mm-130 >> If one throws a die 100,000 times without getting a 1 even once, one >> may reasonably conclude that the die throwing is not honest, and >> that the chances of the die coming up 1 in future similar throws is >> negligible.>If one throws a fair die enough times, one should get 100,000>consecutive rolls without a 1. It should take a while. One should have>to roll 50,000+ without a 1 several times in the process.ThatÕs the difference between saying that the die is not honest (which Virgildid not say), and saying that one may reasonably conclude that the die isnot honest (which Virgil did say, and large number of times. The>eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now,>if that holds for, say 100.000 throws, how is the chance>of {1} coming up in the 100.001:st throw?It would have to be a pretty complicated dice to remember how it hadcome up in the past. Perhaps an intricate mechanism could be devisedthat recorded the results and shifted some internal weights around todo it. It would certainly be a marvel of engineering. Let me know ifyou find have to be a pretty complicated dice to remember how it had> come up in the past. Perhaps an intricate mechanism could be devised> that recorded the results and shifted some internal weights around to> do it. It would certainly be a marvel of engineering. Let me know if> you find one.The ßawed reasoning is perhaps as follows: As the probability ofgetting no ones in 100k throws is quite close to zero, the probabilityof getting at least one one is close to one. Therefore, it shouldbe pretty certain that the next throw should give you a one.This kind of reasoning you can see in Norwegian papers where lottodrawings are discussed and recommendations are given for the nextweekÕs lottery. Watch out for number 17! It has been hiding in thepast weeks, but seems to be reappearing now. But avoid number 3, fourweeks in a row would make it unlikely for a fifth. (Or something.)-- Jon papers where lotto> drawings are discussed and recommendations are given for the next> weekÕs lottery. Watch out for number 17! It has been hiding in the> past weeks, but seems to be reappearing now. But avoid number 3, four> weeks in a row would make it unlikely for a fifth. (Or something.)Actually, while this type of logic is ßawed, it does give one theonly true opportunity to increase the expected pay-off by selectingnumbers: pick a set of numbers kind of reasoning you can see in Norwegian papers where lotto> drawings are discussed and recommendations are given for the next> weekÕs lottery. Watch out for number 17! It has been hiding in the> past weeks, but seems to be reappearing now. But avoid number 3, four> weeks in a row would make it unlikely for a fifth. (Or something.)IÕd say (obviously errorfully) that itÕs right, as well. Why isnÕt thatmore likely to get 17 provided the information above?I use to stake out at Liseberg* for numbers that havenÕt come up fora while and that pays up in chocklad, the most real currency ever. :)* http://www.liseberg.se/start.asp?tree_id=422-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be This kind of reasoning you can see in Norwegian papers where lotto> drawings are discussed and recommendations are given for the next> weekÕs lottery. Watch out for number 17! It has been hiding in the> past weeks, but seems to be reappearing now. But avoid number 3, four> weeks in a row would make it unlikely for a fifth. (Or something.) IÕd say (obviously errorfully) that itÕs right, as well. Why isnÕt that> more likely to get 17 provided the information above? I use to stake out at Liseberg* for numbers that havenÕt come up for> a while and that pays up in chocklad, the most real currency ever. :)Suppose that 17 has not appeared for the last 6 weeks.There are two common fallacious arguments:17 has not appeared for 6 weeks, so it probably will appear this week. [tobalance things up]17 has not appeared for 6 weeks, so it probably wonÕt appear this week.[it somehow continually hides]Assuming the lottery is fair, both arguments are wrong. The probability of17 appearing is constant from week to lottery is fair, both arguments are wrong. The> probability of 17 appearing is constant from week to week.I believe the theory. The thing that makes me not fully at easeis that i canÕt see any difference between these two cases:1) The die is thrown 100.000 times and then 1 time again (nohold up or so there between).2) The die is thrown 100.001 times.Since the dice has no memory, it should be the same chanceto get something in the 100.001:st throw. Obviously itÕs not(provided everybody on this NG isnÕt just living to basicallyset me up) so i must be missing something. WhatÕs that?-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be Assuming the lottery is fair, both arguments are wrong. The>>probability of 17 appearing is constant from week to week.> I believe the theory. The thing that makes me not fully at ease> is that i canÕt see any difference between these two cases:1) The die is thrown 100.000 times and then 1 time again (no> hold up or so there between).> 2) The die is thrown 100.001 times.Since the dice has no memory, it should be the same chance> to get something in the 100.001:st throw. Obviously itÕs not> (provided everybody on this NG isnÕt just living to basically> set me up) so i must be missing something. WhatÕs that?> Compare these:1) The die is thrown 100.000 times (with results being ignored) and then 1 time again (last result recorded).2) The die is thrown 1 time and is thrown a large number of times. The> eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now,> if that holds for, say 100.000 throws, how is the chance> of {1} coming up in the 100.001:st throw?If one sees those two actions (100k throws and the single> throw respectively) as separate, the probability would be> 1/6 for {1}.> Otherwise, one would expect {1} to come up far more likely> than that.Some people would. The 1 is overdue. Personally, I would say thedie is defective and has no 1 on it at all! So the probability thenext time is 0.> Which is correct statistically times. The> eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now,> if that holds for, say 100.000 throws, how is the chance> of {1} coming up in the 100.001:st throw? If one sees those two actions (100k throws and the single> throw respectively) as separate, the probability would be> 1/6 for {1}.> Otherwise, one would expect {1} to come up far more likely> than that.> Which is correct statistically speaking? Kindly> KonradIf itÕs a fair die, the chance is 1/6. (However, IÕd certainly suspectthe die was loaded in your experiment, so IÕd bet it wonÕt come up atall.)--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over the chance is 1/6.Would you care to elaborate, please? I find it very oddthat itÕs 1/6 chance. WouldnÕt you expect {1} pretty soonand any time now if {2,..,6} came up all the time?I know what the theory says but i find it hard to live with.> IÕd... suspect the die was loaded...Yes, thatÕs kind of what iÕd do too in practice.-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be If itÕs a fair die, the chance is 1/6.> Would you care to elaborate, please? I find it very odd> that itÕs 1/6 chance. WouldnÕt you expect {1} pretty soon> and any time now if {2,..,6} came up all the time?> I know what the theory says but i find it hard to live with.> LetÕs say you had just rolled the die and *not* gotten a {1} 100 times. Then someone else walks into the room. If you told the person This is a fair die, what is the probability that it will roll a 1?, what would that person say?Now, does the die itself know any more about what just happened in the last 100 rolls than the person who walked into the room?> >>IÕd... suspect the die was loaded...> Yes, thatÕs kind of what iÕd do too in chance is 1/6. Would you care to elaborate, please? I find it very odd> that itÕs 1/6 chance. WouldnÕt you expect {1} pretty soon> and any time now if {2,..,6} came up all the time?> I know what the theory says but i find it hard to live with. > IÕd... suspect the die was loaded... Yes, thatÕs kind of what iÕd do too in practice.Multiple Choice coin 100 times and record the number of heads andtails. After the first ten tosses the counts are Heads=6,Tails=4.What is the most likely pair of counts after all 100 tosses?1) Heads=51,Tails=49 [The subsequent 90 tosses are most likely to be split45/45.]2) Anything from Heads=40,Tails=60 to Heads=60,Tails=40 is equally likely[The two numbers are likely to be close, but you cannot tell what they willbe.]3) Heads=50,Tails=50 [Things have a tendency to even out.]4) Heads=60,Tails=40 [Things tend to carry on as they have been.]5) There is no most likely pair of counts [You are tossing a coin.Anything could happen.]-- Clive for a very long time and whilewe jumped between 1, 2, 3 and 4 we finally concludedthat it must be either 1 or 3.Since one is easier to spell than three for us Swedes,(weÕre friendly but slow people) we go with number 1. :)We are picking #1 since the dice thrown before the 90remaining throws does not affect the probability of theoutcome of the big throwing sequence.So, we THINK we got it now. Gratefull KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be My girlfriend and i disputed for a very long time and while> we jumped between 1, 2, 3 and 4 we finally concluded> that it must be either 1 or 3. Since one is easier to spell than three for us Swedes,> (weÕre friendly but slow people) we go with number 1. :) We are picking #1 since the dice thrown before the 90> remaining throws does not affect the probability of the> outcome of the big throwing sequence. who tried not to confuse. :)#1 is the correct answer! If itÕs a fair die, the chance is 1/6.Would you care to elaborate, please? I find it very odd>that itÕs 1/6 chance. WouldnÕt you expect {1} pretty soon>and any time now if {2,..,6} came up all the time?>I know what the theory says but i find it hard to live with.Why do you find this hard to live with?Do you really believe that the die has a memory?If not, then it seems completely obvious that the past behaviour of thedie cannot affect its future behaviour. This is not a mathematical orstatistical observation, it just common sense.What is strange is that many people will happily use their theoreticalknowledge of probability theory to (erroneously) deduce and believe acompletely absurd C. BondSuppose a dice is thrown a large number of times. The> eyes that come up are {2, 3, 4, 5, 6} but not {1}. Now,> if that holds for, say 100.000 throws, how is the chance> of {1} coming up in the 100.001:st throw?> > If one sees those two actions (100k throws and the single> throw respectively) as separate, the probability would be> 1/6 for {1}.> Otherwise, one would expect {1} to come up far more likely> than that.> Which is correct statistically speaking?If itÕs a fair die, the chance is 1/6. (However, IÕd certainly suspect> the die was loaded in your experiment, so IÕd bet it wonÕt come up at> all.)And in any case, what the original question is about, does the first100k throws push the probability to be the original question is about, does the first> 100k throws push the probability to be _larger_ than 1/6.Larger, smaller or not affecting at all. I gueesed that IF it does, itwill be up but iÕm certainly open for suggestions. :)-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be And in any case, what the original question is about, does the first>> 100k throws push the probability to be _larger_ than 1/6.Larger, smaller or not affecting at all. I gueesed that IF it does, it>will be up but iÕm certainly open for suggestions. :)No - if the die is a fair one, then the fact that you have had a runwithout a particular result, does not mean that the result becomesmore probable on the next throw - a computer programmer and I need to solve arecurrence relation for a model IÕm trying to build. This isNOT homework. I saw a few recurrence relations while I wasstudying maths at university (e.g. the Fibonacci numbers) butnever anything like this: a_{n} = A + Bsum_{i=1}^{n-1}a_{i}For two constants A and B. So we get a_{1} = A a_{2} = A + AB a_{3} = A + 2AB +AB^{2}and so on. IÕve written out a few more terms and canÕt seea pattern that I can use to guess at the solution. IÕd guessI could prove it by induction if I knew what the solution was.I had a look on wikipedia(http://www.wikipedia.org/wiki/Recurrence_relation) and therewas some good stuff about solving problems like the Fibonaccinumbers, but nothing that gave me any idea how to solve myproblem.Anyone have any ideas?Best wishes, sum^n a_i2) in the resulting equation, rewrite sum^n a_i as a_n + sum^{n-1} a_i3) write out a_n as a function of A,B and sum^{n-1} a_i4) solve the previous equation for sum^{n-1} a_i5) substitute the expression obtained in (4) into the equation in (2)6) solve for a_{n+1}| I am a computer programmer and I need to solve a| recurrence relation for a model IÕm trying to build. This is| NOT homework. I saw a few recurrence relations while I was| studying maths at university (e.g. the Fibonacci numbers) but| never anything like this:|| a_{n} = A + Bsum_{i=1}^{n-1}a_{i}|| For two constants A and B. So we get|| a_{1} = A| a_{2} = A + AB| a_{3} = A + 2AB +AB^{2}|| and so on. IÕve written out a few more terms and canÕt see| a pattern that I can use to guess at the solution. IÕd guess| I could prove it by induction if I knew what the solution was.| I had a look on wikipedia| (http://www.wikipedia.org/wiki/Recurrence_relation) and there| was some good stuff about solving problems like the Fibonacci| numbers, but nothing that gave me any idea how to solve my| problem.|| Anyone have any ideas?|| Best wishes,| a_i, s_0 = 0, thenrewrite as s_n - s_{n-1} = A + B s_{n-1},i.e. s_n = A + (B+1) s_{n-1} (n >= 1).This is easily solved for s_n = A*(1 + (B+1) + ... + (B+1)^(n-2)) + (B+1)^(n-1) s[1], after which a_n is recovered as s_n - s_{n-1} = A*(B+1)^(n-1).But surely the OP should have guessed this answer by factoring a_1,a_2, a_3 and a_4.--Ron Bruck> 1) write out a_{n+1} as a function of A,B and sum^n a_i> 2) in the resulting equation, rewrite sum^n a_i as a_n + sum^{n-1} a_i> 3) write out a_n as a function of A,B and sum^{n-1} a_i> 4) solve the previous equation for sum^{n-1} a_i> 5) substitute the expression obtained in (4) into the equation in (2)> 6) solve for a_{n+1}| I am a computer programmer and I need to solve a> | recurrence relation for a model IÕm trying to build. This is> | NOT homework. I saw a few recurrence relations while I was> | studying maths at university (e.g. the Fibonacci numbers) but> | never anything like this:> |> | a_{n} = A + Bsum_{i=1}^{n-1}a_{i}> |> | For two constants A and B. So we get> |> | a_{1} = A> | a_{2} = A + AB> | a_{3} = A + 2AB +AB^{2}> |> | and so on. IÕve written out a few more terms and canÕt see> | a pattern that I can use to guess at the solution. IÕd guess> | I could prove it by induction if I knew what the solution was.> | I had a look on wikipedia> | (http://www.wikipedia.org/wiki/Recurrence_relation) and there> | was some good stuff about solving problems like the Fibonacci> | numbers, but nothing that gave me any idea how to solve my> | problem.> |> | Anyone have any ideas?> |> | guessed this answer by factoring a_1,> a_2, a_3 and a_4.Yes, I should have. What can I say? IÕm pretty rustyand not as confident as I used to be (so I give up quicker).Best wishes, recurrence relation for a model IÕm trying to build. This is> NOT homework. I saw a few recurrence relations while I was> studying maths at university (e.g. the Fibonacci numbers) but> never anything like this: a_{n} = A + Bsum_{i=1}^{n-1}a_{i}For two constants A and B. So we geta_n=A(B+1)^(n-1).How to get that? Use generating functions. The idea is thatwe begin by writing a(x)=sum_{i=1}^infinite a_i x^i.Your recurrence relation can be rewritten as a(x) (Bx+Bx^2+Bx^3+...)= a(x)- (Ax+Ax^2+Ax^3+...),which is equivalent to (using the sum of am a computer programmer and I need to solve a> recurrence relation for a model IÕm trying to build. This is> NOT homework. I saw a few recurrence relations while I was> studying maths at university (e.g. the Fibonacci numbers) but> never anything like this:a_{n} = A + Bsum_{i=1}^{n-1}a_{i}So, let s_n = a_1 + a_2 + ... + a_n. Then s_0 = 0,a_n = s_n - s_{n-1} and your recurrence iss_n - s_{n-1} = A + B s_{n-1},i.e.,s_n - (B+1) s_{n-1} = A.This is an inhomogenous linear recurrence and can be solvedby standard methods. (General solution -A/B + D(B+1)^n for B =/= 0).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The problem has been studied? Are therewell-known algorithms to solve it in reasonable time? I am sure thatit is NP-complete...Suppose we are given a basic set M with |M|=m (may be quite large, say10^5 in order of magnitude) and a subset X of P(M) such that all setsin X have the same cardinality k. The task is to find all subsets S ofM such that:- |S| = 2k- all subsets of S with size k lie in X.Assume that, to give some numbers, k<=25 so that k< well-known algorithms to solve it in reasonable time? I am sure that> it is NP-complete...> Suppose we are given a basic set M with |M|=m (may be quite large, say> 10^5 in order of magnitude) and a subset X of P(M) such that all sets> in X have the same cardinality k. The task is to find all subsets S of> M such that:- |S| = 2k> - all subsets of S with size k lie in X.Assume that, to give some numbers, k<=25 so that k< anything about X, maybe be empty, may also be that |X|=(m choose k).For each disjoint pair x,y in X, count the number of zÕs in X with z subset (x union y). If this count equals (2k choose k), add x union y to your output (maybe checking that it hasnÕt been output already from some other disjoint union).Time = O(|X|^3) or so, so this is clearly not NP-complete.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, disjoint pair x,y in X, count the number of zÕs in X with z > subset (x union y). If this count equals (2k choose k), add x union y > to your output (maybe checking that it hasnÕt been output already from > some other disjoint union).Time = O(|X|^3) or so, so this is clearly not NP-complete.Well, wasjust plain stupid and I have to admit that I did not really thinkabout it. But it is obvious.Nevertheless, O(|X|^3) is too slow. Because of k is quite smallwhereas |X| may be large (but the greater k the smaller |X|) that isabout, say, O(|X| log |X|) for constant k might be best even if thealgorithmÕs complexity grows very fast with quite small>whereas |X| may be large (but the greater k the smaller |X|) that is>about, say, O(|X| log |X|) for constant k might be best even if the>algorithmÕs complexity grows very fast with k.Well, you canÕt hope for that because the number of SÕs in the output can be of order |X|^2. For example, suppose X consists of all (m choose k) subsets of M with cardinality k. Then the outputconsists of all (m choose 2k) subsets of M with cardinality 2k, and(m choose 2k) ~ const. (m choose k)^2 as m -> infinity withk constant. Note also that to specify a single member of M takes log(m) bits, so the size of the output is of order|X|^2 log(m).Now for an algorithm of complexity O(|X|^2 log(m) log |X|) for constant k,you can examine pairs of members of k (using David EppsteinÕs ideato avoid duplication), and then we just need to check in time O(log(m) log |X|) whether every k-subset of our candidate S is a member of X. Since weÕre keeping k constant, the number (2k choose k) of such subsets is no problem, we just have to be able to check in time O(log(m) log |X|) whether a single k-subset is a member of X. And this can be done by standard techniques (e.g. constructing and searching an AVL tree).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of O(|X|^3) or so, so this is clearly not NP-complete.> Well, was> just plain stupid and I have to admit that I did not really think> about it. But it is obvious.It may be easy in retrospect, but I donÕt think itÕs obvious, or that not getting it is stupid. I suspect if I gave this as a question on an algorithms class exam that very few students would find a solution, especially if phrased the way you posted it so that itÕs unclear whether to look for an efficient algorithm or an NP-completeness proof. I posed it at lunch today to another skilled algorithms researcher and this person didnÕt solve it.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, disjoint pair x,y in X, count the number of zÕs in X with z > subset (x union y). If this count equals (2k choose k), add x union y > to your output (maybe checking that it hasnÕt been output already from > some other disjoint union).Time = O(|X|^3) or so, so this is clearly not NP-complete....and Robert Israel suggested basically the same thing.HereÕs a faster randomized algorithm:- build a data structure (e.g. hash table or balanced search tree) for quickly looking up whether a set belongs to X- randomly permute the elements of the union of X- for each disjoint pair x,y in X, if every element of x is earlier than every element of y in the random permutation, test all k-element subsets of x union y for membership in X. If all subsets belong to X, output x union y.When a pair x,y has its subsets tested, this results in (2k choose k) membership tests, however the chance that a pair x,y has everything in x earlier than everything in y is 1/(2k choose k), so the expected number of membership tests is O(|X|^2). This method also avoids the need to check for duplicates in the output: each output set is produced exactly once.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer thing.> HereÕs a faster randomized algorithm:- build a data structure (e.g. hash table or balanced search tree) for > quickly looking up whether a set belongs to X- randomly permute the elements of the union of X- for each disjoint pair x,y in X, if every element of x is earlier than > every element of y in the random permutation, test all k-element subsets > of x union y for membership in X. If all subsets belong to X, output x > union y.When a pair x,y has its subsets tested, this results in (2k choose k) > membership tests, however the chance that a pair x,y has everything in x > earlier than everything in y is 1/(2k choose k), so the expected number > of membership tests is O(|X|^2). This method also avoids the need to > check for duplicates in the output: each output set is produced exactly > once.This sounds quite ok, and in comparison to my own idea(clique-detecting) it is very easily implemented and problem has been studied? Are there>well-known algorithms to solve it in reasonable time? I am sure that>it is NP-complete...>Suppose we are given a basic set M with |M|=m (may be quite large, say>10^5 in order of magnitude) and a subset X of P(M) such that all sets>in X have the same cardinality k. The task is to find all subsets S of>M such that:>- |S| = 2k>- all subsets of S with size k lie in X.>Assume that, to give some numbers, k<=25 so that k<anything about X, maybe be empty, may also be that |X|=(m choose k).Note that |X| must be at least (2k choose k) in order for any S toexist. So the size of the input must be at least exponential in k.There is a polynomial-time algorithm. Note that every S must be theunion of two disjoint members of X. Start with an empty list.For each pair A,B of members of X if A and B are disjoint Let S = A union B. Count the number of members of X that are subsets of S. If the count is (2k choose k), add S to the list.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia following problem has been studied? Are there> well-known algorithms to solve it in reasonable time? I am sure that> it is NP-complete...> Suppose we are given a basic set M with |M|=m (may be quite large, say> 10^5 in order of magnitude) and a subset X of P(M) such that all sets> in X have the same cardinality k. The task is to find all subsets S of> M such that:- |S| = 2k> - all subsets of S with size k lie in X.Therefore no subset S contains an element outsideof X. Therefore all elements of S are finite unions ofmembers of X. That suggests an algorithm anyway: - form (recursively) all finite unions of members of X - for each, determine the cardinality and discard any which do not have cardinality 2k.There may of course be a more efficient approach.How many finite unions? These are the results of expressions like X1 U X2 U ... U Xp. So there are2^|X|-1 such expressions in one-to-one correspondencewith the subsets of X (except the empty set). Hmm.I think the end result is that this algorithm isexponential. I forget what exactly the definition ofNP-complete is, and obviously this is not a completeanalysis, but isnÕt it worse than exponential? - of X. Therefore all elements of S are finite unions of> members of X. That suggests an algorithm anyway:> - form (recursively) all finite unions of members of X> - for each, determine the cardinality and discard any> which do not have cardinality 2k.There may of course be a more efficient approach.> Yes, and I think something of the following kind might work:construct the sets S layer by layer from k to 2k. So, in a first step,find all sets S(k+1) with the second property, but with |S(k+1)|=k+1,use these then to get all corresponding sets S(k+2) with |S(k+2)|=k+2all the way to |S|=2k.For each step, it may be helpful to determine all (m-1)-subsets of theS(m) to get the S(m+1). For each (m-1)-subset, mark of which S(m)-setsit is a subset. So we have the following graph:- each S(m)-set as a node- two nodes share an (undirected) edge if they differ in only oneelement <=> they have a (m-1)-subset in commonFinding all S(m+1) then corresponds to enumerating all (m+1)-cliques(subgraphs that are isomorphic to the complete graph on (m+1) nodes).I do absolutely not know what to do about this.> I think the end result is that this algorithm is> exponential. I forget what exactly the definition of> NP-complete is, and obviously this is not a complete> analysis, but isnÕt it worse than exponential?Well, it was too unprecise in my OP: we have to consider thecorresponding recognition problem, which is: given X, is there a set Sthat fulfills the conditions?I do not know much about complexity theory, so please correct me if Iam wrong:That this problem is in NP means that we can check a given S forfeasibility in polynomial time. This is true because we just need togo through X once. NP-completeness then means: suppose we have apolynomial algorithm for the problem, every other NP-problem can besolved in polynomial time because it amounts to calling our problemalgorithm a polynomial number of times.The problem reminded me of the k-clique-problem (see above), whoserecognition version (is there a k-clique in the graph?) is known to beNP-complete.It has nothing to do with wether there is an exponential algorithm forthe problem or not. Up to date it is not even known (but widelyconjectured) if there are NP-complete problems for which no polynomialalgorithm difference betweenmean and median has cropped up. I defended the claim that a for a strictly convex function, themean is provably strictly larger than the median. I assumed thefunction to be real-valued, continuous, defined on a closed intervalof real numbers. Since my proof would be similar for arbitrary intervals [a,b], Ionly proved it for the [0,1] interval. This is mostly high-school calculus level math, but bear with meand read this simple proof:----------- ItÕs a bit awkward to write definite integrals in this medium, butIÕll try.Let S(a,b)((F(x))dx mean the definite integral of F(x) on the domain[a,b]. If, to simplify things, a strictly convex function F has the domain[0,1], then we have to prove that the median --F(1/2)-- is strictlysmaller than the mean, which in this case is S(0,1)(F(x))dx (since1-0=1). Now we have successively: S(0,1)(F(x))dx - F(1/2) =S(0,1)(F(x)-F(1/2))dxwhich is further S(0,1/2)(F(x)-F(1/2))dx + S(1/2,1)(F(x)-F(1/2))dx orafter a change of variable S(0,1/2)(F(y)-F(1/2))dy +S(0,1/2)(F(1-y)-F(1/2))dy which is againS(0,1/2)(F(y)+F(1-y)-2F(1/2))dy. But F(x) is strictly convex so we have that F(y)+F(1-y)-2F(1/2) isstrictly positive for y in [0,1/2]. Therefore the last definiteintegral mentioned above is strictly positive, hence S(0,1)(F(x))dx isstrictly larger than F(1/2) -- that means the mean is different thanthe median.----------- Other than the slight inaccuracy that I should have saidF(y)+F(1-y)-2F(1/2) is strictly positive on [0,1/2) instead of[0,1/2], which doesnÕt really affect the proof, do you find this proofto be correct? I am asking because my interlocutor wants me to get the opinion subject of the difference between> mean and median has cropped up.> I defended the claim that a for a strictly convex function, the> mean is provably strictly larger than the median.> ----------- ItÕs a bit awkward to write definite integrals in this medium, but> IÕll try.> Let S(a,b)((F(x))dx mean the definite integral of F(x) on the domain> [a,b].ThatÕs fine. Sometimes people write integral(a,b) f(x)dx. Sometimespeople write the limits at the end. If, to simplify things, a strictly convex function F has the domain> [0,1], then we have to prove that the median --F(1/2)-- is strictly> smaller than the mean, which in this case is S(0,1)(F(x))dx (since> 1-0=1).Is everybody in your argument agreed that F(1/2) is themedian? Now we have successively: S(0,1)(F(x))dx - F(1/2) => S(0,1)(F(x)-F(1/2))dx> which is further S(0,1/2)(F(x)-F(1/2))dx + S(1/2,1)(F(x)-F(1/2))dx or> after a change of variable S(0,1/2)(F(y)-F(1/2))dy +> S(0,1/2)(F(1-y)-F(1/2))dyThe change was to y = 1-x in the second integral, andjust a relabeling from x to y in the first?> which is again> S(0,1/2)(F(y)+F(1-y)-2F(1/2))dy.> But F(x) is strictly convex so we have that F(y)+F(1-y)-2F(1/2) is> strictly positive for y in [0,1/2]. Therefore the last definite> integral mentioned above is strictly positive, hence S(0,1)(F(x))dx is> strictly larger than F(1/2) -- that means the mean is different than> the median.Not just different, provide. Get a stat book and look up definitions of mean and median.>Recently, in a usenet debate, the subject of the difference between>mean and median has cropped up.> I defended the claim that a for a strictly convex function, the>mean is provably strictly larger than the median. I assumed the>function to be real-valued, continuous, defined on a closed interval>of real numbers.> Since my proof would be similar for arbitrary intervals [a,b], I>only proved it for the [0,1] interval. This is mostly high-school calculus level math, but bear with me>and read this simple proof:----------- ItÕs a bit awkward to write definite integrals in this medium, but>IÕll try.>Let S(a,b)((F(x))dx mean the definite integral of F(x) on the domain>[a,b]. If, to simplify things, a strictly convex function F has the domain>[0,1], then we have to prove that the median --F(1/2)-- is strictly>smaller than the mean, which in this case is S(0,1)(F(x))dx (since>1-0=1). Now we have successively: S(0,1)(F(x))dx - F(1/2) =>S(0,1)(F(x)-F(1/2))dx>which is further S(0,1/2)(F(x)-F(1/2))dx + S(1/2,1)(F(x)-F(1/2))dx or>after a change of variable S(0,1/2)(F(y)-F(1/2))dy +>S(0,1/2)(F(1-y)-F(1/2))dy which is again>S(0,1/2)(F(y)+F(1-y)-2F(1/2))dy.> But F(x) is strictly convex so we have that F(y)+F(1-y)-2F(1/2) is>strictly positive for y in [0,1/2]. Therefore the last definite>integral mentioned above is strictly positive, hence S(0,1)(F(x))dx is>strictly larger than F(1/2) -- that means the mean is different than>the median.----------- Other than the slight inaccuracy that I should have said>F(y)+F(1-y)-2F(1/2) is strictly positive on [0,1/2) instead of>[0,1/2], which doesnÕt really affect the proof, do you find this proof>to be correct? I am asking because my interlocutor wants me to get the opinion of>experts on the difference between>mean and median has cropped up.> I defended the claim that a for a strictly convex function, the>mean is provably strictly larger than the median. I assumed the>function to be real-valued, continuous, defined on a closed interval>of real numbers.> Since my proof would be similar for arbitrary intervals [a,b], I>only proved it for the [0,1] interval. This is mostly high-school calculus level math, but bear with me>and read this simple proof:----------- ItÕs a bit awkward to write definite integrals in this medium, but>IÕll try.>Let S(a,b)((F(x))dx mean the definite integral of F(x) on the domain>[a,b].General advice is to read a few posts when you visit a new group:in fact int_a^b F(x) dx is standard notation for this here.Your notation is also fine, since you explained what it means.> If, to simplify things, a strictly convex function F has the domain>[0,1], then we have to prove that the median --F(1/2)-- is strictly>smaller than the mean, which in this case is S(0,1)(F(x))dx (since>1-0=1).Are you also assuming that F is _increasing_? (Ie, non-decreasing:F(s) >= F(t) whenever s > t.) If F is not increasing I donÕt see whyF(1/2) deserves to be called the median.> Now we have successively: S(0,1)(F(x))dx - F(1/2) =>S(0,1)(F(x)-F(1/2))dx>which is further S(0,1/2)(F(x)-F(1/2))dx + S(1/2,1)(F(x)-F(1/2))dx or>after a change of variable S(0,1/2)(F(y)-F(1/2))dy +>S(0,1/2)(F(1-y)-F(1/2))dy which is again>S(0,1/2)(F(y)+F(1-y)-2F(1/2))dy.> But F(x) is strictly convex so we have that F(y)+F(1-y)-2F(1/2) is>strictly positive for y in [0,1/2]. Therefore the last definite>integral mentioned above is strictly positive, hence S(0,1)(F(x))dx is>strictly larger than F(1/2) -- that means the mean is different than>the median.----------- Other than the slight inaccuracy that I should have said>F(y)+F(1-y)-2F(1/2) is strictly positive on [0,1/2) instead of>[0,1/2], which doesnÕt really affect the proof, do you find this proof>to be correct?ItÕs a correct proof that S(0,1) F is larger than F(1/2). (Or at leastessentially correct, I didnÕt read every detail carefully). Regardingwhether this shows that mean is larger than the median see thequestion above.> I am asking because my interlocutor wants me to get the opinion of>experts on >-----------> > ItÕs a bit awkward to write definite integrals in this medium, but>IÕll try.>Let S(a,b)((F(x))dx mean the definite integral of F(x) on the domain>[a,b].General advice is to read a few posts when you visit a new group:> in fact int_a^b F(x) dx is standard notation for this here.Your notation is also fine, since you explained what it means. Ok, simplify things, a strictly convex function F has the domain>[0,1], then we have to prove that the median --F(1/2)-- is strictly>smaller than the mean, which in this case is S(0,1)(F(x))dx (since>1-0=1).Are you also assuming that F is _increasing_? (Ie, non-decreasing:> F(s) >= F(t) whenever s > t.) If F is not increasing I donÕt see why> F(1/2) deserves to be called the median.> You are correct. The context was the distribution of income, andthe function represents the income of the population, where thepopulation is ordered by income to begin with. I claimed that under certain assumptions, (continuity and strictconvexity) the mean income is strictly larger than the median income. I also gave a short paragraph describing the proof in the discretecase --assuming first and second differences strictly increasing-- butthat proof was not challenged. ItÕs a correct proof that S(0,1) F is larger than F(1/2). (Or at least> essentially correct, I didnÕt read every detail carefully). Regarding> whether this shows that mean is larger than the median see the> question above.> Ok, thank you for Ôconvex functionto describe afunction with the second derivative positive. My interlocutor claimedthat the Ôcorrectusage is Ôconcave up.IÕm not in doubt that my usage is correct, but IÕm curious: doesanyone know the history of using Ôconcave upinstead of Ôconvexandwhat are, in your experience, the most likely places or times whereÕconcave upcould be the norm for naming such Ôconvex functionto describe a> function with the second derivative positive. My interlocutor claimed> that the Ôcorrectusage is Ôconcave up.IÕm not in doubt that my usage is correct, but IÕm curious: does> anyone know the history of using Ôconcave upinstead of Ôconvexand> what are, in your experience, the most likely places or times where> Ôconcave upcould be the norm for naming such functions?My original training was in physics. Convex in that arenais convex in the geometric sense, like a convex lens. The mathbackground for that was fairly elementary: calculus, vectorcalculus, differential equations, complex analysis. IÕmpretty sure concave up is the way I would have heardy = x^2 described, with y = -ax^2 concave down andy = x neither convex nor concave.It wasnÕt until taking a graduate optimization course that I wasexposed to convex in the mathematical sense, and it tooka while to get used to the idea that a curve that looksconcave is called convex. It seemed backward. It helpedto connect it with other properties, such as theconvexity of the level sets. - that arena> is convex in the geometric sense, like a convex lens. The math> background for that was fairly elementary: calculus, vector> calculus, differential equations, complex analysis. IÕm> pretty sure concave up is the way I would have heard> y = x^2 described, with y = -ax^2 concave down and> y = x neither convex nor concave.It wasnÕt until taking a graduate optimization course that I was> exposed to convex in the mathematical sense, and it took> a while to get used to the idea that a curve that looks> concave is called convex. It seemed backward. It helped> to connect it with other properties, such as the> convexity of the level sets. - Randy This seems to be relevant to my discussion since my interlocutoralso seems to have a background in physics. I want to thank everyone who replied to my question, of Ôconvex functionto describe a> function with the second derivative positive. My interlocutor claimed> that the Ôcorrectusage is Ôconcave up.IÕm not in doubt that my usage is correct, but IÕm curious: does> anyone know the history of using Ôconcave upinstead of Ôconvexand> what are, in your experience, the most likely places or times where> Ôconcave upcould be the norm for naming such functions?AlexMathematicians say convex and concave. Calculus students foundthat confusing. So textbook writers invented concave up and concavedown, hoping it would be less confusing to beginners. (Did it work?)Mathematicians at the research level still say convex and concave.Much as mathematicians at the research level stll write log for thenatural logarithm...-- G. A. Edgar challenged on my use of Ôconvex functionto describea> function with the second derivative positive. My interlocutorclaimed> that the Ôcorrectusage is Ôconcave up.IÕm not in doubt that my usage is correct, but IÕm curious: does> anyone know the history of using Ôconcave upÕ instead of ÔconvexÕand> what are, in your experience, the most likely places or times where> Ôconcave up could be the norm for naming such functions?>Try you really arguing about this with your interlocutor?Do you know the Feynman story about the name of the bird?See Ôconvex functionto describe a>function with the second derivative positive. My interlocutor claimed>that the Ôcorrectusage is Ôconcave up.ÕYou should suggest to him that while itÕs easy to know that oneterm is correct itÕs very hard to know for sure that another termfor the same concept is not correct. The word convex is notused this way in calculus books, but itÕs a perfectly standardterm in mathematics, meaning the same as concave up.> IÕm not in doubt that my usage is correct, but IÕm curious: does>anyone know the history of using Ôconcave upinstead of ÔconvexÕ and>what are, in your experience, the most likely places or times where>Õconcave upcould be the norm for naming such functions?I donÕt know about the history, but regarding where the term isused, see any text on real analysis. (Royden, Rudin, Folland...)>Alex************************David C. checked up to a verylarge number and not have a counterexample ( if one exists at all ). The Challenge is to make a list of primes and then find the largesteven number that can be created out of the list without skipping anysmaller even numbersEX.L= {2,3}4 = 2+26 = 3+3L = {2,3,5}4 = 2+26 = 3+38 = 5+310 = 5+5L = {2,3,5,7}4 = 2+26 = 3+38 = 5+310 = 5+512 = 5+714 = 7+7L = {2,3,5,7,11}4 = 2+26 = 3+38 = 5+310 = 5+512 = 5+714 = 7+716 = 11+518=11+720 = no wayNotice I stopped, even with the fact that 22 can be produced.Other examples.With the first 100 primes the largest even number possible (withoutskipping) is 966the 100th prime is 541 but there was no way to create 968 with any ofthe first 100 primes so the program had to stopWith the first 1,000,000 primes the largest even number possible is30,970,934(this will be a test case for your programs)With the first 298,538,262 primes the largest even number possible is12,856,609,956So your job is to write an efficient program (C++) that can calculatethe top even number.-----Construct a functionunsigned long long Gold[your last name](unsigned long long n)that has as input the number of primes, and outputs the top evennumber-----Expect to have numbers for input larger than 10,000,000. As you cansee from above IÕve solved the problem up to about n = 300M, soefficient calculations for the making primes are needed. Either postsubject name of GoldbachI will compile them on a PowerPC G4 with 768 Mb of memory running at1GHzI will rank the programs be speed on 9-15-03 English textbook I find a claim that contains the following:let x be a distance from MIs is a typo or is it correct? I am not a native English speaker but shouldnot it belet x be at distance from M?Zdenek---Odchoz.92 zpr.87va neobsahuje viry.Zkontrolov.87no antivirovm syst.8emem AVG quotation too short. The Ôxin the quoted text isa vector and ÔMis a subspace. Perhaps this makes things clearer now.Z.H.Zden.93k Hur.87k p.92e v diskusn.92m p.92sp.93vku> In an English textbook I find a claim that contains the following: let x be a distance from M Is is a typo or is it correct? I am not a native English speaker butshould> not it be let x be at distance from M? Zdenek> ---> Odchoz.92 zpr.87va neobsahuje viry.> Zkontrolov.87no antivirovm syst.8emem AVG (http://www.grisoft.cz).---Odchoz.92 zpr.87va neobsahuje viry.Zkontrolov.87no antivirovm syst.8emem AVG claim that contains the following:let x be a distance from MIs is a typo or is it correct? I am not a native English speaker but should>not it belet x be at distance from M?Neither of those makes much sense to me. Are you sureyouÕre quoting the entire phrase correctly?>Zdenek>--->Odchoz.92 zpr.87va neobsahuje viry.>Zkontrolov.87no antivirovm syst.8emem AVG (http://www.grisoft.cz).>************************David C. p.92se v diskusn.92m pr.92spevku> Neither of those makes much sense to me. Are you sure> youÕre quoting the entire phrase correctly?You are right, I omitted something by mistake. Please have a look at myanswer to Guy Corrigall. I am sorry.Zdenek---Odchoz.92 zpr.87va neobsahuje viry.Zkontrolov.87no antivirovm syst.8emem AVG depends very much on context. One would need toread the whole paragraph to answer your question correctly.Nevertheless on the limited evidence you provide, one would say :let x the the distance from M. Now, IÕm assuming that the quantity x is alength or distance in metres or some suitable unit of length. IÕm alsoassuming that M is a defined point in space, and so the distance x metresfrom M is meaningful in the context of the paragraph in your textbook.hthGuy> In an English textbook I find a claim that contains the following: let x be a distance from M Is is a typo or is it correct? I am not a native English speaker butshould> not it be let x be at distance from M? Zdenek> ---> Odchoz.92 zpr.87va neobsahuje viry.> Zkontrolov.87no antivirovm syst.8emem AVG depends very much on context. One would need to> read the whole paragraph to answer your question correctly.I am pretty much sorry for the confusing question. Fuller two alternative senteces are:a vector $x$ is a distance $d$ from a subspace $M$ora vector $x$ is at distance $d$ from a subspace $M$ question. Fuller two alternative >senteces are:a vector $x$ is a distance $d$ from a subspace $M$ora vector $x$ is at distance $d$ from a subspace $M$ The Oxford English Dictionary entry for distance includes(9b). Also used without preposition as an adverbial adjunct of measure. It includes several quotes, dating back as far as 1577. So your firstversion is well established in English usage. Nevertheless, I mightprefer to say a vector $x$ is at a distance $d$ from a subspace $M$for fear that a naive reader might misinterpret your statement assaying that the vector _is_ the distance.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of agree with Robert Israel - now you should begin to see how redundantEnglish is. I wouldnÕt have read that the vector is the distance, but thesentence could be construed that way quite logically.Moving along a little, does it make sense to say that a vector lies at acertain distance from a point or a space? IsnÕt there some ambiguity aboutthe location of a vector? To say nothing about the location of a space!Guy Corrigall >I am pretty much sorry for the confusing question. Fuller two alternative>senteces are:> >a vector $x$ is a distance $d$ from a subspace $M$> >or> >a vector $x$ is at distance $d$ from a subspace $M$ The Oxford English Dictionary entry for distance includes (9b). Also used without preposition as an adverbial adjunct of measure. It includes several quotes, dating back as far as 1577. So your first> version is well established in English usage. Nevertheless, I might> prefer to say a vector $x$ is at a distance $d$ from a subspace $M$> for fear that a naive reader might misinterpret your statement as> saying that the vector _is_ the distance. Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T context. One would need to>> read the whole paragraph to answer your question correctly.I am pretty much sorry for the confusing question. Fuller two alternative >senteces are:a vector $x$ is a distance $d$ from a subspace $M$ora vector $x$ is at distance $d$ from a subspace $M$ That makes more sense. Either one is correct - I think thatthe second is probably better.>Zdenek ************************David C. context. One would need to>> read the whole paragraph to answer your question correctly.I am pretty much sorry for the confusing question. Fuller two alternative >senteces are:a vector $x$ is a distance $d$ from a subspace $M$ora vector $x$ is at distance $d$ from a subspace $M$ Both sentences are acceptable idiomatic English (thefirst one is more idiomatic, and probably less acceptableto many listeners; the role of a before distance isa bit hard to describe--but see my note below), and theymean the same thing as the more precise sentences The distance between the vector x and the subspace M is d.and The distance from the vector x to the subspace M is d.Note: _A Comprehensive Grammar of the English Language_by Quirk, Greenbaum, Leech, and Svartvik, is the standarddescriptive grammar. ItÕs about 1800 pages long and I havenÕtturned up anything directly relevant to your question in quicklook through it. My personal grammatically judgments abovewere just that, personal; the stuff in Quirk et al. is backedby study of large corpora of actual written and spoken English,so if something relevant to your question *does* turn up in thatbook (which IÕll keep thumbing through) it will be the closestyouÕll get to a definitive answer about depends very much on context. One would needto> read the whole paragraph to answer your question correctly. I am pretty much sorry for the confusing question. Fuller two alternative> senteces are: a vector $x$ is a distance $d$ from a subspace $M$ or a vector $x$ is at distance $d$ from a subspace $M$Both sound OK to me. IÕd prefer the first.-- Bob same idea, butmathematically they are quite distinct concepts. OTOH the veryconstruction on Lebesgue measure in R^n shows an explicit connectionbetween them: loosely speaking we find the volume of boxes multiplyingthe lenghts of their sides... but are there similar connections inmuch more abstract settings?Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on intuitively recall the same idea, but>mathematically they are quite distinct concepts. OTOH the very>construction on Lebesgue measure in R^n shows an explicit connection>between them: loosely speaking we find the volume of boxes multiplying>the lenghts of their sides... but are there similar connections in>much more abstract settings?Yes. Look up Hausdorff measure somewhere - thatÕs a constructionof measures on a metric space thatÕs analogous to Lebesgue measure.>Michele************************David C. recall the same idea, but>>mathematically they are quite distinct concepts. OTOH the very>>construction on Lebesgue measure in R^n shows an explicit connection>>between them: loosely speaking we find the volume of boxes multiplying>>the lenghts of their sides... but are there similar connections in>>much more abstract settings?Yes. Look up Hausdorff measure somewhere - thatÕs a construction>of measures on a metric space thatÕs analogous to Lebesgue measure.And if you *really* want to get into it (in *much* more abstractsettings), look up what Mikhail Gromov has been doing distance between two points on a torus?You have to specify the metric. For example you might have the torus> embedded in R^3 with the inherited metric, you might be thinking of a> ßat torus, you might be thinking of the intrinsic (geodesic) metric,> or you might be thinking of a ßat torus.Sorry, I am another silly question but here goes :)For my own education and amusement I was considering what happens when a person has an initial amount of money, invests this money for a period of time, collects the interest and reinvests the original principle plus the interest plus an additional amount smaller than the original principle.Just so you know that IÕve done some work on my own :), this is what I have so far:v=final valuea=additional amounti=interest rate payable per periodp=initial amount investedt=number of time periodst=0: v=pt=1: v=ip+at=2: v=i(ip+a)+a =i^2p+ai+at=3: v=i(i^2p+ai+a)+a =i^3p+ai^2+ai+a...t=6: v=i^6p+ai^5+ai^4+ai^3+ai^2+ai+aLooking at that, it seems to me that the ai terms are a geometrical progression and so the function for any number of time periods should be (if I have this bit right)v=(a(1-i)^t/(1-i))+p(i^t)However, this doesnÕt seem to work ... I have calculated it manually a few times and graphed it using a graphical calculator and a graphing programme (gtkgraph) and it completely falls over :(The numbers I used are:p=1000i=1.0125 (1.25%+1 because itÕs increasing)a=100t=25and I got a result of 1364.19If someone would put me on the right track IÕd be very here goes :) For my own education and amusement I was considering what happens when a> person has an initial amount of money, invests this money for a period of> time, collects the interest and reinvests the original principle plus the> interest plus an additional amount smaller than the original principle. Just so you know that IÕve done some work on my own :), this is what Ihave> so far: v=final value> a=additional amount> i=interest rate payable per period> p=initial amount invested> t=number of time periods t=0: v=p> t=1: v=ip+a> t=2: v=i(ip+a)+a> =i^2p+ai+a t=3: v=i(i^2p+ai+a)+a> =i^3p+ai^2+ai+a> ...> t=6: v=i^6p+ai^5+ai^4+ai^3+ai^2+ai+a Looking at that, it seems to me that the ai terms are a geometrical> progression and so the function for any number of time periods should be> (if I have this bit right) v=(a(1-i)^t/(1-i))+p(i^t)Replace (1-i)^t by (1-i^t). This gives us v=a*(i^t-1)/(i-1)+p*i^t[Writing i^t-1 rather than 1-i^t since i>1.]> However, this doesnÕt seem to work ... I have calculated it manually a few> times and graphed it using a graphical calculator and a graphing programme> (gtkgraph) and it completely falls over :( The numbers I used are:> p=1000> i=1.0125 (1.25%+1 because itÕs increasing)> a=100> t=25 and I got a result of 1364.19-- Clive silly question but here goes :)>> v=(a(1-i)^t/(1-i))+p(i^t)Replace (1-i)^t by (1-i^t). This gives us v=a*(i^t-1)/(i-1)+p*i^t[Writing i^t-1 rather than 1-i^t since i>1.]Clive,Works like a charm - you saved my following question came to mind and I am at a loss for an answer. Does anyone know a solution?Mathematical QueryGiven a square containing an even 2 by 2 matrix similar to a very small crossword puzzle, how many unique combinations of filled and unfilled squares are possible?If the squares are numbered 1 4, then there are 16 possible combinations:1 no squares filled.2-5 one square filled.6 11 two squares filled:a. 1&2, 1&3, 1&4.b. 2&3, 2&4.c. 3&4.12 15 three squares filled.16 all four squares filled.HOWEVER: if the squares are unnumbered, the viewpoint from each of the 4 sides is indistinguishable from any other side view. Then the possible uniquely identifiable combinations are reduced to 6:1 no squares filled.2 one square filled.3 4 two squares filled.a. Two adjacent squares.b. Two diagonally opposite squares.5 three squares filled.6 all four squares filled.QUESTION? Is there an equation that calculates the latter? If so:1. Does it expand to cover grids of more than 2 x 2?2. Does it expand to cover 3 dimensional grids, which have 6 sides?3. Does it expand to shapes other than squares & cubes (e.g. triangles & pyramids (either 3 or 4 sided))?4. Does it expand to (hypothetical) more than three dimensions & shapes?As a three dimensional example:A 2 x 2 x2 grid cube has only 24 possible uniquely identifiable combinations (as best I can determine)if the internal divisions (cubettes?) are un-numbered and the cube can be rotated to any viewpoint (23 if mirroring is also allow, but that is a separate question).1 all 8 cubettes empty.2 1 cubette filled.3 5 two cubettes filled:a. Two adjacent.b. Two diagonally on the same plane.c. Two diagonally across the three dimensions.6 8 three cubettes filled:a. All on the same plane.b. Two adjacent on one plane, one on the other plane. c. Two diagonally on one plane, the third on the other plane but not adjacent to either of the first two.9 17 four cubettes filled:a. All on the same plane.b. 3 on one plane, one on the other plane:1. Adjacent to the middle cubette on plane one.2. Adjacent to a side cube on plane one (two possible configurations without mirroring).3. Diagonally across the three dimensions from the middle cubette on plane one.c. Two on one plane, two on the other plane:1. Two adjacent on plane one.2. Two adjacent on plane two at 90 degrees from the two on plane one (two possible configurations without mirroring).3. Two adjacent on plane one at the opposite diagonal of the cube.4. Two adjacent on plane one, two diagonally opposite each other on plane two.5. Two diagonally opposed on plane one, two on the opposite diagonal of plane two.18 20 - five cubettes filled (same number of combinations as 3 5 ).21 22 six cubettes filled (same number of combinations as 2 4).23 seven cubettes [-1,1] as a subspace of R(the reals). Which of thefollowing sets are open in Y? Which are open in R?A = {x : 1/2 < |x| < 1}B = {x : 1/2 < |x| <=1}C = {x : 0 < |x| < 1 and 1/x not in the integers greater than 0}My confusion is as follows. I must consider the subspace topology of Y as asubspace of R, which by definition isT_y = {Y intersect U : U is in T} where T is a topology on R. Is T just anytopology on R? Can I pick any one?Because if I pick the collection of open intervals (a,b), then A is openin Y since it equals Y intersect (1/2,1), and B would notbe open in Y. However, if I pick the topology of all half-open intervals(a,b], then A is not open, however B is.WhatÕs the deal here?Does the question assume the the set Y = [-1,1] as a subspace of R(the reals). Which of the>following sets are open in Y? Which are open in R?A = {x : 1/2 < |x| < 1}>B = {x : 1/2 < |x| <=1}>C = {x : 0 < |x| < 1 and 1/x not in the integers greater than 0}My confusion is as follows. I must consider the subspace topology of Y as a>subspace of R, which by definition is>T_y = {Y intersect U : U is in T} where T is a topology on R. Is T just any>topology on R? Can I pick any one?No, if someone says something about topology on R without specifyingwhich topology is intended then the standard topology is meant.ThatÕs what standard means...>Because if I pick the collection of open intervals (a,b), then A is open>in Y since it equals Y intersect (1/2,1), and B would not>be open in Y. However, if I pick the topology of all half-open intervals>(a,b], then A is not open, however B is.WhatÕs the deal here?The deal is you donÕt quite have the definition of the subspacetopology straight. Or maybe youÕre missing something else,but if we give R the standard topology and then consider thesubspace topology on Y = [-1,1] then B _is_ open in Y.WhatÕs the definition of the subspace topology?>Does the question assume the standard topology on R? (i.e. the >Consider the set Y = [-1,1] as a subspace of R(the reals). Which of the>following sets are open in Y? Which are open in R?> >A = {x : 1/2 < |x| < 1}>B = {x : 1/2 < |x| <=1}>C = {x : 0 < |x| < 1 and 1/x not in the integers greater than 0}> >My confusion is as follows. I must consider the subspace topology of Yas a>subspace of R, which by definition is>T_y = {Y intersect U : U is in T} where T is a topology on R. Is T justany>topology on R? Can I pick any one? No, if someone says something about topology on R without specifying> which topology is intended then the standard topology is meant.> ThatÕs what standard means... >Because if I pick the collection of open intervals (a,b), then A isopen>in Y since it equals Y intersect (1/2,1), and B would not>be open in Y. However, if I pick the topology of all half-open intervals>(a,b], then A is not open, however B is.> >WhatÕs the deal here? The deal is you donÕt quite have the definition of the subspace> topology straight. Or maybe youÕre missing something else,> but if we give R the standard topology and then consider the> subspace topology on Y = [-1,1] then B _is_ open in Y. WhatÕs the definition of the subspace topology? >Does the question assume the standard topology on R? (i.e. the open>intervals?)> > ************************ line and let f be increasing functionthen f(max(x,y))= max(f(x),f(y)).My proof:Let x My theorem:> Let x,y be points in real line and let f be increasing function> then f(max(x,y))= max(f(x),f(y)).> My proof:> Let x LHS:max(x,y)=y,f(max(x,y))=f(y)> RHS:f(x) Is this correct? Any comments?> I wonder if the converse is also true i.e.> If f(max(x,y))=max(f(x),f(y)) then f is increasing.> Is there someone who Ôd like to prove it?> Is the condition f is increasing the only characterization or most general> one such that equation f(max(x,y))=max(f(x),f(y)) is still let f be increasing function> then f(max(x,y))= max(f(x),f(y)).> My proof:> Let x LHS:max(x,y)=y,f(max(x,y))=f(y)> RHS:f(x) Is this correct? Any comments?You canÕt just let x=y.I would write the proof like this, x>=y.If x=y:f(max(x,y))=f(x)=max(f(x),f(y)), since f(x)>=f(y).Hence, in all cases f(max(x,y))= if the converse is also true i.e.> If f(max(x,y))=max(f(x),f(y)) then f is increasing.> Is there someone who Ôd like to prove f(max(x,y))=max(f(x),f(y))Then, if x>y, we have:f(x)=f(max(x,y))=max(f(x),f(y))Therefore, f(x)>=f(y)Hence, the function is f is increasing the only characterization or most general> one such that equation f(max(x,y))=max(f(x),f(y)) is still valid?I am not sure that that question is quite meaningful.-- x,y be points in real line and let f be increasing function| > then f(max(x,y))= max(f(x),f(y)).| > My proof:| > Let x LHS:max(x,y)=y,f(max(x,y))=f(y)| > RHS:f(x) Is this correct? Any comments?|| You canÕt just let x=y.No need. The hypothesis is symmetric in x and y, and the case x=y is trivial.|| I would write the proof like this, perhaps:|| x>=y.| If x=y:| f(max(x,y))=f(x)=max(f(x),f(y)), since f(x)>=f(y).| Hence, in all cases f(max(x,y))= max(f(x),f(y)). QED| is also true i.e.| > If f(max(x,y))=max(f(x),f(y)) then f is increasing.| > Is there someone who Ôd like to prove it?|| f(max(x,y))=max(f(x),f(y))|| Then, if x>y, we have:| f(x)=f(max(x,y))=max(f(x),f(y))| Therefore, f(x)>=f(y)| Hence, the function is increasing.| increasing the only characterization or most general| > one such that equation f(max(x,y))=max(f(x),f(y)) is still valid?|| I am not sure that that question is quite meaningful.|| --| Clive Tooth| x,y be points in real line and let f be increasing function> | > then f(max(x,y))= max(f(x),f(y)).> | > My proof:> | > Let x | > LHS:max(x,y)=y,f(max(x,y))=f(y)> | > RHS:f(x) | > Is this correct? Any comments?> |> | You canÕt just let x | for all y. So you have only proved half of it. You must also cover thecase> | x>=y. No need. The hypothesis is symmetric in x and y, and the case x=y istrivial.I agree with your second sentence. But the original poster had not madeeither of those points and I felt that a bald let x> | > then f(max(x,y))= max(f(x),f(y)).>> | > My proof:>> | > Let x> | > LHS:max(x,y)=y,f(max(x,y))=f(y)>> | > RHS:f(x)> | > Is this correct? Any comments?>> |>> | You canÕt just let xand>> | for all y. So you have only proved half of it. You must also cover the>case>> | x>=y.>> No need. The hypothesis is symmetric in x and y, and the case x=y is>trivial.I agree with your second sentence. But the original poster had not made>either of those points and I felt that a bald let xenough.Saying WLOG x:>> It looks like your spell checker changed my memoized to memorized.>> The term is actually memoized. ItÕs a CS thing.Uh, that was me. There was no CS when I was in school... Memoize makes> sense (I had to look up a formal definition), and I would definitely use the> term for a static table, but in the case where you build a dynamic set of> return values, you are in a sense Ômemorizingthe return values in order to> memoize the function. So weÕre both right.> >>Note that the input value is the highest value that the program can>>evaluate. ItÕs equal to 0xFFFE0000 which is not the maximum 32 bit> unsigned>>int. IÕm not exactly sure why if fails at 0xFFFE0001. This is true both> in>>Borland C++ and g++ so itÕs not a compiler issue...>> Probably some computation exceeds 2^32, e.g. if S2 calls JSHpi with a>> value of xin greater than 2^32.IÕm sure thatÕs whatÕs happenning, and itÕs probably the i*i term in the for> loop in S2. That will cause premature termination of the loop on overßow.> So, maybe that sqrt needs to go back in, which will also slow things down,> but maybe if itÕs only used for large input values like over 0xFFFE0000?> >> When I replace unsigned with unsigned>> long long I get pi(10,000,000,000) = 105097565. I have no idea if thatÕs>> correct since I know of no other prime counting program for which I have>> enough patience to wait for it to finish. Using long long makes the>> program twice as slow, by the way.Some sample values of pi(x) for big numbers> (from http://www.trnicely.net/pi/tabpi.html)> 1.0000e+09 50847534> 2.0000e+09 98222287> 3.0000e+09 144449537> 4.0000e+09 189961812> 5.0000e+09 234954223> 6.0000e+09 279545368> 7.0000e+09 323804352> 8.0000e+09 367783654> 9.0000e+09 411523195> 1.0000e+10 455052511The JSH program variants produce the same answers through 4*10^9. So your> answer for 10^10 is way off. I wish I still had access to a native 64bit> machine... Last time was in 1975. I have a feeling that hacked 64 bit> integer math on a fast P3 or P4 class machine is still way faster.fun.DonÕt be so sure. :-) An anonymous submitter memoized my legendrephialgorithm and so far that oneÕs in third place. :-)--Stan Gula> -- #191, ewill3@earthlink.netItÕs still legal to go submitter memoized my legendrephi> algorithm and so far that oneÕs in third place. :-)Oh, sure, if youÕre going to cheat Jimenez:>>The JSH program variants produce the same answers through 4*10^9. So your>>answer for 10^10 is way off. Hmm.>>fun.Yes, IÕve been having fun, too. IÕve got one more modification (sorry,> I wonÕt promise to stop, but IÕll try :-) Notice that the loop in S2 only> needs to iterate over the primes. I peeled off the first iteration of> the loop for i=2, then propagated a bunch of constants to simplfy things.> I moved the code inside the loop to a function called compute_sum, then> had the inner loop iterate over values of i of the form 6*k+1 and 6*k+5> where k is an integer; all primes except for 2 and 3 take this form.> The result is 30% faster for large values on a Pentium 4 and about 7%> (thereÕs a lot of noise in the measurement) on an Athlon. HereÕs the code:> [submission snipped for brevity]-- #191, ewill3@earthlink.netItÕs still legal to go arenÕt nice powers of 10> (which are available in published tables). Could you time> PrimePi[4294500000]? On the slow machine: 0.37 seconds.> On the fast machine: 0.15 .8ecrit dans le message> > Wow. I wonder is it is as fast for values that arenÕt nice powers of10> > (which are available in published tables). Could you time> > PrimePi[4294500000]?> > On the slow machine: 0.37 seconds.> the effort. No surprise, they use a *good* algorithm :)----------------------------------------------------------- --PrimePi and Prime use sparse caching and sieving for n lessthan about 10^8. For large n the Lagarias-Miller-Odlyzkoalgorithm is used for PrimePi, based on asymptotic estimatesof the density of primes, and is inverted to give Prime[n].---------------------------------------------------- ---------Taken fromhttp://support.wolfram.com/mathematica/kernel/Symbols/ the FBI, the ASPCA and whoever else> he had on the mathematicianÕs trail.Nobody called me on this but itÕs been irritating me. Ihate misplaced apostrophes but this one got away from me.... on the mathematicianstrail. - whoever else>>he had on the mathematicianÕs trail.> Nobody called me on this but itÕs been irritating me. I> hate misplaced apostrophes but this one got away from me.... on the mathematicianstrail.> - RandyHmm. I would have thought you were speaking of the genericmathematician, the proverbial some great math results on-line> > trusting that when I found that great math, someone in the world would> > recognize it.> > That faith in the broad populace was mislaid, and in fact, mostly I> > got insults, and lies.> > My paper Advanced Polynomial Factorization is being published by the> > Mega Foundation, a society of the highest I.Q.Õs in the country.>Yeah, IÕve known people who could write stuff like what I found in the MegaFoundation, but they had to smoke a lot of weed to get to this level.Congratulations, Edwards. I am stuck on> a formula on page 9.Int(0,oo) (e^-nx)(x^s-1) dx = Gamma (s)/n^s. (s>0,n=1,2,3)Ugh, ItÕs x^{s-1} not x^s-1 :-( (and e^{-nx} not e^-nx).integral_0^infinity exp(-nx) x^s dx/x = Gamma(s)/n^sfollows by replacing x by nx in the definition of Gamma.> >Riemann sums> this over n and uses Sigma(n=1,oo)1/n^s=(r-1)^-1 to obtainInt(0,oo)(x^(s-1))/(e^x)-1 dx = Gamma(s)Sigma(n=1,oo)1/n^s.Ugh squared. Make thatintegral_0^infinity x^s/(exp(x) - 1) dx/x = Gamma(s) zeta(s).This follows from the geometric seriessum_{n=1}^infinity exp(-nx) = exp(-x)/[1 - exp(-x)] = 1/[exp(x) - 1].I donÕt see the connection. He also states Convergence of the> improper integral on the left and the validity of the interchange of> summation and integration are not difficult to establishAs long as the real part of s is > 1. Use the monotone and dominated convergence theorems.backtrack a bit. On page 9, Edwards states (hope I have the notationright)integral_0^infinity exp(-nx)x^(s-1) dx = Gamma(s)/n^sHow does n^s wind up in the denominator, based states (hope I have the notation> right)integral_0^infinity exp(-nx)x^(s-1) dx = Gamma(s)/n^sHow does n^s wind up in the denominator, based on substitution of xThis is really just a simple substitution. Start with Gamma(s) = int_0^infinity exp(-x) x^s dx/xand put x = ny.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of in an optimal number of transformations) algorithm is used in almost every mathematical field there is. However, IÕm interested to know why itÕs used so widely? What are its time and space complexities (rough ideas)? What makes is better than other algorithms that exist? (even computer --A and B are sets.q and p are members.Option 1: q and p are members of A, but then q is not equal to p .Option 2: q is a member of A , p is a member of B .D = Discreteness = q XOR p = a localized element = {.}C = Continuum = q to p correspondence = a non-localized element = {.___.}In the Common Math 0^0 is not well defined,because each member is D.Let us say that power 0 is the simplest levelof existence of some setÕs content.Because there are no Ds in C, its base value = 0,but because it exists (unlike the emptiness),its cardinality = 0^0 = 1.There are now 3 kinds of cardinality:|{}| = 0 = the cardinality of the Empty set.|{._.}| = 0^0 = 1 = the cardinality of C.|{.}| = 1^0 = 1 = the cardinality of D.Any point is a D element. Any line a C element.It means that there is a XOR ratio between LINES to POINTS.XOR ratio between LINES to POINTS---------------------------------0(LINE) 0(POINT) -> 0-(No information) -> no conclusion. -> conclusions on points.1(LINE) 0(POINT) -> 1-(Clear Wave-like information) -> conclusions on lines.1(LINE) 1(POINT) -> 0-(No clear information) -> no see:http://www.geocities.com/complementarytheory/ matrices of same rank and AA=A and BB=B then A is similar to then A is similar to B.>(Assumed: they are also of the same BB=B then A is similar toB.(Assumed: they are also of the donÕt know a better one.I canÕt figure out how to program the following in matlab:(iow, to fill hyper-matrices of arbitrary size)function x = fill(N = positive integer)H(1,1, ..., 1).matrix = 1 % (H has N arguments)% end function The the best newsgroup, but I donÕt know a better one.> HereÕs a good one: comp.soft-sys.matlab> I canÕt figure out how to program the following in matlab:(iow, to fill hyper-matrices of arbitrary size)function x = fill(N = positive integer)H(1,1, ..., 1).matrix = 1 % (H has N arguments)% end function> > The problem is that you need N for-loops.WilbertItÕs not clear whether you wish to fill in elements individually,or simply to initialize a matrix.ALL ZEROS: H = zeros( [n1,n2,n3,...,nN] );or H(n1,n2,...nN) = 0;ALL ONES: H = ones( [n1,n2,n3,...,nN] );ALL NaNs (Not a Number): H = NaN*zeros( [n1,n2, ..., nN] );If you have a vector L of length ell: ell = prod([n1, n2, ..., nN] );then H = reshape( L, [n1, n2, ..., nN] );or H(:) = L;will fill H by running down the indices 1 through N in order.That is, if you have a 2x2x3 matrix, and the vector [1:12],you get this: H(1,1,1) = 1 H(2,1,1) = 2; H(1,2,1) = 3; H(2,2,1) = 4; H(1,1,2) = 5; H(2,1,2) = 6; H(1,2,2) = 7; H(2,2,2) = 8; H(1,1,3) = 9; H(2,1,3) = 10; H(1,2,3) = 11; H(2,2,3) = 12;Also, check out these functions, which I found by entering thecommand lookfor(ÔindexÕ): NUM2IND finds the index for element in array given the number of the element SUBSINDEX Subscript index. IND2SUB Multiple subscripts from linear index. SUB2IND Linear index from multiple subscripts.Folks at comp.soft-sys.matlab may (as in almost surely will)have better/faster/cuter ways of filling in an n-dimensionalarray. Generally speaking, the greater control you wish tohave over the placement of elements, the greater amount ofwork youÕll have to do to accomplish that, but itÕs notalways (or even typically) the case that youÕll need I donÕt know a better one.I canÕt figure out how to program the following in matlab:(iow, to fill hyper-matrices of arbitrary size)function x = fill(N = positive integer)H(1,1, ..., 1).matrix = 1 % (H has N arguments)% end function> > The problem is that you need N for-loops.WilbertTry donÕt know a better one.I canÕt figure out how to program the following in matlab:(iow, to fill hyper-matrices of arbitrary size)function x = fill(N = positive integer)H(1,1, ..., 1).matrix = 1 % (H has N arguments)% end function> > The problem is that you need N for-loops.Wilbertviable solution in that particular language.Create an array of triples with values that would be fed into a for-loop: (start_value, end_value, step). This array could/would be generated by a for-loop counting for the number of for-loops you are trying to emulate.Create a second array of scalar values that is the same length as the array of triples with each value set to the start values from their respective triples. This array now contains all of the dynamically created for-loop index counters.Loop all the counters at once with one big while loop (pseudo-C notation):typedef enum {start_val, end_val, step_val} triple_index;for_triples = {{0,5,1} , {1,5,1} , {7,19,2}, {0,5,5}}; // for examplefor(i=0;i 0 && index_vars[i] <= for_triples[i][end_val]) ||(for_triples[i][step_val] < 0 && index_vars[i] >= for_triples[i][end_val]) ) break; // Reset this level to the beginning index_vars[i] = for_triples[i][start_val] // If we drop out of this loop here, we also need to drop out // of the while because weÕve done with the dynamic-fors cond = 0; } // end for} // end whilecode is here: http://www.cyreksoft.yorks.com/euphoria/nfor.e (full of bugs, but the example included at the bottom works know a better one.>> >> I canÕt figure out how to program the following in matlab:>> >> (iow, to fill hyper-matrices of arbitrary size)>> >> function x = fill(N = positive integer)>> >> H(1,1, ..., 1).matrix = 1 % (H has N arguments)>> >> % end function>> >> The problem is that you need N for-loops.>> >> Wilbert> viable solution in that particular language.The question was about MATLAB, not MathCad.You might try comp.soft-sys.matlab.-- Dave SeamanJudge YohnÕs mistakes revealed in Mumia Abu-Jamal ruling. The question was about MATLAB, not MathCad.My bad. But since the pseudo-code I provided wasnÕt for any language in particular, the concept and instructions should still work, providing the OP knows enough about Mat*LAB* to be able to translate or use them.> You might try comp.soft-sys.matlab.The OP perhaps. I was only interested in providing some generic help since it was a coding problem I have a question about the fourier transformation (probably a reallystupid question since IÕm new to this...):Let f(x) be an infinite train of delta (dirac) impulses:f(x) = SUM(DELTA(x-v*T), v=-INFTY...+INFTY)According to my textbook, the fourier transform of f(x) is F(w) =2*PI/T * SUM(DELTA(w-v*2*PI/T), v=-INFTY..+INFTY), another train ofimpulses with a different frequency and a different amplitude.My question is: Where does the first 2*PI/T factor (the one outside theSUM) come from? This factor obviously changes the amplitude. I dounderstand why the frequency changes, but why the amplitude?I tried calculating the fourier transform myself by using the fourierintegral formula F(w)=int(f(t)*exp(iwt))dt - that got me F(w) =SUM(exp(-iwvT),v=-INFTY..+INFTY) which unfortunately didnÕt help mewith my problem.If any of you had a formal or intuitive explanation, IÕd be verythankful ;-)TIA, bye,Florian-- eMail: GoldbachÕs conjecture is>>> undecidable,> >>>>but suppose it is. It would then be true, even though unprovable,>>> because> >>>>there could be no exception to it. The existence of any even number>>>>>greater>>>>>than 2 that is not the sum of two odd primes would disprove the>>> conjecture> >>>>and therefore contradict its undecidability.>>>>>>Is this correct? Gell-Man suggests there could be no exception. Why>>>>>not?>>>>>I believe there could be, but with no hope of finding it, should it>>>>>exist.>>>>Why? If an exception exists, you can in principal find it, by testing all>>>even numbers until you reach the exception.>>>>BTW, I reckon Goldbach is an excellent candidate for true but unprovable.>>>There are very strong probabalistic arguments for believeing it to be>> true> >>>because larger numbers have lots of smaller primes to select from in>> forming> >>>the sum, and it has been tested right out to some very large numbers. The>>>probability of a given number being an exception drops fast enough (as>> the> >>>size of the number increases) to make it very likely that there are no>>>exceptions. But despite 200 years of effort no actual proof has been>> found.> >>>Goldbach may just happen to be true for probabilistic reasons, but there>> is> >>>no actual proof for all even numbers.>>>IÕm assuming it canÕt be proven to be unprovable (since that would prove> it to> >>be true)No. It could be false; that would be a good reason why we canÕt prove it.> The problem lies more with the converse statement there exists an even> number which is NOT the sum of two primes. This cannot be true and> unprovable, as if it is true it is easy to prove by finding the even number.It sure can be true and unprovable. Suppose the even number is very large. Then it is not possible to find the number let alone is easyto prove by finding the even number. How do you find a number which is so large that it is not possible to even contain it in any recorded formother than it is a counterexample for GBC. A small number, say, in the range of factorial(googolplex) or 10^10^100! cannot have all its digits recorded. One could not even know all the possible primes less than the number, though one only needs the primes greater than that the argument of the factorial function. Thus finding a counterexample is not a reasonable method for deciding GBC if the counterexample is not small enough.Larry>>but is it possible that it could be something like GoodsteinÕs sequence>>(i.e. we can prove that itÕs undecidable, but not within Peano> arithmetic)?Yes, good point. It is provable (or unprovable) only in some axiom system.> Its certainly provable in the axiom system PA + GC, where GC is the Goldbach> conjecture. Trouble is, I donÕt know if PA + GC is a consistent system ....>>Or>>am I babbling?Not at all. Or There is no reason to believe that GoldbachÕs conjecture is undecidable,> but suppose it is. It would then be true, even though unprovable, because> there could be no exception to it. The existence of any even number greater> than 2 that is not the sum of two odd primes would disprove the conjecture> and therefore contradict its undecidability.Is this correct? No. If PA does not prove GoldbachÕs conjecture, then there is a model of PA in which GoldbachÕs conjecture is false -- this is a consequence of the so-called Ôcompleteness theoremwhich says that something is provable in a theory if and only if it is true in every model of that theory. Wait 5 hours and ten other posts will explain what I mean.The difficulty you are having is that you are thinking of the Ôstandardmodel of PA, and asking ÔIf Goldbach is undecidable, is it true in the standard model of PA?This is a more subtle question. First, you need to believe that the standard model exists, so that youcan ask whether it has a certain property. If you believe the standard model exists, then you will probably believe that ÔIf Goldbach is undecidable, it is true in the standard model of PAÕ.But that is not the statement you asked thinking of the Ôstandard>model of PA, and asking ÔIf Goldbach is undecidable, is it true in the >standard model of PA?This is a more subtle question. First, you >need to believe that the standard model exists, so that you >can ask whether it has a certain property. If you believe the >standard model exists, then you will probably believe that >ÕIf Goldbach is undecidable, it is true in the standard model of PAÕ. >But that is not the statement you asked about. This is another fine example of the awful effect that anacquaintance with logic can have on peopleÕs thinking in mathematics.There is nothing subtle whatever about the question whetherGoldbachÕs conjecture is undecidable in PA implies GoldbachÕsconjecture is true: the answer to the question is, trivially, yes.Weird misgivings about the existence of the standard model, whichmean absolutely nothing, mathematically speaking, are irrelevant.That GoldbachÕs conjecture is true means, as it does in any mathematicalcontext, that every even number greater than 2 is 4^3--------------------------- C(52,5)is the probability of getting an exclusive pair(a a b c d, with unique a, b, c, d) in poker.The stupid book dares to disagree. Whosright and why?-- V.8anligenKerstin K.8all (The x C(4,2) x 4^3> ---------------------------> C(52,5)is the probability of getting an exclusive pair> (a a b c d, with unique a, b, c, d) in poker.> The stupid book dares to x C(4,2) x 4^3>> --------------------------->> C(52,5)>> and why?> you firstYou pick 4 denominations out of 13 possible (a, b, c, d).For every such thing there are different colorifications.If we regard [ a1 a2 b c d ] we can see that the colorof d doesnÕt affect the color of c which doesnÕt affectthe color of b which doesnÕt affect the colors of aÕs.So - d can be colored in 4 ways. So can c and b, whileaÕs can be colored in 6 ways (well, 4_C_2, actually).So, now your turn!-- V.8anligenKerstin K.8all (The 4^3>>>--------------------------->>> C(52,5)>>>>and why?>>you first> You pick 4 denominations out of 13 possible (a, b, c, d).Ok.> For every such thing there are different colorifications.The way you count the colorings is different for the pair and the non-pairs. Also, you have to count the ways to choose *which* of a,b,c,d will be the pair.> If we regard [ a1 a2 b c d ] we can see that the color> of d doesnÕt affect the color of c which doesnÕt affect> the color of b which doesnÕt affect the colors of aÕs.Except youÕve made an assumption that a will be the pair. Why canÕt the pair be in b,c, or d?So - d can be colored in 4 ways. So can c and b, while> aÕs can be colored in 6 ways (well, 4_C_2, actually).So, now your turn!> You should be off by a factor of 4=C(4,1)= number of ways to choose *which* value is the --------------------------->> C(52,5)>> You pick 4 denominations out of 13 possible (a, b, c, d).> For every such thing there are different colorifications.> If we regard [ a1 a2 b c d ] we can see that the color> of d doesnÕt affect the color of c which doesnÕt affect> the color of b which doesnÕt affect the colors of aÕs. So - d can be colored in 4 ways. So can c and b, while> aÕs can be colored in 6 ways (well, 4_C_2, actually).IÕm sorry, 52 cards with millions of possible hands are too much formy puny brain. Could you kindly work out a smaller example? LetÕs sayweÕre playing with an 8-card deck: the aces, kings, queens, and jacksin the black suits. Now, if you deal 5 cards, whatÕs the probabilityof getting a one-pair hand?IÕm guessing your going to say that the answer isC(4,4)*C(2,2)*(2^3)/C(8,5) = 8/56. Is that right?No, that canÕt be right, because I started counting them by hand andgot:1) As Ac Ks Qs Js2) As Ac Ks Qs Jc3) As Ac Ks Qc Js4) As Ac Ks Qc Jc5) As Ac Kc Qs Js6) As Ac Kc Qs Jc7) As Ac Kc Qc Js8) As Ac Kc Qc Jc9) As Ks Kc Qs JsOops, thatÕs already more than 8 one-pair hands, who knows how manythere will be by the time IÕm done counting them.What am I doing hand from your eight-carddeck. In how many ways could you do that? One way is to first decide to form the pair. There are four suits tochoose from. Once the suit for the pair is chosen there is only one way topick the two cards that will form the pair. Then you must choose the three oddcards. Since there are only three ranks left there is only one way to pickthose three ranks. Finally, for each of the odd-card ranks, there are two waysto pick the card from each rank. Therefore, the number of one-pair hands fromyour depleted deck is 4*1*1*2^3 = 32. Dividing by C(8,5) gives the probabilityof getting a one-pair hand of 32/56, or 4/7. In this type of counting problem it is easy to make a mistake, so a checkmight be to determine the number of two-pair hands that could be formed. Thereare C(4,2) ways of picking the ranks that will form the pairs. In each ofthese there is only one pair that can be formed. Then there are two ranksremaining that might supply the odd card, and for the chosen rand there are twochoices for the suit. So the number of two-pair hands is C(4,2)*2*2=24. Theprobability of a two-pair hand is 24/56 = 3/7. Notice that your depleted deck can only form either a one-pair hand or atwo-pair hand, and that the sum of the probabilities of these is unity. The probability of getting a one-pair hand for a 52-card deck can bedetermined in the same radius R. Given that the massdensity of the plate varies directly as the distance from a point P on theboundary of the plate, locate the center of massHow can I get delta, the density function, in terms of one variable which issomewhat friendly? Can I assume that P = (R,0) ? If P = (R, 0) then I gotdelta (x) = k sqrt (2R^2-2XR). Then things get extremely messy from there.Any hints would be has the form of a disc of radius R. Given that the> mass density of the plate varies directly as the distance from a point> P on the boundary of the plate, locate the center of massHow can I get delta, the density function, in terms of one variable> which is somewhat friendly? Can I assume that P = (R,0) ? If P = (R,> 0) then I got delta (x) = k sqrt (2R^2-2XR). Then things get extremely> messy from there. Any hints would be appreciated.> StevenDue to symmetry considerations, the center of mass will be on the (Ox) axis.You first have to compute the mass. Given a point M=(x, y) on the plate, you have the mass distribution function :?(M) = a * MP = a * sqrt((R-x)^2 + y^2)Then, you have to calculate the following integral : m = iint(?(M)dxdy)for -R <= x <= R and -sqrt(R^2-x^2) <= y <= sqrt(R^2-x^2) It gives you m = a*iint(sqrt((R-x)^2 + y^2) dxdy) that Maple can give you (I hope).I think however that a polar parametrization of the circle with rho=2Rcos(theta) could work much better.I keep on having a look on this.-- Alexandre CharitopoulosEm6 / factorial integer seeds in the Collatz conjecture?Factorials > 5! where the seed is a factorial, then any integercreated in its sequence will never be larger than its seed?Any counter examples for factorials >5!?There are others like --Where n= 0,1,2,3,4,5, . . n, then where 2^n is a seed and terminatingthe loop @ 1 then any integer created in its sequence will never belarger than its seed.Also seeds where (2^n *10) where n=>1.Is there a sequential list of these integers that have this property?If this list exists then Is this true about factorial integer seeds in the Collatz conjecture?Factorials > 5! where the seed is a factorial, then any integer> created in its sequence will never be larger than its seed?Very likely. In order for the excursion (the maximum value in the sequence) to be greater than the seed, you need a lot of 3x+1iterations. With factorials, you end up with a lot of factors of2. For instance, 992! starts out with 987 factors of 2 (in binary,this would appear as 987 contiguous least significant 0s).The sequence can exceed the seed only if it can muster up enoughcontiguous 1s at the least significant end of the binary number after the 0s have been stripped away. This doesnÕt appear very likely (but I canÕt prove it wonÕt ever happen).Any counter examples for factorials >5!?In a quick test up to 1000!, the most consecutive 1s next to theleast significant 0s barely got past 10:240! had 10 1s followed by 236 0s688! had 11 1s followed by 684 0s990! had 11 1s followed by 982 0s992! had 12 1s followed by 987 0s> There are others like --> Where n= 0,1,2,3,4,5, . . n, then where 2^n is a seed and terminating> the loop @ 1 then any integer created in its sequence will never be> larger than its seed.Well, yeah, powers of 2 donÕt ever have any iterations of 3x+1 so theexcursion can never be greater than the seed.> Also seeds where (2^n *10) where n=>1.This will give you exactly 1 iteration of 3x+1, so basically for thesame reason as powers of 2, you donÕt have enough iterations of 3x+1when n gets large.> Is there a sequential list of these integers that have this property?> If this list exists then each integer in this list has a maximum> divisor of 4.DanHereÕs another question: how small can the sequence shrink from the seedbefore turning around and eventually getting larger than the seed?If the excursion is the maximum point, call the minimum value that occurs_before_ the excursion the anti-excursion. HereÕs an example: seed 0000020370359763344860862684456884093781610514683936659 36250636139181703781071533935209480192000anti-excursion 0000000000000000000000000000000000016069380442589902755 41962092341162602522202993782792835301375 excursion 5312279777517495386775626440715592536584669053067889919 49149923478184981802604365988769398088000The sequence shrank to 30 orders of magnitude less than the seed beforerallying and eventually growing 5 orders of magnitude larger than the seed.Of course, these numbers can be constructed. The formula for the above isseed = (2**i - 1) * 2**(i/2) where i is an even integer. For i=200, we get a binary number with 100contiguous 1s followed by 100 contiguous 0s. This means there will be100 consecutive iterations of x/2 before the true about factorial integer seeds in the Collatz conjecture?>> >> Factorials > 5! where the seed is a factorial, then any integer>> created in its sequence will never be larger than its seed?Very likely. In order for the excursion (the maximum value in the >sequence) to be greater than the seed, you need a lot of 3x+1>iterations. With factorials, you end up with a lot of factors of>2. For instance, 992! starts out with 987 factors of 2 (in binary,>this would appear as 987 contiguous least significant 0s).The sequence can exceed the seed only if it can muster up enough>contiguous 1s at the least significant end of the binary number >after the 0s have been stripped away. This doesnÕt appear very >likely (but I canÕt prove it wonÕt ever happen).> >> Any counter examples for factorials >5!?In a quick test up to 1000!, the most consecutive 1s next to the>least significant 0s barely got past 10:240! had 10 1s followed by 236 0s>688! had 11 1s followed by 684 0s>990! had 11 1s followed by 982 0s>992! had 12 1s followed by 987 0s>> >> >> There are others like -->> Where n= 0,1,2,3,4,5, . . n, then where 2^n is a seed and terminating>> the loop @ 1 then any integer created in its sequence will never be>> larger than its seed.Well, yeah, powers of 2 donÕt ever have any iterations of 3x+1 so the>excursion can never be greater than the seed.> Also seeds where (2^n *10) where n=>1.This will give you exactly 1 iteration of 3x+1, so basically for the>same reason as powers of 2, you donÕt have enough iterations of 3x+1>when n gets large.> >> >> Is there a sequential list of these integers that have this property?>> If this list exists then each integer in this list has a maximum>> divisor of 4.>> >> DanHereÕs another question: how small can the sequence shrink from the seed>before turning around and eventually getting larger than the seed?If the excursion is the maximum point, call the minimum value that occurs>_before_ the excursion the anti-excursion. HereÕs an example: seed 0000020370359763344860862684456884093781610514683936659> 36250636139181703781071533935209480192000>anti-excursion 0000000000000000000000000000000000016069380442589902755> 41962092341162602522202993782792835301375> excursion 5312279777517495386775626440715592536584669053067889919> 49149923478184981802604365988769398088000The sequence shrank to 30 orders of magnitude less than the seed before>rallying and eventually growing 5 orders of magnitude larger than the seed.Of course, these numbers can be constructed. The formula for the above isseed = (2**i - 1) * 2**(i/2) where i is an even integer. For i=200, we get a binary number with 100>contiguous 1s Oops, that should be 200 1s and 100 0s. That ratio is chosen because the growthrate for an m-bit all 1s pattern is m*1.585 so with m=200 after the original100s are removed, the excursion ends up being 317 bits making it slightlylarger than the seed.>followed by 100 contiguous 0s. This means there will be>100 consecutive iterations of x/2 before the first 3x+1 is encountered.>3x+1, x/2 causing the sequence to grow, eventually passing the seed>before the 1s run out.--Mensanator2 of Clubs =A solution to problem 7.4 in SL DixonÕs Fluid Mechanics andThermodynamics of Turbomachinery 4th Ed. has me stumped and I wouldappreciate any help understanding the steps that were left out.The compressible ßow relation between mass ßow rate, speed ofrotation and the ßow parameters at the eye tip is given by eqn.(7.11),0.6598 x 10^-8 x (mdot x rads ^2)/k = ...IÕve not figured out yet how weÕre to get 0.6598 x of the eqn ismdot*Omega^2/[pi*k*gamma*Po1*sqrt(gamma*R*To1)]. If you take outmdot*omega^2/k, whatÕs left,1/[pi*gamma*Po1*sqrt(gamma*R*To1)], is equal to 0.6598e-8. Did you actuallytry to plug in numbers?Pete K.> A solution to problem 7.4 in SL DixonÕs Fluid Mechanics and> Thermodynamics of Turbomachinery 4th Ed. has me stumped and I would> appreciate any help understanding the steps that were left out. The compressible ßow relation between mass ßow rate, speed of> rotation and the ßow parameters at the eye tip is given by eqn.> (7.11), 0.6598 x 10^-8 x (mdot x rads ^2)/k = ... IÕve not figured out yet how weÕre to get 0.6598 Fluid Mechanics and> Thermodynamics of Turbomachinery 4th Ed. has me stumped and I would> appreciate any help understanding the steps that were left out. The compressible ßow relation between mass ßow rate, speed of> rotation and the ßow parameters at the eye tip is given by eqn.> (7.11), 0.6598 x 10^-8 x (mdot x rads ^2)/k = ... IÕve not figured out yet how weÕre to get 0.6598 x 10^-8I donÕt know. But, if nobody else seems to know, I would suggest that youre-post, quoting the whole of the question and the whole of the answer.-- Clive passed on. What i mean by this is bestillustrated in the following ex.T= 2*pi (elongation/g)where g is gravity (numerical constant)and where elongaton is 2.5 cmwith a standard deviation of .01cm. after i find the value of T is itpossible for the standard deviation of T to be calculated/ passed onfrom the standard deviation of elongation by plugging-it in with thesame equation??If its no bother can standard deviation be passed on. What i mean by this is best>illustrated in the following ex.>T= 2*pi (elongation/g)>where g is gravity (numerical constant)and where elongaton is 2.5 cm>with a standard deviation of .01cm. after i find the value of T is it>possible for the standard deviation of T to be calculated/ passed on>from the standard deviation of elongation by plugging-it in with the>same equation??>If its no bother can you please reply to firedreamer83@aol.comSomewhere in your textbook there should be a mention of some of the properties of standard deviation. One of the main ones is thatif c is a positive constant, the standard deviation of c X is c times the standard deviation of X.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of prove that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 whenever n is apositive integer using Mathematical Induction. Could ... n*n! = (n + 1)! - 1 whenever n is a>positive integer using Mathematical Induction. Could someone please take me>First one verifies that the formula is valid for the smallest valueof n, i.e. n=1. Which as usual is quite simple.Then you have to prove that if the formula is true for a given n,then the formula in which you have replaced (everywhere)n by n+1 is also true (for the same value of n). How does theresult of this replacement look like ? In this case, the specialaspect of the left hand side makes it that the replacementhas the effect that juxt one more term is included in the sum.Using the so called induction hypothesis, which is by defin. ourassumption that the formula is valid for the originally given n,you transform the new left hand size (with n replaced by n+1)in a sum of just two terms. To prove that this is equal to theright hand side (also with n replaced by n+1) is then easy.(Remember the def. of n! - which is a def. by math. that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 whenevern is a> positive integer using Mathematical Induction. Could someone pleasetake meFirst verify for n = 1: 1*1! + 2*2! = 5 = 6 - 1 = 3! - 1 = (2+1)! - 1 QED the statement for n = 1Now suppose the statement is true for n, i.o.w. suppose that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1and try to prove that the statement is true for n+1, i.o.w. prove that 1*1! + 2*2! + ... n*n! + (n+1)(n+1)! = (n+1 + 1)! - 1LetÕs prove this by examining the left hand side:LHS = 1*1! + 2*2! + ... n*n! + (n+1)(n+1)! use the statement for n = (n + 1)! - 1 + (n+1)(n+1)! rearrange = (n + 1)! + (n+1)(n+1)! - 1 isolate common factor (n+1)! = (1 + n+1)(n+1)! - 1 rearrange first factor = (n+1 + 1)(n+1)! - 1 = RHS QED statement for n+1, if we are allowed to use the statement for n.QED statement for all nNow itÕs your turn that 1*1! + 2*2! + ... n*n! = (n + 1)! - 1 whenever n is a>positive integer using Mathematical Induction. Could someone please take meMight I suggest attending office hours with your instructor?In any case: you want to prove something by induction, you need to dotwo things: (1) Prove that it is true for the first case; in this case, for n=1. The basis of the induction. (2) Prove that IF it is true when n=k, then it is also true for n=k+1. The inductive step.1*1! + 2*2! + ... + k*k! = (k+1)!-1.You want to show that it will also be true for k+1. That is, that ifyou calculate1*1! + 2*2! + ... + k*k! + (k+1)*(k+1)!and you calculate ((k+1)+1)! - 1,you will get the same answers.Usually, what you want to do is think of the case k+1 as being thecase n=k, plus something extra.Here, for instance, you should think of adding1*1! + 2*2! + ... + k*k! + (k+1)*(k+1)!asFirst do 1*1! + 2*2! + ... + k*k!, and then add to that(k+1)*(k+1)!.By thinking of it this way, you can use the fact that you supposedlyalready know what 1*1! + 2*2! + ... + k*k! is equal to: it is equal to(k+1)!-1.So1*1! + 2*2! + ... + k*k! + (k+1)*(k+1)! = (1*1! + 2*2! + ... + k*k!) + (k+1)*(k+1)! = (k+1)!-1 + (k+1)*(k+1)!Now just manipulate this formula to see if you can show it to be equalto (k+2)!-1. It should not be what I accept as reality. --- Calvin (Calvin and to the there is no such thing as a dumbquestion rule, but is there an algorithmn or computer program thatwill predict the spherical packing density of a distribution ofdensity of theoretical spherical sand grains having a sizedistribution 2-dimensional lattice random walk with balanced probabilities,> i.e. four of 1/4 each, starting at <0,0>; what is the probabil;ity> that it gets to <0,1> before returning to <0,0> ?The answer was given as exactly 1/2, and my crude numerical estimations> seem to bear this out. At the time, I think a full proof was given,> but it seemed a rather long one, and it struck me that for such a simple> result, there ought to be a simple proof. But I never found one.So, can anyone find a quick proof of the result?Or failing that, can anyone recall the earlier longish proof?I donÕt feel like repeating the entire proof (which doestake a little work), but let me whet your appetite by statingthe more general fact: If z is a point in the plane otherthan the origin, then the probability that the random walk S_jreaches z before returning to the origin is 1 ------- 2 a(z)where a(z) denotes the potential kernel defined by a(z) = lim_{N rightarrow infty} sum_{j=0}^N [P(S_j = 0) - P(S_j = z)]For nearest neighbors of the origin, a(z) = 1.For a discussion of the potential kernel you can seethe classic book on random walk, Spitzer, Principlesof Random Walk or a number of other books, so its expectation is: E(Z) = (1-p) sum_k [ k p^(k-1) (1-p) ]> = (1-p) E Geom(p) = (1-p)/p = 1/p - 1as claimed. IÕm sorry, but I still donÕt get it. You haveE[Z] = (1-p)^2 Sum_k kp^(k-1)right? But that sum is equal to (1-p)^(-2), so you just get E[Z]=1.Oops again. The above should have read E(Z) = (1-p) E Geom(1-p) whichdoes indeed just give E(Z)=1. IÕll ask Vlada about it, but perhaps weboth made the same mistake and think that this approach to a proof will not work, at least notwithout some significant new idea added. HereÕs why: you can see inthis second way of calculating the expected number of visits to (1,0)before (0,0) that we never actually used the fact that the point inquestion was (1,0). Thus the expected number of visits to any pointbefore returning to (0,0) is 1! This surprised me at first, and Ithought I had made a mistake, but whatÕs happening is that if a pointis way out than the number of visits before returning is 0 with highprobability, but if we do actually hit the point then weÕre likely tohit it a bunch of times before we return to (0,0). So, anyway, itseems that this approach, if it worked, would show that theprobability is 1/2 no matter what point we pick, and this is clearlyfalse. I will say, though, that I have no idea how it do it. I thoughtabout it for a while, and got nowhere. I do like this problem a lot,though.> > and so its expectation is:> > E(Z) = (1-p) sum_k [ k p^(k-1) (1-p) ]> > = (1-p) E Geom(p) = (1-p)/p = 1/p - 1> > as claimed.> > IÕm sorry, but I still donÕt get it. You haveE[Z] = (1-p)^2 Sum_k kp^(k-1)right? But that sum is equal to (1-p)^(-2), so you just get E[Z]=1.Oops again. The above should have read E(Z) = (1-p) E Geom(1-p) which> does indeed just give E(Z)=1. IÕll ask Vlada about it, but perhaps we> both made the same difference between Lenene Test and F-test?Anybody can give me and F-test?> Anybody can give me some hints?royGoogle for let this affect your ego,remember, we are only using our IMAGINATIONS here...) So being God, you are able to look at every single number thatexists all at once. The whole infinite of infinite spectum of numbersin one glance. Now, lets say that you picked a number completely atrandom (God has a VERY good random number generator, duh!) What are the properties of this number, N?This is what I imagine they are...1. N is very very very VERY large, because smaller numbers are muchless common than larger numbers.2. N is probably complex, because reals fill only a line, while thecomplex numbers fill an entire plane.3. N is most likely trancsendental, because...um...ah... Thats justthe way it is, okay? I donÕt really know why, this is a hunch.4. N is probably of a class of numbers we cannot even imagine, muchless put into words. There are probably universes of different typesof numbers that us mere mortals will never see.5. The probability is N being a plain old vanilla integer isinfitessimally(is that a word?) close to pretend that you are like God. (DonÕt let this affect your ego,> remember, we are only using our IMAGINATIONS here...)> So being God, you are able to look at every single number that> exists all at once. The whole infinite of infinite spectum of numbers> in one glance. Now, lets say that you picked a number completely at> random (God has a VERY good random number generator, duh!)> What are the properties of this number, N?> This is what I imagine they are...> 1. N is very very very VERY large, because smaller numbers are much> less common than larger numbers.> 2. N is probably complex, because reals fill only a line, while the> complex numbers fill an entire plane.> 3. N is most likely trancsendental, because...um...ah... Thats just> the way it is, okay? I donÕt really know why, this is a hunch.> 4. N is probably of a class of numbers we cannot even imagine, much> less put into words. There are probably universes of different types> of numbers that us mere mortals will never see.> 5. The probability is N being a plain old vanilla integer is> infitessimally(is that a word?) close to 0.Any other properties you can imagine?> -PaulI am not sure which numbers you mean here, perhaps we mean some sortof field that includes the integers.To start, lets talk about algebraically closed fields characteristic0.Here is a rephrasing and generalisation of the question. I hope youagree it is similar.Given some base wisdom W, that allows us to partition algebraicallyclosed fields into subsets. We now pick an elements that is only inlarge sets.How do we define what a large set is? How do we know that we canalways distinguish between a set and its complement? These aredifficult questions, unfortunately. If W is large enough, then we canjust pick partitions isolating each point in every algebraicallyclosed field in a set of size 1, which must be small.In this way one might say that God, could not be random.If we take some sort of mini god though, or think of the choices thatwould look random to us, we can do something;Suppose that W knows the names of every element in some largealgebraically closed field M (char(M)=0). Also W knows about algebraicoperations, and W can use first order logic. (Thus in particular, Wcan not name the integers as a set explicitly). Then exery set that Wcan name in M is either finite or cofinite (see for example HodgesModel Theory and look under strong minimality). Of course everyelement of M will occur in a singleton, so there is no random, Ishall call it generic element over M. However, if we look in anyextension N (also an acf) then evcery element of NM is generic.Conclusion, if W only knows first order logic and field operations andthe names of an acf M then any trancendental element is generic(relative to W).This is actually a useful idea. Now suppose that we are dealing with a real closed field K, then it isnot clear that we can do the same thing. Certainly if W knows aboutthe ordering, then it knows to much, as it can easily split up R intosay positive and negative elements, and which of these sets islarger?As I said, the notion of a generic element (generic type) is useful instrongly minimal structures such as algebraically closed fields. Thereare other places where one can develop a similar notion of a uniquegeneric type, relative to first order logic. One needs to know when adefinable set is large or small.My solution to the above quandry, which again proved to be ratheruseful, is to consider two generic types, the left generic is thetrancendental elements which are less than every element in K, theright generic type is that of pretend that you are like God. (DonÕt let this affect your ego,> remember, we are only using our IMAGINATIONS here...)> So being God, you are able to look at every single number that> exists all at once. The whole infinite of infinite spectum of numbers> in one glance. Now, lets say that you picked a number completely at> random (God has a VERY good random number generator, duh!)> What are the properties of this number, N?I have a similar question. What if you were God and could pick aBaire-typical real number?What would be the properties of that number?1. It would not be normal in any base.2. Any attempt to estimate the fraction of its digits that are anygiven digit in any given base would not pretend that you are like God. (DonÕt let this affect your ego,> remember, we are only using our IMAGINATIONS here...)> So being God, you are able to look at every single number that> exists all at once. The whole infinite of infinite spectum of numbers> in one glance. Now, lets say that you picked a number completely at> random (God has a VERY good random number generator, duh!)> What are the properties of this number, N?> This is what I imagine they are...> 1. N is very very very VERY large, because smaller numbers are much> less common than larger numbers.> 2. N is probably complex, because reals fill only a line, while the> complex numbers fill an entire plane.> 3. N is most likely trancsendental, because...um...ah... Thats just> the way it is, okay? I donÕt really know why, this is a hunch.> 4. N is probably of a class of numbers we cannot even imagine, much> less put into words. There are probably universes of different types> of numbers that us mere mortals will never see.> 5. The probability is N being a plain old vanilla integer is> infitessimally(is that a word?) close to 0.Any other properties you can imagine?> -PaulSo this god knows at once the output of each rng he ever wouldand takes it modulo 10...The number is *so much* ...ehmm... *transcendental*... <*pheww*>(After this is done he stops doing things that even he cannot do...for a (DonÕt let this affect your ego,> remember, we are only using our IMAGINATIONS here...)> So being God, you are able to look at every single number that> exists all at once. The whole infinite of infinite spectum of numbers> in one glance. Now, lets say that you picked a number completely at> random (God has a VERY good random number generator, duh!)I assume that this god is approximately the traditional christian god.It is not clear to me that such a god could do *anything* at random.immediately says Yeah, I *knew* it would pick that that you are like God. (DonÕt let this affect your ego,> remember, we are only using our IMAGINATIONS here...)> So being God, you are able to look at every single number that> exists all at once. The whole infinite of infinite spectum of numbers> in one glance. Now, lets say that you picked a number completely at> random (God has a VERY good random number generator, duh!)I assume that this god is approximately the traditional christian god.It is not clear to me that such a god could do *anything* at random.> immediately says Yeah, I *knew* it would pick that one.So, are you saying that God does not play with spherical dice??...;)Can God pick a number *SO* uncalculable/uncomputable that even HEcannot calculate/compute it???...;) -againAs for the OPÕs questions, I would guess that all of PaulÕs propertiesof the God-picked number would, for all practical purposes, betrue...except there might have an issue with property #1.I am definitely no expert on this topic, but I wonder if God wouldhave any problem picking a number COMPLETELY at random, without anyrestrictions, when the number is by-definition finite, but hasabsolutely NO chance of being within *ANY* arbitrary finite range ofpre-determined numbers.So, obviously this problem vanishes if we allow God to restrict hischoices to some finite-sized range. But then that would be our hubris,would that such a god could do *anything* at random.> immediately says Yeah, I *knew* it would pick that one.You know, that is a good point for the philosophers. God being allBut then again God being all powerful should be able to create aperfect random number generator. So maybe He creates a random numbergenerator that He can say, Yeah, I *knew* it would pick that one, butboy, am I surprised by the results!Hey God, can you create a mountain so big you cannot lift it? But anyway, this thread isnÕt about the powers of God, but about theproperties of a truly random number, and the properties of numbers ingeneral, I guess.If all real numbers fit on a line, and all complex numbers fit on aplane, what does it mean if a number is in some sort of 3D space?Can it be possible for there to be numbers we cannot even imagine, ordescribe?it took man thousands of years to discover the (to us) basic number 0,and then negative numbers, and then irrationals, and complex, andtranscendentals, and on and on... What are we still missing? Probablylots.Hey James clear to me that such a god could do *anything* at random.> immediately says Yeah, I *knew* it would pick that one.You know, that is a good point for the philosophers. God being all> But then again God being all powerful should be able to create a> perfect random number generator. So maybe He creates a random number> generator that He can say, Yeah, I *knew* it would pick that one, but> boy, am I surprised by the results!Hey God, can you create a mountain so big you cannot lift it? But anyway, this thread isnÕt about the powers of God, but about the> properties of a truly random number, and the properties of numbers in> general, I guess.If all real numbers fit on a line, and all complex numbers fit on a> plane, what does it mean if a number is in some sort of 3D space?Can it be possible for there to be numbers we cannot even imagine, or> describe?it took man thousands of years to discover the (to us) basic number 0,> and then negative numbers, and then irrationals, and complex, and> transcendentals, and on and on... What are we still missing? Probably> lots.> Can the undirected reciprocal of 0 be counted as a number? Or at leasta numerical concept? Maybe we should give it a real existence. Afterall, when we type out 1/x and say x = 0 the universe doesnÕt blow upjust because of this, so it ought to have a fundamental existence asat least an abstract numerical concept.> Hey James Harris, you got any insights about this?> -Paul(...Starblade Riven computer science and I am considering>> going back to school to pursue an MS degree. Should I work towards an>> MS in mathematics taking many (if not all) electives in Comp Theory or>> should I pursue an MS in Comp Sci taking relevant electives in math?>> IÕve noticed that most people involved in Theoretical comp science>> hold degrees in mathematics. As an example, most of the individuals>> listed in the book People and Ideas in Theoretical Computer Science>> have PhDs in math. Maybe one reason is that there was no Computer>> Science department back when they were students. Now that many>> colleges and universities offer MS and PhDs in Comp Sci does it make>> more sense to pursue a graduate degree in Comp Sci instead of>> Mathematics for someone interested in Theory? Comments please!piggybacki have some bad news for you.unless you really love math and or comp.sci for their own sake, i advice you> keep away from them. advanced degrees in either of those subjects, especially> mathematics, are bound to lead you nowhere careerwise.I find this quite harsh... My long term goal is to either join areputable company in their research department or join academia as aprofessor. I disagree that a graduate degree in really love math and or comp.sci for their own sake, i > advice you keep away from them. advanced degrees in either of those > subjects, especially mathematics, are bound to lead you nowhere > careerwise.I find this quite harsh... My long term goal is to either join a> reputable company in their research department or join academia as a> professor. I disagree that a graduate degree in Comp Sci or> Mathematics will lead me nowhere.If thatÕs your goal, you should probably be looking into Ph.D. programs rather than Masters.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer increase your earning> potential in the real world, it doesnÕt matter what you pursue (math> or cs) as long as you get the degree. What you learn in school as a> result of your pursuing the MS degree in theoretical computer science> will almost certainly have nothing to do with your job; itÕs only the> diploma that counts. So just take whatever classes interest you.given that monkeys are running the show in the private and publicsectors practically everywhere, i call the above assertions intoquestion. could you tell us how a math degree increases earningpotential in the real world? mind you, i really mean the REAL world.try avoid re-stating the obvious observation that legitimate knowledgeand understanding of mathematics is fair indicator of competence forpractically any job in an advanced society. instead, focus onpresenting an argument that establishes (got that? ESTABLISHES) that amath degree stand-alone will in fact increase earning potential in theREAL world. when you accomplish that feat, endevour to present anaccurate estimate of math majors that actually realise your utopicideas in the REAL world...> If you are doing it to learn more about theoretical computer science,> then getting a degree in computer science or math will certainly help> you learn more, but it is not necessary to do such. ItÕs much more> efficient to just pick up a book and teach yourself. Everything they> will teach you will come from some book textbooks is being dealt with.Sure. Just not by you.[Giving an unresponsive and off-tangent response to a posting,to which you have nothing of substance or value to contribute,just to use as a hook for ßogging your bent obsession, thenthrowing in a tired old list of things youÕve posted identicallya hundred times before, is paradigmatic net-kook behavior,Arthur. I guess you really do see it as your destiny to displaceJames Harris. Otherwise, throw away your cut and paste postinginserts. Sorry for the a formula for the number of strings containing m ofeach of k > 1 distinct characters (so the strings are of length mk)that are cyclically different. What i mean by this is canÕt berotated to make the same string...so the strings ABC BCA and CAB areall the same cyclically, but they are all different to, for example,the string ACB.I reckon the number of strings with m occurences of each of kcharacters is simply(mk)!-----(m!)^kas there are (mk)! permutations of the characters in a string mk inlength,but the m occurences of each of the k characters can be permutedwithout making a difference.However, I canÕt tell what I need to divide the above number by to getthe number of cyclically different strings.An example might help to explain. Imagine i wanted to get allcyclically different strings containing two As and two Bs (so oßength 4). The formula above gives me the answer six, correspondingto the six stringsi/ AABBii/ ABABiii/ ABBAiv/ BABAv/ BAABvi/ BBAAHowever there are only two cyclically different strings AABB and ABABasi/,iii/,v/ and vi/ are all in the same class as AABB, and ii/ and iv/are both in the same class as ABAB.How do I tell how many cyclic equivalence classes there are (so I knowwhat to divide my formula by)? Or how do I count the number ofcyclically different strings of length mk containing m occurrences ofeach of my k characters for the number of strings containing m of> each of k > 1 distinct characters (so the strings are of length mk)> that are cyclically different. What i mean by this is canÕt be> rotated to make the same string...so the strings ABC BCA and CAB are> all the same cyclically, but they are all different to, for example,> the string ACB. I reckon the number of strings with m occurences of each of k> characters is simply (mk)!> -----> (m!)^k as there are (mk)! permutations of the characters in a string mk in> length,> but the m occurences of each of the k characters can be permuted> without making a difference. However, I canÕt tell what I need to divide the above number by to get> the number of cyclically different strings. An example might help to explain. Imagine i wanted to get all> cyclically different strings containing two As and two Bs (so of> length 4). The formula above gives me the answer six, corresponding> to the six strings> i/ AABB> ii/ ABAB> iii/ ABBA> iv/ BABA> v/ BAAB> vi/ BBAA However there are only two cyclically different strings AABB and ABAB> as> i/,iii/,v/ and vi/ are all in the same class as AABB, and ii/ and iv/> are both in the same class as ABAB. How do I tell how many cyclic equivalence classes there are (so I know> what to divide my formula by)? Or how do I count the number of> cyclically different strings of length mk containing m occurrences of> each of my k characters in some other way? IÕm stuck!>Apparently youÕve thought about this enough to realizewhat the central difficulty in this sort of problem is.These are sometimes referred to as bead problemsor necklace problems because of the equivalenceof an arrangement of colored beads on a necklaceunder rotation.George Polya developed a theory for counting thesearrangements. You might be interested in this nicelywritten introduction to those methods:http://www.geometer.org/mathcircles/polya.pdfX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: (Gregory L. Hansen)said:>I was just reading about Riemann spaces as a particular type of>affine spaces, which as I understand it means the space has a>metric,Riemann space, yes. There is an affine connection uniquely determinedby the metric, which gives the associated affine space.>a rule for finding distances between points. No, only for finding the distances[1] along a curve, or,alternatively, finding the lengths of tangent vectors.>But what does it mean for a space to be affine but non-Riemannian? It means that you are only given a connection, a means for paralleltransport of tensors and thus[2] for differentiating them.[1] Actually a bit more.[2] More precisely, for taking covariant t derivatives. There areimportant notions of derivatives that do not require a connection.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>In the context of differentiable manifolds, an affine space is one>with a connection, i.e., a field of coefficients with well-defined>transformation properties that allows one to do covariant>differentiation, i.e., parallel transport of tensors, which in turn>allows one to compare tensors defined at different points of the>manifold.Not unless the curvature vanishes. Otherwise the parallel transportbetween two points will depend on the path chosen.BTW, I would not expect a contemporary class in Differential Geometryto define a connection in terms of components and their transformationproperties.-- Shmuel (Seymour J.) Metz, SysProg <3f608294$2$fuzhry+tra$mr2ice@news.patriot.net>:> at 12:12 PM, Jim Heckman said: >In the context of differentiable manifolds, an affine space is one>with a connection, i.e., a field of coefficients with well-defined>transformation properties that allows one to do covariant>differentiation, i.e., parallel transport of tensors, which in turn>allows one to compare tensors defined at different points of the>manifold. Not unless the curvature vanishes. Otherwise the parallel transport> between two points will depend on the path chosen.Yes, I probably should have noted that. I certainly didnÕt meanto imply the comparison was unique.> BTW, I would not expect a contemporary class in Differential Geometry> to define a connection in terms of components and their transformation> properties.Yeah, a better way is to say the connection coefficients andtheir transformation properties follow from the existence of acovariant derivative operator satisfying the relevant axioms*.In particular, itÕs the Ônicebehavior wrt linearity of vectorfields multiplied by functions that distinguishes a covariantderivative from the Lie derivative you can get from just being adifferentiable manifold.* Type Preservation, Linearity, Commutation with Contractions,Leibniz, and Action : > In the context of differentiable manifolds, an affine space is> one with a connection, i.e., a field of coefficients with> well-defined transformation properties that allows one to do> covariant differentiation, i.e., parallel transport of tensors,> which in turn allows one to compare tensors defined at different> points of the manifold. While I completely agree with this description of an affine connection,> is the term affine space normally used for a manifold with a connection?> I would have thought this was quite a dangerous usage,> as the term affine space is also used for the underlying space> in affine geometry, ie the homogeneous space associated to a vector space.IÕm not sure about normally, but IÕm pretty sure IÕveencountered this usage, and went along with it here because itÕsclearly what the OP was referring to. I agree that affinemanifold is less prone to confusion, and would be :> In the context of differentiable manifolds, an affine space is>> one with a connection, i.e., a field of coefficients with>> well-defined transformation properties that allows one to do>> covariant differentiation, i.e., parallel transport of tensors,>> which in turn allows one to compare tensors defined at different>> points of the manifold.>> While I completely agree with this description of an affine connection,>> is the term affine space normally used for a manifold with a connection?>> I would have thought this was quite a dangerous usage,>> as the term affine space is also used for the underlying space>> in affine geometry, ie the homogeneous space associated to a vector space.IÕm not sure about normally, but IÕm pretty sure IÕve>encountered this usage, and went along with it here because itÕs>clearly what the OP was referring to. I agree that affine>manifold is less prone to confusion, and would be the term IÕd>usually use.I went with the terminology in an old book on general relativity, Adler, B___, and someone. IÕve been told that physicists can be sloppy about that sort of thing.-- When the fool walks through the street, in his lack of understanding he calls everything foolish. -- Ecclesiastes this description of an affine connection,|is the term affine space normally used for a manifold with a connection?|I would have thought this was quite a dangerous usage,|as the term affine space is also used for the underlying space|in affine geometry, ie the homogeneous space associated to a vector space.I also thought I remembered it being used in Auslander and MacKenziefor a continuous manifold where the transition maps are affine maps. IcouldnÕt think of a good example that wasnÕt a Riemann manifold, though.Keith an affine connection,||is the term affine space normally used for a manifold with a connection?||I would have thought this was quite a dangerous usage,||as the term affine space is also used for the underlying space||in affine geometry, ie the homogeneous space associated to a vector space.||I also thought I remembered it being used in Auslander and MacKenzie|for a continuous manifold where the transition maps are affine maps. I|couldnÕt think of a good example that wasnÕt a Riemann manifold, though.whatÕs the matter with the example of the circle seen as the orbitspace of the [affine connection]-preserving but not metric-preservingoperator x->x*2 on the open half-line?by the way there are some funny potential confusions going on in thisthread, some related to the very general confusions that gerald edgaralluded to in the vector space example thread, and some moreparticular confusions related to the fact that the de-origined vectorspace concept of affine space is the geodesically completeconnected simply connected special case of auslander and mackenzieÕsconcept of affine space that you mentioned which is in turn the ßattorsion-free special case of the space with affine connectionconcept. (or something like circle seen as the orbit> space of the [affine connection]-preserving but not metric-preserving> operator x->x*2 on the open half-line?I didnÕt understand this.What exactly is the orbit of an operator?The question, as I understand it,is whether every torsion-free connectionarises from a metric --or what comes to the same thing,given a connection is there necessarily a tensor field g_{ij}constant with respect to that connection?The following dimensional argument seems to me to show thatthe answer is in the negative even locally,Two torsion-free connections D,Don a manifold differ by a tensor field t^i_{jk} symmetric in j,k.Conversely if D is one such connectionand t such a tensor then D + t is a second torsion-free connection.Thus the space of connections on a manifold of dimenson nhas dimension n^2(n+1)/2 while the space of metrics has dimension n(n+1)/2,Since there is a unique torsion-free connection corresponding to each metricit follows that if n > 1 there are torsion-free connectionswhich do not arise from a metric, even locally.-- Timothy Murphy tel: +353-86-233 6090 X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>What exactly is the orbit of an operator?operator. If F is an operator and O is a non-null subset of itÕsdomain such that F: O->O and for all x and y in O there is an integern (which may be negative) such that F^n(x)=y, then we say that O is anorbit of F. The set off all such orbits is the orbit space.Note that the same nomenclature is used with groups of operators.>The question, as I understand it,>is whether every torsion-free connection>arises from a metricYour understanding is correct.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot orbit of an operator?operator. If F is an operator and O is a non-null subset of itÕs> domain such that F: O->O and for all x and y in O there is an integer> n (which may be negative) such that F^n(x)=y, then we say that O is an> orbit of F.First you say there is no such thing,Then you define it.Strange.> The set off all such orbits is the orbit space.How did the orbit space of the operator x->x^2 on the open half-lineturn out to be a circle?-- Timothy Murphy tel: operator?>> >> operator. If F is an operator and O is a non-null subset of itÕs>> domain such that F: O->O and for all x and y in O there is an integer>> n (which may be negative) such that F^n(x)=y, then we say that O is an>> orbit of F.First you say there is no such thing,>Then you define it.>Strange.> The set off all such orbits is the orbit space.How did the orbit space of the operator x->x^2 on the open half-line>turn out to be a circle?Well, you have to be careful to take the *correct* open half-line,for one thing; I highly recommend the one consisting of real numbersbigger than 1 (and I highly disrecommend the more usual one, consisting of real numbers bigger than 0, on which the operator hasa fixed point, which plays merry hell with the topology of the orbitspace). That being done, you find a natural coordinate in whichthe open half-line looks like the line, and the operator space bythe relation x is equivalent to y iff x and y are in the sameorbit), as a topological space (or as a smooth manifold), iscertainly a circle. But its *affine* structure (which comesfrom the original coordinate, not the new one) is *not* theusual affine structure on the circle.Whatever an affine the operator x->x^2 on the open half-line>>turn out to be a circle?Well, you have to be careful to take the *correct* open half-line,> for one thing; I highly recommend the one consisting of real numbers> bigger than 1 (and I highly disrecommend the more usual one,> consisting of real numbers bigger than 0, on which the operator has> a fixed point, which plays merry hell with the topology of the orbit> space). That being done, you find a natural coordinate in which> the open half-line looks like the line, and the operator x->x^2> becomes translation. Now the orbit space (=quotient space by> the relation x is equivalent to y iff x and y are in the same> orbit), as a topological space (or as a smooth manifold), is> certainly a circle. But its *affine* structure (which comes> from the original coordinate, not the new one) is *not* the> usual affine structure on the circle.OK, so you take the half-line {x in R: x > 1}and then you change coordinates by y = loglog(x)and the operator becomes y -> y + log2 on R.Then apparently you take R mod log2 and you get a circle.But why not take the circle to start with?The construction seems ludicrously complicated.Anyway, as far as I can understand the argument,the standard connection on the half-linegives a connection on the circlewith respect to which the standard metric on the circleis not constant.So what?The question is, is _any_ metric on the circleconstant under this connection(ie under the covariant derivative defined by the connection)?IÕm not too sure of the accuracy of what follows,but tensor analysis on a 1-dimensional manifold is pretty trivial.A connection on a manifold is defined bythe action of the corresponding covariant derivative on differentials.Thus in the case of the circle with coordinate t (0 < t < 2pi)a connection D takes the form D(dt) = f(t)dt^2,while a metric is of the form g(t)dt^2.The action of D on this metric is (gÕ(t) + 2f(t))dt^3.So we are looking for a function g(t) such thatgÕ(t) = -2f(t).It is obvious that there is a local solution to this;but in general there is no global solution,eg if f(t) = 1 then g(t) = -2t is not periodic.So the connection defined by D(dt) = dt^2is not Riemannian,though it is locally Riemannian,as is any connection on a 1-dimensional manifold.(Nb in dimension 1 every connection is torsion-free.) -- Timothy Murphy tel: operator x->x^2 on the open half-line|>>turn out to be a circle?|> |> Well, you have to be careful to take the *correct* open half-line,|> for one thing; I highly recommend the one consisting of real numbers|> bigger than 1 (and I highly disrecommend the more usual one,|> consisting of real numbers bigger than 0, on which the operator has|> a fixed point, which plays merry hell with the topology of the orbit|> space). That being done, you find a natural coordinate in which|> the open half-line looks like the line, and the operator x->x^2|> becomes translation. Now the orbit space (=quotient space by|> the relation x is equivalent to y iff x and y are in the same|> orbit), as a topological space (or as a smooth manifold), is|> certainly a circle. But its *affine* structure (which comes|> from the original coordinate, not the new one) is *not* the|> usual affine structure on the circle.||OK, so you take the half-line {x in R: x > 1}|and then you change coordinates by y = loglog(x)|and the operator becomes y -> y + log2 on R.|Then apparently you take R mod log2 and you get a circle.|But why not take the circle to start with?|The construction seems ludicrously complicated.perhaps lee was making fun of your habit of misquoting people. theconstruction is ludicrously complicated only because of yourmisquotation.|Anyway, as far as I can understand the argument,|the standard connection on the half-line|gives a connection on the circle|with respect to which the standard metric on the circle|is not constant.|So what?and faster and in fact makes an infinite number of revolutions andthen [SZ]ies off the track in a finite amount of time, much unlike thecase of a riemannian connection. (an affine connection gives a goodconcept of inertial motion even Well, you have to be careful to take the *correct* open half-line,> |> for one thing; I highly recommend the one consisting of real numbers> |> bigger than 1 (and I highly disrecommend the more usual one,> |> consisting of real numbers bigger than 0, on which the operator has> |> a fixed point, which plays merry hell with the topology of the orbit> |> space). That being done, you find a natural coordinate in which> |> the open half-line looks like the line, and the operator x->x^2> |> becomes translation. Now the orbit space (=quotient space by> |> the relation x is equivalent to y iff x and y are in the same> |> orbit), as a topological space (or as a smooth manifold), is> |> certainly a circle. But its *affine* structure (which comes> |> from the original coordinate, not the new one) is *not* the> |> usual affine structure on the circle.> perhaps lee was making fun of your habit of misquoting people. the> construction is ludicrously complicated only because of your> misquotation.I didnÕt misquote anyone, nor do I make a habit of misquoting people,and I rather resent the suggestion that I do.Perhaps you could either give an example of a misquotation,or else withdraw your imputation.If I quote what anyone says, I copy their exact words,as I did above.If the aim was to define a non-Riemannian connection on a circlethen it _was_ a very complicated way of doing it.I may be suffering from a misunderstanding,but a connection on a 1-dimensional manifoldseems completely trivial to me.If dt is a local differentialthen the connection must simply be defined by a function f(t),and is given by D(dt) = f(t) dt^2, In the case of a circle, if t is the anglethen the connection given by D(dt) = dt^2does not arise from any metric.> and faster and in fact makes an infinite number of revolutions and> then [SZ]ies off the track in a finite amount of time, much unlike the> case of a riemannian connection. (an affine connection gives a good> concept of inertial motion even though it doesnÕt give a good concept> of distance.)I take it inertial motion on a circleis defined by a locally constant vector field(ie having zero covariant derivative with respect to the connection).I would have thought that the speed would increase by the same amounton each revolution,and I donÕt quite see why it should [SZ]y off the track.But it is perfectly possible that I misunderstand.-- Timothy Murphy your habit of misquoting people. the>> construction is ludicrously complicated only because of your>> misquotation.I didnÕt misquote anyone, nor do I make a habit of misquoting people,>and I rather resent the suggestion that I do.I do have a habit of making fun of people and those of theirhabits which are not my habits, but neither was I making fun ofTimothy Murphy (or any of his habits) nor do I hold the opinion that he either has a habit of misquoting people or was misquotinganyone in this thread. I also revel in ludicrous complication, and I rather resent thesuggestion that any putative misquotation was necessary to inspiremy ludicrously complicated attempt at clearing up some of the :I was just reading about Riemann spaces as a particular type of affine> spaces, which as I understand it means the space has a metric, a rule for> finding distances between points. But what does it mean for a space to be> affine but non-Riemannian? What are some examples of that?IÕm surprised no oneÕs answered this yet, so IÕll take a quick> shot at it.but IÕd hoped for more of a response given some of the high-powereddenizens here.In the context of differentiable manifolds, an affine space is> one with a connection, i.e., a field of coefficients with> well-defined transformation properties that allows one to do> covariant differentiation, i.e., parallel transport of tensors,> which in turn allows one to compare tensors defined at different> points of the manifold. Note that being affine is an additional> level of structure beyond just being a differentiable manifold.> (Without a connection, the best one can do is something like the> Lie derivative, which doesnÕt quite have all the nice properties> one wants of a covariant derivative.)A metric, which is an additional level of structure beyond a> connection, is a tensor field which allows one to take dot> products of vectors in the tangent space at each point in the> manifold. With a metric, one can always define a (torsion-free,> metric-compatible) connection in terms of derivatives of the> metric, as is done in General Relativity -- but the converse> isnÕt true.I pretty much know the formal definitions, but they donÕt really meanmuch to me.I know of the existence of non-metric theories of gravity. Knowanything about non-metric theories of gravity? What would it mean forit not to have a metric? If itÕs a theory of gravity youÕre lookingat relationships between objects in space and time, so IÕd think itwould pretty much have to have a metric, because distances have apretty clear physical know of the existence of non-metric theories of gravity. Know> anything about non-metric theories of gravity? What would it mean for> it not to have a metric? If itÕs a theory of gravity youÕre looking> at relationships between objects in space and time, so IÕd think it> would pretty much have to have a metric, because distances have a> pretty clear physical meaning.This is a question for sci.physics.relativity, or even better,sci.physics.research. There are lots of people in both forumswho know a lot more about this kind of thing than most peoplehere.-- Jim Heckman X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>Stranger, even, is that my teaching texts>for my HS classes make a real point of saying Ôno even roots for>integersexcept those texts that go into complex numbers, who allow>sqrt(-1), but still maintian Ôno even roots for negative>integersÕ.ObFeynman Those books are wrong, which is SOP for HS texts.>But while weÕre on it (and negative bases to reducible rational powers?The same as any exponential. You must at all times distinguishbewtween single valued expressions and expressions that representmultiple values.> For example, IÕve>been teaching the advanced algebra kids that (-16)^(2/4) does not>exist,Which is wrong. It can refer to two distinct values.>as it is not well-definedIt is up to factors of cos Pi/2 + i sin Pi/2.> (-16)^(2/4) should equal (-16)^(1/2) = 4i,ThatÕs only one of the four possible values.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>I donÕt understand your question. Of course you can take even roots>of negative integers. For example sqrt(sqrt(-1)) is a well-defined>complex number.No. It could refer to any of 4 different numbers.>Its value is equal to sqrt(1/2) + i*sqrt(1/2).Nu, so what is sqrt(1/2) - i*sqrt(1/2)? What are the negatives ofthose two? -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to following:> at 04:38 PM, Joona I Palaste said:>>I donÕt understand your question. Of course you can take even roots>>of negative integers. For example sqrt(sqrt(-1)) is a well-defined>>complex number.> No. It could refer to any of 4 different numbers.>>Its value is equal to sqrt(1/2) + i*sqrt(1/2).> Nu, so what is sqrt(1/2) - i*sqrt(1/2)? What are the negatives of> those two? Point taken. I stand corrected.sqrt(sqrt(-1)) has four values:sqrt(1/2) + i*sqrt(1/2)-sqrt(1/2) + i*sqrt(1/2)sqrt(1/2) - i*sqrt(1/2)-sqrt(1/2) - i*sqrt(1/2)But then, sqrt(-1) also has two values, i and -i, so having multiplevalues does not distinguish sqrt(sqrt(-1)) from sqrt(-1).-- /-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.fi/~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/You could take his life and... - Mirja following:> at 04:38 PM, Joona I Palaste said:>>I donÕt understand your question. Of course you can take even roots>>of negative integers. For example sqrt(sqrt(-1)) is a well-defined>>complex number.> > No. It could refer to any of 4 different numbers.> >>Its value is equal to sqrt(1/2) + i*sqrt(1/2).> > Nu, so what is sqrt(1/2) - i*sqrt(1/2)? What are the negatives of> those two? Point taken. I stand corrected.> sqrt(sqrt(-1)) has four values:> sqrt(1/2) + i*sqrt(1/2)> -sqrt(1/2) + i*sqrt(1/2)> sqrt(1/2) - i*sqrt(1/2)> -sqrt(1/2) - i*sqrt(1/2)> But then, sqrt(-1) also has two values, i and -i, so having multiple> values does not distinguish sqrt(sqrt(-1)) from sqrt(-1).Or rather, x^(1/2) has two branches. When x is non-negativereal, we all agree that the symbol sqrt refers to thenon-negative real branch. When x is negative real, I guessyou could say that thereÕs a convention that sqrt refersto the branch containing positive imaginary values.ThereÕs no convention for sqrt for non-real arguments,though I suppose it wouldnÕt be too hard to make one upwhen one needed it. - arguments,> though I suppose it wouldnÕt be too hard to make one up> when one needed it.But you canÕt do it in such a way thatsqrt(zw) = sqrt(z) sqrt(w)is always true :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of GentlemenX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: numbers, is there any reason that we cannot>take even roots of negative integers, other than the square root of>-1?Of course not; the Fundamental Theorem of Algebra says that anypolynomial equation has solutions. The roots, of course, will not beunique. If r is an nth root of x, then so are r*e^2Pi*i/n. . . .r*e^2(n-1)Pi*i/n. This doesnÕt depend on the sign of x.-- Shmuel (Seymour J.) Metz, SysProg and some of the mathematical induction problems, I tried toremember why it is necessary to replace the n with k in the induction step.Can somebody help me? Is it that k in N is supposed to be a particularelement; whereas, n in N is meant as a general element. How is thisdescribed in reading some of the mathematical induction problems, I tried to> remember why it is necessary to replace the n with k in the inductionstep.> Can somebody help me? Is it that k in N is supposed to be a particular> element; whereas, n in N is meant as a general element. How is this> described in logic?> induction problems, I tried to>remember why it is necessary to replace the n with k in the induction step.>Can somebody help me? Is it that k in N is supposed to be a particular>element; whereas, n in N is meant as a general element. ItÕs not necessary to replace n by k, depending on how you phrasethings. But there are common incoherent phrasings that can befixed by using two variables. Say weÕre proving that(*) 1 + ... + n = n(n+1)/2.Students sometimes say something about assume (*) with nand then prove (*) with n = n+1 and people cringe. If you feltlike saying that you could say assume (*) with n = k and prove(*) with n = k+1 - that phrasing makes sense.But you could just as well say assume 1 + ... + n = n(n+1)/2and prove 1 + ... + (n+1) = (n+1)(n+1+1)/2. No need for a kif you put it that way.Or you could say assume (*), and then prove (*) with n+1in place of n. ThatÕs acceptable, but I wish you wouldnÕtput it that way because hearing people talk that way leadsstudents to speak in the incoherent way they sometimes do.>How is this>described in logic?With only one variable, for example(P(1) & An(P(n) -> P(n+1)) -> AnP(n).>TIA,Charlie>************************David induction problems, I tried to> remember why it is necessary to replace the n with k in the inductionstep.> Can somebody help me? Is it that k in N is supposed to be a particular> element; whereas, n in N is meant as a general element. How is this> described in logic?>I think the short answer is, to avoid the appearance of circular argument.It really doesnÕt matter what variable, n or k or some other variable isused,so long as one is careful about the (often implied) quantification of thesevariables.LetÕs review the outline of mathematical induction and point out why itmight appear to an unfamiliar eye that the principle espouses a form ofcircular proof that could be used to prove anything at all.We refer to a basis step and an induction step as the ingredientsthat make up a proof by mathematical induction. Although some maydisagree, I hold with zero as being the least natural number (it makes Nboth an additive as well as multiplicative monoid). Thus to prove by(weak) induction for some property P of natural numbers that:for all n in N, P(n)we should show both ingredients:(basis step) P(0)(induction step) for all k in N, P(k) implies P(k+1)The second ingredient, the induction step, is often explained inhigh school algebra by a somewhat dubious notion that we areassuming the truth of P(k) in order to show that P(k+1) is true.While this does amount to the procedural reasoning involved,the bright student will begin to suspect that by assuming thetruth of P(k), we are indulging in a circular argument.And of course, we _almost_ are. ItÕs a delicate point. The changeof variable from n to k is really a distraction, designed to suppresswhat is really an intelligent but difficult to answer question.Your note hits pretty close to the mark in raising the issue ofwhether n may be a general element, while k is to be a particularelement. This is precisely what quantification is about in thetheory of symbolic logic. When asserting that:for all n, P(n)the meaning of n is intuitively a general element.The meaning of k in:for all k, P(k) implies P(k+1)is also that of a general element. However, what happens isthat once we begin to assume the induction hypothesis, P(k),a logical procedure whose correctness is justified by somethingin the theory of symbolic logic called the Deduction Theorem,we no longer have complete generality for k. At that point themeaning of k has been limited to a particular element forwhich P(k) happens to be true.Hope this somewhat wordy reading some of the mathematical induction problems, I tried to>> remember why it is necessary to replace the n with k in the induction>step.>> Can somebody help me? Is it that k in N is supposed to be a particular>> element; whereas, n in N is meant as a general element. How is this>> described in logic?>I think the short answer is, to avoid the appearance of circular argument.It really doesnÕt matter what variable, n or k or some other variable is>used,>so long as one is careful about the (often implied) quantification of these>variables.LetÕs review the outline of mathematical induction and point out why it>might appear to an unfamiliar eye that the principle espouses a form of>circular proof that could be used to prove anything at all.We refer to a basis step and an induction step as the ingredients>that make up a proof by mathematical induction. Although some may>disagree, I hold with zero as being the least natural number (it makes N>both an additive as well as multiplicative monoid). Thus to prove by>(weak) induction for some property P of natural numbers that:for all n in N, P(n)we should show both ingredients:(basis step) P(0)(induction step) for all k in N, P(k) implies P(k+1)The second ingredient, the induction step, is often explained in>high school algebra by a somewhat dubious notion that we are>assuming the truth of P(k) in order to show that P(k+1) is true.While this does amount to the procedural reasoning involved,>the bright student will begin to suspect that by assuming the>truth of P(k), we are indulging in a circular argument.And of course, we _almost_ are. ItÕs a delicate point. The change>of variable from n to k is really a distraction, designed to suppress>what is really an intelligent but difficult to answer question.Your note hits pretty close to the mark in raising the issue of>whether n may be a general element, while k is to be a particular>element. This is precisely what quantification is about in the>theory of symbolic logic. When asserting that:for all n, P(n)the meaning of n is intuitively a general element.The meaning of k in:for all k, P(k) implies P(k+1)is also that of a general element. However, what happens is>that once we begin to assume the induction hypothesis, P(k),>a logical procedure whose correctness is justified by something>in the theory of symbolic logic called the Deduction Theorem,>we no longer have complete generality for k. At that point the>meaning of k has been limited to a particular element for>which P(k) happens to be true.Hmm. Of course I agree about the particular element bit -where I come from we do two things: (i) prove P(1), (ii)assume P(n) and prove P(n+1). We say three times thatwe do _not_ assume P(n) for all n, one or two studentsdo anyway, and we say it again...But I donÕt follow your distaste for using the DeductionTheorem. I wouldnÕt call that something in the theoryof symbolic logic, although of course it is that. Seemsto me that the simplest _definition_ of a correct proofof if A then B is A proof that looks like the following:Assume A....Hence B.Leaving aside induction for a moment. Howwould you have them prove, say, if n iseven then n^2 is even.? Seems to me acorrect proof is this:[1] Pf: Assume n is even. Then n = 2k, son^2 = 2(2k^2), hence n^2 is even,while IÕd call[2] Pf Since n = 2k, n^2 = 2(2K^2), so n^2is even._wrong_, because we donÕt know that n = 2k,because we never assumed that n was even.(While [2] _is_ a correct proof if what is to beproved was phrased Suppose n is even. Thenn^2 is even.Seems to me to give a correct proof _without_explicitly assuming the hypotheses are trueone has to begin every sentence withif n is even then..., which is awkward ina longer proof.(Explictly stating what weÕre assuming seemsmore important if, say, weÕre proving If n is even then n^2 is even, while if n is odd then n^2 is odd.)>Hope this somewhat wordy discussion helps to clarify the>matter for you...************************David C. induction problems, I tried to> remember why it is necessary to replace the n with k in the induction step.> Can somebody help me? Is it that k in N is supposed to be a particular> element; whereas, n in N is meant as a general element. How is this> described in logic?> .... ItÕs not necessary, but it can be helpful in teaching induction tobeginners who find it difficult. Understandably they can be confused if aproof of a statement p(n) begins by saying Assume p(n). I never use theword assume there, but many people do. This is a case where some careful logical notation can be a real aidto clear thinking. What needs to be proved has a quantifier:(for every natural number n) p(n).The induction proof establishes p(1) and also(for every natural number k) (p(k) implies p(k+1)).That quantifier *outside* the bracketed implication is the key to whatinduction is all about, and what students often have trouble with. IÕmsure that any successful explanation of induction must embody thatformula, explicitly or implicitly. The words Assume p(k) may herald alogically legitimate use of the Deduction Theorem but itÕs very confusingto beginners. who told his students>>> that two decimal places were sufficient for ALL applications,>>> since that was the accuracy used for _money_. My world must>>> be THE world.>>> >>> BartThatÕs not even good advice in the money world. Banks hold a few more than two decimal places when figuring things like interest. A famous bank hack was when an employee programmed the computer to truncate the amount less than one cent and added it to his own account. That particular financial institution was doing so many transactions per day that it amounted to millions of dollars in his account before he was caught.>> >> >> Thats a riot - I hope they didnÕt have any students who became nuclear>> engineers, missile trajectory anaylists or medical equipment>> designers, (among other things), and stuck to the money decimal>> rule.I might add that part of this same rule was that one _always_>rounded up, since thatÕs what happened to one in stores....>merchants, to this guy, evidently always represented the>real world.BartI remember Tom Thumb advertised sandwiches for Under A Dollar! They were 99.5 cents each. Of course the customer still payed $1 for one, but only $1.99 for two.-- When the fool walks through the street, in his lack of understanding he calls everything foolish. -- Ecclesiastes 10:3, New American Bible tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>I had a teacher who once explained in great detail how there was no>general solution for a cubicDid one of her students set her straight?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to of a linear transform is for T(x)1.T(x+y)=T(x)+T(y)2.T(kx)=kT(x)We are wondering whether there is any example that only condition 1 is satisfied but not condition 2.-- Shi the definition of a linear transformNo, they transform is for T(x)>1.T(x+y)=T(x)+T(y)>2.T(kx)=kT(x)>We are wondering whether there is any example that only condition 1 is >satisfied but not condition 2. If x, y, k are allowed to be complex, consider complex conjugation.-- Wanted: Experts at choosing the best of 100+ applicants for a position.Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San ask on what sets you are operating? Does T : R --> R ? Or, issomething else? Are you studying Vector spaces?Lurch Everybody knows the definition of a linear transform is for T(x)> 1.T(x+y)=T(x)+T(y)> 2.T(kx)=kT(x)> We are wondering whether there is any example that only condition 1 is> satisfied but not condition 2.> -- > Shi Jin> the definition of a linear transform is for T(x)>1.T(x+y)=T(x)+T(y)>2.T(kx)=kT(x)>We are wondering whether there is any example that only condition 1 is >satisfied but not condition 2.See e.g. the thread Additive but not linear, without AC? from November 2002, or Difficult problem from February 1996.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia by 20 equilateral triangles. This constructionapproximates a sphere, more or less. What is a god book to study in order tobe able to answer questions like WhatÕs a better approximation to a sphere,still using only equilateral triangles? or what are the angles in awire-frame construction using triangles with sides of different equilateral triangles. Thisconstruction> approximates a sphere, more or less. What is a god book to study in orderto> be able to answer questions like WhatÕs a better approximation to asphere,> still using only equilateral triangles? or what are the angles in a> wire-frame construction using triangles with sides of different length?> etc.A polyhedron whose faces are all congruent equilateral triangles is called adeltahedron. There are exactly 8 convex deltahedra.http://mathworld.wolfram.com/Deltahedron.htmlThree of those 8 are platonic solids (the tetrahedron, the octahedron andthe icosahedron). Among the other 5 the snub disphenoid is (in my opinion)the most interesting.http://mathworld.wolfram.com/ SnubDisphenoid.htmlThe edges at either end of a snub disphenoid are perpendicular and itsgreatest dihedral angle is about 166.44 degrees.Among all the deltahedra, there is no better approximation to a sphere thanthe icosahedron.If you restrict yourself to polyhedra with triangular faces, not necessarilyequilateral, one approach is to take any polyhedron you like and, on itsnon-triangular faces, build little pyramids of some suitable height. I donot know a book on this particular topic.-- Clive called snub disphenoid? anyway,itÕs my favorite shape. note that thereÕs no 18-deltahedron(that isnÕt somewhat degenerate). the generalization of these is dueto Alain Lobel; check his site! > http://mathworld.wolfram.com/Deltahedron.htmlThree of those 8 are platonic solids (the tetrahedron, the octahedron and> the icosahedron). Among the other 5 the snub disphenoid is (in my opinion)> the most interesting.> http://mathworld.wolfram.com/SnubDisphenoid.htmlThe edges at either end of a snub disphenoid are perpendicular and its> greatest dihedral angle is about 166.44 degrees.--still not cause my math teacher ended up confusing me when shewent over this---= 19 East/West-Coast Specialized Servers - Total cause my math teacher ended up confusing me when she> went over this> ---= 19 East/West-Coast Specialized Servers - not 3*x^3?---= 19 East/West-Coast Specialized Servers - Total 3*x^3?Because 27*x^6 = (3*x^2)^3> News==----> http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 > ---= 19 East/West-Coast Specialized Servers it 3*x^2 and not 3*x^3?Because 27*x^6 = (3*x^2)^3IIRC, we didnÕt get much practice with the power rules whenI was in high school. It looks like the OP may have missedthat one class. The way I did these problems, when I wasnÕtsure, was to not use the shorthand. E.g., x^6 was writtenas xxxxxx. If I wanted to decube it, IÕd write it out 1) a .8ecrit dans le message de plz help me factor this cause my math teacher ended up confusing me whenshe> went over thisNews==----> http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000> ---= 19 East/West-Coast Specialized Servers - Total Privacy via to factor 27*X^6 + 1.-- Wanted: Experts at choosing the best of 100+ applicants for a position.Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California a .8ecrit dans le>(3*X^2 - 1)*(9*X^4 + 3*X^2 + 1)> Now try to factor 27*X^6 + 1.>(3*X^2 a .8ecrit:>> plz help me factor this cause my math teacher ended up confusing>> me when she went over this> (3*X^2 - 1)*(9*X^4 + 3*X^2 + 1)That answers the question what is 27X^6-1, factored, which is notthe same a .8ecrit dans le message de> a .8ecrit: >> plz help me factor this cause my math teacher ended up confusing>> me when she went over this > (3*X^2 - 1)*(9*X^4 + 3*X^2 + 1) That answers the question what is 27X^6-1, factored, which is not> the same question. xanthian.>Ok. let me restate it :hint: 6 = 2*3 .8ecrit:>>> a .8ecrit:>>>> plz help me factor this cause my math teacher ended up confusing>>>> me when she went over this>>> (3*X^2 - 1)*(9*X^4 + 3*X^2 + 1)>> That answers the question what is 27X^6-1, factored, which is not>> the same question.> Ok. let me restate it :> hint: 6 = 2*3 and 27 = 3^3.Okay, but why then does the OP not decide the answer with which to replyis: (sqrt(27)*x^3 + 1)*(sqrt(27)*x^3 - 1)rather than your original answer?That is, what motivates the OP to find _your_ answer?The original confusion, I would guess, is in the why do I go from hereto there part, as much as in the how do I go from here to there part.xanthian, no pedagogue, so not ready to supply an decide the answer with which to reply> is: (sqrt(27)*x^3 + 1)*(sqrt(27)*x^3 - 1) rather than your original answer?>Sure, you can start factorizing it that way, but I supposed the expectedanswerwas a factorization over Z[X] and not R[X] nor following problem.We have 20 ekls in a wood and we catch 5 of them and thenrelease them back again after they are being marked. For noreason what so ever we catch 4 of the 20 elks once again.What is the probability that 2 of them are marked?IÕm not sure how to reason here and reading the chapter in thebook gave me much but nothing i ca use. I understand that ineed to refrain the problem in terms of something less elk-ish.If i only pick one elk, the chance of it being marked is 0.25.What if i pick two? ItÕs not (0.25)^2 chance of getting twomarked because (0.25)^20 would be 1 (picking up every elkmakes it sure to get one marked)...How should i approach this kind of probability?-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be I have a following problem.We have 20 ekls in a wood and we catch 5 of them and then> release them back again after they are being marked. For no> reason what so ever we catch 4 of the 20 elks once again.> What is the probability that 2 of them are marked?IÕm not sure how to reason here and reading the chapter in the> book gave me much but nothing i ca use. I understand that i> need to refrain the problem in terms of something less elk-ish.If i only pick one elk, the chance of it being marked is 0.25.> What if i pick two? ItÕs not (0.25)^2 chance of getting two> marked because (0.25)^20 would be 1 (picking up every elk> makes it sure to get one marked)...Right. These arenÕt independent picks. This issampling without replacement so every pick changesthe population remaining.How should i approach this kind of probability?There are many equivalent approaches. You just have tobe careful about undercounting or overcounting. Hereare a few:1. Assume I number my catches in order of catching. Numerator:total number of arrangements of 2 Marked and 2 Unmarked inall orders. Denominator: Total number of arrangements of4 elk. That is: Numerator: My marked elk could be in positions 1 and 2, or 1 and 3, or 1 and 4... there are 4C2 positions they could be in. For each of these possibilities, there are 4*3 ways to fill the marked positions (Take any of the 4 marked elk, then any of the 3 remaining). How many ways are there to fill the unmarked positions? Denominator: 20 ways to fill 1st position, 19 ways to fill second one, ...2. Ignore ordering right from the start. Numerator: Totalnumber of possible catches with a pair of marked elk (how many distinct pairs of marked elk are there, if you donÕtcare about order?) and pairs of unmarked elk (how many distinct pairs of those are there?). Denominator: Totalnumber of ways to grab 4 elk from among 20, i.e. 20C4.3. Write down the 6 possible events (catches in differentorders) and work out the probabilities:For instance: P(MMUU) = (4/20)(3/19)(16/18)(15/17)for catching marked, marked, unmarked, unmarked in thatorder. The first term is 4/20 because there are 20 elkand 4 are marked. The second one is 3/19 because thereare only 19 elk and 3 of them are marked, etc.P(MUMU) = (4/20)(16/19)(3/18)(15/17)After writing down a few of these youÕll probablynotice a pattern (what is the following problem.We have 20 ekls in a wood and we catch 5 of them and then>release them back again after they are being marked. For no>reason what so ever we catch 4 of the 20 elks once again.>The real reason is that eventually youÕre going to do this problem with n elk, where n is unknown. Then, when 2 come back marked, youÕre going endangered species.Jon and then > release them back again after they are being marked. For no > reason what so ever we catch 4 of the 20 elks once again. > What is the probability that 2 of them are marked?How many different set of 4 elks are there? Amongst these, how many have exactly two marked elks? (Or at least 2 if that is what you want.)A literal question: Does this help? If not, ask for more hints.-- Stephen J. Herschkorn The best kind of help is when itÕs much enoughto get you going but just little enough to make you think youmade it on your own. :)-- KindlyKonrad------------------------------------------------- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be I have a following problem. We have 20 ekls in a wood and we catch 5 of them and then> release them back again after they are being marked. For no> reason what so ever we catch 4 of the 20 elks once again.> What is the probability that 2 of them are marked? IÕm not sure how to reason here and reading the chapter in the> book gave me much but nothing i ca use. I understand that i> need to refrain the problem in terms of something less elk-ish. If i only pick one elk, the chance of it being marked is 0.25.> What if i pick two? ItÕs not (0.25)^2 chance of getting two> marked because (0.25)^20 would be 1 (picking up every elk> makes it sure to get one marked)... How should i approach this kind of probability?>Check out the Hypergeometric Chess? Time, 19 May 1997.)He continues: The gap between the human and the>surrogate is permanent and will never be closed.>Machines will continue to make life easier, healthier,>richer, and more puzzling. And humans will continue to>care, ultimately, about the same things they always>have: about themselves, about one another, and many of>them, about God.Eugene WignerÕs 1961 paper in The Logic of Personal Knowledge, thetitle of which paper has been translated into at least two Englishversions, making finding the book containing the paper the bestmethod, esentially proves that for a simple model of life, quantummechanics proves life cannot reproduce.By now, I think we realize a little different model is needed, butWignerÕs model includes the salient features we still use: life iscomplex, life lives in an ecology; a nutrient sea, and lifereplicates itself. Wigner proved this canÕt happen. So a new model isneeded, and WignerÕs paper may provide the line by which the new modelis separated from the old. wignerÕs paper means that life must be insome way more complex than the life modeled in his paper, since thenutrient sea can have pretty much any properties we wish to give it,and still replication is impossible.So I would say that until computers can exceed the complexity ofWignerÕs model, and we can provide them with the appropriate nutrientsea, we donÕt have to worry that(I believe from TWF Week 198, by John Baez)>>perhaps philosophers need to be convinced >> that the business of mathematics is still a >> mysterious process, not yet easily automated.>Yale professor David Gelertner was notThe idea that Deep Blue has a mind is absurd. How can>an object that wants nothing, fears nothing, enjoys>nothing, needs nothing, and cares about nothing have a>mind?Well, if we are created by God, then perhaps our most noble purpose isto create such new life as we can, in addition to trivially creatingcopies of ourselves; while if we are not, then we have the power ofGod and may create such life as we choose. I do not believe thesecopies are even functionally interchageable and are certainly notidentical, not even to their progenitors.I do not believe our most noble purpose is to reproduce, while about 6billion of my cousins appear to believe this. Nor do I claim to knowwhat our most noble purpose is. This is the realm of the philosopher.I do know that for me, to take non living machine tools and animatethem through my own labor, without the use of computers, to createnearly indistinguishible, and certainly functionally interchangeablecopies of themselves, feels like a noble purpose. So that is what Ido.Yours,Doug Goncz (at aol dot com)Replikon Research Replikon Research researches replikons, which are self-reproducingconfigurations of non-living matter in environments that support replication,analogous to organisms living in interesting problems,including some relating to the fixed points of the hyperpower sequencex^x^x^...Turns out there is another bonus in there, thatÕs again using theLambertW, this time to define the analytic continuation of the realhyperpower function F(x) = x^x^x^...investigation, but the idea that it could be used as the analyticcontinuation of F(x) hadnÕt occured to me, until two days ago.There is also a nice investigation of the two components of the Hopfbifurcation using Maple, and some code to sketch the branches in(0,e^(-e)),in my math section:Enjoy.-- Ioannishttp://users.forthnet.gr/ath/jgal/____________________ _______________________Eventually, _everything_ is power of a number to come up with an answer.In this case I already know the answer (see below), but if I didnÕt, how would I find out the value for n in this scenario?2n - 1 = 4095 (I know that from trial and error, that n = 12) Please realize that the n is 2 to the power of n and not 2 x nSo Okay folks...I need to know how to find the power of a number to come up with an > answer.In this case I already know the answer (see below), but if I didnÕt, how > would I find out the value for n in this scenario?2n - 1 = 4095 (I know that from trial and error, that n = 12) Please realize that the n is 2 to the power of n and not 2 x nSo any help in figuring this out would be appreciated.GlennStandard newsnet for writing powers is ^, so that 2^3 means 2 times 2 times 2, or 8.Your problem, finding n such that 2^n - 1 = 4095, or, equivalently,2^n = 4096 can be done by simply writing out the powers of 2 until you get there:2^0 = 12^1 = 22^2 = 42^3 = 82^4 = 162^5 = 322^6 = 642^7 = 1282^8 = 2562^9 = 5122^10 = 10242^11 = 20482^12 = 4096And, as a short cut, one might memorize that 2^10 = 1024 is one kilo in binary based things like computer memory (kilobytes are actually 1024 to find the power of a number to come up with an > answer.In this case I already know the answer (see below), but if I didnÕt, how > would I find out the value for n in this scenario?2n - 1 = 4095 (I know that from trial and error, that n = 12) Please realize that the n is 2 to the power of n and not 2 x nSo any help in figuring this out would be appreciated.GlennWell, you have 2^n = 4094 and 2^2 = 4, 2^3 = 8, ... 2^12 = 4094.Or, you could write n = (ln(4094))/(ln(2)) = 11.999295.ThereÕs a moral there someplace.-- Paul power of a number to come up with an> answer.> In this case I already know the answer (see below), but if I didnÕt, how> would I find out the value for n in this scenario?> 2n - 1 = 4095 (I know that from trial and error, that n = 12)> Please realize that the n is 2 to the power of n and not 2 x n Well, you have 2^n = 4094 and 2^2 = 4, 2^3 = 8, ... 2^12 = 4094. Or, you could write n = (ln(4094))/(ln(2)) = 11.999295.>Adds and subs apart, log base 2, as well as common logarithm (base 10 -notation log) or natural log (base e - notation ln) also have their specialnotation, more widely known in computer scientists circles. The aboveproblem can be need to know how to find the power of a number to come up with an >> answer.>> >> In this case I already know the answer (see below), but if I didnÕt, how >> would I find out the value for n in this scenario?>> >> 2n - 1 = 4095 (I know that from trial and error, that n = 12) >> >> Please realize that the n is 2 to the power of n and not 2 x n>> >> So any help in figuring this out would be appreciated.>> >> GlennWell, you have 2^n = 4094 and 2^2 = 4, 2^3 = 8, ... 2^12 = 4094.Or, you could write n = (ln(4094))/(ln(2)) = 11.999295.ThereÕs a moral there someplace.Something like: all the fancy logarithms donÕt do you any good if you canÕtadd.2n - 1 = 40952n = 4096-- >Paul Sperry>Columbia, SC (USA)--Mensanator2 of Clubs => >> Okay folks...>> >> I need to know how to find the power of a number to come up with an >> answer.>> >> In this case I already know the answer (see below), but if I didnÕt, how >> would I find out the value for n in this scenario?>> >> 2n - 1 = 4095 (I know that from trial and error, that n = 12) >> >> Please realize that the n is 2 to the power of n and not 2 x n>> >> So any help in figuring this out would be appreciated.>> >> Glenn> >Well, you have 2^n = 4094 and 2^2 = 4, 2^3 = 8, ... 2^12 = 4094.> >Or, you could write n = (ln(4094))/(ln(2)) = 11.999295.> >ThereÕs a moral there someplace.Something like: all the fancy logarithms donÕt do you any good if you canÕt> add.2n - 1 = 40952n = 4096> -- Paul SperryColumbia, SC (USA)X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Darksquall) said:sci.logic? Your question doesnÕt belong there.Ask again in 200 years; right now we donÕt know whether there is sucha thing, much less what it is.>I want to learn about string theory and loop quantum gravity, and the>mathematics used in it. YouÕve got a lot of background to pick up first. Ask again when youÕvelearned some Functional Analysis, some Differential Topology and a fewother odds and ends.>I know that they use similar mathematical formalisms,Do you? How? Similar in what way?>They are not too different to be unified,Things can be very different and still unified.>So IÕm wondering if anybody can tell me the mathematics used, Off the top of my head Algebraic topology Complex Variables Differential Geometry Differential Topology Functional Analysis Group Theory>and the type of logic, and such, and where I can learn them.Plain old ordinary Logic, such as you need for any real Mathematicscourse.>and where I can learn them.Well, if youÕre bright enough, have access to a good library and workat it, you can learn them on your own. If you try that, work theexercises.If you have a good college nearby, find out whether you would beallowed to audit any classes. Plan on doing the homework and takingthe exams. When the kindly instructor allows you to take a test homefor the weekend, that doesnÕt mean that heÕs giving the class a break;it means that the exam will take longer than he wants to stay aroundfor. But such exams can be fun when you understand the material andthe questions are well chosen.>meaning I have to take calculus and>physics again in college, or Jr. College.DonÕt expect to find any real Mathematics or Physics classes in ajunior college. Look for a university that has a good graduate school.Attend[1] colloquia and seminars. Talk to the faculty and the graduatestudents.>I have a lot of good ideas IÕd like to use, but I want to be able to >develop the mathematics to back them up.DonÕt be upset if some of them turn out to be not so good. YouÕllprobably develop new ideas when you have more background.Why donÕt you give some information on what you already know beyondCalculus so that people can suggest reading material and/or courses?[1] Some of the material will be over your head. DonÕt worry about it;just absorb what you can.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to Poe<585ab5d8.0309090829.52bbcea9@ :>> >> Sire, there is no royal road to geometry.>> >> But the _is_ the Harris road.>> >> Gib>> >> >> Oooooooooh....youÕll take the Harris road and IÕll take>> the standard road and IÕll get the correct result before thee..... :-)>> >> (No, you donÕt want to hear me in person.)DoonÕt forrrget yerrr bad Scottish brrrrogue, laddie.For prrractice: The dilithiooom crrrystals are fyoooosinÕ,> Keptin! - RandyWe canna take any mair power, Keptin! The dilithiooomcrrrrystals have vanished into thin air!:-)-- #191, ewill3@earthlink.netItÕs still legal to go .sigless.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: hebl9901@stcloudstate.edu (Blake Hegerle) said:>Quick question: can anyone give an example of a vector space thatÕsSure; any vector space without a metric. A vector space with a metricthat is not[1] positive definite. An inner product space that is[1] Important in Physics both for Relativity and for Quantum FieldTheories that include ghosts.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <3f5bd98d$3$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>No,we are in l_1, Whoops!-- Shmuel (Seymour J.) Metz, <3F5642D1.30304@comcast.net> <3f54ff5f$1_4@newsfeed.slurp.net> <3f552de4$1_3@newsfeed.slurp.net> <3F56FF82.5070502@tcs.inf.tu-dresden.de> <3F572D7D.7090108@tcs.inf.tu-dresden.de>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>I wonder if the students of history have the same perception when >arriving at universtiy.I suspect so. Likewise students of English.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <3Z16b.2042$_26.498@newsread2.news.atl.earthlink.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: (Herman Rubin) said:>In probability>courses, many students will claim that exp(-|x|) cannot be>integrated, as they cannot come up with a simple formula for the>anti-derivative.IÕd like to believe that youÕre joking.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to legislation, teachers with great experience are> being asked to take demeaning examinations in the subject matter(s)> they teach. They are exempt from examination if they have MasterÕs> degrees in their subjects. Check this with teachers or principals or> headmasters.> I donÕt know why you think these examinations are demeaning. IÕve heardof several cases when the teachers fail and the teachers union puts up agreat fuss. Seems to me that if these tests are showing gaps in theteachersmathematical understanding, then the revelation of this may bedemeaning (or rather, very embarassing) to the teachers, but demeaningis in the eye of the beholder especially here.I donÕt know about this masterÕs degree exemption, but different stateshave tried implementing this kind of thing before (with no exemptions),and a majority of teachers fail the test, the union gets upset, and thestate quits this testing business. I recall this happened withMassachusetts a number of years ago.> In fact a solid BachelorÕs degree from a good math department is good> enough to teach high school, but ...> I donÕt agree with this. ItÕs easy enough to get abachelors, even from a good place, without having learned what a proof is(most likely the person will have a rather artificial, precise idea ofproof, which is as bad as having no idea of proof), or without having seen asimple concept taken to a very deep level, or without having had a glimpseof the unity of mathematics. Of course, the issue really is good math teachers, and no piece of paper,no matter how fancy, can certify or guarantee that a teachers with great experience are> being asked to take demeaning examinations in the subject matter(s)> they teach. They are exempt from examination if they have MasterÕs> degrees in their subjects. Check this with teachers or principals or> headmasters.> I donÕt know why you think these examinations are demeaning. IÕve heard> of several cases when the teachers fail and the teachers union puts up a> great fuss. Seems to me that if these tests are showing gaps in the> teachersmathematical understanding, then the revelation of this may be> demeaning (or rather, very embarassing) to the teachers, but demeaning> is in the eye of the beholder especially here.The *teachers* consider these examinations to be demeaning. They havetaught their subjects successfully for many years, apparently withoutsignificant complaints about their performance, and are then asked tosubject themselves to an authority that is newly formed and holdspower of authority over them.I donÕt know about this masterÕs degree exemption, but different states> have tried implementing this kind of thing before (with no exemptions),> and a majority of teachers fail the test, the union gets upset, and the> state quits this testing business. I recall this happened with> Massachusetts a number of years ago.In fact a solid BachelorÕs degree from a good math department is good> enough to teach high school, but ...> I donÕt agree with this. ItÕs easy enough to get a> bachelors, even from a good place, without having learned what a proof is> (most likely the person will have a rather artificial, precise idea of> proof, which is as bad as having no idea of proof), or without having seen a> simple concept taken to a very deep level, or without having had a glimpse> of the unity of mathematics. > Whoever has no idea what constitues a proof perhaps didnÕt get adegree from a good math department. > Of course, the issue really is good math teachers, and no piece of paper,> no matter how fancy, can certify or guarantee that a person is one.Yes. That must be the point at issue. And it is extremely probablethat some top-notch third-semester calculus students could teachfirst- and second-semester calculus without having degrees orcredentials.David Ames tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>The *teachers* consider these examinations to be demeaning. They>have taught their subjects successfullyFSVO successfully. Maybe your expectations are set too low.>apparently without significant complaints about their performance,Apparent to whom? Determined how?>and are then asked to>subject themselves to an authority that is newly formed and holds>power of authority over them.IÕm more concerned about the kids that they have power of authorityover.>And it is extremely probable>that some top-notch third-semester calculus students could teach>first- and second-semester calculus without having degrees or>credentials.ItÕs also extremely probable that some of the teachers complainingabout the tests took Mathematics for dummies[1] instead of realMathematics classes. Those top notch 3rd semester student wouldprobably do a better job than the certified teachers.[1] IÕm suspicious of any class that is not accepted for departmentalcredit. ThatÕs usually a red ßag indicating that the contents havebeen watered down.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to *teachers* consider these examinations to be demeaning. They have>taught their subjects successfully for many years, apparently without>significant complaints about their performance, and are then asked to>subject themselves to an authority that is newly formed and holds>power of authority over them.Try to complain about the low level of teaching in a publicschool. These complaints are ignored. The teacher willsay that this is all the children can learn, and untilrecently, that was essentially the end. Anyone who is capable of getting a good education is denied access to it.>> I donÕt know about this masterÕs degree exemption, but different states>> have tried implementing this kind of thing before (with no exemptions),>> and a majority of teachers fail the test, the union gets upset, and the>> state quits this testing business. I recall this happened with>> Massachusetts a number of years ago.If we tested the teachers to honest standards, most teachersWOULD fail. The public schools have brought it on themselves,by insisting on age grouping. >> In fact a solid BachelorÕs degree from a good math department is good>> enough to teach high school, but ...>> I donÕt agree with this. ItÕs easy enough to get a>> bachelors, even from a good place, without having learned what a proof is>> (most likely the person will have a rather artificial, precise idea of>> proof, which is as bad as having no idea of proof), or without having seen a>> simple concept taken to a very deep level, or without having had a glimpse>> of the unity of mathematics. It might be that a school like Harvard or Princeton,which has a surplus of strong applicants, can afford todemand that those who get degrees understand somemathematics. If a school demanded that the prospectiveteachers of mathematics they turn out qualify in thismethod, the result will be that the appropriate courseswill be moved to the schools of education, or that thosecandidates will go to other schools with far lowerstandards. Unless there is an outside examination by abody which can resist pressure to lower standards, thiswill not happen.A half century ago, most going to college had decentEuclid geometry courses and decent college algebracourses, including induction. Now, a fair proportionof high schools do not even have a Euclid course, andmany which do just have the students memorize proofs.>Whoever has no idea what constitues a proof perhaps didnÕt get a>degree from a good math department.Any math department which insists on this is likely tolose much of its funding, unless it is at a school whichcan afford to be highly selective. >> Of course, the issue really is good math teachers, and no piece of paper,>> no matter how fancy, can certify or guarantee that a person is one.>Yes. That must be the point at issue. And it is extremely probable>that some top-notch third-semester calculus students could teach>first- and second-semester calculus without having degrees or>credentials.Of course; degrees, and even grades, are misleading. Only comprehensive examinations, without multiple choicequestions, can test on what should be tested.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, result of Federal legislation, teachers with great experience are>>>being asked to take demeaning examinations in the subject matter(s)>>>they teach. They are exempt from examination if they have MasterÕs>>>degrees in their subjects. Check this with teachers or principals or>>>headmasters.>>I donÕt know why you think these examinations are demeaning. IÕve heard>>of several cases when the teachers fail and the teachers union puts up a>>great fuss. Seems to me that if these tests are showing gaps in the>>teachersmathematical understanding, then the revelation of this may be>>demeaning (or rather, very embarassing) to the teachers, but demeaning>>is in the eye of the beholder especially here.The *teachers* consider these examinations to be demeaning. They have> taught their subjects successfully for many years, apparently without> significant complaints about their performance, and are then asked to> subject themselves to an authority that is newly formed and holds> power of authority over them.These same teachers that get high marks for teaching successfully for many years sometimes have students who consider them to be ineffective teachers who completely fail to explain concepts. The issue is: by whose standards are the teachers successful? Some of these teachers are getting local or state awards yet have students who do not have respect for them because they want to learn but are not learning.> >>I donÕt know about this masterÕs degree exemption, but different states>>have tried implementing this kind of thing before (with no exemptions),>>and a majority of teachers fail the test, the union gets upset, and the>>state quits this testing business. I recall this happened with>>Massachusetts a number of years ago.>>>In fact a solid BachelorÕs degree from a good math department is good>>>enough to teach high school, but ...>>>I donÕt agree with this. ItÕs easy enough to get a>>bachelors, even from a good place, without having learned what a proof is>>(most likely the person will have a rather artificial, precise idea of>>proof, which is as bad as having no idea of proof), or without having seen a>>simple concept taken to a very deep level, or without having had a glimpse>>of the unity of mathematics. Whoever has no idea what constitues a proof perhaps didnÕt get a> degree from a good math department.If you get CÕs through junior/senior courses, and then donÕt do any proofs for a while, you are likely to have a shaky notion of proof.> > >>Of course, the issue really is good math teachers, and no piece of paper,>>no matter how fancy, can certify or guarantee that a person is one.> Yes. That must be the point at issue. And it is extremely probable> that some top-notch third-semester calculus students could teach> first- and second-semester calculus without having degrees or> credentials.David AmesPerhaps, but teaching ability and math ability are not the same.-- Will get high marks for teaching successfully for >many years sometimes have students who consider them to be ineffective >teachers who completely fail to explain concepts.How do they get such high marks? Mostly by getting thestudents to do a good job of regurgitation on the exams.Where do our college students get the idea that there isno point to learning the concepts, but just to learn thealgorithms to be able to do the exercises, and to answerthe multiple choice questions on the mass finals? Theonly way to judge teaching is by the results, not theperception of students before the end of the term on howwell they have been taught.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman ....................>>These same teachers that get high marks for teaching successfully for >>many years sometimes have students who consider them to be ineffective >>teachers who completely fail to explain concepts.> How do they get such high marks? Mostly by getting the> students to do a good job of regurgitation on the exams.Precisely. Meanwhile, the students know they donÕt understand so the instructor looks good to the system, while the students wander around wondering how to do math.Where do our college students get the idea that there is> no point to learning the concepts, but just to learn the> algorithms to be able to do the exercises, and to answer> the multiple choice questions on the mass finals? The> only way to judge teaching is by the results, not the> perception of students before the end of the term on how> well they have been taught.> IÕm worried about the high-school/middle-school students. If they enter college with poor preparation, they are going to have problems catching up in college.-- Will TwentymanX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>As a result of Federal legislation, teachers with great experience>are being asked to take demeaning examinations in the subject>matter(s) they teach. How do you define great experience and in what way are theexaminations demeaning?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <3F575754.7050107@pacbell.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>The epithet Top Number Theorist in the World TodayBut in what world?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: said:>IÕve seen the occasional immaturity and name calling by JSH, but I am>hoping to present him with another point of viewIn a lost cause there are no failures. In the end, I suspect that youwill do what many others have done and add him to your twit list. Butwho knows, the horse might learn to sing.>I donÕt know about the rest of you,>but I think that if he came to me and said I donÕt see why people>are hung up on this part of my paper, IÕd be willing to say Well,>you could rewrite this as..... and maybe they would understand this>better.That would only help if it were correct. What if the people rejectinghis paper already understand it and are rejecting it because itÕswrong?>It seems like he ismaking up his own language as he goesA person lacking in trust might believe that it was deliberateobfuscation, in an attempt to make his errors less obvious.Fortunately nobody here is so suspicious. But if you succeed inbringing him to see the light of Reason, I have a few more cranks youmight want to take a stab at reforming ;-)-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to colleges, academia, and sciences will encounter. Whyare thereso many disproportionate number of jews in sciences? Is it becausethey area superior race and conversely lending truth to racism and thusShockleyÕsfallacy that blacks are inferior? Do the jews have greater staminathanothers? Who is Benjamin Disraeli? Who is Marx, Who is Freud (Fraud)? Who isFelix Klein and which country did he live a life of great freedom in?How did USA got dragged into WW1? Who are neturei karta? visitwww.nkusa.org, and sites pointed from there to find more truth withsolid evidence. These questions are all answers. View the documentary, QueenVictoriaÕs England.The worst enemy is the doctor who misdiagnoses a problem andprescribes awrong medicine. Medication that not only ignores the existing ailmentbut also adds side-effects for a malady that does not exist.If you find anything offensive, refute it and give us a chance tomodify these views -- but only based on solid supporting arguments.Internet is not censored by vested interests.-----> Prime Minister SharonÕs visit to India is> unbelievable. No matter what we may think> about him, we must acknowledge that> contributions of Jewish people from the time> of antiquities to the prsent day in all walks> of human endeavour are extraordinary. This is a false conclusion of a mind blinded by a dog bone thrown by Sharon. Not one jew in Indiaor the middle-east made any significant contribution.For thousands of years jews failed to invent thepositional number system despite their fierce work in money-lending. They failed to invent algebra. Certainly,they learnt a lot from the people they lived among andback-stabbed them. For example, Germany, the land ofAryan race was the leading country in Europe. Theylearnt much from it, yet they back-stabbed a winningGermany of WW1 to get the land of Palestine andmaneouvered US into WW1 to defeat Germany. Notonly this, they got a lot of practical hands-on training in precision building of large monuments from pharaonic egyptians. In the past few centuries they are motivated solely by weapons for armageddon.> Look at their names in religious life, arts,> literature, philosophy, technology and science> etc.. They have shown they are second to none> in their love and sacrifice in blood for their> country and society.Much of the European science is financed on themoney of the colonies and jews were not motivatedby science but the power it gives to fight their wars.As a matter of fact a jew disclosed to me the shockingtruth, the purpose of a university is not to educate,but to prepare for war. And he said that in a mostserious and a deep manner, So it is the Armageddonthat Technion, Weizmann and others are involved inwith enticing coverup stories.> Sure, they have made few mistakes during the> past decades when dealing with Palestanians.> It will pass it will only be a footnote in history. There is no possibility of this in sight. Thisone is not going to end like South Africa withtruth and reconciliation, and those south africanwhites not wanting to stay, leaving for US/Australia/Europe.Sharon is in India to prepare for the final solution which is going to be the Armageddon. Not for nothing,it is in every religious scripture so much aboutjews. He is going to bring mischief to India as Jewsbrought misery, war and revolutions to Germany, Russia and every place that gave them refuge and prosperity.THEIR ORIGINS ARE DEEPLY TAINTED WITH FRATRICIDE AND BACK-STABBING. JUST STUDY WHAT JACOBÕS SONS DID TO THEIR OWN BROTHER AND FOR WHAT REASON!!!> Palestanians will be better off with Israelies > as friends and neighbours than other Arabs who have > not given them even a refuge when it was most needed.If they do give refuge, Israel will only expand.This one is an expansionist beast and has no realintent on making peace. Israel cannot acceptdemocracy to make peace like South Africa. Itsaxiom is apartheid. Religious equality is basicto a democracy.> Indians should learn something from Israelies> how to build a formidable little modern> society out of desert sand by sheer hardwork.India should learn not to build a society thatcan be held together only by hate and fear but basedon principles of Vedas and Ghandhi. One thing wecan learn from jewish scriptures is that devil alwaysbrings evil in very fine wrappings. The devilÕs rep hasactually shown the sign of this right below:-----Prime Minister SharonÕs visit to India is after 50> years of being spit on by Europe and the US > (in a lesser way) , have taken the logical turn> East to find people with similar interest. Can you see what he considers billions of US aidand veto usage in every security council resolutioncondemnation as spitting. These are a bunch ofungrateful people who will NEVER BE SATISFIED,no matter what you do for them. Newyorkers get thetitle of rudeness from the jews that are in them.Most Italian newyorkers are actually very polite.Only jews are generally rude with characteristicnewengland accent.And while Irgun talks of east, China did not trustthem nor have they done well with moslem easterners -I understand Jew+Moslem is another story of fratricide ofthese evil children of abram!!!!???? Not only this,their First fore-father abram was the biggest haterof hinduism and statues. I must conclude that thejews are the biggest enemies of hindus. The mainaccomplishment on abramÕs resume was breaking idols.They need us because they are actually weak. Theyare weak because of the presence of Europe and Christian. They need our demographic strength to counter the Christiansand that is why they are making alliance with NRIs.They will help using their inßuence to multiply theirpower by snatching from the mouth of the christians, impoverish them, and give us - for a while - but where will it ultimately lead to!!!???Consider their history and any alliance as a math problem. Open your eyes and look from any side, you find terrible problems. But the wrappingsare deceptively beautiful indeed. I am sure Advani ishaving orgasms just thinking about phalcon.> As the great master of diplomacy, the Frenchman> Talleyrand put it, Nations have no permanent> friends, only permanent interestsAnd Irgun has made this his own modus operandi and indeed if it is in their interest to go to US and drag it into WW1 to defeat the Germany that gave them refuge, so be it for the land of Palestine, from Britian. If Russianscan be brought under control by a false communist lullaby,that is right too. If Britian can be brought under controlby pretense of workers rights, that is fair enough fora Benjamin Disraeli and if IndiaÕs wealth is necessaryto be looted for planned palestinian conquest, Disraeliwould go and buy Suez to ensure that India is securelycolonized.In yesteryears, the interest of the zionists outsideGermany was to send jewish brothers and mothers to Aushwitzand they did it. Neturei Karta site will give you all theevidence you want.Just make sure that your path does not cross withjewsÕ. India better had enough cannon fodder tosacrifice at their altar when the time comes!!!> irgun43> Even their own god elohim cursed them and sent them toan exile. Their god gotta be the biggest academia, and sciences will> encounter. Why are there so many disproportionate number of jews in> sciences?< huge snip >Judging from the other groups on the list, the OP obviously thinks that Cprogramming is a science.The clc community is focussed on technical issues and political agitationhas no place here, particularly when it is of a type calculated to incitehatred against people on the basis of their ethnic or academia, and sciences will encounter. Why> are there> so many disproportionate number of jews in sciences? Is it because> they are> a superior race and conversely lending truth to racism and thus> ShockleyÕs> fallacy that blacks are inferior? Do the jews have greater stamina> than> others?They have Jewish mothers. Read PortnoyÕs Complaint by Philip Roth. Chinese mothers are equally lethal and man-hating. That Jews from theoff the bottom 90% of the bell curve for European Jews doesnÕt hurteither - evolution is a hoot if you are one of the survivors. Thebest of the best emigrated to the New World and were scythed again byanti-Semitism. The survivors are intellectually superior in every way(on the average).Selective breeding works the other way, too. The absolutely moststupid Blacks in West Africa were captured by their marginally moreintelligent and vicious brethren, then sold to Arabs who in turn soldthem to the Triangular Trade and New World slavery. Uppity Darkieswere weeded out. The result is US Inner Cities as celebrations ofethnic Alhttp://www.mazepath.com/uncleal/eotvos.htm (Do something right now we are covering vectors inplanes and space. The only problem is the homework the professor assigns isso much different from high school.In high school we only had problems that tested the concept, so if you knowthe dot product, for example, you just use it. But in college, heÕs makingus do all these problems with proofs, eg show that this equals that, and allthese other problems that you have to actually think in variables andletters in order to solve. :)So my question is - is there a program to assist or some way to help meunderstand these things better. I get the general concept; itÕs just that hegoes into great depth with course and right now we are covering vectors in> planes and space. The only problem is the homework the professor assigns is> so much different from high school.In high school we only had problems that tested the concept, so if you know> the dot product, for example, you just use it. But in college, heÕs making> us do all these problems with proofs, eg show that this equals that, and all> these other problems that you have to actually think in variables and> letters in order to solve. :)So my question is - is there a program to assist or some way to help me> understand these things better. I get the general concept; itÕs just that he> goes into great depth with these really hard problems. : /> D. Solow, HOW TO right now we are covering vectors in> planes and space. The only problem is the homework the professor assigns is> so much different from high school.In high school we only had problems that tested the concept, so if you know> the dot product, for example, you just use it. But in college, heÕs making> us do all these problems with proofs, eg show that this equals that, and all> these other problems that you have to actually think in variables and> letters in order to solve. :)So my question is - is there a program to assist or some way to help me> understand these things better. I get the general concept; itÕs just that he> goes into great depth with these really hard problems. : /Most likely there is a textboook (or a number of textbooks) for thecourse. IÕd suggest that go very carefully over the proofs youÕvelearnt so far, questioning every detail until youÕre sure you know whyevery step is true.After you do this try proving some of the simpler facts yourself andthen let yourself loose on here and I, or someone else in this NG, will tryand help you. What kind of proofs are they?Lurch> IÕm taking a calculus 3 course and right now we are covering vectors in> planes and space. The only problem is the homework the professor assignsis> so much different from high school. In high school we only had problems that tested the concept, so if youknow> the dot product, for example, you just use it. But in college, heÕs making> us do all these problems with proofs, eg show that this equals that, andall> these other problems that you have to actually think in variables and> letters in order to solve. :) So my question is - is there a program to assist or some way to help me> understand these things better. I get the general concept; itÕs just thathe> goes snippage>Having just muddled my way through this myself, I am going to take a shot atexplaining what I have learned as it applies to your proposal.> Just budding in, but I have found few people come up with the following> solution. Before you open the envelope in your hand, think of a number.> If the amount in your envelope is smaller then that number ask to> switch. If it is greater donÕt switch.>Your proposal would work (I think) if it just so happened that over time, thefrequency of each possible amount was the same - that is, that the distributionof amounts was uniform. However, there is no reason to think that is the case.Illustrating that it is a false assumption is the real purpose of the paradox.> There are three possibilities. 1) Both envelopes contain an amount under your number.> You will always switch and on average gain nothing.#1 is false, because the on average is assuming a uniform distribution ofamounts, and that it will all average out over time. There just is nothingthat makes that so. 2) Both envelopes contain an amount above your number.> You will never switch and on average gain nothing.#2 works, since you are not changing anything.> 3) Your number is between the amounts in the envelopes.> You will keep the higher amount but will swicth> with the lower amount, so you will gain some.#3 might happen, and if it did, you might make a gain. Might not. There is noway to tell, since no one knows the true distribution of amounts that will becarried around in pairs of envelopes by generous persons with an interest inlogic problems, for the rest of eternity ... So in conclusion with this procedure you will in general be> better of than if you always switch or never switch. -- > Antoon PardonIn fact, the distribution is uniform only if something, or someone, makes ituniform, as happens with a roulette wheel, or a deck of cards. But there is nosense in which you can say that in general the number 136 occurs just as oftenin the universe as the number 27. In fact, it sounds kind of that you should switch when the envelope you opened contains After posting, I mulled it around in my head some more, and I> saw most of what you have pointed out.Bummer :-). I hate it when I come to that conclusion just _after_ postingsomething.> I agree that it does pay off in this situation. But this is precisely the> calculation that is being done in the original situation. Somehow, this is> the right way to calculate in my puzzle, but not the right way to> calculate in the original situation.Right, hence the paradox of the original puzzle. Why is the simpleprobability calculation correct for your formulation, but somehow wrong inthe original puzzle?> The mathematicianÕs web site treatment may be correct, but it is not> satisfying to me to say something about expectation values going to infinity.ThatÕs actually a subtle wrinkle on the basic insight. Most philosophersor statisticians who look at it quickly realize that the problem with theoriginal puzzle is figuring out what probability distribution you areexpecting. You canÕt calculated expected value without knowing how likelyany particular value is.The original puzzle said roughly there is an equal chance of each envelopecontaining _any_ (positive) amount of money. If there was a range, like theintegers from 1 to 100, you would interpret this English statement as sayingthat each value should have the same chance (1%), and that the sum of thechances should be 100%.But the original puzzle didnÕt specify a range. The naive thing to do is tosay well, itÕs some probability distribution such that every integer (orreal number) has the same chance of showing up.Unfortunately, that doesnÕt work, because there doesnÕt exist _any_distribution that gives every integer the same chance of appearing, and yetthe sum adds up to 100%, since there are an infinite number of integers.Once you realize that, you see the trick in the puzzle. The question is,where to go from there? Some folks invent ranges by talking about the totalmoney supply in the world. Some folks invent distributions, where numbersbecome less likely as they get bigger. (These can have either finite orinfinite expected values.)But the main insight is that, for the original puzzle, you canÕt tell me thelikelihood that the first envelope contains $20. ThereÕs really nothing forcedby the puzzle statement, that is also consistent with mathematics, that cangive you a number for the probability of finding $20.> What is the rule for avoiding the error in calculated payoff? IsnÕt there> a principle here? What else does on average mean in the original> situation, if not the result of my puzzle?Perhaps my analysis can get you started in the right direction. I havenÕtfully resolved the puzzle for you (namely, how should you act if confrontedwith the situation described in the puzzle), but IÕve at least done half ofit: why is the simplistic probability calculation wrong? (Answer: becausethere is no initial distribution such that if you find any amount, say $20,in the first envelope, you can actually conclude that the second contains$10 or $40 with equal chances, and have this be true no matter what youfind in the first envelope.) -- Don__________________________________________________________ _____________________Don Geddis http://don.geddis.org/ don@geddis.orgSeen on the door to a light-wave lab:Do not excellent analysis:>> http://www.u.arizona.edu/~chalmers/papers/envelope.html> Just budding in, but I have found few people come up with the following> solution.You arenÕt proposing a solution. The paradox is, what is wrong with thesimple probabilistic reasons that proves you should always switch? YouhavenÕt explained that; youÕve merely come up with a different action totake.> Before you open the envelope in your hand, think of a number.> If the amount in your envelope is smaller then that number ask to> switch. If it is greater donÕt switch.Read the Chalmer(s) papers. YouÕll see that (depending on someassumptions), thereÕs a different argument you can make that suggests youreally will be better off if you switch every time. It all has to do withthe expected value in the second envelope. If youÕre in a situation wherethe expected value is infinite, but any specific value is finite, thenperhaps you should switch to the second envelope _no_matter_ what you findin the first one. -- Don__________________________________________________________ _____________________Don Geddis http://don.geddis.org/ don@geddis.orgI bet when you first start working to shrink a head, itÕs hard to focus on howgood itÕs going to look when itÕs finished. All you can think about is howmuch work is was curious on the probability on something. Let us say that I have 2dice. They are normal 6 sided dice. How many times would I have to throwthem to GUARANTEE that I will, at least once, come up with 2 and 3.I believe that the probability of 2 events occurring simultaneously is P1 *P2. In which case, the chance, per throw, of getting 2 and 3 is 1/6 * 1/6,which is 1/36.The only catch is that throws are always unique. 1 and 2, or 2 and 1, willonly occur once.I believe that under these circumstances, the probability after X throwswould be P = X*P1*P2. Which for guarantee, 100%, the equation would looklike:1 = X * 1/6 * 1/6X = 36Is there something I am missing here? Also, could anyone tell me the law,axiom, theorem that covers these ( I remember them hazily from something. Let us say that I have 2>dice. They are normal 6 sided dice. How many times would I have to throw>them to GUARANTEE that I will, at least once, come up with 2 and 3.Infinity.On an infinite sequence of throws, with probability 1 every particular outcome will come up infinitely many times.On any finite sequence of throws, anything thatÕs physically possiblehas a strictly positive probability of happening.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2