mm-85 === In a fifty-year-old math paper about integrals of hermite polynomialsthe author uses this strange symbol (-m)_{k}, i.e. lower indexnotation, 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 coef?ients if my own calculations arecorrect. But it's de?ition for integers appears to be some quotientof factorials.Heiko Gimperlein === I would guess that (x)_k denotesx(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, mto 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 kfactors m, m - 1, ..., m - k + 1. For k zero, the empty product is 1as usual; for k negative, de?e so as to keep certain simple laws ofexponents. Factorial of the natural number m is either 1 to m risingor 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 SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === If I did not make a mistake calculating the ?st few terms by hand,here is a recursively-de?ed 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=0And, 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 (ienonrecursive) representation?Also, other sequences can be based on the same idea: a partial sumsomehow involving earlier terms, BOTH in the general term and in thelimit of the indexes used in the sum.Leroy Quet === > No, this is your mistake. If both A and A-B are in?ite, 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 de?e is not a subgroup.> For example, A = integers, B = positive integers, sigma(x) = x+1.> Ahh....thanks. I forgot about in?ite sets. Good to hear (from theother poster) that my proof works for ?ite ones at least. === msk===Bob, your explanation shows why the common presentation ofmultiplication of complex numbers as a free convention : (a,b)*(c,d) =(ac-bd, ad+bc) is insuf?ient because also (a,b)*(c,d) = (ac-2bd,ad+bc) will conform to the laws of algebra. As you showed itsnecessary 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 Vthen 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 youin 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 ofcounterexample against my work would work against *any* non-unitalgebraic 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 polynomialwith integer coef?ients, you will have a non-monic irreducible overQ, which proves that none of its roots are algebraic integers.That fact comes from noting that to get an integer with f, you'll needthe others factors of 5 that some of you apparently presume mustexist.That is, they form a factor ring, so unless f is a simple radical,like sqrt(5), you need the other factors, so algebraic manipulationswon't do it.So then, you have that the polynomial f b^3 + 12b^2 - 65/f^2 = 0which you might presume has one algebraic integer root, has been shownto have none because you can't form a polynomial with integercoef?ients, let alone one that is monic with integer coef?ientsusing less than an in?ite number of algebraic manipulations.That problem results with irreducibles over Q, and is the problem I'vebeen talking about for a while.Posters in arguing with me, were helped by my focus on f=sqrt(5), sopeople were being directed to focus on that case but then failed torealize 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 analgebraic integer.For some of you the question of algebraic manipulations may seemesoteric, so I've presented the question of the polynomial y^3 + sy^2 + ty - 13that would result if you divided factors in common with 5 from theroots.For those of you still not convinced by the proof above, perhapstrying to ?d ?s' and ?t' will \ help you.So yes in fact the odd result is that the three roots r_1, r_2, andr_3, of a^3 + 12a^2 - 65are in fact *each* coprime to 5 ***in the ring of algebraicintegers***.A rough analogy is the case of 2 and 6 in the ring of evens, as 6 isprime in the ring of evens as it is coprime to 2.However, the ring of evens is still complete, while the problem I'vedescribed in the ring of integers leads to the contradiction thatthough r_1 r_2 r_3 = 65r_1, r_2 and r_3 are each coprime to 5.The mathematics is here strange enough to confuse, but it needs to beconsidered carefully to see if any large errors have snuck into coremathematics along with it.After all, mathematics is extremely sensitive to ?ven a verysmall error can have huge repercussions.James Harris === ESB Consultancy is pleased to announce the latest release of ESBCalc Pro - a full featured, scienti? calculator for Win32 Platforms.http://www.esbconsult.com/esbcalc/ esbcalcpro.htmlESBCalc Pro is an Enhanced Scienti? Calculator for Win32 Platforms with In? Notation, Full Exponential Notation support, Brackets, Scienti? 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.comHome of ESBPCS, ESBStats, ESBPDF Analysis & ESBCalcKalgoorlie-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 getattention for themselves out of the problems of others. Get help.You need to see a psychiatrist.-- Michelle Malkin (Mickey)http://questioner.www2.50megs.comatheist/agnostic list ordainerEAC Bible thumper thumperBAAWA 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 dif?ult, this whole town of 100,000 people are all in on it.>> No one responded to your ?st post because you are a COMPLETE FUCKING> LOON, and it made no sense.>> the others have clari?d 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'.> HercHerc,What you say if the following townsville numbers didn't have a cluewhat on earth you were talking about:4750 0700 4031 1107 4753 44444779 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 Of?er - Research> 4031 1107 NuShape Foods Outlet> 4753 4444Australian Institute of Marine Science> 4779 0233Townsville Vet Clinic> 4760 1380Alamo Rent-A-Car - Townsville Airport> 4781 4182James Cook University - School of Psychology> 4781 4301James Cook University - Distance EducationInteresting 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?re?=mitaka === > What you say if the following townsville numbers didn't have a clue what on earth you weretalking about:>> Let's take a look at these numbers (just out of curiosity):> 4750 0700> Great Barrier Reef Marine Park Authority - ermits Of?er - 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 Townsvilleand Yes we can hear him. I can post up more random numbers fromthe phone book or scan a few pages and upload that if its not enoughwitnesses. 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> 47831124The comments you will get will be exactly like the posts to aus.tv,the only people admitting they were from townsville ALL shared thesame sarchasm approach, direct con?mation.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 con?mation.> >>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 unjusti?d. People might not speak up here becauseof 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 isabout.Could you please take your time for a minute or so and tell me what is it that you're trying tosay.Ok, this is what I know so far, you claim that there are camera's on you, and people fromTownsville 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 workedyour psychological assesment back from that.There's only one alias showing on the google search phrase i am the truman. Its a 5 billionto one, but had to be someone.I have an online proof at www.adamskingdom.com I have half dozen proofs butyou've yet to pull your ?gers out of your ears. Take several hours to understandthe website proof if you're smart and open minded, several minutes to contact the100,000 witnesses. They all listen to the truman company interrogate me anddiscuss the myriad of evidence, so coming from an authorative source theyare open enough to believe it, noone in Townsville disputes it. Likely you willonly 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 LoewenHad 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 thepeople 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. Wouldsomeone set this up as a criterion-based set of standards (based on whattheory of how mathematical ability develops?) or would one simplyempirically observe age norms and then report those? How could we be surethat presentation bypaper worksheets in vertical format,paper worksheets in horizontal format,?ards 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 towhat is the age norm in another country?Part of the premise that underlies the question here, I fear, is the usualassumption of lock-step age-grading in schoolhttp://learninfreedom.org/age_grading_bad.htmland part of it is the assumption that facility with math facts is IN ALLINSTANCES necessary for further advancement in mathematical reasoning, forwhich the famous story of Ernst Eduard Kummerhttp://dirphys.harvard.edu/Humour/Science/sciencejokes/ sciencejokes.10of9.txtprovides a counterexample.Just wondering . . . .P. S. The EPGY program at Stanford Universityhttp://epgy.stanford.edu/would have data in their ?es about math facts performance of aself-selected group of gifted children who use a computer-baseddistance-learning math course. I don't know how much of the researchspeci?ally related to their math races program has been published. Myson used those materials to learn to solve conditional probability wordproblems mentally at fairly high speed while he was of ?th-grade age. Ihave no idea what is a reasonable expectation for the general population forsolving that kind of problem, which is probably insoluble to most collegegraduates in the United States.-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn 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 ?ite. There are an> in?ite number of times each ball is removed and an in?ite number> of times it is put back (albeit with a new label)...a classic in?ity> minus in?ity 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 ?ite> 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 noballs. It seems clear to me that any ball with an initial label notdivisible by 10 will never be removed. So at noon, there will stillbe an in?ite number of (unlabeled) balls remaining. It's notcontradictory that the balls are labeled before noon but not afternoon, just as in the original problem it's not contradictory that thebucket 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 speci? ball remains is 0. I don't *think* I have made an > unjusti?ble 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> speci?d 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 withthis more straightforward variant:You have a bucket, initially with no coins in it. For n>0, at time1-2^(-n), you put a coin in the bucket (say coin n), ?l thecoins in the bucket, and remove all the coins that show heads. What'sthe probability distribution for the number of coins in the bucket attime 1?It feels like the bucket should be empty with probability 1, but Ihaven'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 in?ite 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!) istherefore valid. Unless you can show me a natural number n such that10n is not a natural number, I think the labeless ball hypothesiscannot 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), ?l 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 thecoin ?ails is 1/2, and thus the probability that it will nevergo 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 withprobability 1. === > The balls-in-the-bucket problem de?es what happens at every ?ite> 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 in?ityeach room has one ball and one bucket, and is responsible for placing theball in the bucket and then removing it at the prescribed times. At noonthe occupants are to phone their results to the front desk, reportingwhether each ball is in its bucket or not.How many balls are in the buckets at noon?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > The balls-in-the-bucket problem de?es what happens at every ?ite>> 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 in?ity>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 itmakes the most physical sense, and I have never disagreed that with thatassumption there will be no balls at noon. I have said this numeroustimes. However, it is not necessary that this type of convergence beassumed, and by assuming another type of convergence, you might get adifferent limit.I am not saying that yours is not the answer I would give if asked thisquestion. However, given that the question was asked about some otherpossible limits, I tried as best I could to give topologies that wouldlead to those limits.Rob Johnson take out the trash before replying === > The balls-in-the-bucket problem de?es what happens at every ?ite> 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 in?ity>>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 mentionidea 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 atopology on the space of functions a_n is irrelevant to the problem.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > The balls-in-the-bucket problem de?es what happens at every ?ite>> 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 in?ity>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 seeshis bucket (his point) converging to an empty bucket (under the discretetopology, where convergence is rather monotonous). Considering all thebuckets, 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 timebefore noon, then that bucket is empty at noon. This is the assumptionof pointwise convergence. If you don't assume some sort of continuity,you cannot deduce the state at noon from the states before noon and allwe 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 in?ity>>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* pointwiseconvergence holds, or are you saying that pointwise convergence holds*because* it describes the fact (independently veri?ble via ZF) thatno balls are left at noon?If the former, then you have not explained *why* pointwise convergenceholds.> 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 equallynecessary for solving the ?e 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 oneattempted solution fails. Particularly not when you have already beenshown an entirely different solution that does not have that defect.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >>Let's try this version. Instead of a single bucket, we have an in?ity>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 veri?ble via ZF) that>no balls are left at noon?>>If the former, then you have not explained *why* pointwise convergence>holds.|De?ition:|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 whetherthat 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 isabout the only topology we can give to the range of f_n. To converge inthe discrete topology, all terms must be constant after some point.|De?ition:|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 tothe constancy of discrete convergence, it may not appear that there isany convergence at all, the terms are all just the same.Translating the de?ition of pointwise convergence to the language ofbuckets 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 restatementof pointwise convergence. This is why I have said that you are assumingpointwise convergence. Perhaps I should say you are taking pointwiseconvergence 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 ?e apples, take away two problem.No, because there is no in?ite sequence in the ?e apples, take awaytwo problem. Convergence deals with limits of in?ite 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 describehow to derive the state at noon from the states at previous times.The problem only tells us what happens before noon; we need to havesome way of determining what happens at noon. Assigning the topologyof pointwise convergence to the bucket has a ball function seems themost 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, Iam trying to show that the solution depends on an added assumption andthat 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 veri?ble via ZF) that>>no balls are left at noon?>>If the former, then you have not explained *why* pointwise convergence>>holds.>|De?ition:>|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 thatmay *deduce* that pointwise convergence holds, but I do not need thatfact to solve the problem. I need only set theory.>|De?ition:>|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 morefully and completely, and all will become clear. I know what convergencemeans.> Translating the de?ition 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 ?st law, does it not? Convergence ismerely the language you have chosen to describe what we observe; it isnot 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 ?st law and ZF for granted. Pointwise convergenceis a *consequence* of those assumptions, not an assumption in its ownright.> 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 ?e apples, take away two problem.> No, because there is no in?ite sequence in the ?e apples, take away> two problem. Convergence deals with limits of in?ite sequences.You have ?e apples at 11:55. I take away two at 11:59. How many doyou have at noon?We need to assume that Newton's ?st law applies, just as we do with thebuckets and balls.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >>> The balls-in-the-bucket problem de?es what happens at every ?ite>> 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 in?ity>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 isan 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 confusescard(lim S_n) with lim(card S_n). Where is it stated thatthe sequence of cardinalities *must* have a limit equal to thecardinality of the set at noon? You have merely assumed this,in your solution 1. Unfortunately, as Dave Seaman has pointedout, your assumption contradicts the axiom of separation, i.e.is invalid in ZF. So it *cannot* be taken as an unstated givenof the problem; in other words, the problem statement is notambiguous 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 inthe argument.Btw I cannot see how your assumption could be considered a matterof topology anyway. You are asserting that card S is a continuousfunction of t at t=0, and proper topology (e.g. nondiscrete) is ofcourse necessary for that assertion to hold. However, that is amere triviality; the real question is suf?iency. === [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 suf?iency.And I should have added, I'm just a beginner at this so I couldbe wrong. My last comment wasn't quite right; I meant rather thattopological arguments are insuf?ient to establish what the valueof f is at 0; they can only decide, given the value that f has at0, 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, andeven then, that still won't give you the value of f at 0, which iswhat 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 ?e apples. I take>>away two. How many are left? Are you claiming that every ?st-grade>>arithmetic problem is an exercise in topology?> No, the apples problem is a two step problem. There is no in?ite set> of steps which requires a topology to say what happens in the limit.There is not an in?ite set of steps in the balls problem either. Foreach ball there are exactly two steps that change the state of that ball.Viewing the total aggregate of transitions for all the balls is simplynot productive; there is another way to the solution that does notconsider 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 de?es what happens at every ?ite> 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 nnot 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 preciselythe assumption that you can cancel the 6's.The fact that you can look at a problem in two different ways and get thesame answer is not evidence that the two methods are indistinguishable,or even that both methods are correct. I can conclude that the bucket isempty at noon, or that subtracting two apples from ?e apples leavesthree apples, without using pointwise convergence. It is not enough toshow that your pointwise convergence approach gets the right answer; youalso have to justify the introduction of a topology by referring to theoriginal problem statement. You can't. There is nothing there about atopology. I base my answer on the axiom schema of separation, not onpointwise 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, whenwhat we really have is the states at noon (one for each ball). Yes,the problem absolutely does give us enough information to derive each andevery one of those states at noon, without using topology. In preciselythe way that the apples problem gives us enough information to derive all?e 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 simplychosen to consider each ball as an individual object, independent of theothers. You cannot solve the apple problem without making the identicalassumption.> 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 SeamanJudge 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 groupG_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 veri?d 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 otherrelations for S_n. \ I want to show that G_n = S_n. Since phi: G_n -> S_n issurjective, 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.htmlgives the distribution of |det| if the elements of the matrix are within