. Then you must think the assumption of pointwise convergence is equally necessary for solving the five apples, take away two problem. >If you don't assume some sort of continuity, > you cannot deduce the state at noon from the states before noon and all > we are given is how to know the states of the buckets before noon. For which bucket are we unable to deduce the state at noon? You can't prove that no solution exists by merely pointing out where one attempted solution fails. Particularly not when you have already been shown an entirely different solution that does not have that defect. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Let's try this version. Instead of a single bucket, we have an infinity >each room has one ball and one bucket, and is responsible for placing the >ball in the bucket and then removing it at the prescribed times. At noon >the occupants are to phone their results to the front desk, reporting >whether each ball is in its bucket or not. How many balls are in the buckets at noon? > Here again you are assuming the pointwise convergence, and I agree it >> makes the most physical sense, and I have never disagreed that with that >> assumption there will be no balls at noon. I have said this numerous >> times. However, it is not necessary that this type of convergence be >> assumed, and by assuming another type of convergence, you might get a >> different limit. No, I am not assuming convergence of any sort. If you were to mention >idea what you were talking about. All he sees is one ball and one bucket. > Yes, and he sees his bucket always empty after a certain time, so he sees >> his bucket (his point) converging to an empty bucket (under the discrete >> topology, where convergence is rather monotonous). Considering all the >> buckets, we get pointwise convergence of the bucket indicator functions. Are you saying we have no balls left at noon *because* pointwise >convergence holds, or are you saying that pointwise convergence holds >*because* it describes the fact (independently verifiable via ZF) that >no balls are left at noon? If the former, then you have not explained *why* pointwise convergence >holds. |Definition: |Suppose {f_n} is a sequence of functions with domain D and that f is a |function also with domain D. The sequence {f_n} converges pointwise to |f iff for each x in D, f_n(x) converges to f(x). Let D be the set of buckets, and f_n(x) be 1 or 0 depending on whether that bucket has a ball in it or not at time 12 - 2^{1-n}. Since f_n(x) can take only two values, 1 or 0, the discrete topology is about the only topology we can give to the range of f_n. To converge in the discrete topology, all terms must be constant after some point. |Definition: |Suppose {a_n} is a sequence in a space with the discrete topology and a |is another point in that same space. The sequence {a_n} converges to a |iff there is an N so that a_n = a for all n > N. This is why I called discrete convergence monotonous. Perhaps, due to the constancy of discrete convergence, it may not appear that there is any convergence at all, the terms are all just the same. Translating the definition of pointwise convergence to the language of buckets and balls and using discrete convergence at each point, we get |A bucket is empty at noon iff it has always been empty since some time |before noon. A bucket has a ball in it at noon iff it has always had a |ball in it since some time before noon. This last quoted statement is something I believe you take for granted, but others take it is an assumption. Whatever, it is just a restatement of pointwise convergence. This is why I have said that you are assuming pointwise convergence. Perhaps I should say you are taking pointwise convergence for granted. >> Your assumption is that if a bucket is always empty after a certain time >> before noon, then that bucket is empty at noon. This is the assumption >> of pointwise convergence. Then you must think the assumption of pointwise convergence is equally >necessary for solving the five apples, take away two problem. No, because there is no infinite sequence in the five apples, take away two problem. Convergence deals with limits of infinite sequences. >>If you don't assume some sort of continuity, >> you cannot deduce the state at noon from the states before noon and all >> we are given is how to know the states of the buckets before noon. For which bucket are we unable to deduce the state at noon? We cannot deduce the state of any bucket at noon if we don't describe how to derive the state at noon from the states at previous times. The problem only tells us what happens before noon; we need to have some way of determining what happens at noon. Assigning the topology of pointwise convergence to the bucket has a ball function seems the most reasonable when asking which balls are left. >You can't prove that no solution exists by merely pointing out where one >attempted solution fails. Particularly not when you have already been >shown an entirely different solution that does not have that defect. I am not trying to show that no solution exists. Quite the contrary, I am trying to show that the solution depends on an added assumption and that depending on that added assumption, different answers arise. Rob Johnson take out the trash before replying ==== >>Are you saying we have no balls left at noon *because* pointwise >>convergence holds, or are you saying that pointwise convergence holds >>*because* it describes the fact (independently verifiable via ZF) that >>no balls are left at noon? >>If the former, then you have not explained *why* pointwise convergence >>holds. >|Definition: >|Suppose {f_n} is a sequence of functions with domain D and that f is a >|function also with domain D. The sequence {f_n} converges pointwise to >|f iff for each x in D, f_n(x) converges to f(x). > Let D be the set of buckets, and f_n(x) be 1 or 0 depending on whether > that bucket has a ball in it or not at time 12 - 2^{1-n}. > Since f_n(x) can take only two values, 1 or 0, the discrete topology is > about the only topology we can give to the range of f_n. To converge in > the discrete topology, all terms must be constant after some point. I do not *assume* that convergence of any type holds. I assume that may *deduce* that pointwise convergence holds, but I do not need that fact to solve the problem. I need only set theory. >|Definition: >|Suppose {a_n} is a sequence in a space with the discrete topology and a >|is another point in that same space. The sequence {a_n} converges to a >|iff there is an N so that a_n = a for all n > N. > This is why I called discrete convergence monotonous. Perhaps, due to > the constancy of discrete convergence, it may not appear that there is > any convergence at all, the terms are all just the same. You keep thinking that all you need to do is explain convergence more fully and completely, and all will become clear. I know what convergence means. > Translating the definition of pointwise convergence to the language of > buckets and balls and using discrete convergence at each point, we get >|A bucket is empty at noon iff it has always been empty since some time >|before noon. A bucket has a ball in it at noon iff it has always had a >|ball in it since some time before noon. And this follows from Newton's first law, does it not? Convergence is merely the language you have chosen to describe what we observe; it is not in any way a *cause* of what we observe. > This last quoted statement is something I believe you take for granted, > but others take it is an assumption. Whatever, it is just a restatement > of pointwise convergence. This is why I have said that you are assuming > pointwise convergence. Perhaps I should say you are taking pointwise > convergence for granted. I am taking Newton's first law and ZF for granted. Pointwise convergence is a *consequence* of those assumptions, not an assumption in its own right. > Your assumption is that if a bucket is always empty after a certain time > before noon, then that bucket is empty at noon. This is the assumption > of pointwise convergence. >>Then you must think the assumption of pointwise convergence is equally >>necessary for solving the five apples, take away two problem. > No, because there is no infinite sequence in the five apples, take away > two problem. Convergence deals with limits of infinite sequences. You have five apples at 11:55. I take away two at 11:59. How many do you have at noon? We need to assume that Newton's first law applies, just as we do with the buckets and balls. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > > The balls-in-the-bucket problem defines what happens at every finite >> step, but does not specify how to pass to the limit. Therein lies the >> ambiguity. Let's try this version. Instead of a single bucket, we have an infinity >each room has one ball and one bucket, and is responsible for placing the >ball in the bucket and then removing it at the prescribed times. At noon >the occupants are to phone their results to the front desk, reporting >whether each ball is in its bucket or not. How many balls are in the buckets at noon? > > Here again you are assuming the pointwise convergence, and I agree it > makes the most physical sense, Nothing about this problem makes *physical* sense. It is an abstract set theory problem. and I have never disagreed that with that > assumption there will be no balls at noon. I have said this numerous > times. However, it is not necessary that this type of convergence be > assumed, and by assuming another type of convergence, you might get a > different limit. As Jonathan Hoyle has pointed out, your solution 1 confuses card(lim S_n) with lim(card S_n). Where is it stated that the sequence of cardinalities *must* have a limit equal to the cardinality of the set at noon? You have merely assumed this, in your solution 1. Unfortunately, as Dave Seaman has pointed out, your assumption contradicts the axiom of separation, i.e. is invalid in ZF. So it *cannot* be taken as an unstated given of the problem; in other words, the problem statement is not ambiguous at all. If you disagree, you have only to answer Dave Seaman's question -- viz., which n is still in the set at noon? -- to prevail in the argument. Btw I cannot see how your assumption could be considered a matter of topology anyway. You are asserting that card S is a continuous function of t at t=0, and proper topology (e.g. nondiscrete) is of course necessary for that assertion to hold. However, that is a mere triviality; the real question is sufficiency. ==== [snip] > Btw I cannot see how your assumption could be considered a matter > of topology anyway. You are asserting that card S is a continuous > function of t at t=0, and proper topology (e.g. nondiscrete) is of > course necessary for that assertion to hold. However, that is a > mere triviality; the real question is sufficiency. And I should have added, I'm just a beginner at this so I could be wrong. My last comment wasn't quite right; I meant rather that topological arguments are insufficient to establish what the value of f is at 0; they can only decide, given the value that f has at 0, whether the function *is* continuous there or not. That is, you can't start with topology and argue from there that the function *has* to be continuous. (Except of course in the trivial case, and even then, that still won't give you the value of f at 0, which is what you really are asking for here.) ==== >>It is an unstated fact that if a ball is removed from the bucket before >>noon and is not subsequently returned, then that ball is not in the bucket >>at noon. This is so without regard to what may happen to other balls in >>the meantime. > That is precisely the assumption for pointwise convergence. If you take > that as an unstated fact, then yes, you do get the limit you claim. It > is the fact that that fact is not given that makes the problem ambiguous > to start with. >>The same assumption operates if I ask, You have five apples. I take >>away two. How many are left? Are you claiming that every first-grade >>arithmetic problem is an exercise in topology? > No, the apples problem is a two step problem. There is no infinite set > of steps which requires a topology to say what happens in the limit. There is not an infinite set of steps in the balls problem either. For each ball there are exactly two steps that change the state of that ball. Viewing the total aggregate of transitions for all the balls is simply not productive; there is another way to the solution that does not consider that entity at all. >>The balls-in-the-bucket problem is no more ambiguous than the apples >>problem. There may be more balls than there are apples, but if you focus >>your attention on just one ball or one apple at a time, exactly the same >>principle applies. > The balls-in-the-bucket problem defines what happens at every finite > step, but does not specify how to pass to the limit. Therein lies the > ambiguity. You have not answered my question. For which n is the state of ball n not known at noon? > Would you agree that a ball is in the bucket at noon if and only if that > ball was put in at some time before noon and then never taken out after > that; and that a ball is not in the bucket if it is either never put in > the bucket or was taken out some time before noon and never put back in? > What I have just described is precisely the assumption of pointwise > convergence of the indicator functions of whether the balls are in the > bucket or not. Perhaps this is the only thing that makes sense to you. > In any case, under that assumption, I agree that there will be no balls > in the bucket at noon (and I have said so before). Do you agree that 16/64 = 1/4? What I have just described is precisely the assumption that you can cancel the 6's. The fact that you can look at a problem in two different ways and get the same answer is not evidence that the two methods are indistinguishable, or even that both methods are correct. I can conclude that the bucket is empty at noon, or that subtracting two apples from five apples leaves three apples, without using pointwise convergence. It is not enough to show that your pointwise convergence approach gets the right answer; you also have to justify the introduction of a topology by referring to the original problem statement. You can't. There is nothing there about a topology. I base my answer on the axiom schema of separation, not on pointwise convergence. > The problem is that without this (or another) assumption, the state at > noon is not derivable from the states before noon. Perhaps you think > that this is part of the statement of the problem. If so, then I can > see why you are so convinced that this is the only way to look at the > problem. Your error is using the term the state at noon in the singular, when what we really have is the states at noon (one for each ball). Yes, the problem absolutely does give us enough information to derive each and every one of those states at noon, without using topology. In precisely the way that the apples problem gives us enough information to derive all five states after the taking without topology. > This seems to be our difference, that you assume this as given and I do > not. If so, I have no disagreement with you over the outcome of the > process; I just don't see this as necessarily an assumption for the > problem. Since replacing that assumption yields differing results, I > wanted to point out that the cause for the ambiguity noted by the OP is > the choice of these assumptions, and that each assumption relates to its > own topology. I have not assumed anything different than you have. I have simply chosen to consider each ball as an individual object, independent of the others. You cannot solve the apple problem without making the identical assumption. > If you still think I am full of crap, then so be it. I don't think that > there is much point to us arguing about this further. Are we agreed on > this at least? We can agree to disagree. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== I am given a set of generators (s_1, ..., s_{n-1}) and relations for a group G_n: s_i^2 = 1 (i=1...n-1) s_i s_j = s_j s_i ( |i-j|> 1) s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1} (i=1...n-2) I have verified that these properties hold for transpositions in S_n, e.g. for (1,2), (2,3), (n-1,n), but I don't know that there can't be other relations for S_n. I want to show that G_n = S_n. Since phi: G_n -> S_n is surjective, I was thinking of using Todd-Coxeter and induction to show that |G_n| = |S_n|. I tried this, but did not get very far. Any hints? John ==== http://mathworld.wolfram.com/Determinant.html gives the distribution of |det| if the elements of the matrix are within